PHSC 121 - INTRODUCTION TO ASTRONOMY Kepler's Laws of Planetary Motion Question Bank Set 4

PHSC 121 - INTRODUCTION TO

ASTRONOMY - Kepler’s Laws of

Planetary Motion

Question Bank - Set 4

Liberty University

Question 1

Question

Suppose a planet orbits a star with a semi-major axis of 2 AU. If the period of

this planet’s orbit is 1 year, determine the eccentricity of the planet’s orbit.

Solution

Let’s use Kepler’s Third Law of Planetary Motion, which relates the period of

a planet’s orbit to the semi-major axis of its elliptical orbit:

T2=4π2

G(m1+m2)a3

where: - Tis the period of the orbit, - ais the semi-major axis of the orbit, and

-Gis the gravitational constant.

Given that T= 1 year, a= 2 AU, and assuming the star’s mass is much

greater than the planet’s mass (m2≈0), we have:

12=4π2

Gm1

(2)3

1 = 8π2

Gm1×8

Gm1= 64π2

Now, let’s determine the eccentricity of the planet’s orbit using the formula:

Eccentricity (e) =√1−(b

a)2

where: - bis the semi-minor axis of the orbit.

From the definition of eccentricity: eccentricity =√1+b2/a2

Let’s first determine busing the formula:

b=a√1−e2

b= 2√1−e2

We know that the area enclosed by the ellipse is the same as a circle with a

radius of 2 AU:

πab =π(2)2

π×2√1−e2×2 = π×4

4√1−e2= 4

√1−e2= 1

1−e2= 1

e2= 0

e= 0

Therefore, the eccentricity of the planet’s orbit is 0, indicating a circular

orbit.

Question 2

Question

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

planet is proportional to the cube of its semi-major axis. Consider two planets,

Planet A and Planet B, orbiting a star. Planet A has a semi-major axis of 2

AU and a period of 4 years, while Planet B has a semi-major axis of 4 AU.

Calculate the period of Planet B.

Solution

Step 1: We can express Kepler’s Third Law mathematically as:

(TA

TB)2

=(aA

aB)3

where: - TAand TBare the periods of Planet A and Planet B, respectively. -

aAand aBare the semi-major axes of Planet A and Planet B, respectively.

Step 2: Substituting the known values into the equation, we get:

TB)2

=(2

4)3

Step 3: Simplifying the equation gives:

TB)2

=(1

2)3

Step 4: Solving for TBby taking the square root of both sides, we have:

=√(1

2)3

Step 5: Simplifying further yields:

√2

Step 6: Multiplying both sides by TBgives:

TB= 4√2

Therefore, the period of Planet B is 4√2years.

Question 3

Question

An asteroid is orbiting the Sun in an elliptical orbit. The semi-major axis of its

orbit is 2.5 AU and the eccentricity of the orbit is 0.6. Determine the periapsis

and apoapsis distances of the asteroid from the Sun.

Solution

Let’s denote the periapsis distance as rpand the apoapsis distance as ra. The

periapsis distance occurs at the point where the asteroid is closest to the Sun,

while the apoapsis distance occurs at the point where the asteroid is farthest

from the Sun.

Step 1: Calculate the periapsis distance using the formula:

rp=a(1 −e)

where ais the semi-major axis and eis the eccentricity.

Substitute a= 2.5AU and e= 0.6into the formula:

rp= 2.5(1 −0.6) = 2.5×0.4 = 1 AU

Therefore, the periapsis distance of the asteroid from the Sun is 1 AU.

Step 2: Calculate the apoapsis distance using the formula:

ra=a(1 + e)

Substitute a= 2.5AU and e= 0.6into the formula:

ra= 2.5(1 + 0.6) = 2.5×1.6 = 4 AU

Therefore, the apoapsis distance of the asteroid from the Sun is 4 AU.

Question 4

Question

In a distant planetary system, the semi-major axis of a certain planet’s orbit is

3.5 AU. The period of the planet’s orbit around its star is 9.2 years. Determine

the mass of the star (in solar masses) given that the gravitational constant

G= 6.67 ×10−11 m3kg−1s−2.

