PHSC 121 - INTRODUCTION TO ASTRONOMY Kepler's Laws of Planetary Motion Question Bank Set 3

PHSC 121 - INTRODUCTION TO

ASTRONOMY - Kepler’s Laws of

Planetary Motion

Question Bank - Set 3

Liberty University

Question 1

Question

A planet orbits a star in an elliptical orbit. The closest distance of the planet

to the star (perihelion) is 0.3 AU, and the farthest distance of the planet to

the star (aphelion) is 0.7 AU. If the period of the planet’s orbit is 0.5 years,

calculate the eccentricity of the planet’s orbit.

Solution

Step 1: Recall Kepler’s Third Law of Planetary Motion, which states that the

square of the period of an orbit is proportional to the cube of the semi-major

axis of the orbit, i.e., T2∝a3, where Tis the period of the orbit and ais the

semi-major axis of the orbit.

Step 2: The semi-major axis of an elliptical orbit is the average of the peri-

helion and aphelion distances, a=rmin +rmax

Step 3: We can first calculate the semi-major axis a:

a=0.3AU + 0.7AU

2=1.0AU

2= 0.5AU

Step 4: Next, we can use Kepler’s Third Law to find the constant of propor-

tionality. Since the period is given as 0.5 years, we write:

T2=k·a3

(0.5)2=k·(0.5)3

0.25 = k·0.125

k=0.25

0.125 = 2

Step 5: Now, we can find the period of the planet’s orbit when it is at its

perihelion and aphelion distances. Let’s denote the distances as rmin = 0.3AU

(perihelion) and rmax = 0.7AU (aphelion).

Step 6: For the perihelion distance:

(0.5)2= 2 ·(0.3)3

0.25 = 2 ·0.027

0.25 = 0.054

(Not true)

Step 7: For the aphelion distance:

(0.5)2= 2 ·(0.7)3

0.25 = 2 ·0.343

0.25 = 0.686

(Not true)

Step 8: Since both calculations are not true, this indicates that the eccen-

tricity of the orbit must be between 0 and 1. To calculate the eccentricity, we

use the formula e=rmax −rmin

rmax +rmin .

Step 9: Calculate the eccentricity:

e=0.7−0.3

0.7+0.3=0.4

1.0= 0.4

Step 10: Therefore, the eccentricity of the planet’s orbit is 0.4.

Question 2

Question

According to Kepler’s laws of planetary motion, the square of the orbital period

(T) of a planet is proportional to the cube of the semi-major axis (a) of its orbit.

The Earth’s average distance from the Sun is approximately 1 astronomical unit

(AU) and its orbital period is about 1 year. Suppose an exoplanet has an orbital

period of 8 years. What is the semi-major axis of its orbit in AU?

Solution

Step 1: Let’s first write down Kepler’s third law in symbolic form:

T2∝a3

Step 2: Since the Earth’s average distance from the Sun is 1 AU and its orbital

period is 1 year, we can write this relationship as:

(1)2=k(1)3

where kis the constant of proportionality. Step 3: Now, we can set up a similar

equation for the exoplanet with an orbital period of 8 years and an unknown

semi-major axis in AU:

(8)2=k(a)3

Step 4: To find the semi-major axis of the exoplanet’s orbit, we can set up a

ratio using the two equations:

13=82

Step 5: Solving for agives:

a=3

√82= 4 AU

Therefore, the semi-major axis of the exoplanet’s orbit is 4 astronomical units.

Question 3

Question

Suppose a new planet is discovered in a distant solar system with an orbital

period of 200 Earth days. If the semi-major axis of its orbit is measured to be 3

AU (astronomical units), determine the mass of the star around which this new

planet orbits.

(Given: the mass of the sun is 1.989×1030 kg and the gravitational constant

is 6.674 ×10−11 N m2/kg2.)

Solution

Step 1: Calculate the orbital period of the planet in seconds. The formula

relating the orbital period (T) of a planet to its semi-major axis (a) is Kepler’s

third law:

T2=(4π2

G(M1+M2))·a3

where: T= orbital period, G= gravitational constant, M1= mass of the star,

M2= mass of the planet, and a= semi-major axis of the planet’s orbit.

Given T= 200 Earth days and a= 3 AU, we convert these values to seconds

and meters:

T= 200 ×24 ×60 ×60 seconds

a= 3 ×1.496 ×1011 meters

Step 2: Substitute the known values into the formula to solve for the mass

of the star (M1).

T2=(4π2

G(M1+M2))·a3

(200 ×24 ×60 ×60)2=(4π2

6.674 ×10−11 ×M1)·(3 ×1.496 ×1011)3

Step 3: Solve for M1.

M1=4π2·a3

G·T2−M2

Substitute the given values (G,M2) into the equation.

M1=4π2·(3 ×1.496 ×1011)3

6.674 ×10−11 ·(200 ×24 ×60 ×60)2−M2

Calculate M1to find the mass of the star.

Question 4

Question

Suppose a hypothetical planet, Planet X, has an orbital period around its star

of 200 days and an average distance from the star of 0.08 AU. Calculate the

orbital speed of Planet X in km/s. (Hint: Use Kepler’s Third Law of Planetary

Motion)

Solution

Step 1: First, we need to calculate the star’s mass in solar masses. We can use

Kepler’s Third Law of Planetary Motion, which states:

P2=4π2

G(M1+M2)

where: a= average distance between the planet and the star in AU, P=

orbital period of the planet in years, G= gravitational constant (6.67430 ×

10−11 m3kg−1s−2), M1= mass of the star in kg, M2= mass of the planet in

kg.

Step 2: We are given that the average distance, a, is 0.08 AU and the orbital

period, P, is 200 days. Converting the orbital period to years:

P=200 days

365.25 days/year ≈0.547 years

Step 3: Plugging in the known values into Kepler’s Third Law, we have:

0.083

0.5472=4π2

G(M1+M2)

Step 4: Solving for M1+M2:

M1+M2=4π2

G×0.083

0.5472≈2.3×1029 kg

Step 5: Since the mass of the planet is negligible compared to the mass of

the star, we can approximate M1≈M2≈2.3×1029 kg.

Step 6: Next, we can calculate the orbital speed of Planet X using the

formula for orbital speed:

v=√GM1

where: v= orbital speed in m/s, G= gravitational constant, M1= mass of the

star in kg, a= average distance between the planet and the star in m.

Step 7: Converting the average distance, a, to meters:

0.08 AU = 0.08 ×1.496 ×1011 m

Step 8: Plugging in the known values into the orbital speed formula:

v=√6.67430 ×10−11 ×2.3×1029

0.08 ×1.496 ×1011

Step 9: Calculating the orbital speed:

v≈3.63 km/s

Therefore, the orbital speed of Planet X is approximately 3.63 km/s.

Question 5

Question

An asteroid, located at a distance of 2.5 AU from the Sun, takes 12 years to

complete one full orbit. Determine the mass of the Sun using Kepler’s Third

Law of Planetary Motion.

Solution

To find the mass of the Sun using Kepler’s Third Law, we need to relate the

period of the asteroid’s orbit to the distance from the Sun and the mass of the

Sun.

Step 1: Recall Kepler’s Third Law, which states that the square of the

period of revolution of a planet is proportional to the cube of the semi-major

axis of its orbit: T2

=T2

where: - T1and a1are the period and semi-major axis of one orbit, - T2and a2

are the period and semi-major axis of another orbit.

Step 2: We can express the relationship in terms of the asteroid and the

Sun: Given: - T= 12 years is the period of the asteroid’s orbit, - a= 2.5AU is

the distance of the asteroid from the Sun, - M⊙is the mass of the Sun.

We can write the equation as:

a3=4π2

G(M⊙+Masteroid)

where Gis the gravitational constant, Masteroid is the mass of the asteroid which

we assume to be negligible compared to the Sun.

Step 3: Solve for the mass of the Sun: Plugging in the values:

(12 years)2

(2.5AU)3=4π2

G·M⊙

144

(2.5)3=4π2

G·M⊙

144

15.625 =4π2

G·M⊙

9.216 = 4π2

G·M⊙

Step 4: Calculate the mass of the Sun:

M⊙=4π2

9.216 ·G

M⊙=4π2

9.216 ·6.674 ×10−11

M⊙≈39.478

6.168 ×10−10

M⊙≈6.4×1010 kg

Therefore, the mass of the Sun is approximately 6.4×1010 kg.

Question 6

Question

Consider a hypothetical solar system with a star of mass Mat the center. A

planet of mass morbits the star in an elliptical orbit. The semi-major axis of

the orbit is a, and the period of the planet is T. The eccentricity of the orbit is

Assume the planet is at its closest point to the star (perihelion) and its

furthest point from the star (aphelion).

Determine the speed of the planet when it is at perihelion in terms of the

constants and variables provided.

Solution

Step 1: Find the distance from the star to the planet at perihelion (rmin) and

at aphelion (rmax): At perihelion, the distance rmin is equal to the semi-major

axis minus the eccentricity times the semi-major axis:

rmin =a(1 −e)

At aphelion, the distance rmax is equal to the semi-major axis plus the ec-

centricity times the semi-major axis:

rmax =a(1 + e)

Step 2: Using Kepler’s third law, we can express the period of the planet T

in terms of the semi-major axis a:

T2=4π2a3

G(M+m)

where Gis the gravitational constant.

Step 3: Use conservation of angular momentum to relate the angular mo-

mentum of the planet in circular orbit to its angular momentum in elliptical

orbit:

mvr =m√G(M+m)r

At perihelion, the velocity of the planet vmin can be expressed as:

vmin =√G(M+m)(1 + e)

a(1 −e)

Therefore, the speed of the planet at perihelion is √G(M+m)(1 + e)

a(1 −e).

Question 7

Question

According to Kepler’s third law of planetary motion, the square of the period of

revolution of a planet around the sun is proportional to the cube of its average

distance from the sun. Suppose a newly discovered planet has an average dis-

tance from the sun of 2.5 astronomical units (AU). If Earth’s average distance

from the sun is 1 AU and its period of revolution is 1 year, find the period of

revolution of the newly discovered planet.

Solution

Step 1: Calculate the ratio of the cube of the average distance of the newly

discovered planet to that of Earth: Let Tbe the period of revolution of the

newly discovered planet. According to Kepler’s third law:

(Tnew

TEarth )2

=(dnew

dEarth )3

Given that dnew = 2.5AU, dEarth = 1 AU, and TEarth = 1 year:

(Tnew

1)2

=(2.5

1)3

new

1= 2.53

new = 2.53

Tnew =√2.53(years)

Step 2: Calculate the period of revolution of the newly discovered planet:

Tnew =√2.53

Tnew =√15.625

Tnew ≈3.95 years

Therefore, the period of revolution of the newly discovered planet is approx-

imately 3.95 years.

Question 8

Question

For a certain planet orbiting the sun, its average distance from the sun is 1.5

AU and its orbital period is 3 years. Determine the mass of the sun in terms of

Earth’s mass.

Solution

Let’s denote the mass of the sun as M⊙, the distance of the planet from the sun

as r, the orbital period of the planet as T, and the mass of Earth as M⊕. We

will use Kepler’s third law of planetary motion, which states that the square of

the period of any planet is proportional to the cube of the semimajor axis of its

orbit. This can be mathematically represented as:

=T2

Step 1: Given that the average distance from the sun is 1.5 AU and the

orbital period is 3 years, and Earth’s values are r2= 1 AU and T2= 1 year, we

can set up the following equation:

(1.5)3=12

Step 2: Simplify the equation:

3.375 = 1

9= 1

Step 3: Therefore, we now have an equation in terms of masses:

M⊙

(M⊕)3=32

Step 4: Rearrange the equation to solve for the mass of the sun in terms of

Earth’s mass:

M⊙=32

9×(M⊕)3

Step 5: Since we are looking for the mass of the sun in terms of Earth’s

mass, no numerical calculation is needed. The mass of the sun in terms of

Earth’s mass is 32

9M⊕.

