PHSC 121 - INTRODUCTION TO ASTRONOMY Kepler's Laws of Planetary Motion Question Bank Set 6

PHSC 121 - INTRODUCTION TO

ASTRONOMY - Kepler’s Laws of

Planetary Motion

Question Bank - Set 6

Liberty University

Question 1

Question

An exoplanet orbits a star in an elliptical orbit such that the closest approach to

the star is 0.25 astronomical units (AU) and the farthest distance from the star

is 0.85 AU. If the exoplanet takes 400 days to complete one full orbit, determine

the average speed of the exoplanet in its orbit.

Solution

Step 1: First, we need to find the semi-major axis of the elliptical orbit of the

exoplanet. Given that the closest approach to the star is 0.25 AU (perihelion

distance) and the farthest distance from the star is 0.85 AU (aphelion distance),

the semi-major axis is the average of these two distances.

Semi-major axis (a) =0.25 + 0.85

2= 0.55 AU

Step 2: Next, we use Kepler’s Third Law of Planetary Motion to find the

period of the exoplanet’s orbit. The period of an orbiting body squared is

proportional to the semi-major axis of its orbit cubed.

T2=k×a3

where Tis the period of the orbit and ais the semi-major axis. Since we are

given the period T= 400 days, and the semi-major axis a= 0.55 AU, we can

now solve for the constant of proportionality k.

k=T2

a3=(400)2

(0.55)3≈7127.27

Step 3: Finally, we can calculate the average speed of the exoplanet in its

orbit using the formula for orbital speed. The formula for orbital speed is:

v=√k

Substitute the values of kand ainto the formula to find the average speed of

the exoplanet.

v=√7127.27

0.55 ≈√12958.76 ≈113.89 km/s

Therefore, the average speed of the exoplanet in its orbit is approximately

113.89 km/s.

Question 2

Question

A planet orbiting a star has an orbital period of 8 years and an average distance

of 2 AU from the star. Determine the semi-major axis of the planet’s elliptical

orbit.

Solution

Step 1: Recall Kepler’s third law of planetary motion, which states that the

square of the orbital period of a planet is proportional to the cube of the semi-

major axis of its orbit:

P2=a3

where: - Pis the orbital period of the planet in years, - ais the semi-major axis

of the planet’s elliptical orbit in astronomical units (AU).

Step 2: Given that the orbital period of the planet is 8 years, we can sub-

stitute this value into the equation:

82=a3

Step 3: Solve for the semi-major axis aby taking the square root of both

sides of the equation: √82=√a3

8 = a3/2

Step 4: Cube both sides to isolate a:

83=a3

512 = a3

Step 5: Take the cube root of both sides to find the semi-major axis a:

√512 = a

a= 8 AU

Therefore, the semi-major axis of the planet’s elliptical orbit is 8 AU.

Question 3

Question

Consider a hypothetical planetary system where the semi-major axis of a planet’s

orbit is 2.5 AU. If the period of the planet’s orbit is 4 years, determine the mass

of the star around which the planet orbits. Assume the star is stationary.

Solution

To find the mass of the star, we can use Kepler’s Third Law of Planetary Motion,

which states that the ratio of the cube of the semi-major axis of an orbit to the

square of the period of the orbit is constant for all planets orbiting the star.

Step 1: Determine the constant of proportionality using known values. Let

a= 2.5AU be the semi-major axis of the planet’s orbit and T= 4 years be the

period of the planet’s orbit. According to Kepler’s Third Law, we have:

a3=k·T2where kis the constant of proportionality.

Plugging in the values:

(2.5)3=k·(4)2

15.625 = 16k

k=15.625

16 = 0.9765625

Step 2: Use the constant of proportionality to find the mass of the star.

Now, we’ll apply the formula for the constant kto find the mass of the star.

The formula is:

M=4π2

G·a3

where Mis the mass of the star, Gis the gravitational constant, and ais the

semi-major axis of the planet’s orbit.

Substitute the known values:

M=4π2

G·(2.5)3

M=4π2

6.674 ×10−11 ·15.625

M≈5.917 ×1029 kg

Therefore, the mass of the star around which the planet orbits is approxi-

mately 5.917 ×1029 kg.

Question 4

Question

Describe Kepler’s third law of planetary motion and explain how it relates to

the distance of a planet from the sun.