Solution

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

of an orbit (T) and the semi-major axis of the orbit (a), as follows:

T2=(4π2

G·M)a3,

where Mis the mass of the star.

Step 2: We are given a= 3.5AU and T= 9.2years. We will convert the

semi-major axis from astronomical units (AU) to meters using the conversion

factor 1AU = 1.496 ×1011 m.

a= 3.5×1.496 ×1011 m= 5.236 ×1011 m.

Step 3: Substitute aand Tinto Kepler’s Third Law to solve for the mass of

the star M:

(9.2years)2=(4π2

6.67 ×10−11 m3kg−1s−2·M)(5.236 ×1011 m)3.

Step 4: Simplify and solve the equation for M:

9.22=4π2

6.67 ×10−11 ·M·(5.236 ×1011)3.

84.64 = 4π2

6.67 ×10−11 ·M·1.190 ×1035.

M=4π2

6.67 ×10−11 ·1.190 ×1035 .

Step 5: Calculate the mass of the star M:

M=4π2

6.67 ×10−11 ·1.190 ×1035 ≈39.48

7.95 ×1024 ≈4.96 ×1023 kg.

Step 6: Finally, convert the mass of the star from kilograms to solar masses

using the conversion factor 1solar mass = 1.989 ×1030 kg:

Mstar =4.96 ×1023 kg

1.989 ×1030 kg/solar mass ≈2.49 ×10−7solar masses.

Therefore, the mass of the star is approximately 2.49 ×10−7solar masses.

Question 5

Question

According to Kepler’s laws 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 3.5 astronomical units (AU). If another planet has a period of revolution

around the Sun that is twice as long as the first planet, what is the average

distance of the second planet from the Sun?

Solution

Step 1: Let Tbe the period of revolution of the first planet, dbe the average

distance from the Sun of the first planet, and T′and d′be the corresponding

quantities for the second planet. We know that according to Kepler’s Third

Law: T2∝d3. Therefore, for the first planet: T2=k·d3, where kis the

constant of proportionality.

Step 2: Since the second planet has a period of revolution that is twice as

long as the first planet, we have T′= 2T, which implies that T′= 2√k·d3.

Step 3: We now solve for the average distance of the second planet from

the Sun by appealing to Kepler’s Third Law. For the second planet, we have:

T′2=k·(d′)3. Substituting T′= 2√k·d3, we get: (2√k·d3)2=k·(d′)3.

Step 4: Simplifying, we have: 4·k·d3=k·(d′)3. Dividing both sides by k,

we have: 4·d3= (d′)3.

Step 5: Taking the cube root of both sides, we find: d′=3

√4·d. Given

that the average distance of the first planet from the Sun is 3.5 AU, we have:

d′=3

√4·3.5AU.

Step 6: Calculating the result, we find: d′= 3.5·3

√4≈3.5·1.587 =

5.557 AU. Therefore, the average distance of the second planet from the Sun is

approximately 5.557 astronomical units.

Question 6

Question

At perihelion, a planet is at its closest point to the Sun in its elliptical orbit.

Consider a planet with a semi-major axis of 2.5 AU and an eccentricity of 0.3:

1. Calculate the distance of the planet from the Sun at perihelion.

2. Calculate the distance of the planet from the Sun at aphelion.

Solution

1. To calculate the distance of the planet from the Sun at perihelion, we can

use the following formula based on the properties of an elliptical orbit:

rperi =a(1 −e)

where: - rperi is the distance at perihelion, - ais the semi-major axis, - eis the

eccentricity.

Substitute a= 2.5AU and e= 0.3into the formula:

rperi = 2.5×(1 −0.3)

rperi = 2.5×0.7 = 1.75 AU

Therefore, the planet’s distance from the Sun at perihelion is 1.75 AU.