Question 9

Question

According to Kepler’s third law of planetary motion, the square of the period

(T) of a planet’s orbit is proportional to the cube of its average distance from

the sun (r). Suppose planet A has a period of 2.5 years and an average distance

from the sun of 1.2 AU (astronomical units), while planet B has a period of

4.8 years and an average distance from the sun of 1.8 AU. Which planet has a

greater average orbital speed in m/s?

Solution

Let’s denote the period as Tin years and the average distance from the sun as

rin astronomical units (AU). We can use Kepler’s third law to find the average

orbital speed for each planet.

Step 1: Calculate the orbital speed of planet A - For planet A: Given

TA= 2.5years and rA= 1.2AU Kepler’s third law states:

A∝r3

We can write this proportionality as:

A=k·r3

Where kis a constant. To find the value of k, we can use the data for planet A:

2.52=k·1.23

6.25 = k·1.728

k≈3.62

Now, we can find the average orbital speed for planet A:

vA=2πrA

vA=2π·1.2

2.5·365.25 ·24 ·3600 m/s

Step 2: Calculate the orbital speed of planet B - For planet B: Given

TB= 4.8years and rB= 1.8AU Using the same constant k≈3.62, we can find

the average orbital speed for planet B:

vB=2π·1.8

4.8·365.25 ·24 ·3600 m/s

Step 3: Compare the orbital speeds Now, we can compare the orbital

speeds of planet A and planet B to determine which planet has a greater average

orbital speed in m/s.

Question 10

Question

According to Kepler’s third law of planetary motion, the square of the period of

revolution of a planet around the Sun is proportional to the cube of its average

distance from the Sun. Suppose a planet has an average distance from the Sun

of 2 astronomical units (AU). If the period of revolution of the planet is 3 years,

what is the average distance of another planet from the Sun if its period of

revolution is 6 years?

Solution

Let T1be the period of revolution for the first planet and D1be its average

distance from the Sun. Similarly, let T2be the period of revolution for the second

planet and D2be its average distance from the Sun. According to Kepler’s third

law, we have the following relationship:

=T2

Given that D1= 2 AU and T1= 3 years, we can solve for D2when T2= 6

years.

Step 1: Substitute the given values into Kepler’s third law equa-

tion.

Plugging in the values we have:

23=62

Step 2: Solve for D2.

Solving for D2:

8=36

Cross multiply to solve for D2:

9D3

2= 288 ⇒D3

2= 32 ⇒D2= 2 3

√16 ⇒D2= 2 ×2 = 4 AU

Therefore, the average distance of the second planet from the Sun is 4 as-

tronomical units.

Question 11

Question

Consider a planet with a semi-major axis of 2.5 AU and an orbital eccentricity

of 0.4. Calculate the closest (perihelion) and farthest (aphelion) distances of

the planet from the Sun in AU.

Solution

Step 1: Calculate the perihelion distance of the planet from the Sun. The

perihelion distance is given by:

rmin =a(1 −e)

where: - rmin is the perihelion distance, - ais the semi-major axis, and - eis

the eccentricity.

Substitute a= 2.5AU and e= 0.4into the formula:

rmin = 2.5(1 −0.4)

rmin = 2.5(0.6)

rmin = 1.5AU

Step 2: Calculate the aphelion distance of the planet from the Sun. The

aphelion distance is given by:

rmax =a(1 + e)

where: - rmax is the aphelion distance, - ais the semi-major axis, and - eis the

eccentricity.

Substitute a= 2.5AU and e= 0.4into the formula:

rmax = 2.5(1 + 0.4)

rmax = 2.5(1.4)

rmax = 3.5AU

Thus, the perihelion distance (closest distance) of the planet from the Sun

is 1.5 AU and the aphelion distance (farthest distance) is 3.5 AU.

Question 12

Question

Consider a planet with a semi-major axis of 2.5 AU orbiting a star with a mass

of 1.5×1030 kg. Calculate the period of the planet’s orbit around the star (in

years) using Kepler’s Third Law.

Solution

To find the period of the planet’s orbit, we can use Kepler’s Third Law, which

states that the square of the period of an orbiting object is proportional to the

cube of its semi-major axis.

Step 1: Convert the semi-major axis to meters. Given: 1 Astronomical

Unit (AU) = 1.496 ×1011 meters.

2.5AU ×1.496 ×1011 meters/AU = 3.74 ×1011 meters

Step 2: Calculate the period using Kepler’s Third Law formula:

T2=4π2

G(M1+M2)a3

Where: T= period of the planet’s orbit G= gravitational constant (6.67430×

10−11 m3kg−1s−2)M1= mass of the star (1.5×1030 kg) M2= mass of the planet

(assumed negligible compared to star) a= semi-major axis of the planet’s orbit

(3.74 ×1011meters)

Plugging in the values:

T2=4π2

6.67430 ×10−11 m3kg−1s−2

(1.5×1030 kg)

3.74 ×1011 m

Step 3: Calculate the period T:

T2=4π2

6.67430 ×10−11 m3kg−1s−2

1.5×1030 kg

(3.74 ×1011 m)3

T2= 1.22372 ×107s2

T=√1.22372 ×107s=3496.40 s≈3.50 years

Therefore, the period of the planet’s orbit around the star is approximately

3.50 years.

Question 13

Question

Consider a hypothetical planetary system where a planet orbits a star in a

perfectly circular orbit. The planet takes 200 days to complete one orbit around

the star. If the distance between the planet and the star is 0.6 AU, determine

the mass of the star.

Solution

Step 1: First, let’s convert the orbital period of the planet from days to years.

Since there are approximately 365.25 days in a year:

Orbital period (years) =200 days

365.25 days/year

Step 2: Calculate the orbital period in years:

Orbital period (years) ≈200

365.25 ≈0.547 years

Step 3: Next, let’s use Kepler’s third law:

Orbital period2=4π2

G(M+Mplanet)×Semi-major axis3

where Gis the gravitational constant, Mis the mass of the star, Mplanet is the

mass of the planet (which we assume to be negligible compared to the star), and

the semi-major axis is the average distance between the star and the planet.

Step 4: Since the planet’s orbit is perfectly circular, the semi-major axis is

the same as the distance between the planet and the star. We are given that

the distance is 0.6 AU, so we can substitute the known values into the equation:

0.5472=4π2

G(M)×(0.6)3

Step 5: Solve for the mass of the star M:

M=4π2×(0.6)3

G×0.5472

Step 6: Now, calculate the mass of the star using the value of the gravita-

tional constant G≈6.67 ×10−11 m3kg−1s−2:

M=4π2×(0.6)3

(6.67 ×10−11)×0.5472

Step 7: Finally, calculate the mass of the star:

M≈4π2×0.216

6.67 ×0.5472≈8.494

2.266 ×10−10 ≈3.741 ×1010 kg

Therefore, the mass of the star in the hypothetical planetary system is ap-

proximately 3.741 ×1010 kg.

Question 14

Question

Explain Kepler’s Third Law of Planetary Motion in detail, and then use it to

calculate the period of a hypothetical planet orbiting a star with a mass of

2×1030 kg, at an average distance of 5 AU.

Solution

Kepler’s Third Law of Planetary Motion states that the square of the period of

revolution (T) of a planet is directly proportional to the cube of the semi-major

axis (a) of its elliptical orbit. Mathematically, it can be expressed as:

T2=k×a3

where kis a constant that depends on the system of measurement used.

Step 1: Calculate the period of the hypothetical planet using Kepler’s Third

Law. Assume k= 1 for simplicity. Since the mass of the star is given as 2×1030

kg and the average distance from the star is 5 AU (1 AU = 1.496 ×1011 m), we

need to find the semi-major axis (a) in meters:

a= 5 ×1.496 ×1011

Step 2: Substitute the value of ainto Kepler’s Third Law equation to solve

for T:

T2= 1 ×(5 ×1.496 ×1011)3

T2= (7.48 ×1011)3

T2= 3.3455 ×1035

Step 3: Take the square root of both sides to find T:

T=√3.3455 ×1035

T≈1.83 ×1017 seconds

Therefore, the period of the hypothetical planet orbiting the star would be

approximately 1.83 ×1017 seconds.

Question 15

Question

According to Kepler’s Laws of Planetary Motion, the square of the orbital period

of a planet is proportional to the cube of its semi-major axis. Suppose a planet

has an orbital period of 8 years and a semi-major axis of 3 AU (astronomical

units). Determine the orbital period of another planet with a semi-major axis

of 6 AU.

Solution

Step 1: Let P1be the orbital period of the first planet, a1be its semi-major

axis, P2be the orbital period of the second planet, and a2be its semi-major

axis. Kepler’s Third Law states that

(P1

P2)2

=(a1

a2)3

Step 2: Substituting the given values, we have

P2)2

=(3

6)3

Step 3: Simplifying the equation gives

P2)2

=(1

2)3

Step 4: Solving for P2yields

P2)2

=(1

64 = P2

Therefore, the orbital period of a planet with a semi-major axis of 6 AU is

64 years.

Question 16

Question

According to Kepler’s third law of planetary motion, the square of the period of

a planet’s orbit is proportional to the cube of the semi-major axis of its orbit.

Consider a planet with a semi-major axis of 2.5 AU (astronomical units) that

takes 6 years to complete one orbit. Calculate the period of another planet in

years if it has a semi-major axis of 3.5 AU.

Solution

Let’s denote the period of the second planet as Tyears. According to Kepler’s

third law,

(T1

T2)2

=(a1

a2)3

where T1= 6 years, a1= 2.5AU, and a2= 3.5AU.

Step 1: Rewrite Kepler’s Third Law

6)2

=(2.5

3.5)3

Step 2: Solve for T

6)2

=(5

7)3

36 =125

343

Step 3: Find T

T2= 36 ×125

343

T2=4500

343

T=√4500

343

T≈4.44 years

Therefore, the period of the second planet is approximately 4.44 years.

Question 17

Question

A planet orbits the Sun in an elliptical path such that its closest distance to the

Sun (perihelion) is 0.3 AU and its farthest distance from the Sun (aphelion) is

0.7 AU. If the time taken by the planet to travel from perihelion to aphelion

is 24 days, determine the time taken by the planet to travel from aphelion to

perihelion.

Solution

Step 1: To find the speed of the planet at perihelion and aphelion, we can use

Kepler’s second law which states that a planet sweeps out equal areas in equal

times. This implies that the planet moves faster when it is closer to the Sun

(perihelion) and slower when it is farther from the Sun (aphelion).

Step 2: Let r1be the distance of the planet from the Sun at perihelion (0.3

AU) and r2be the distance at aphelion (0.7 AU). The area swept out by the

planet at perihelion is A1=1

2r2

1θ1and at aphelion is A2=1

2r2

2θ2, where θ1and

θ2are the angles swept by the planet.

Step 3: Since the areas are equal, we have A1=A2. This gives us 1

2r2

1θ1=

2r2

2θ2. We can simplify this to r2

1θ1

2θ2= 1.

Step 4: We are given that the planet takes 24 days to travel from perihelion

to aphelion, so the time taken to travel from aphelion to perihelion is the same.

This implies that the angle swept in each case is the same, i.e., θ1=θ2.

Step 5: Substituting θ1=θ2into our equation, we get r2

= 1. Substituting

the given values of r1= 0.3AU and r2= 0.7AU, we obtain 0.32

0.72= 1.

Step 6: Solving this equation, we find that the time taken by the planet to

travel from aphelion to perihelion is 24 days as well. Therefore, the time taken

by the planet to travel from aphelion to perihelion is 24 days.

Question 18

Question

An exoplanet is discovered orbiting a distant star with a semi-major axis of 1.5

AU. The star has a mass of 3×1030 kg. Determine the period of the exoplanet’s

orbit using Kepler’s third law of planetary motion.