Solution

Step 1: Kepler’s third law of planetary motion states that the square of the

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

axis of its orbit. Mathematically, it can be expressed as:

=T2

where T1and T2are the orbital periods of two planets, and a1and a2are the

semi-major axes of their orbits.

Step 2: The orbital period of a planet refers to the time it takes for the planet

to complete one full orbit around the sun. On the other hand, the semi-major

axis of a planet’s orbit is half the longest diameter of the elliptical orbit.

Step 3: According to Kepler’s third law, planets that are farther away from

the sun have longer orbital periods compared to planets that are closer to the

sun. This is because the semi-major axis of the orbit (which represents the

distance of the planet from the sun) is in the denominator of the equation.

Step 4: Therefore, the farther a planet is from the sun (larger semi-major

axis), the longer it takes to complete one orbit around the sun (larger orbital

period). This relationship helps us understand how the distance of a planet

from the sun affects its orbital motion.

Step 5: In summary, Kepler’s third law of planetary motion relates the

orbital period of a planet to the distance of the planet from the sun. The law

provides a mathematical relationship that helps explain the observed motions

of planets in our solar system and beyond.

Question 5

Question

Suppose a planet completes one orbit around the sun in 500 days. If the distance

between the planet and the sun is 2.5 AU (astronomical units), determine the

time period (in days) for a planet located at 4 AU to complete one orbit around

the sun. Use Kepler’s third law of planetary motion.

Solution

Step 1: Kepler’s Third Law states that the square of the period of revolution of

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

This can be mathematically represented as:

T2=k×a3

where Tis the time period of revolution in days, ais the semi-major axis in

AU, and kis a constant of proportionality.

Step 2: Let’s first calculate the constant of proportionality, k, using the

given information for the planet at 2.5 AU:

5002=k×2.53

250000 = k×15.625

k=250000

15.625 = 16000

Step 3: Now, we can use the derived value of kto find the time period for a

planet located at 4 AU:

T2= 16000 ×43

T2= 16000 ×64

T2= 1024000

T=√1024000

T≈1012.4days

Therefore, a planet located at 4 AU would take approximately 1012.4 days

to complete one orbit around the sun.

Question 7

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 its average distance from the Sun

(r). Suppose a newly discovered planet has an orbital period of 8 years and an

average distance from the Sun of 6 astronomical units (AU). If another planet

has an orbital period of 15 years, what is its average distance from the Sun in

astronomical units?

Solution

Step 1: Let’s denote the orbital period of the second planet as T2and its average

distance from the Sun as r2. According to Kepler’s third law, we have:

=T2

where T1= 8 years and r1= 6 AU.

Step 2: Substitute the known values into the equation:

63=152

Simplify:

216 =225

Step 3: Cross multiply to solve for r2:

64 ·r3

2= 225 ·216

Step 4: Divide by 64 to isolate r3

2=225 ·216

Step 5: Calculate r2by taking the cube root of the value obtained in the

previous step:

r2=3

√225 ·216

Therefore, the average distance from the Sun for the second planet is ap-

proximately 13.5 astronomical units (AU).

Question 8

Question

In a distant solar system, a planet is orbiting a star with a semi-major axis of

4.5 Astronomical Units (AU). The planet takes 9.5 Earth years to complete one

full orbit around the star. Calculate the orbital period of a hypothetical moon

that orbits this planet at a distance of 0.75 AU.

Solution

Step 1: Determine the mass of the star using Kepler’s third law. Kepler’s third

law states: T2

=T2

, where Tis the orbital period and ais the semi-major axis.

Let T1= 9.5years, a1= 4.5AU, T2=Tmoon (unknown), and a2= 0.75 AU.

9.52

4.53=T2

moon

0.753

90.25

91.125 =T2

moon

0.421875

0.99065 = T2

moon

0.421875

moon = 0.99065 ×0.421875

moon = 0.41849478

Tmoon ≈√0.41849478 ≈0.646 years

Therefore, the orbital period of the hypothetical moon that orbits the planet

at a distance of 0.75 AU is approximately 0.646 years.

Question 9

Question

A planet is in an elliptical orbit around a star with an eccentricity of 0.3. If the

planet is closest to the star at a distance of 0.5 AU and farthest from the star

at a distance of 1.5 AU, determine the semi-major axis of the orbit.

Solution

Step 1: Recall that the semi-major axis of an ellipse is the average of the dis-

tances from the center to the nearest and farthest points on the ellipse.