2. To calculate the distance of the planet from the Sun at aphelion, we can

use a similar formula:

rapo =a(1 + e)

where: - rapo is the distance at aphelion, - ais the semi-major axis, - eis the

eccentricity.

Substitute a= 2.5AU and e= 0.3into the formula:

rapo = 2.5×(1 + 0.3)

rapo = 2.5×1.3 = 3.25 AU

Therefore, the planet’s distance from the Sun at aphelion is 3.25 AU.

Question 7

Question

Consider a planet orbiting a star in an elliptical orbit with the star located at

one of the foci. If the semi-major axis of the orbit is 3 AU and the eccentricity

of the orbit is 0.4, calculate the semi-minor axis of the orbit.

Solution

Step 1: Recall the relationship between the semi-major axis a, semi-minor axis

b, and eccentricity efor an elliptical orbit:

b=a√1−e2

Step 2: Given that the semi-major axis a= 3 AU and the eccentricity

e= 0.4, we can now substitute these values into the equation:

b= 3√1−0.42

Step 3: Calculate 1−0.42:

1−0.42= 0.84

Step 4: Substitute the result back into the equation:

b= 3√0.84

Step 5: Calculate √0.84:

√0.84 ≈0.917

Step 6: Finally, multiply the semi-major axis by the calculated value:

b= 3 ×0.917 = 2.751 AU

Therefore, the semi-minor axis of the orbit is approximately 2.751 AU.

Question 8

Question

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

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

has an orbital period of 5 years and an average distance from the sun of 3

astronomical units (AU), what would be the orbital period of another planet

with an average distance from the sun of 12 AU?

Solution

To solve this problem, we will first express Kepler’s third law mathematically

and then use the given information to find the orbital period of the second

planet.

Step 1: Write Kepler’s third law mathematically: Kepler’s third law states

that:

T2∝a3

where Tis the orbital period (in years) and ais the semi-major axis (in AU).

Step 2: Use the given values for the first planet: For the first planet, we

have:

T1= 5 years

a1= 3 AU

This gives us:

1∝a3

52∝33

25 ∝27

Step 3: Set up the proportion to find the orbital period of the second planet:

We can now set up a proportion using the information from the first planet and

the average distance from the sun of the second planet:

=T2

Step 4: Substitute the values and solve for the orbital period of the second

planet: Substitute the values we know:

27 =T2

123

27 =T2

1728

Now, solve for T2:

2=25

27 ×1728

2= 1600

T2=√1600

T2= 40 years

Therefore, the orbital period of the second planet with an average distance

from the sun of 12 AU would be 40 years.

Question 9

Question

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

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

Given that Earth has an orbital period of approximately 365.25 days and a

semi-major axis of approximately 1 astronomical unit (AU), calculate the orbital

period of a planet with a semi-major axis of 4 AU.

Solution

Step 1: Let T1be the orbital period of Earth and a1be the semi-major axis of

Earth’s orbit. Let T2be the orbital period of the unknown planet and a2be

the semi-major axis of the unknown planet’s orbit. According to Kepler’s third

law:

=T2

Step 2: Substitute the values for Earth’s orbital period (T1= 365.25 days)

and semi-major axis (a1= 1 AU) into the equation:

(365.25)2

13=T2

Step 3: Simplify the equation to solve for T2.

365.252=T2

2×43

2=365.252

2= 365.252×1

2=365.252

T2=√365.252

T2=365.25

8= 45.65625 days

Step 4: Therefore, the orbital period of a planet with a semi-major axis of 4

AU is approximately 45.66 days.

Question 10

Question

An exoplanet with a semi-major axis of 0.4 AU orbits a star with a mass of

1.5×1030 kg. If the exoplanet takes 100 Earth days to complete one full orbit

around the star, calculate the period of the exoplanet’s orbit (in Earth days)

according to Kepler’s third law.

Solution

Step 1: Convert the given values to SI units: The semi-major axis of the exo-

planet’s orbit a= 0.4AU = 0.4×1.496 ×1011 m= 5.984 ×1010 m. The mass

of the star M= 1.5×1030 kg. The period of the orbit T= 100 Earth days =

100 ×24 ×3600 seconds = 8.64 ×106seconds.