Solution

Step 1: Determine the period of the exoplanet’s orbit using Kepler’s third law

of planetary motion, which states:

=T2

where T1and T2are the periods of the orbits, and a1and a2are the semi-

major axes of the orbits.

Step 2: Substitute the known values into the equation:

(1.5)3=T2

1.53=T2

3.375 =⇒T2= 3.375

Step 3: Solve for T:

T=√3.375 = 1.837 years

Therefore, the period of the exoplanet’s orbit is approximately 1.837 years.

Question 19

Question

In a distant star system, a planet with a semi-major axis of 2.5×1011 meters

takes 3.6 Earth years to complete one full orbit around its star. Calculate the

mass of the star if the gravitational constant G= 6.67430 ×10−11 m3kg−1s−2.

Solution

Step 1: Firstly, we need to calculate the orbital period of the planet using

Kepler’s third law: T2=(4π2a3

GM ), where Tis the orbital period, ais the semi-

major axis, Gis the gravitational constant, and Mis the mass of the star.

Given: Semi-major axis, a= 2.5×1011 meters Orbital period, T= 3.6years

Converting the orbital period to seconds: 1year = 365.25 days ≈365.25×24

hours ≈365.25 ×24 ×3600 seconds = 3.1536 ×107seconds

Therefore, T= 3.6×3.1536 ×107= 1.1356 ×108seconds

Step 2: Substituting the known values into Kepler’s third law equation:

T2=4π2a3

(1.1356 ×108)2=4π2(2.5×1011)3

G×M

Step 3: Solving for the mass of the star M:

M=4π2(2.5×1011)3

(1.1356 ×108)2×G

M=4π2(2.5×1011)3

1.2875 ×1016 ×6.67430 ×10−11

M=4π2(1.9531 ×1034)

8.6045 ×105

M≈3.04 ×1029 kg

Therefore, the mass of the star in the distant star system is approximately

3.04 ×1029 kg.

Question 20

Question

Calculate the period of revolution of a planet in the outer region of our solar

system if its semi-major axis is 4.5×109km. Assume that the planet is in a

circular orbit around the Sun.

Solution

Step 1: Recall Kepler’s Third Law, which states that the square of the period of

revolution (T) of a planet is directly proportional to the cube of its semi-major

axis (a), while their ratio remains constant. This relationship can be expressed

as:

T2=k·a3

where kis the constant of proportionality.

Step 2: To find the period of revolution, we need to determine the value of the

constant k. We can do this by using the known values for the Earth’s period of

revolution and semi-major axis. The Earth’s semi-major axis is approximately

1 astronomical unit (AU), which is about 1.496 ×108km, and its period of

revolution is one year.

Step 3: Substituting the values of the Earth’s period and semi-major axis

into Kepler’s Third Law equation, we get:

(1 year)2=k·(1.496 ×108km)3

Step 4: Solve for k:

k=(1 year)2

(1.496 ×108km)3

Step 5: Now that we have the value of k, we can find the period of revolution

for the planet with a semi-major axis of 4.5×109km. Substitute the values

into Kepler’s Third Law:

T2=k·(4.5×109km)3

Step 6: Calculate the period of revolution of the planet:

T=√k·(4.5×109km)3

Question 21

Question

According to Kepler’s Laws of Planetary Motion, what is the relationship be-

tween the square of the orbital period of a planet (T) and the cube of its semi-

major axis (a)? Explain the significance of this relationship in terms of a planet’s

distance from the Sun.

Solution

Step 1: Kepler’s Third Law of Planetary Motion states that the square of the

orbital period of a planet (T) is proportional to the cube of its semi-major axis

(a), mathematically represented as:

T2∝a3

Step 2: This proportionality relationship can be expressed as an equation

with a constant of proportionality:

T2=k·a3

Step 3: By rearranging the equation, we can solve for the constant k:

k=T2

Step 4: The significance of this relationship is that it allows us to compare

the orbital periods and average distances of different planets from the Sun.

Planets farther from the Sun have larger semi-major axes and hence longer

orbital periods, while planets closer to the Sun have shorter orbital periods due

to their smaller semi-major axes. This relationship provides a systematic way to

understand how the distance of a planet from the Sun affects its orbital period.

Question 23

Question

Star X has an orbital period around a black hole of 2.6 years and a semi-major

axis of 14.8 AU. Calculate the mass of the black hole. (Hint: Use Kepler’s Third

Law)

Solution

Step 1: Convert the period of Star X from years to seconds:

1year = 365.25 days ×24 hours/day ×60 minutes/hour ×60 seconds/minute

1year = 31,557,600 seconds

Therefore, the period of Star X in seconds is:

T= 2.6years ×31,557,600 seconds/year

Step 2: Calculate the period of Star X in seconds:

T= 2.6×31,557,600 = 82,009,760 seconds

Step 3: Use Kepler’s Third Law to find the mass of the black hole:

T2=4π2

G(M1+M2)a3

where: T= orbital period of Star X in seconds, G= gravitational constant

(6.67430 ×10−11 m3kg−1s−2), M1= mass of Star X, M2= mass of the black

hole, a= semi-major axis of Star X in meters.

Step 4: Convert the semi-major axis of Star X from AU to meters:

1AU = 1.496 ×1011 m

Therefore, the semi-major axis of Star X in meters is:

a= 14.8AU ×1.496 ×1011 m/AU

Step 5: Calculate the semi-major axis of Star X in meters:

a= 14.8×1.496 ×1011 = 22.2368 ×1011 m

Step 6: Substitute the known values into Kepler’s Third Law equation:

(82009760)2=4π2

6.67430 ×10−11 ×(M1+M2)

22.2368 ×1011

Step 7: Solve for the mass of the black hole, M2:

M2=(4π2×(22.2368 ×1011)3

6.67430 ×10−11 ×(82009760)2−M1)

Step 8: Substitute M1= 0 (since the mass of Star X is negligible compared

to the black hole) and calculate M2.

Question 24

Question

According to Kepler’s Second Law of Planetary Motion, a line segment joining

a planet and the Sun sweeps out equal areas during equal intervals of time.

Suppose a planet takes 60 days to complete an orbit around the Sun with a

semi-major axis of 2 AU. Determine the planet’s orbital speed when it is 1 AU

away from the Sun.

Solution

Step 1: Calculate the orbital period of the planet using Kepler’s Third Law of

Planetary Motion. Step 2: Calculate the planet’s mean orbital speed. Step 3:

Determine the planet’s orbital speed when it is 1 AU away from the Sun.

Step 1: The orbital period (T) of a planet can be calculated using Kepler’s

Third Law of Planetary Motion:

T2=4π2a3

G(M⊙+m)

where: T= orbital period of the planet, a= semi-major axis of the planet’s

orbit, G= gravitational constant, M⊙= mass of the Sun, and m= mass of the

planet (assumed negligible compared to Sun’s mass).

Given that a= 2 AU and the orbital period is 60 days, we can solve for T:

T2=4×(π)2×(2)3

G×M⊙

T2=32π2

T=√32π2

Step 2: The mean orbital speed (v) of a planet can be calculated using the

formula:

v=2πa

where: v= mean orbital speed of the planet, a= semi-major axis of the planet’s

orbit, and T= orbital period of the planet.

Given that a= 2 AU and Twas calculated in Step 1, we can solve for v.

Step 3: To determine the planet’s orbital speed when it is 1 AU away from

the Sun, we will use the concept of conservation of angular momentum. Since

the area of a sector of a circle is constant, the product of orbital radius and

orbital speed remains constant. Thus, when the planet is 1 AU away from the

Sun, its orbital speed would be:

v′=a·v

Question 25

Question

According to Kepler’s Third Law of Planetary Motion, the square of the period

of revolution of a planet around the sun is directly proportional to the cube of

its average distance from the sun. A planet has an average distance from the

sun of 2.5×108miles. If another planet has an average distance from the sun

of 1.2×108miles, what is the ratio of their periods of revolution?

Solution

To find the ratio of the periods of revolution of the two planets, we can use

Kepler’s Third Law of Planetary Motion, which states that the square of the

period of revolution of a planet around the sun is directly proportional to the

cube of its average distance from the sun.

Let T1be the period of revolution of the first planet and T2be the period of

revolution of the second planet.

From Kepler’s Third Law, we have:

(T1

T2)2

=(r1

r2)3

Given that the average distance from the sun for the first planet is r1=

2.5×108miles and for the second planet is r2= 1.2×108miles, we can

substitute these values into the equation:

(T1

T2)2

=(2.5×108

1.2×108)3

(T1

T2)2

=(2.083

1.2)3

(T1

T2)2

= 1.7363

(T1

T2)2

= 5.93

Taking the square root of both sides, we get:

=√5.93 ≈2.435

Therefore, the ratio of their periods of revolution is approximately 2.435 .

k=0.25

0.125 = 2

Step 5: Now, we can find the period of the planet’s orbit when it is at its

perihelion and aphelion distances. Let’s denote the distances as rmin = 0.3AU

(perihelion) and rmax = 0.7AU (aphelion).

Step 6: For the perihelion distance:

(0.5)2= 2 ·(0.3)3

0.25 = 2 ·0.027

0.25 = 0.054

(Not true)

Step 7: For the aphelion distance:

(0.5)2= 2 ·(0.7)3

0.25 = 2 ·0.343

0.25 = 0.686

(Not true)

Step 8: Since both calculations are not true, this indicates that the eccen-

tricity of the orbit must be between 0 and 1. To calculate the eccentricity, we

use the formula e=rmax −rmin

rmax +rmin .

Step 9: Calculate the eccentricity:

e=0.7−0.3

0.7+0.3=0.4

1.0= 0.4

Step 10: Therefore, the eccentricity of the planet’s orbit is 0.4.

Question 2

Question

According to Kepler’s laws of planetary motion, the square of the orbital period

(T) of a planet is proportional to the cube of the semi-major axis (a) of its orbit.

The Earth’s average distance from the Sun is approximately 1 astronomical unit

(AU) and its orbital period is about 1 year. Suppose an exoplanet has an orbital

period of 8 years. What is the semi-major axis of its orbit in AU?

Solution

Step 1: Let’s first write down Kepler’s third law in symbolic form:

T2∝a3

Step 2: Since the Earth’s average distance from the Sun is 1 AU and its orbital

period is 1 year, we can write this relationship as:

(1)2=k(1)3

where kis the constant of proportionality. Step 3: Now, we can set up a similar

equation for the exoplanet with an orbital period of 8 years and an unknown

semi-major axis in AU:

(8)2=k(a)3

Step 4: To find the semi-major axis of the exoplanet’s orbit, we can set up a

ratio using the two equations:

13=82

Step 5: Solving for agives:

a=3

√82= 4 AU

Therefore, the semi-major axis of the exoplanet’s orbit is 4 astronomical units.

Question 3

Question

Suppose a new planet is discovered in a distant solar system with an orbital

period of 200 Earth days. If the semi-major axis of its orbit is measured to be 3

AU (astronomical units), determine the mass of the star around which this new

planet orbits.

(Given: the mass of the sun is 1.989×1030 kg and the gravitational constant

is 6.674 ×10−11 N m2/kg2.)

Solution

Step 1: Calculate the orbital period of the planet in seconds. The formula

relating the orbital period (T) of a planet to its semi-major axis (a) is Kepler’s

third law:

T2=(4π2

G(M1+M2))·a3

where: T= orbital period, G= gravitational constant, M1= mass of the star,

M2= mass of the planet, and a= semi-major axis of the planet’s orbit.

Given T= 200 Earth days and a= 3 AU, we convert these values to seconds

and meters:

T= 200 ×24 ×60 ×60 seconds

a= 3 ×1.496 ×1011 meters

Step 2: Substitute the known values into the formula to solve for the mass

of the star (M1).