Step 2: The formula to find the semi-major axis of an ellipse is given by

a=rmin +rmax

2, where rmin is the minimum distance and rmax is the maximum

distance from the center.

Step 3: In this case, rmin = 0.5AU and rmax = 1.5AU.

Step 4: Substitute the values into the formula to find the semi-major axis:

a=0.5+1.5

Step 5: Calculate the average:

a=2

2= 1 AU

Step 6: Therefore, the semi-major axis of the planet’s orbit around the star

is 1 AU.

Question 10

Question

Consider a hypothetical planetary system with two planets, Planet A and Planet

B. Planet A has a semi-major axis of 2 AU and an orbital period of 2 years,

while Planet B has a semi-major axis of 4 AU. Determine the orbital period of

Planet B around the star in years.

Solution

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

square of the orbital period of a planet is directly proportional to the cube of

the semi-major axis of its orbit. Mathematically, this can be expressed as:

=a3

where TAand TBare the orbital periods of Planet A and Planet B respectively,

and aAand aBare the semi-major axes of their orbits.

Step 2: Substitute the given values into the equation. For Planet A, aA= 2

AU and TA= 2 years. For Planet B, aB= 4 AU and TB=? years. Substituting

these values into the equation gives:

=23

Step 3: Simplify the equation to solve for TB. This gives:

Step 4: Simplify further to find the value of TB.

Step 5: Cross multiply to solve for TB.

B= 32 =⇒TB=√32

Step 6: Finally, calculate the value of TB.

TB≈5.66 years

Therefore, the orbital period of Planet B around the star is approximately

5.66 years.

Question 11

Question

The eccentricity of a planet’s orbit around the Sun is 0.2. If the planet’s closest

distance to the Sun (perihelion) is 0.6 AU, what is the planet’s farthest distance

from the Sun (aphelion) in AU?

Solution

Step 1: We can use the formula for calculating the aphelion distance of an orbit

given the eccentricity (e) and perihelion distance (rp):

ra=rp

1−e

where rais the aphelion distance.

Step 2: Given that e= 0.2and rp= 0.6AU, we can substitute these values

into the formula:

ra=0.6AU

1−0.2

Step 3: Calculating the denominator:

1−0.2 = 0.8

Step 4: Now, we can calculate the aphelion distance:

ra=0.6AU

0.8

ra= 0.75 AU

Step 5: Therefore, the planet’s farthest distance from the Sun (aphelion) is

0.75 AU.

Question 12

Question

Which of Kepler’s laws of planetary motion states that a planet moves in an

ellipse with the sun at one focus?

Solution

To determine the correct law of planetary motion that describes a planet moving

in an ellipse with the sun at one focus, we need to review Kepler’s three laws:

Kepler’s First Law (Law of Ellipses): The orbit of a planet is an ellipse

with the sun at one of the two foci.

Kepler’s Second Law (Law of Equal Areas): A line segment joining a

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

Kepler’s Third Law (Law of Harmonies): The square of the period of

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

Therefore, the law that describes a planet moving in an ellipse with the sun

at one focus is Kepler’s First Law (Law of Ellipses).

Question 13

Question

A planet is orbiting a star with a semi-major axis of 5 AU. If the planet’s orbital

eccentricity is 0.4, calculate the aphelion and perihelion distances of the planet

from the star.

Solution

Step 1: Recall Kepler’s First Law, which states that all planets move in elliptical

orbits with the star at one of the foci. The semi-major axis (a) is the average of

the distances from the center to the farthest and closest points of the planet’s

orbit.

Given: Semi-major axis, a = 5 AU Eccentricity, e = 0.4

Step 2: The aphelion (farthest distance) distance can be calculated using

the formula:

Aphelion distance =a(1 + e)

Substitute the given values to find the aphelion distance:

Aphelion distance = 5(1 + 0.4) = 5(1.4) = 7 AU

Therefore, the aphelion distance of the planet from the star is 7 AU.

Step 3: The perihelion (closest distance) distance can be calculated using

the formula:

Perihelion distance =a(1 −e)

Substitute the given values to find the perihelion distance:

Perihelion distance = 5(1 −0.4) = 5(0.6) = 3 AU

Therefore, the perihelion distance of the planet from the star is 3 AU.