Step 2: Use Kepler’s third law to calculate the period of the exoplanet’s

orbit: Kepler’s third law states that T2=4π2

G(M+m)a3, where Tis the period of

the orbit, Gis the gravitational constant, Mis the mass of the star, mis the

mass of the exoplanet, and ais the semi-major axis of the exoplanet’s orbit.

Since the mass of the exoplanet is much smaller compared to the star’s mass,

we can neglect it in the equation.

Substitute the given values into the equation:

T2=4π2

6.67430×10−11 ×(1.5×1030 )×(5.984 ×1010)3

Step 3: Solve for the period of the exoplanet’s orbit: T2=4π2

1.001145×1020 ×

2.149052 ×1032

T2= 8.587107 ×104

T=√8.587107 ×104

T≈293 Earth days

Therefore, the period of the exoplanet’s orbit around the star is approxi-

mately 293 Earth days.

Question 11

Question

According to Kepler’s third law 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 2.5 astronomical

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

axis of 4.2 AU.

Solution

Step 1: 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 of planetary motion, the following

relationship holds:

=T2

Step 3: Substituting the given values for the first planet (orbital period = 8

years, semi-major axis = 2.5 AU) and the unknown values for the second planet,

we have: 82

2.53=T2

4.23

Step 4: Solving for T2, we get:

T2= 82×4.23

2.53

T2= 64 ×74.088

15.625

T2≈304.56

Step 5: Taking the square root of both sides to find the orbital period T, we

get:

T≈√304.56

T≈17.45 years

Step 6: Therefore, the orbital period of the second planet with a semi-major

axis of 4.2 AU is approximately 17.45 years.

Question 12

Question

A planet orbits around the sun in an elliptical path according to Kepler’s Laws

of Planetary Motion. The planet’s closest approach to the sun (perihelion) is

0.3 astronomical units (AU) and its farthest point from the sun (aphelion) is

0.7 AU. If the planet takes 250 days to complete one full orbit, find the average

speed of the planet in its orbit.

Solution

Step 1: First, we find the semi-major axis of the planet’s elliptical orbit, denoted

by a, which is the average of the perihelion and aphelion distances:

a=0.3+0.7

2= 0.5AU

Step 2: We calculate the eccentricity of the orbit, denoted by e, as the ratio

of the distance between the foci of the ellipse to the length of the major axis.

e=0.7−0.5

0.5= 0.4

Step 3: To find the average speed of the planet in its orbit, we can use

Kepler’s Third Law which relates the period of orbit (T) and the semi-major

axis (a) of the orbit: T2= 4π2a3/(G(M+m)), where Gis the gravitational

constant, Mis the mass of the sun, and mis the mass of the planet.

Step 4: Since we are interested in the average speed of the planet in its orbit,

we can find the total distance traveled by the planet in one complete orbit. This

distance is equal to the circumference of the ellipse, which can be approximated

by 2πa.

Step 5: Using the total distance traveled and the period of orbit, we can

calculate the average speed of the planet:

Average speed =Total distance

Total time

Step 6: Substituting the relevant values into the equation, we find:

Average speed =2π×0.5AU

250 days

Step 7: Converting the average speed from AU/day to km/s (1 AU = 1.496×

108km):

Average speed =2π×0.5×1.496 ×108

250 ×24 ×3600

Step 8: Calculating the final value gives us the average speed of the planet

in its orbit.

Question 13

Question

Consider a planet orbiting the Sun in an elliptical orbit. The planet’s closest

distance to the Sun is 0.3 AU and its farthest distance is 0.6 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 relates the period

of an orbit (T) to the semi-major axis of the orbit (a) through the equation:

T2=k·a3

where kis a constant specific for the system of measurements being used.