T2=(4π2

G(M1+M2))·a3

(200 ×24 ×60 ×60)2=(4π2

6.674 ×10−11 ×M1)·(3 ×1.496 ×1011)3

Step 3: Solve for M1.

M1=4π2·a3

G·T2−M2

Substitute the given values (G,M2) into the equation.

M1=4π2·(3 ×1.496 ×1011)3

6.674 ×10−11 ·(200 ×24 ×60 ×60)2−M2

Calculate M1to find the mass of the star.

Question 4

Question

Suppose a hypothetical planet, Planet X, has an orbital period around its star

of 200 days and an average distance from the star of 0.08 AU. Calculate the

orbital speed of Planet X in km/s. (Hint: Use Kepler’s Third Law of Planetary

Motion)

Solution

Step 1: First, we need to calculate the star’s mass in solar masses. We can use

Kepler’s Third Law of Planetary Motion, which states:

P2=4π2

G(M1+M2)

where: a= average distance between the planet and the star in AU, P=

orbital period of the planet in years, G= gravitational constant (6.67430 ×

10−11 m3kg−1s−2), M1= mass of the star in kg, M2= mass of the planet in

kg.

Step 2: We are given that the average distance, a, is 0.08 AU and the orbital

period, P, is 200 days. Converting the orbital period to years:

P=200 days

365.25 days/year ≈0.547 years

Step 3: Plugging in the known values into Kepler’s Third Law, we have:

0.083

0.5472=4π2

G(M1+M2)

Step 4: Solving for M1+M2:

M1+M2=4π2

G×0.083

0.5472≈2.3×1029 kg

Step 5: Since the mass of the planet is negligible compared to the mass of

the star, we can approximate M1≈M2≈2.3×1029 kg.

Step 6: Next, we can calculate the orbital speed of Planet X using the

formula for orbital speed:

v=√GM1

where: v= orbital speed in m/s, G= gravitational constant, M1= mass of the

star in kg, a= average distance between the planet and the star in m.

Step 7: Converting the average distance, a, to meters:

0.08 AU = 0.08 ×1.496 ×1011 m

Step 8: Plugging in the known values into the orbital speed formula:

v=√6.67430 ×10−11 ×2.3×1029

0.08 ×1.496 ×1011

Step 9: Calculating the orbital speed:

v≈3.63 km/s

Therefore, the orbital speed of Planet X is approximately 3.63 km/s.

Question 5

Question

An asteroid, located at a distance of 2.5 AU from the Sun, takes 12 years to

complete one full orbit. Determine the mass of the Sun using Kepler’s Third

Law of Planetary Motion.

Solution

To find the mass of the Sun using Kepler’s Third Law, we need to relate the

period of the asteroid’s orbit to the distance from the Sun and the mass of the

Sun.

Step 1: Recall Kepler’s Third Law, which states that the square of the

period of revolution of a planet is proportional to the cube of the semi-major

axis of its orbit: T2

=T2

where: - T1and a1are the period and semi-major axis of one orbit, - T2and a2

are the period and semi-major axis of another orbit.

Step 2: We can express the relationship in terms of the asteroid and the

Sun: Given: - T= 12 years is the period of the asteroid’s orbit, - a= 2.5AU is

the distance of the asteroid from the Sun, - M⊙is the mass of the Sun.

We can write the equation as:

a3=4π2

G(M⊙+Masteroid)

where Gis the gravitational constant, Masteroid is the mass of the asteroid which

we assume to be negligible compared to the Sun.

Step 3: Solve for the mass of the Sun: Plugging in the values:

(12 years)2

(2.5AU)3=4π2

G·M⊙

144

(2.5)3=4π2

G·M⊙

144

15.625 =4π2

G·M⊙

9.216 = 4π2

G·M⊙

Step 4: Calculate the mass of the Sun:

M⊙=4π2

9.216 ·G

M⊙=4π2

9.216 ·6.674 ×10−11

M⊙≈39.478

6.168 ×10−10

M⊙≈6.4×1010 kg

Therefore, the mass of the Sun is approximately 6.4×1010 kg.

Question 6

Question

Consider a hypothetical solar system with a star of mass Mat the center. A

planet of mass morbits the star in an elliptical orbit. The semi-major axis of

the orbit is a, and the period of the planet is T. The eccentricity of the orbit is

Assume the planet is at its closest point to the star (perihelion) and its

furthest point from the star (aphelion).

Determine the speed of the planet when it is at perihelion in terms of the

constants and variables provided.

Solution

Step 1: Find the distance from the star to the planet at perihelion (rmin) and

at aphelion (rmax): At perihelion, the distance rmin is equal to the semi-major

axis minus the eccentricity times the semi-major axis:

rmin =a(1 −e)

At aphelion, the distance rmax is equal to the semi-major axis plus the ec-

centricity times the semi-major axis:

rmax =a(1 + e)

Step 2: Using Kepler’s third law, we can express the period of the planet T

in terms of the semi-major axis a:

T2=4π2a3

G(M+m)

where Gis the gravitational constant.

Step 3: Use conservation of angular momentum to relate the angular mo-

mentum of the planet in circular orbit to its angular momentum in elliptical

orbit:

mvr =m√G(M+m)r

At perihelion, the velocity of the planet vmin can be expressed as:

vmin =√G(M+m)(1 + e)

a(1 −e)

Therefore, the speed of the planet at perihelion is √G(M+m)(1 + e)

a(1 −e).

Question 7

Question

According to Kepler’s third law of planetary motion, the square of the period of

revolution of a planet around the sun is proportional to the cube of its average

distance from the sun. Suppose a newly discovered planet has an average dis-

tance from the sun of 2.5 astronomical units (AU). If Earth’s average distance

from the sun is 1 AU and its period of revolution is 1 year, find the period of

revolution of the newly discovered planet.

Solution

Step 1: Calculate the ratio of the cube of the average distance of the newly

discovered planet to that of Earth: Let Tbe the period of revolution of the

newly discovered planet. According to Kepler’s third law:

(Tnew

TEarth )2

=(dnew

dEarth )3

Given that dnew = 2.5AU, dEarth = 1 AU, and TEarth = 1 year:

(Tnew

1)2

=(2.5

1)3

new

1= 2.53

new = 2.53

Tnew =√2.53(years)

Step 2: Calculate the period of revolution of the newly discovered planet:

Tnew =√2.53

Tnew =√15.625

Tnew ≈3.95 years

Therefore, the period of revolution of the newly discovered planet is approx-

imately 3.95 years.

Question 8

Question

For a certain planet orbiting the sun, its average distance from the sun is 1.5

AU and its orbital period is 3 years. Determine the mass of the sun in terms of

Earth’s mass.

Solution

Let’s denote the mass of the sun as M⊙, the distance of the planet from the sun

as r, the orbital period of the planet as T, and the mass of Earth as M⊕. We

will use Kepler’s third law of planetary motion, which states that the square of

the period of any planet is proportional to the cube of the semimajor axis of its

orbit. This can be mathematically represented as:

=T2

Step 1: Given that the average distance from the sun is 1.5 AU and the

orbital period is 3 years, and Earth’s values are r2= 1 AU and T2= 1 year, we

can set up the following equation:

(1.5)3=12

Step 2: Simplify the equation:

3.375 = 1

9= 1

Step 3: Therefore, we now have an equation in terms of masses:

M⊙

(M⊕)3=32

Step 4: Rearrange the equation to solve for the mass of the sun in terms of

Earth’s mass:

M⊙=32

9×(M⊕)3

Step 5: Since we are looking for the mass of the sun in terms of Earth’s

mass, no numerical calculation is needed. The mass of the sun in terms of

Earth’s mass is 32

9M⊕.

Question 9

Question

According to Kepler’s third law of planetary motion, the square of the period

(T) of a planet’s orbit is proportional to the cube of its average distance from

the sun (r). Suppose planet A has a period of 2.5 years and an average distance

from the sun of 1.2 AU (astronomical units), while planet B has a period of

4.8 years and an average distance from the sun of 1.8 AU. Which planet has a

greater average orbital speed in m/s?

Solution

Let’s denote the period as Tin years and the average distance from the sun as

rin astronomical units (AU). We can use Kepler’s third law to find the average

orbital speed for each planet.

Step 1: Calculate the orbital speed of planet A - For planet A: Given

TA= 2.5years and rA= 1.2AU Kepler’s third law states:

A∝r3

We can write this proportionality as:

A=k·r3

Where kis a constant. To find the value of k, we can use the data for planet A:

2.52=k·1.23

6.25 = k·1.728

k≈3.62

Now, we can find the average orbital speed for planet A:

vA=2πrA

vA=2π·1.2

2.5·365.25 ·24 ·3600 m/s

Step 2: Calculate the orbital speed of planet B - For planet B: Given

TB= 4.8years and rB= 1.8AU Using the same constant k≈3.62, we can find

the average orbital speed for planet B:

vB=2π·1.8

4.8·365.25 ·24 ·3600 m/s

Step 3: Compare the orbital speeds Now, we can compare the orbital

speeds of planet A and planet B to determine which planet has a greater average

orbital speed in m/s.

Question 10

Question

According to Kepler’s third law of planetary motion, the square of the period of

revolution of a planet around the Sun is proportional to the cube of its average

distance from the Sun. Suppose a planet has an average distance from the Sun

of 2 astronomical units (AU). If the period of revolution of the planet is 3 years,

what is the average distance of another planet from the Sun if its period of

revolution is 6 years?

Solution

Let T1be the period of revolution for the first planet and D1be its average

distance from the Sun. Similarly, let T2be the period of revolution for the second

planet and D2be its average distance from the Sun. According to Kepler’s third

law, we have the following relationship:

=T2

Given that D1= 2 AU and T1= 3 years, we can solve for D2when T2= 6

years.

Step 1: Substitute the given values into Kepler’s third law equa-

tion.

Plugging in the values we have:

23=62

Step 2: Solve for D2.

Solving for D2:

8=36

Cross multiply to solve for D2:

9D3

2= 288 ⇒D3

2= 32 ⇒D2= 2 3

√16 ⇒D2= 2 ×2 = 4 AU

Therefore, the average distance of the second planet from the Sun is 4 as-

tronomical units.

Question 11

Question

Consider a planet with a semi-major axis of 2.5 AU and an orbital eccentricity

of 0.4. Calculate the closest (perihelion) and farthest (aphelion) distances of

the planet from the Sun in AU.

Solution

Step 1: Calculate the perihelion distance of the planet from the Sun. The

perihelion distance is given by:

rmin =a(1 −e)

where: - rmin is the perihelion distance, - ais the semi-major axis, and - eis

the eccentricity.

Substitute a= 2.5AU and e= 0.4into the formula:

rmin = 2.5(1 −0.4)

rmin = 2.5(0.6)

rmin = 1.5AU

Step 2: Calculate the aphelion distance of the planet from the Sun. The

aphelion distance is given by:

rmax =a(1 + e)

where: - rmax is the aphelion distance, - ais the semi-major axis, and - eis the

eccentricity.

Substitute a= 2.5AU and e= 0.4into the formula:

rmax = 2.5(1 + 0.4)

rmax = 2.5(1.4)

rmax = 3.5AU

Thus, the perihelion distance (closest distance) of the planet from the Sun

is 1.5 AU and the aphelion distance (farthest distance) is 3.5 AU.

Question 12

Question

Consider a planet with a semi-major axis of 2.5 AU orbiting a star with a mass

of 1.5×1030 kg. Calculate the period of the planet’s orbit around the star (in

years) using Kepler’s Third Law.

Solution

To find the period of the planet’s orbit, we can use Kepler’s Third Law, which

states that the square of the period of an orbiting object is proportional to the

cube of its semi-major axis.

Step 1: Convert the semi-major axis to meters. Given: 1 Astronomical

Unit (AU) = 1.496 ×1011 meters.