Question 14

Question

Consider a planetary system where the semi-major axis of Planet X’s orbit is 2.5

AU. If Planet Y has an orbit with a semi-major axis 3 times that of Planet X,

what is the period of Planet Y’s orbit in terms of Planet X’s period, according

to Kepler’s Third Law?

Solution

To find the relationship between the periods of the two planets, we can use

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

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

the orbit.

Step 1: Let’s denote the period of Planet X as TXand the period of Planet

Y as TY. Similarly, the semi-major axis of Planet Y’s orbit is aY= 3 ×aX.

From Kepler’s Third Law:

(TY

TX)2

=(aY

aX)3

Step 2: Substituting the given values, where aX= 2.5AU and aY= 3aX=

3×2.5 = 7.5AU, we have:

(TY

TX)2

=(7.5

2.5)3

= 33= 27

Step 3: Solving for TY

TX:

(TY

TX)2

= 27 =⇒TY

= 3√3

Therefore, the period of Planet Y’s orbit in terms of Planet X’s period is

3√3×TX.

Question 15

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 its average distance from the

sun. Suppose a planet has a period of Tyears and an average distance from

the sun of rastronomical units (AU). If a different planet has a period 9 times

longer and an average distance from the sun 27 times larger, what is the ratio

of the new planet’s period to its average distance from the sun in AU?

Solution

Step 1: Let’s denote the period of the second planet as T′and its distance from

the sun as r′. We are given that:

T′= 9T

r′= 27r

Step 2: According to Kepler’s third law, we have the proportion:

T′2

r′3=T2

Step 3: Substituting the given values for T′and r′, we get:

(9T)2

(27r)3=T2

Step 4: Simplifying the left side of the equation:

81T2

19683r3=T2

Step 5: Cross multiplying to solve for the ratio of the new planet’s period

to its average distance from the sun in AU:

81T2·r3=T2·19683r3

81T2r3= 19683T2r3

Step 6: Dividing both sides by T2r3:

81 = 19683

Step 7: Thus, there seems to be a mistake in the calculations. Let’s reeval-

uate our steps to find and correct the error.

Question 16

Question

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

perfectly circular orbit. The planet takes 490 days to complete one orbit around

the star. If the planet’s semi-major axis is 2.5 AU, calculate the mass of the

star in terms of the mass of the Sun.

Solution

Step 1: Recall Kepler’s Third Law of Planetary Motion which relates the orbital

period of a planet (T) to the semi-major axis of its orbit (a) and the mass of

the central star (M):

T2=(4π2

G(M1+M2))a3

Step 2: We are given that the orbital period T= 490 days and the semi-

major axis a= 2.5AU. Our goal is to find the mass of the star Min terms of

the mass of the Sun.

Step 3: First, let’s convert the orbital period from days to seconds. There

are 24 hours in a day, each with 3600 seconds, so:

T= 490 days ×24 hours/day ×3600 seconds/hour = 42336000 seconds

Step 4: Plugging in the known values for Tand ainto Kepler’s Third Law

equation:

(42336000)2=(4π2

G(MSun +M))(2.5)3

Step 5: Next, we know that MSun is the mass of the Sun, which is approxi-

mately 1.989 ×1030 kg. Solving for Min terms of MSun:

(42336000)2=(4π2

G(1.989 ×1030 +M))(2.5)3

Step 6: Now, solve for Mby rearranging the equation and plugging in the

known values for Gand π:

M=4π2(2.5)3

G−1.989 ×1030

Step 7: Calculating the final result gives the mass of the star in terms of the

mass of the Sun.

Question 17

Question

For a certain planet in the solar system, the semi-major axis of its orbit around

the Sun is 3.84 AU. If the orbital period of this planet is 4.6 years, calculate the

mass of the Sun. (Hint: Use Kepler’s Third Law)

Solution

Step 1: Recall Kepler’s Third Law, which states:

a3=4π2

G(M1+M2)

where: - Tis the orbital period of the planet, - ais the semi-major axis of the

planet’s orbit, - Gis the gravitational constant, - M1is the mass of the Sun,

and - M2is the mass of the planet.

Step 2: Rearrange the formula to solve for the mass of the Sun:

M1=4π2a3

GT 2−M2

Step 3: Since the mass of the planet is usually much smaller than the mass

of the Sun, we can neglect it in this calculation. Therefore:

M1≈4π2a3

GT 2

Step 4: Now, substitute the given values into the formula:

M1≈4π2(3.84 AU)3

G(4.6years)2

Step 5: Convert the semi-major axis from AU to meters (1 AU = 1.496×1011

meters):

M1≈4π2(3.84 ×1.496 ×1011 m)3

G(4.6×365.25 ×24 ×3600 s)2

Step 6: Simplify the expression and calculate the mass of the Sun. The final

answer will be in kilograms.

Question 18

Question

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

of a planet is proportional to the cube of its average distance from the Sun.

Suppose a planet has an average distance from the Sun of 2.5×108km and

a period of 3 years. Calculate the average distance of another planet from the

Sun if it has a period of 5 years.

Solution

Step 1: Calculate the constant of proportionality using the data given for the

first planet. Given: r1= 2.5×108km, T1= 3 years

Using Kepler’s Third Law: T2

1=k·r3

Solving for the constant of proportionality k:

k=T2

=(3)2

(2.5×108)3

Step 2: Calculate the average distance of the second planet from the Sun

using the period of 5 years. Given: T2= 5 years

Using Kepler’s Third Law with the constant of proportionality k:

r2=3

√k·T2

1=3

√k·(5)2

Substitute the value of kcalculated in Step 1 into the equation:

r2=3

√(3)2/(2.5×108)3·(5)2

Simplify the equation to find the average distance of the second planet from

the Sun.

Question 19

Question

The semi-major axis of a planet’s elliptical orbit around the Sun is 2.7 AU. If

the planet’s period of revolution is 3.5 years, determine the mass of the Sun in

terms of the mass of the Earth.

Solution

Step 1: Calculate the period of revolution of the Earth using Kepler’s Third

Law. The period of revolution of a planet, T, is related to the semi-major axis

of its orbit, a, by the formula:

T2=k·a3

where kis a constant.

Substitute T= 1 year (the period of revolution of Earth) and a= 1 AU (the

semi-major axis of Earth’s orbit) to find k:

12=k·13

k= 1

Now, plug in the values T= 3.5years (the period of revolution of the given

planet) and a= 2.7AU (the semi-major axis of the planet’s orbit) into the

formula:

(3.5)2= 2.73

12.25 = 19.683

Step 2: Use the ratio of the cubes of the semi-major axes to find the ratio

of the masses of the Sun and the Earth. According to Kepler’s Third Law:

m1+m2

mSun

=(a1

a2)3

Substitute a1= 1 AU, a2= 2.7AU, and m1=mEarth:

1 + mEarth

mSun

=(1

2.7)3

Step 3: Solve for the mass of the Sun in terms of the mass of the Earth.

1 + mEarth

mSun

=(1

2.7)3

1 + mEarth =(1

2.7)3

·mSun

1 + mEarth =(1

2.7)3

·mEarth

mSun =1 + mEarth

2.7)3

mSun =1+1

2.7)3

mSun =2

2.7)3

mSun =2·2.73

mSun = 32.706 mEarth

Therefore, the mass of the Sun is 32.706 times the mass of the Earth.

Question 20

Question

Consider a hypothetical solar system with a central star of mass Mand a planet

of negligible mass in a circular orbit around the star. The planet orbits at a

distance rfrom the star with a period of revolution T.

Prove Kepler’s third law of planetary motion: The square of the period of

revolution of a planet in a circular orbit around a star is directly proportional

to the cube of the semi-major axis of the orbit.

Solution

Step 1: Let’s consider the gravitational force between the star and the planet.

According to Newton’s law of gravitation, the gravitational force between two

objects of masses Mand mseparated by a distance ris given by:

F=GMm

where Gis the gravitational constant.

Step 2: This gravitational force provides the centripetal force required for

the planet to move in a circular orbit. The centripetal force is given by:

F=mv2

where vis the velocity of the planet in its orbit.

Step 3: Equating the gravitational force to the centripetal force, we have:

GMm

r2=mv2

Step 4: Simplifying the equation further, we find:

v2=GM

Step 5: We know that the velocity of an object in circular motion is given

by:

v=2πr

where Tis the period of revolution.

Step 6: Substituting the expression for velocity into our equation, we get:

(2πr

T)2

=GM

Step 7: Simplifying the equation, we find:

4π2r=GM

rT2

Step 8: Rearranging the terms, we arrive at Kepler’s third law:

T2=4π2

GM r3

Therefore, the square of the period of revolution of a planet in a circular

orbit around a star is directly proportional to the cube of the semi-major axis

of the orbit, as stated by Kepler’s third law of planetary motion.