2.5AU ×1.496 ×1011 meters/AU = 3.74 ×1011 meters

Step 2: Calculate the period using Kepler’s Third Law formula:

T2=4π2

G(M1+M2)a3

Where: T= period of the planet’s orbit G= gravitational constant (6.67430×

10−11 m3kg−1s−2)M1= mass of the star (1.5×1030 kg) M2= mass of the planet

(assumed negligible compared to star) a= semi-major axis of the planet’s orbit

(3.74 ×1011meters)

Plugging in the values:

T2=4π2

6.67430 ×10−11 m3kg−1s−2

(1.5×1030 kg)

3.74 ×1011 m

Step 3: Calculate the period T:

T2=4π2

6.67430 ×10−11 m3kg−1s−2

1.5×1030 kg

(3.74 ×1011 m)3

T2= 1.22372 ×107s2

T=√1.22372 ×107s=3496.40 s≈3.50 years

Therefore, the period of the planet’s orbit around the star is approximately

3.50 years.

Question 13

Question

Consider a hypothetical planetary system where a planet orbits a star in a

perfectly circular orbit. The planet takes 200 days to complete one orbit around

the star. If the distance between the planet and the star is 0.6 AU, determine

the mass of the star.

Solution

Step 1: First, let’s convert the orbital period of the planet from days to years.

Since there are approximately 365.25 days in a year:

Orbital period (years) =200 days

365.25 days/year

Step 2: Calculate the orbital period in years:

Orbital period (years) ≈200

365.25 ≈0.547 years

Step 3: Next, let’s use Kepler’s third law:

Orbital period2=4π2

G(M+Mplanet)×Semi-major axis3

where Gis the gravitational constant, Mis the mass of the star, Mplanet is the

mass of the planet (which we assume to be negligible compared to the star), and

the semi-major axis is the average distance between the star and the planet.

Step 4: Since the planet’s orbit is perfectly circular, the semi-major axis is

the same as the distance between the planet and the star. We are given that

the distance is 0.6 AU, so we can substitute the known values into the equation:

0.5472=4π2

G(M)×(0.6)3

Step 5: Solve for the mass of the star M:

M=4π2×(0.6)3

G×0.5472

Step 6: Now, calculate the mass of the star using the value of the gravita-

tional constant G≈6.67 ×10−11 m3kg−1s−2:

M=4π2×(0.6)3

(6.67 ×10−11)×0.5472

Step 7: Finally, calculate the mass of the star:

M≈4π2×0.216

6.67 ×0.5472≈8.494

2.266 ×10−10 ≈3.741 ×1010 kg

Therefore, the mass of the star in the hypothetical planetary system is ap-

proximately 3.741 ×1010 kg.

Question 14

Question

Explain Kepler’s Third Law of Planetary Motion in detail, and then use it to

calculate the period of a hypothetical planet orbiting a star with a mass of

2×1030 kg, at an average distance of 5 AU.

Solution

Kepler’s Third Law of Planetary Motion states that the square of the period of

revolution (T) of a planet is directly proportional to the cube of the semi-major

axis (a) of its elliptical orbit. Mathematically, it can be expressed as:

T2=k×a3

where kis a constant that depends on the system of measurement used.

Step 1: Calculate the period of the hypothetical planet using Kepler’s Third

Law. Assume k= 1 for simplicity. Since the mass of the star is given as 2×1030

kg and the average distance from the star is 5 AU (1 AU = 1.496 ×1011 m), we

need to find the semi-major axis (a) in meters:

a= 5 ×1.496 ×1011

Step 2: Substitute the value of ainto Kepler’s Third Law equation to solve

for T:

T2= 1 ×(5 ×1.496 ×1011)3

T2= (7.48 ×1011)3

T2= 3.3455 ×1035

Step 3: Take the square root of both sides to find T:

T=√3.3455 ×1035

T≈1.83 ×1017 seconds

Therefore, the period of the hypothetical planet orbiting the star would be

approximately 1.83 ×1017 seconds.

Question 15

Question

According to Kepler’s Laws of Planetary Motion, the square of the orbital period

of a planet is proportional to the cube of its semi-major axis. Suppose a planet

has an orbital period of 8 years and a semi-major axis of 3 AU (astronomical

units). Determine the orbital period of another planet with a semi-major axis

of 6 AU.

Solution

Step 1: Let P1be the orbital period of the first planet, a1be its semi-major

axis, P2be the orbital period of the second planet, and a2be its semi-major

axis. Kepler’s Third Law states that

(P1

P2)2

=(a1

a2)3

Step 2: Substituting the given values, we have

P2)2

=(3

6)3

Step 3: Simplifying the equation gives

P2)2

=(1

2)3

Step 4: Solving for P2yields

P2)2

=(1

64 = P2

Therefore, the orbital period of a planet with a semi-major axis of 6 AU is

64 years.

Question 16

Question

According to Kepler’s third law of planetary motion, the square of the period of

a planet’s orbit is proportional to the cube of the semi-major axis of its orbit.

Consider a planet with a semi-major axis of 2.5 AU (astronomical units) that

takes 6 years to complete one orbit. Calculate the period of another planet in

years if it has a semi-major axis of 3.5 AU.

Solution

Let’s denote the period of the second planet as Tyears. According to Kepler’s

third law,

(T1

T2)2

=(a1

a2)3

where T1= 6 years, a1= 2.5AU, and a2= 3.5AU.

Step 1: Rewrite Kepler’s Third Law

6)2

=(2.5

3.5)3

Step 2: Solve for T

6)2

=(5

7)3

36 =125

343

Step 3: Find T

T2= 36 ×125

343

T2=4500

343

T=√4500

343

T≈4.44 years

Therefore, the period of the second planet is approximately 4.44 years.

Question 17

Question

A planet orbits the Sun in an elliptical path such that its closest distance to the

Sun (perihelion) is 0.3 AU and its farthest distance from the Sun (aphelion) is

0.7 AU. If the time taken by the planet to travel from perihelion to aphelion

is 24 days, determine the time taken by the planet to travel from aphelion to

perihelion.

Solution

Step 1: To find the speed of the planet at perihelion and aphelion, we can use

Kepler’s second law which states that a planet sweeps out equal areas in equal

times. This implies that the planet moves faster when it is closer to the Sun

(perihelion) and slower when it is farther from the Sun (aphelion).

Step 2: Let r1be the distance of the planet from the Sun at perihelion (0.3

AU) and r2be the distance at aphelion (0.7 AU). The area swept out by the

planet at perihelion is A1=1

2r2

1θ1and at aphelion is A2=1

2r2

2θ2, where θ1and

θ2are the angles swept by the planet.

Step 3: Since the areas are equal, we have A1=A2. This gives us 1

2r2

1θ1=

2r2

2θ2. We can simplify this to r2

1θ1

2θ2= 1.

Step 4: We are given that the planet takes 24 days to travel from perihelion

to aphelion, so the time taken to travel from aphelion to perihelion is the same.

This implies that the angle swept in each case is the same, i.e., θ1=θ2.

Step 5: Substituting θ1=θ2into our equation, we get r2

= 1. Substituting

the given values of r1= 0.3AU and r2= 0.7AU, we obtain 0.32

0.72= 1.

Step 6: Solving this equation, we find that the time taken by the planet to

travel from aphelion to perihelion is 24 days as well. Therefore, the time taken

by the planet to travel from aphelion to perihelion is 24 days.

Question 18

Question

An exoplanet is discovered orbiting a distant star with a semi-major axis of 1.5

AU. The star has a mass of 3×1030 kg. Determine the period of the exoplanet’s

orbit using Kepler’s third law of planetary motion.

Solution

Step 1: Determine the period of the exoplanet’s orbit using Kepler’s third law

of planetary motion, which states:

=T2

where T1and T2are the periods of the orbits, and a1and a2are the semi-

major axes of the orbits.

Step 2: Substitute the known values into the equation:

(1.5)3=T2

1.53=T2

3.375 =⇒T2= 3.375

Step 3: Solve for T:

T=√3.375 = 1.837 years

Therefore, the period of the exoplanet’s orbit is approximately 1.837 years.

Question 19

Question

In a distant star system, a planet with a semi-major axis of 2.5×1011 meters

takes 3.6 Earth years to complete one full orbit around its star. Calculate the

mass of the star if the gravitational constant G= 6.67430 ×10−11 m3kg−1s−2.

Solution

Step 1: Firstly, we need to calculate the orbital period of the planet using

Kepler’s third law: T2=(4π2a3

GM ), where Tis the orbital period, ais the semi-

major axis, Gis the gravitational constant, and Mis the mass of the star.

Given: Semi-major axis, a= 2.5×1011 meters Orbital period, T= 3.6years

Converting the orbital period to seconds: 1year = 365.25 days ≈365.25×24

hours ≈365.25 ×24 ×3600 seconds = 3.1536 ×107seconds

Therefore, T= 3.6×3.1536 ×107= 1.1356 ×108seconds

Step 2: Substituting the known values into Kepler’s third law equation:

T2=4π2a3

(1.1356 ×108)2=4π2(2.5×1011)3

G×M

Step 3: Solving for the mass of the star M:

M=4π2(2.5×1011)3

(1.1356 ×108)2×G

M=4π2(2.5×1011)3

1.2875 ×1016 ×6.67430 ×10−11

M=4π2(1.9531 ×1034)

8.6045 ×105

M≈3.04 ×1029 kg

Therefore, the mass of the star in the distant star system is approximately

3.04 ×1029 kg.

Question 20

Question

Calculate the period of revolution of a planet in the outer region of our solar

system if its semi-major axis is 4.5×109km. Assume that the planet is in a

circular orbit around the Sun.

Solution

Step 1: Recall Kepler’s Third Law, which states that the square of the period of

revolution (T) of a planet is directly proportional to the cube of its semi-major

axis (a), while their ratio remains constant. This relationship can be expressed

as:

T2=k·a3

where kis the constant of proportionality.

Step 2: To find the period of revolution, we need to determine the value of the

constant k. We can do this by using the known values for the Earth’s period of

revolution and semi-major axis. The Earth’s semi-major axis is approximately

1 astronomical unit (AU), which is about 1.496 ×108km, and its period of

revolution is one year.

Step 3: Substituting the values of the Earth’s period and semi-major axis

into Kepler’s Third Law equation, we get:

(1 year)2=k·(1.496 ×108km)3

Step 4: Solve for k:

k=(1 year)2

(1.496 ×108km)3

Step 5: Now that we have the value of k, we can find the period of revolution

for the planet with a semi-major axis of 4.5×109km. Substitute the values

into Kepler’s Third Law:

T2=k·(4.5×109km)3

Step 6: Calculate the period of revolution of the planet:

T=√k·(4.5×109km)3

Question 21

Question

According to Kepler’s Laws of Planetary Motion, what is the relationship be-

tween the square of the orbital period of a planet (T) and the cube of its semi-

major axis (a)? Explain the significance of this relationship in terms of a planet’s

distance from the Sun.

Solution

Step 1: Kepler’s Third Law of Planetary Motion states that the square of the

orbital period of a planet (T) is proportional to the cube of its semi-major axis

(a), mathematically represented as:

T2∝a3

Step 2: This proportionality relationship can be expressed as an equation

with a constant of proportionality:

T2=k·a3

Step 3: By rearranging the equation, we can solve for the constant k:

k=T2

Step 4: The significance of this relationship is that it allows us to compare

the orbital periods and average distances of different planets from the Sun.

Planets farther from the Sun have larger semi-major axes and hence longer

orbital periods, while planets closer to the Sun have shorter orbital periods due

to their smaller semi-major axes. This relationship provides a systematic way to

understand how the distance of a planet from the Sun affects its orbital period.