Question 21

Question

In the study of planetary motion, Kepler’s third law states that the square of

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

of its semi-major axis.

Given that Earth’s semi-major axis is approximately 1 astronomical unit

(AU) and its period of revolution around the sun is approximately 1 year, de-

termine the approximate period of revolution for a planet with a semi-major

axis of 2 AU.

Solution

Step 1: Let T1and a1denote the period of revolution and the semi-major axis

of Earth, respectively. Similarly, let T2and a2represent the period of revolution

and the semi-major axis of the other planet.

Step 2: According to Kepler’s third law, we have the equation:

=T2

Substitute T1= 1 year, a1= 1 AU, and a2= 2 AU into the equation:

13=T2

Step 3: Simplify the equation:

1 = T2

Step 4: Solving for T2:

2= 8

T2=√8

T2= 2.83 years

Therefore, a planet with a semi-major axis of 2 AU would have an approxi-

mate period of revolution around the sun of 2.83 years.

Question 22

Question

Explain Kepler’s Third Law of Planetary Motion and how it relates to the period

and distance of a planet from the sun.

Solution

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

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

semi-major axis of the planet’s orbit. Mathematically, this can be written as:

T2=k×a3

where: - Tis the period of revolution of the planet, - ais the semi-major

axis of the planet’s orbit, - kis a constant of proportionality.

Step 1: Let’s express Kepler’s Third Law in terms of Earth’s period and

distance from the sun.

For Earth, the period of revolution (1 year) is denoted by TEarth and the

average distance from the sun (1 Astronomical Unit) is denoted by aEarth.

So, Kepler’s Third Law can be written for Earth as:

Earth =k×a3

Earth

Step 2: Since the period and distance of Earth are known, we can use them

to find the constant of proportionality, k.

Given: TEarth = 1 year and aEarth = 1 AU, we have:

12=k×13⇒k= 1

Thus, Kepler’s Third Law for Earth simplifies to:

Earth =a3

Earth

Step 3: Now, let’s consider a different planet, denoted as P, with a period

of revolution TPand a semi-major axis aP.

The relationship for planet Pcan be expressed as:

P=a3

Step 4: Comparing the equations for Earth and planet P, we find:

=T2

Earth

= 1

This demonstrates that Kepler’s Third Law holds for any planet in the solar

system - the square of the period of revolution divided by the cube of the semi-

major axis is always equal to 1.

Question 23

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 4 years and a semi-major axis of 3 astronomical units

(AU). Calculate the orbital period of another planet with a semi-major axis of

2 AU.

Solution

Step 1: First, let’s denote the orbital period of the second planet as T(in years)

and its semi-major axis as a(in AU).

Step 2: According to Kepler’s third law, the following relationship holds:

=T2

Where T1and a1are the orbital period and semi-major axis of the first

planet, and T2and a2are the orbital period and semi-major axis of the second

planet.

Step 3: Substituting the values we know into the equation, we get:

33=T2

Step 4: Simplifying both sides gives us:

27 =T2

Step 5: Cross multiplying, we have:

16 ×8 = 27 ×T2

Step 6: Solving for T:

T2=16 ×8

27 =128

Step 7: Therefore, the orbital period of the second planet is:

T=√128

27 ≈3.794 years

So, the orbital period of the second planet with a semi-major axis of 2 AU

is approximately 3.794 years.

Question 24

Question

In the context of Kepler’s laws of planetary motion, state the difference between

Kepler’s first law and Kepler’s third law. Provide examples to illustrate each

law.

Solution

Step 1: Kepler’s First Law: Kepler’s first law, also known as the law of orbits,

states that each planet moves in an elliptical orbit with the Sun at one of the

two foci.

Example: The orbit of Mars around the Sun is an ellipse with the Sun

located at one of the foci.

Step 2: Kepler’s Third Law: Kepler’s third law, also known as the law of

harmonies, states that the square of a planet’s orbital period is directly propor-

tional to the cube of the semi-major axis of its orbit.

Example: Consider the Earth and Mars. Let TEbe the orbital period of

Earth (365.25 days) and TMbe the orbital period of Mars (686.98 days). Let

aEbe the semi-major axis of Earth’s orbit (1 AU) and aMbe the semi-major

axis of Mars’ orbit (1.52 AU). According to Kepler’s third law, we have:

=a3

365.252

686.982=13

1.523

133225.06

471865.60 ≈1

3.522

Therefore, Kepler’s third law holds true for the Earth and Mars.