Question 23

Question

Star X has an orbital period around a black hole of 2.6 years and a semi-major

axis of 14.8 AU. Calculate the mass of the black hole. (Hint: Use Kepler’s Third

Law)

Solution

Step 1: Convert the period of Star X from years to seconds:

1year = 365.25 days ×24 hours/day ×60 minutes/hour ×60 seconds/minute

1year = 31,557,600 seconds

Therefore, the period of Star X in seconds is:

T= 2.6years ×31,557,600 seconds/year

Step 2: Calculate the period of Star X in seconds:

T= 2.6×31,557,600 = 82,009,760 seconds

Step 3: Use Kepler’s Third Law to find the mass of the black hole:

T2=4π2

G(M1+M2)a3

where: T= orbital period of Star X in seconds, G= gravitational constant

(6.67430 ×10−11 m3kg−1s−2), M1= mass of Star X, M2= mass of the black

hole, a= semi-major axis of Star X in meters.

Step 4: Convert the semi-major axis of Star X from AU to meters:

1AU = 1.496 ×1011 m

Therefore, the semi-major axis of Star X in meters is:

a= 14.8AU ×1.496 ×1011 m/AU

Step 5: Calculate the semi-major axis of Star X in meters:

a= 14.8×1.496 ×1011 = 22.2368 ×1011 m

Step 6: Substitute the known values into Kepler’s Third Law equation:

(82009760)2=4π2

6.67430 ×10−11 ×(M1+M2)

22.2368 ×1011

Step 7: Solve for the mass of the black hole, M2:

M2=(4π2×(22.2368 ×1011)3

6.67430 ×10−11 ×(82009760)2−M1)

Step 8: Substitute M1= 0 (since the mass of Star X is negligible compared

to the black hole) and calculate M2.

Question 24

Question

According to Kepler’s Second Law of Planetary Motion, a line segment joining

a planet and the Sun sweeps out equal areas during equal intervals of time.

Suppose a planet takes 60 days to complete an orbit around the Sun with a

semi-major axis of 2 AU. Determine the planet’s orbital speed when it is 1 AU

away from the Sun.

Solution

Step 1: Calculate the orbital period of the planet using Kepler’s Third Law of

Planetary Motion. Step 2: Calculate the planet’s mean orbital speed. Step 3:

Determine the planet’s orbital speed when it is 1 AU away from the Sun.

Step 1: The orbital period (T) of a planet can be calculated using Kepler’s

Third Law of Planetary Motion:

T2=4π2a3

G(M⊙+m)

where: T= orbital period of the planet, a= semi-major axis of the planet’s

orbit, G= gravitational constant, M⊙= mass of the Sun, and m= mass of the

planet (assumed negligible compared to Sun’s mass).

Given that a= 2 AU and the orbital period is 60 days, we can solve for T:

T2=4×(π)2×(2)3

G×M⊙

T2=32π2

T=√32π2

Step 2: The mean orbital speed (v) of a planet can be calculated using the

formula:

v=2πa

where: v= mean orbital speed of the planet, a= semi-major axis of the planet’s

orbit, and T= orbital period of the planet.

Given that a= 2 AU and Twas calculated in Step 1, we can solve for v.

Step 3: To determine the planet’s orbital speed when it is 1 AU away from

the Sun, we will use the concept of conservation of angular momentum. Since

the area of a sector of a circle is constant, the product of orbital radius and

orbital speed remains constant. Thus, when the planet is 1 AU away from the

Sun, its orbital speed would be:

v′=a·v

Question 25

Question

According to Kepler’s Third Law of Planetary Motion, the square of the period

of revolution of a planet around the sun is directly proportional to the cube of

its average distance from the sun. A planet has an average distance from the

sun of 2.5×108miles. If another planet has an average distance from the sun

of 1.2×108miles, what is the ratio of their periods of revolution?

Solution

To find the ratio of the periods of revolution of the two planets, we can use

Kepler’s Third Law of Planetary Motion, which states that the square of the

period of revolution of a planet around the sun is directly proportional to the

cube of its average distance from the sun.

Let T1be the period of revolution of the first planet and T2be the period of

revolution of the second planet.

From Kepler’s Third Law, we have:

(T1

T2)2

=(r1

r2)3

Given that the average distance from the sun for the first planet is r1=

2.5×108miles and for the second planet is r2= 1.2×108miles, we can

substitute these values into the equation:

(T1

T2)2

=(2.5×108

1.2×108)3

(T1

T2)2

=(2.083

1.2)3

(T1

T2)2

= 1.7363

(T1

T2)2

= 5.93

Taking the square root of both sides, we get:

=√5.93 ≈2.435

Therefore, the ratio of their periods of revolution is approximately 2.435 .

k=0.25

0.125 = 2

Step 5: Now, we can find the period of the planet’s orbit when it is at its

perihelion and aphelion distances. Let’s denote the distances as rmin = 0.3AU

(perihelion) and rmax = 0.7AU (aphelion).

Step 6: For the perihelion distance:

(0.5)2= 2 ·(0.3)3

0.25 = 2 ·0.027

0.25 = 0.054

(Not true)

Step 7: For the aphelion distance:

(0.5)2= 2 ·(0.7)3

0.25 = 2 ·0.343

0.25 = 0.686

(Not true)

Step 8: Since both calculations are not true, this indicates that the eccen-

tricity of the orbit must be between 0 and 1. To calculate the eccentricity, we

use the formula e=rmax −rmin

rmax +rmin .

Step 9: Calculate the eccentricity:

e=0.7−0.3

0.7+0.3=0.4

1.0= 0.4

Step 10: Therefore, the eccentricity of the planet’s orbit is 0.4.

Question 2

Question

According to Kepler’s laws of planetary motion, the square of the orbital period

(T) of a planet is proportional to the cube of the semi-major axis (a) of its orbit.

The Earth’s average distance from the Sun is approximately 1 astronomical unit

(AU) and its orbital period is about 1 year. Suppose an exoplanet has an orbital

period of 8 years. What is the semi-major axis of its orbit in AU?

Solution

Step 1: Let’s first write down Kepler’s third law in symbolic form:

T2∝a3

Step 2: Since the Earth’s average distance from the Sun is 1 AU and its orbital

period is 1 year, we can write this relationship as:

(1)2=k(1)3

where kis the constant of proportionality. Step 3: Now, we can set up a similar

equation for the exoplanet with an orbital period of 8 years and an unknown

semi-major axis in AU:

(8)2=k(a)3

Step 4: To find the semi-major axis of the exoplanet’s orbit, we can set up a

ratio using the two equations:

13=82

Step 5: Solving for agives:

a=3

√82= 4 AU

Therefore, the semi-major axis of the exoplanet’s orbit is 4 astronomical units.

Question 3

Question

Suppose a new planet is discovered in a distant solar system with an orbital

period of 200 Earth days. If the semi-major axis of its orbit is measured to be 3

AU (astronomical units), determine the mass of the star around which this new

planet orbits.

(Given: the mass of the sun is 1.989×1030 kg and the gravitational constant

is 6.674 ×10−11 N m2/kg2.)

Solution

Step 1: Calculate the orbital period of the planet in seconds. The formula

relating the orbital period (T) of a planet to its semi-major axis (a) is Kepler’s

third law:

T2=(4π2

G(M1+M2))·a3

where: T= orbital period, G= gravitational constant, M1= mass of the star,

M2= mass of the planet, and a= semi-major axis of the planet’s orbit.

Given T= 200 Earth days and a= 3 AU, we convert these values to seconds

and meters:

T= 200 ×24 ×60 ×60 seconds

a= 3 ×1.496 ×1011 meters

Step 2: Substitute the known values into the formula to solve for the mass

of the star (M1).

T2=(4π2

G(M1+M2))·a3

(200 ×24 ×60 ×60)2=(4π2

6.674 ×10−11 ×M1)·(3 ×1.496 ×1011)3

Step 3: Solve for M1.

M1=4π2·a3

G·T2−M2

Substitute the given values (G,M2) into the equation.

M1=4π2·(3 ×1.496 ×1011)3

6.674 ×10−11 ·(200 ×24 ×60 ×60)2−M2

Calculate M1to find the mass of the star.

Question 4

Question

Suppose a hypothetical planet, Planet X, has an orbital period around its star

of 200 days and an average distance from the star of 0.08 AU. Calculate the

orbital speed of Planet X in km/s. (Hint: Use Kepler’s Third Law of Planetary

Motion)

Solution

Step 1: First, we need to calculate the star’s mass in solar masses. We can use

Kepler’s Third Law of Planetary Motion, which states:

P2=4π2

G(M1+M2)

where: a= average distance between the planet and the star in AU, P=

orbital period of the planet in years, G= gravitational constant (6.67430 ×

10−11 m3kg−1s−2), M1= mass of the star in kg, M2= mass of the planet in

kg.

Step 2: We are given that the average distance, a, is 0.08 AU and the orbital

period, P, is 200 days. Converting the orbital period to years:

P=200 days

365.25 days/year ≈0.547 years

Step 3: Plugging in the known values into Kepler’s Third Law, we have:

0.083

0.5472=4π2

G(M1+M2)

Step 4: Solving for M1+M2:

M1+M2=4π2

G×0.083

0.5472≈2.3×1029 kg

Step 5: Since the mass of the planet is negligible compared to the mass of

the star, we can approximate M1≈M2≈2.3×1029 kg.

Step 6: Next, we can calculate the orbital speed of Planet X using the

formula for orbital speed:

v=√GM1

where: v= orbital speed in m/s, G= gravitational constant, M1= mass of the

star in kg, a= average distance between the planet and the star in m.

Step 7: Converting the average distance, a, to meters:

0.08 AU = 0.08 ×1.496 ×1011 m

Step 8: Plugging in the known values into the orbital speed formula:

v=√6.67430 ×10−11 ×2.3×1029

0.08 ×1.496 ×1011

Step 9: Calculating the orbital speed:

v≈3.63 km/s

Therefore, the orbital speed of Planet X is approximately 3.63 km/s.

Question 5

Question

An asteroid, located at a distance of 2.5 AU from the Sun, takes 12 years to

complete one full orbit. Determine the mass of the Sun using Kepler’s Third

Law of Planetary Motion.

Solution

To find the mass of the Sun using Kepler’s Third Law, we need to relate the

period of the asteroid’s orbit to the distance from the Sun and the mass of the

Sun.

Step 1: Recall Kepler’s Third Law, which states that the square of the

period of revolution of a planet is proportional to the cube of the semi-major

axis of its orbit: T2

=T2

where: - T1and a1are the period and semi-major axis of one orbit, - T2and a2

are the period and semi-major axis of another orbit.

Step 2: We can express the relationship in terms of the asteroid and the

Sun: Given: - T= 12 years is the period of the asteroid’s orbit, - a= 2.5AU is

the distance of the asteroid from the Sun, - M⊙is the mass of the Sun.

We can write the equation as:

a3=4π2

G(M⊙+Masteroid)

where Gis the gravitational constant, Masteroid is the mass of the asteroid which

we assume to be negligible compared to the Sun.

Step 3: Solve for the mass of the Sun: Plugging in the values:

(12 years)2

(2.5AU)3=4π2

G·M⊙

144

(2.5)3=4π2

G·M⊙

144

15.625 =4π2

G·M⊙

9.216 = 4π2

G·M⊙

Step 4: Calculate the mass of the Sun:

M⊙=4π2

9.216 ·G

M⊙=4π2

9.216 ·6.674 ×10−11

M⊙≈39.478

6.168 ×10−10

M⊙≈6.4×1010 kg

Therefore, the mass of the Sun is approximately 6.4×1010 kg.

Question 6

Question

Consider a hypothetical solar system with a star of mass Mat the center. A

planet of mass morbits the star in an elliptical orbit. The semi-major axis of

the orbit is a, and the period of the planet is T. The eccentricity of the orbit is

Assume the planet is at its closest point to the star (perihelion) and its

furthest point from the star (aphelion).

Determine the speed of the planet when it is at perihelion in terms of the

constants and variables provided.

Solution

Step 1: Find the distance from the star to the planet at perihelion (rmin) and

at aphelion (rmax): At perihelion, the distance rmin is equal to the semi-major

axis minus the eccentricity times the semi-major axis:

rmin =a(1 −e)

At aphelion, the distance rmax is equal to the semi-major axis plus the ec-

centricity times the semi-major axis:

rmax =a(1 + e)

Step 2: Using Kepler’s third law, we can express the period of the planet T

in terms of the semi-major axis a:

T2=4π2a3

G(M+m)

where Gis the gravitational constant.

Step 3: Use conservation of angular momentum to relate the angular mo-

mentum of the planet in circular orbit to its angular momentum in elliptical

orbit:

mvr =m√G(M+m)r

At perihelion, the velocity of the planet vmin can be expressed as:

vmin =√G(M+m)(1 + e)

a(1 −e)

Therefore, the speed of the planet at perihelion is √G(M+m)(1 + e)

a(1 −e).

Question 7

Question

According to Kepler’s third law of planetary motion, the square of the period of

revolution of a planet around the sun is proportional to the cube of its average

distance from the sun. Suppose a newly discovered planet has an average dis-

tance from the sun of 2.5 astronomical units (AU). If Earth’s average distance

from the sun is 1 AU and its period of revolution is 1 year, find the period of

revolution of the newly discovered planet.

Solution

Step 1: Calculate the ratio of the cube of the average distance of the newly

discovered planet to that of Earth: Let Tbe the period of revolution of the

newly discovered planet. According to Kepler’s third law:

(Tnew

TEarth )2

=(dnew

dEarth )3

Given that dnew = 2.5AU, dEarth = 1 AU, and TEarth = 1 year:

(Tnew

1)2

=(2.5

1)3

new

1= 2.53

new = 2.53

Tnew =√2.53(years)

Step 2: Calculate the period of revolution of the newly discovered planet:

Tnew =√2.53

Tnew =√15.625

Tnew ≈3.95 years

Therefore, the period of revolution of the newly discovered planet is approx-

imately 3.95 years.

Question 8

Question

For a certain planet orbiting the sun, its average distance from the sun is 1.5

AU and its orbital period is 3 years. Determine the mass of the sun in terms of

Earth’s mass.

Solution

Let’s denote the mass of the sun as M⊙, the distance of the planet from the sun

as r, the orbital period of the planet as T, and the mass of Earth as M⊕. We

will use Kepler’s third law of planetary motion, which states that the square of

the period of any planet is proportional to the cube of the semimajor axis of its

orbit. This can be mathematically represented as:

=T2

Step 1: Given that the average distance from the sun is 1.5 AU and the

orbital period is 3 years, and Earth’s values are r2= 1 AU and T2= 1 year, we

can set up the following equation:

(1.5)3=12

Step 2: Simplify the equation:

3.375 = 1

9= 1

Step 3: Therefore, we now have an equation in terms of masses:

M⊙

(M⊕)3=32

Step 4: Rearrange the equation to solve for the mass of the sun in terms of

Earth’s mass:

M⊙=32

9×(M⊕)3

Step 5: Since we are looking for the mass of the sun in terms of Earth’s

mass, no numerical calculation is needed. The mass of the sun in terms of

Earth’s mass is 32

9M⊕.

Question 9

Question

According to Kepler’s third law of planetary motion, the square of the period

(T) of a planet’s orbit is proportional to the cube of its average distance from

the sun (r). Suppose planet A has a period of 2.5 years and an average distance

from the sun of 1.2 AU (astronomical units), while planet B has a period of

4.8 years and an average distance from the sun of 1.8 AU. Which planet has a

greater average orbital speed in m/s?

Solution

Let’s denote the period as Tin years and the average distance from the sun as

rin astronomical units (AU). We can use Kepler’s third law to find the average

orbital speed for each planet.

Step 1: Calculate the orbital speed of planet A - For planet A: Given

TA= 2.5years and rA= 1.2AU Kepler’s third law states:

A∝r3

We can write this proportionality as:

A=k·r3

Where kis a constant. To find the value of k, we can use the data for planet A:

2.52=k·1.23

6.25 = k·1.728

k≈3.62

Now, we can find the average orbital speed for planet A:

vA=2πrA

vA=2π·1.2

2.5·365.25 ·24 ·3600 m/s

Step 2: Calculate the orbital speed of planet B - For planet B: Given

TB= 4.8years and rB= 1.8AU Using the same constant k≈3.62, we can find

the average orbital speed for planet B:

vB=2π·1.8

4.8·365.25 ·24 ·3600 m/s

Step 3: Compare the orbital speeds Now, we can compare the orbital

speeds of planet A and planet B to determine which planet has a greater average

orbital speed in m/s.

Question 10

Question

According to Kepler’s third law of planetary motion, the square of the period of

revolution of a planet around the Sun is proportional to the cube of its average

distance from the Sun. Suppose a planet has an average distance from the Sun

of 2 astronomical units (AU). If the period of revolution of the planet is 3 years,

what is the average distance of another planet from the Sun if its period of

revolution is 6 years?

Solution

Let T1be the period of revolution for the first planet and D1be its average

distance from the Sun. Similarly, let T2be the period of revolution for the second

planet and D2be its average distance from the Sun. According to Kepler’s third

law, we have the following relationship:

=T2

Given that D1= 2 AU and T1= 3 years, we can solve for D2when T2= 6

years.

Step 1: Substitute the given values into Kepler’s third law equa-

tion.

Plugging in the values we have:

23=62

Step 2: Solve for D2.

Solving for D2:

8=36

Cross multiply to solve for D2:

9D3

2= 288 ⇒D3

2= 32 ⇒D2= 2 3

√16 ⇒D2= 2 ×2 = 4 AU

Therefore, the average distance of the second planet from the Sun is 4 as-

tronomical units.

Question 11

Question

Consider a planet with a semi-major axis of 2.5 AU and an orbital eccentricity

of 0.4. Calculate the closest (perihelion) and farthest (aphelion) distances of

the planet from the Sun in AU.

Solution

Step 1: Calculate the perihelion distance of the planet from the Sun. The

perihelion distance is given by:

rmin =a(1 −e)

where: - rmin is the perihelion distance, - ais the semi-major axis, and - eis

the eccentricity.

Substitute a= 2.5AU and e= 0.4into the formula:

rmin = 2.5(1 −0.4)

rmin = 2.5(0.6)

rmin = 1.5AU

Step 2: Calculate the aphelion distance of the planet from the Sun. The

aphelion distance is given by:

rmax =a(1 + e)

where: - rmax is the aphelion distance, - ais the semi-major axis, and - eis the

eccentricity.

Substitute a= 2.5AU and e= 0.4into the formula:

rmax = 2.5(1 + 0.4)

rmax = 2.5(1.4)

rmax = 3.5AU

Thus, the perihelion distance (closest distance) of the planet from the Sun

is 1.5 AU and the aphelion distance (farthest distance) is 3.5 AU.

Question 12

Question

Consider a planet with a semi-major axis of 2.5 AU orbiting a star with a mass

of 1.5×1030 kg. Calculate the period of the planet’s orbit around the star (in

years) using Kepler’s Third Law.

Solution

To find the period of the planet’s orbit, we can use Kepler’s Third Law, which

states that the square of the period of an orbiting object is proportional to the

cube of its semi-major axis.

Step 1: Convert the semi-major axis to meters. Given: 1 Astronomical

Unit (AU) = 1.496 ×1011 meters.

2.5AU ×1.496 ×1011 meters/AU = 3.74 ×1011 meters

Step 2: Calculate the period using Kepler’s Third Law formula:

T2=4π2

G(M1+M2)a3

Where: T= period of the planet’s orbit G= gravitational constant (6.67430×

10−11 m3kg−1s−2)M1= mass of the star (1.5×1030 kg) M2= mass of the planet

(assumed negligible compared to star) a= semi-major axis of the planet’s orbit

(3.74 ×1011meters)

Plugging in the values:

T2=4π2

6.67430 ×10−11 m3kg−1s−2

(1.5×1030 kg)

3.74 ×1011 m

Step 3: Calculate the period T:

T2=4π2

6.67430 ×10−11 m3kg−1s−2

1.5×1030 kg

(3.74 ×1011 m)3

T2= 1.22372 ×107s2

T=√1.22372 ×107s=3496.40 s≈3.50 years

Therefore, the period of the planet’s orbit around the star is approximately

3.50 years.

Question 13

Question

Consider a hypothetical planetary system where a planet orbits a star in a

perfectly circular orbit. The planet takes 200 days to complete one orbit around

the star. If the distance between the planet and the star is 0.6 AU, determine

the mass of the star.

Solution

Step 1: First, let’s convert the orbital period of the planet from days to years.

Since there are approximately 365.25 days in a year:

Orbital period (years) =200 days

365.25 days/year

Step 2: Calculate the orbital period in years:

Orbital period (years) ≈200

365.25 ≈0.547 years

Step 3: Next, let’s use Kepler’s third law:

Orbital period2=4π2

G(M+Mplanet)×Semi-major axis3

where Gis the gravitational constant, Mis the mass of the star, Mplanet is the

mass of the planet (which we assume to be negligible compared to the star), and

the semi-major axis is the average distance between the star and the planet.

Step 4: Since the planet’s orbit is perfectly circular, the semi-major axis is

the same as the distance between the planet and the star. We are given that

the distance is 0.6 AU, so we can substitute the known values into the equation:

0.5472=4π2

G(M)×(0.6)3

Step 5: Solve for the mass of the star M:

M=4π2×(0.6)3

G×0.5472

Step 6: Now, calculate the mass of the star using the value of the gravita-

tional constant G≈6.67 ×10−11 m3kg−1s−2:

M=4π2×(0.6)3

(6.67 ×10−11)×0.5472

Step 7: Finally, calculate the mass of the star:

M≈4π2×0.216

6.67 ×0.5472≈8.494

2.266 ×10−10 ≈3.741 ×1010 kg

Therefore, the mass of the star in the hypothetical planetary system is ap-

proximately 3.741 ×1010 kg.

Question 14

Question

Explain Kepler’s Third Law of Planetary Motion in detail, and then use it to

calculate the period of a hypothetical planet orbiting a star with a mass of

2×1030 kg, at an average distance of 5 AU.

Solution

Kepler’s Third Law of Planetary Motion states that the square of the period of

revolution (T) of a planet is directly proportional to the cube of the semi-major

axis (a) of its elliptical orbit. Mathematically, it can be expressed as:

T2=k×a3

where kis a constant that depends on the system of measurement used.

Step 1: Calculate the period of the hypothetical planet using Kepler’s Third

Law. Assume k= 1 for simplicity. Since the mass of the star is given as 2×1030

kg and the average distance from the star is 5 AU (1 AU = 1.496 ×1011 m), we

need to find the semi-major axis (a) in meters:

a= 5 ×1.496 ×1011

Step 2: Substitute the value of ainto Kepler’s Third Law equation to solve

for T:

T2= 1 ×(5 ×1.496 ×1011)3

T2= (7.48 ×1011)3

T2= 3.3455 ×1035

Step 3: Take the square root of both sides to find T:

T=√3.3455 ×1035

T≈1.83 ×1017 seconds

Therefore, the period of the hypothetical planet orbiting the star would be

approximately 1.83 ×1017 seconds.

Question 15

Question

According to Kepler’s Laws of Planetary Motion, the square of the orbital period

of a planet is proportional to the cube of its semi-major axis. Suppose a planet

has an orbital period of 8 years and a semi-major axis of 3 AU (astronomical

units). Determine the orbital period of another planet with a semi-major axis

of 6 AU.

Solution

Step 1: Let P1be the orbital period of the first planet, a1be its semi-major

axis, P2be the orbital period of the second planet, and a2be its semi-major

axis. Kepler’s Third Law states that

(P1

P2)2

=(a1

a2)3

Step 2: Substituting the given values, we have

P2)2

=(3

6)3

Step 3: Simplifying the equation gives

P2)2

=(1

2)3

Step 4: Solving for P2yields

P2)2

=(1

64 = P2

Therefore, the orbital period of a planet with a semi-major axis of 6 AU is

64 years.