Question 25

Question

The period of one complete orbit of a planet around the Sun is known as its

orbital period. Jupiter has an orbital period of approximately 11.86 Earth

years. If Mars has an average distance from the Sun of approximately 1.52

astronomical units (AU), find the approximate orbital period of Mars around

the Sun in Earth years. (Hint: Use Kepler’s third law of planetary motion).

Solution

Step 1: Kepler’s third law of planetary motion relates the orbital period (T) of a

planet around the Sun to its average distance from the Sun (r) by the equation:

planet

=T2

Earth

Step 2: We are given that the orbital period of Jupiter (TJupiter) is 11.86

Earth years and its distance from the Sun (rJupiter) is 5.20 AU.

Step 3: Substituting the values into the formula, we get:

(11.86)2

(5.20)3=T2

Earth

140.6596

140.608 ≈T2

Earth

Step 4: Solving for TEarth, we find:

TEarth ≈√140.6596 ≈11.86 years

Step 5: Finally, to find the orbital period of Mars (TMars) with an average

distance of 1.52 AU, we use the same formula:

Mars

(1.52)3=(11.86)2

(5.20)3

Step 6: Solving for TMars:

TMars ≈√(11.86)2

(5.20)3×(1.52)3

TMars ≈√140.6596 ×1.523

140.608

TMars ≈√140.6596 ×3.5952

140.608

TMars ≈√559.5422 ≈23.66 years

Therefore, the approximate orbital period of Mars around the Sun is 23.66

Earth years.

Step 3: Finally, we can calculate the average speed of the exoplanet in its

orbit using the formula for orbital speed. The formula for orbital speed is:

v=√k

Substitute the values of kand ainto the formula to find the average speed of

the exoplanet.

v=√7127.27

0.55 ≈√12958.76 ≈113.89 km/s

Therefore, the average speed of the exoplanet in its orbit is approximately

113.89 km/s.

Question 2

Question

A planet orbiting a star has an orbital period of 8 years and an average distance

of 2 AU from the star. Determine the semi-major axis of the planet’s elliptical

orbit.

Solution

Step 1: Recall Kepler’s third law of planetary motion, which states that the

square of the orbital period of a planet is proportional to the cube of the semi-

major axis of its orbit:

P2=a3

where: - Pis the orbital period of the planet in years, - ais the semi-major axis

of the planet’s elliptical orbit in astronomical units (AU).

Step 2: Given that the orbital period of the planet is 8 years, we can sub-

stitute this value into the equation:

82=a3

Step 3: Solve for the semi-major axis aby taking the square root of both

sides of the equation: √82=√a3

8 = a3/2

Step 4: Cube both sides to isolate a:

83=a3

512 = a3

Step 5: Take the cube root of both sides to find the semi-major axis a:

√512 = a

a= 8 AU

Therefore, the semi-major axis of the planet’s elliptical orbit is 8 AU.

Question 3

Question

Consider a hypothetical planetary system where the semi-major axis of a planet’s

orbit is 2.5 AU. If the period of the planet’s orbit is 4 years, determine the mass

of the star around which the planet orbits. Assume the star is stationary.

Solution

To find the mass of the star, we can use Kepler’s Third Law of Planetary Motion,

which states that the ratio of the cube of the semi-major axis of an orbit to the

square of the period of the orbit is constant for all planets orbiting the star.

Step 1: Determine the constant of proportionality using known values. Let

a= 2.5AU be the semi-major axis of the planet’s orbit and T= 4 years be the

period of the planet’s orbit. According to Kepler’s Third Law, we have:

a3=k·T2where kis the constant of proportionality.

Plugging in the values:

(2.5)3=k·(4)2

15.625 = 16k

k=15.625

16 = 0.9765625

Step 2: Use the constant of proportionality to find the mass of the star.

Now, we’ll apply the formula for the constant kto find the mass of the star.

The formula is:

M=4π2

G·a3

where Mis the mass of the star, Gis the gravitational constant, and ais the

semi-major axis of the planet’s orbit.

Substitute the known values:

M=4π2

G·(2.5)3

M=4π2

6.674 ×10−11 ·15.625

M≈5.917 ×1029 kg

Therefore, the mass of the star around which the planet orbits is approxi-

mately 5.917 ×1029 kg.