Question 16

Question

According to Kepler’s third law of planetary motion, the square of the period of

a planet’s orbit is proportional to the cube of the semi-major axis of its orbit.

Consider a planet with a semi-major axis of 2.5 AU (astronomical units) that

takes 6 years to complete one orbit. Calculate the period of another planet in

years if it has a semi-major axis of 3.5 AU.

Solution

Let’s denote the period of the second planet as Tyears. According to Kepler’s

third law,

(T1

T2)2

=(a1

a2)3

where T1= 6 years, a1= 2.5AU, and a2= 3.5AU.

Step 1: Rewrite Kepler’s Third Law

6)2

=(2.5

3.5)3

Step 2: Solve for T

6)2

=(5

7)3

36 =125

343

Step 3: Find T

T2= 36 ×125

343

T2=4500

343

T=√4500

343

T≈4.44 years

Therefore, the period of the second planet is approximately 4.44 years.

Question 17

Question

A planet orbits the Sun in an elliptical path such that its closest distance to the

Sun (perihelion) is 0.3 AU and its farthest distance from the Sun (aphelion) is

0.7 AU. If the time taken by the planet to travel from perihelion to aphelion

is 24 days, determine the time taken by the planet to travel from aphelion to

perihelion.

Solution

Step 1: To find the speed of the planet at perihelion and aphelion, we can use

Kepler’s second law which states that a planet sweeps out equal areas in equal

times. This implies that the planet moves faster when it is closer to the Sun

(perihelion) and slower when it is farther from the Sun (aphelion).

Step 2: Let r1be the distance of the planet from the Sun at perihelion (0.3

AU) and r2be the distance at aphelion (0.7 AU). The area swept out by the

planet at perihelion is A1=1

2r2

1θ1and at aphelion is A2=1

2r2

2θ2, where θ1and

θ2are the angles swept by the planet.

Step 3: Since the areas are equal, we have A1=A2. This gives us 1

2r2

1θ1=

2r2

2θ2. We can simplify this to r2

1θ1

2θ2= 1.

Step 4: We are given that the planet takes 24 days to travel from perihelion

to aphelion, so the time taken to travel from aphelion to perihelion is the same.

This implies that the angle swept in each case is the same, i.e., θ1=θ2.

Step 5: Substituting θ1=θ2into our equation, we get r2

= 1. Substituting

the given values of r1= 0.3AU and r2= 0.7AU, we obtain 0.32

0.72= 1.

Step 6: Solving this equation, we find that the time taken by the planet to

travel from aphelion to perihelion is 24 days as well. Therefore, the time taken

by the planet to travel from aphelion to perihelion is 24 days.

Question 18

Question

An exoplanet is discovered orbiting a distant star with a semi-major axis of 1.5

AU. The star has a mass of 3×1030 kg. Determine the period of the exoplanet’s

orbit using Kepler’s third law of planetary motion.

Solution

Step 1: Determine the period of the exoplanet’s orbit using Kepler’s third law

of planetary motion, which states:

=T2

where T1and T2are the periods of the orbits, and a1and a2are the semi-

major axes of the orbits.

Step 2: Substitute the known values into the equation:

(1.5)3=T2

1.53=T2

3.375 =⇒T2= 3.375

Step 3: Solve for T:

T=√3.375 = 1.837 years

Therefore, the period of the exoplanet’s orbit is approximately 1.837 years.

Question 19

Question

In a distant star system, a planet with a semi-major axis of 2.5×1011 meters

takes 3.6 Earth years to complete one full orbit around its star. Calculate the

mass of the star if the gravitational constant G= 6.67430 ×10−11 m3kg−1s−2.

Solution

Step 1: Firstly, we need to calculate the orbital period of the planet using

Kepler’s third law: T2=(4π2a3

GM ), where Tis the orbital period, ais the semi-

major axis, Gis the gravitational constant, and Mis the mass of the star.

Given: Semi-major axis, a= 2.5×1011 meters Orbital period, T= 3.6years

Converting the orbital period to seconds: 1year = 365.25 days ≈365.25×24

hours ≈365.25 ×24 ×3600 seconds = 3.1536 ×107seconds

Therefore, T= 3.6×3.1536 ×107= 1.1356 ×108seconds

Step 2: Substituting the known values into Kepler’s third law equation:

T2=4π2a3

(1.1356 ×108)2=4π2(2.5×1011)3

G×M

Step 3: Solving for the mass of the star M:

M=4π2(2.5×1011)3

(1.1356 ×108)2×G

M=4π2(2.5×1011)3

1.2875 ×1016 ×6.67430 ×10−11

M=4π2(1.9531 ×1034)

8.6045 ×105

M≈3.04 ×1029 kg

Therefore, the mass of the star in the distant star system is approximately

3.04 ×1029 kg.

Question 20

Question

Calculate the period of revolution of a planet in the outer region of our solar

system if its semi-major axis is 4.5×109km. Assume that the planet is in a

circular orbit around the Sun.

Solution

Step 1: Recall Kepler’s Third Law, which states that the square of the period of

revolution (T) of a planet is directly proportional to the cube of its semi-major

axis (a), while their ratio remains constant. This relationship can be expressed

as:

T2=k·a3

where kis the constant of proportionality.

Step 2: To find the period of revolution, we need to determine the value of the

constant k. We can do this by using the known values for the Earth’s period of

revolution and semi-major axis. The Earth’s semi-major axis is approximately

1 astronomical unit (AU), which is about 1.496 ×108km, and its period of

revolution is one year.

Step 3: Substituting the values of the Earth’s period and semi-major axis

into Kepler’s Third Law equation, we get:

(1 year)2=k·(1.496 ×108km)3

Step 4: Solve for k:

k=(1 year)2

(1.496 ×108km)3

Step 5: Now that we have the value of k, we can find the period of revolution

for the planet with a semi-major axis of 4.5×109km. Substitute the values

into Kepler’s Third Law:

T2=k·(4.5×109km)3

Step 6: Calculate the period of revolution of the planet:

T=√k·(4.5×109km)3

Question 21

Question

According to Kepler’s Laws of Planetary Motion, what is the relationship be-

tween the square of the orbital period of a planet (T) and the cube of its semi-

major axis (a)? Explain the significance of this relationship in terms of a planet’s

distance from the Sun.

Solution

Step 1: Kepler’s Third Law of Planetary Motion states that the square of the

orbital period of a planet (T) is proportional to the cube of its semi-major axis

(a), mathematically represented as:

T2∝a3

Step 2: This proportionality relationship can be expressed as an equation

with a constant of proportionality:

T2=k·a3

Step 3: By rearranging the equation, we can solve for the constant k:

k=T2

Step 4: The significance of this relationship is that it allows us to compare

the orbital periods and average distances of different planets from the Sun.

Planets farther from the Sun have larger semi-major axes and hence longer

orbital periods, while planets closer to the Sun have shorter orbital periods due

to their smaller semi-major axes. This relationship provides a systematic way to

understand how the distance of a planet from the Sun affects its orbital period.

Question 23

Question

Star X has an orbital period around a black hole of 2.6 years and a semi-major

axis of 14.8 AU. Calculate the mass of the black hole. (Hint: Use Kepler’s Third

Law)

Solution

Step 1: Convert the period of Star X from years to seconds:

1year = 365.25 days ×24 hours/day ×60 minutes/hour ×60 seconds/minute

1year = 31,557,600 seconds

Therefore, the period of Star X in seconds is:

T= 2.6years ×31,557,600 seconds/year

Step 2: Calculate the period of Star X in seconds:

T= 2.6×31,557,600 = 82,009,760 seconds

Step 3: Use Kepler’s Third Law to find the mass of the black hole:

T2=4π2

G(M1+M2)a3

where: T= orbital period of Star X in seconds, G= gravitational constant

(6.67430 ×10−11 m3kg−1s−2), M1= mass of Star X, M2= mass of the black

hole, a= semi-major axis of Star X in meters.

Step 4: Convert the semi-major axis of Star X from AU to meters:

1AU = 1.496 ×1011 m

Therefore, the semi-major axis of Star X in meters is:

a= 14.8AU ×1.496 ×1011 m/AU

Step 5: Calculate the semi-major axis of Star X in meters:

a= 14.8×1.496 ×1011 = 22.2368 ×1011 m

Step 6: Substitute the known values into Kepler’s Third Law equation:

(82009760)2=4π2

6.67430 ×10−11 ×(M1+M2)

22.2368 ×1011

Step 7: Solve for the mass of the black hole, M2:

M2=(4π2×(22.2368 ×1011)3

6.67430 ×10−11 ×(82009760)2−M1)

Step 8: Substitute M1= 0 (since the mass of Star X is negligible compared

to the black hole) and calculate M2.

Question 24

Question

According to Kepler’s Second Law of Planetary Motion, a line segment joining

a planet and the Sun sweeps out equal areas during equal intervals of time.

Suppose a planet takes 60 days to complete an orbit around the Sun with a

semi-major axis of 2 AU. Determine the planet’s orbital speed when it is 1 AU

away from the Sun.

Solution

Step 1: Calculate the orbital period of the planet using Kepler’s Third Law of

Planetary Motion. Step 2: Calculate the planet’s mean orbital speed. Step 3:

Determine the planet’s orbital speed when it is 1 AU away from the Sun.

Step 1: The orbital period (T) of a planet can be calculated using Kepler’s

Third Law of Planetary Motion:

T2=4π2a3

G(M⊙+m)

where: T= orbital period of the planet, a= semi-major axis of the planet’s

orbit, G= gravitational constant, M⊙= mass of the Sun, and m= mass of the

planet (assumed negligible compared to Sun’s mass).

Given that a= 2 AU and the orbital period is 60 days, we can solve for T:

T2=4×(π)2×(2)3

G×M⊙

T2=32π2

T=√32π2

Step 2: The mean orbital speed (v) of a planet can be calculated using the

formula:

v=2πa

where: v= mean orbital speed of the planet, a= semi-major axis of the planet’s

orbit, and T= orbital period of the planet.

Given that a= 2 AU and Twas calculated in Step 1, we can solve for v.

Step 3: To determine the planet’s orbital speed when it is 1 AU away from

the Sun, we will use the concept of conservation of angular momentum. Since

the area of a sector of a circle is constant, the product of orbital radius and

orbital speed remains constant. Thus, when the planet is 1 AU away from the

Sun, its orbital speed would be:

v′=a·v

Question 25

Question

According to Kepler’s Third Law of Planetary Motion, the square of the period

of revolution of a planet around the sun is directly proportional to the cube of

its average distance from the sun. A planet has an average distance from the

sun of 2.5×108miles. If another planet has an average distance from the sun

of 1.2×108miles, what is the ratio of their periods of revolution?

Solution

To find the ratio of the periods of revolution of the two planets, we can use

Kepler’s Third Law of Planetary Motion, which states that the square of the

period of revolution of a planet around the sun is directly proportional to the

cube of its average distance from the sun.

Let T1be the period of revolution of the first planet and T2be the period of

revolution of the second planet.

From Kepler’s Third Law, we have:

(T1

T2)2

=(r1

r2)3

Given that the average distance from the sun for the first planet is r1=

2.5×108miles and for the second planet is r2= 1.2×108miles, we can

substitute these values into the equation:

(T1

T2)2

=(2.5×108

1.2×108)3

(T1

T2)2

=(2.083

1.2)3

(T1

T2)2

= 1.7363

(T1

T2)2

= 5.93

Taking the square root of both sides, we get:

=√5.93 ≈2.435

Therefore, the ratio of their periods of revolution is approximately 2.435 .

PHSC 121 - INTRODUCTION TO ASTRONOMY Kepler's Laws of Planetary Motion Question Bank Set 3

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