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

and farthest distances from the central body (in this case, the Sun). Therefore,

the semi-major axis ais given by:

a=rclosest +rfarthest

Plugging in the values, we have:

a=0.3+0.6

a= 0.45 AU

Step 3: We are given that the period Tof the orbit is 0.5 years. Substituting

T= 0.5years and a= 0.45 AU into Kepler’s Third Law, we get:

(0.5)2=k·(0.45)3

0.25 = k·0.091125

k≈2.743]

Step 4: Now, we can calculate the eccentricity (e) of the planet’s orbit using

the formula relating eccentricity to the closest and farthest distances:

e=rfarthest −rclosest

rfarthest +rclosest

Substitute rclosest = 0.3AU and rfarthest = 0.6AU into the formula:

e=0.6−0.3

0.6+0.3

e=0.3

0.9

e= 0.333]

Therefore, the eccentricity of the planet’s orbit is 0.333.

Question 14

Question

Let’s consider a hypothetical binary star system with two stars, A and B, orbit-

ing their common center of mass. Star A has a mass of 2.0×1030 kg, while Star

B has a mass of 3.0×1030 kg. If the distance between the two stars is 1.5×1011

m, determine the position of the center of mass of the system from Star A.

Solution

Step 1: Calculate the total mass of the system.

Given: Mass of Star A, ma= 2.0×1030 kg

Mass of Star B, mb= 3.0×1030 kg

The total mass of the system, mtotal, is the sum of the masses of both stars:

mtotal =ma+mb= 2.0×1030 kg + 3.0×1030 kg

mtotal = 5.0×1030 kg

Step 2: Calculate the position of the center of mass from Star A.

Let the position of the center of mass from Star A be x. To find this position,

we use the formula for the center of mass of a two-body system:

x=mb

mtotal

where: mbis the mass of Star B, mtotal is the total mass of the system, and d

is the distance between the two stars.

Substitute the given values into the formula:

x=3.0×1030 kg

5.0×1030 kg ×1.5×1011 m

x= 0.6×1.5×1011 m

x= 0.9×1011 m

Therefore, the position of the center of mass of the system from Star A is

9.0×1010 m.

Question 15

Question

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

of a planet’s orbit (T) is proportional to the cube of the semimajor axis of its

orbit (a). Given that Mars has an average orbital period of 687 days and an

average semimajor axis of 1.524 astronomical units (AU), calculate the orbital

period of Jupiter, whose average semimajor axis is 5.204 AU.

Solution

Step 1: First, let’s set up the proportion using Kepler’s third law:

Jupiter

=T2

Mars

where TJupiter is the orbital period of Jupiter, aJupiter is the semimajor axis of

Jupiter, TMars is the orbital period of Mars, and aMars is the semimajor axis of

Mars.

Step 2: Substitute the given values into the equation:

Jupiter

(5.204)3=(687)2

(1.524)3

Step 3: Solve for TJupiter by cross multiplying:

Jupiter =(687)2

(1.524)3×(5.204)3

Step 4: Calculate the square root to find TJupiter:

TJupiter =√(687)2

(1.524)3×(5.204)3

Step 5: Finally, compute the value of TJupiter:

TJupiter =√(472,369)

(3.276) ×(142.715) ≈4,332 days

Therefore, the orbital period of Jupiter is approximately 4,332 days.

Question 16

Question

The planet Mars orbits the Sun in an elliptical path with an eccentricity of

0.094, where the distance between the foci of the ellipse is 0.194 AU. If the

average distance between Mars and the Sun is about 1.52 AU, determine the

closest and farthest distance between Mars and the Sun during its orbit.

Solution

Step 1: Recall the equation for the distance between a planet and the Sun in

an elliptical orbit:

r=a(1 −e2)

1 + ecos(θ)

where ris the distance from the planet to the Sun, ais the semi-major axis, e

is the eccentricity, and θis the true anomaly.

Step 2: Given that the average distance between Mars and the Sun is 1.52

AU, we have:

a= 1.52 AU

Step 3: The distance between the foci of the ellipse is 2ae = 0.194 AU. Since

we know the eccentricity is 0.094, we can solve for a:

2ae = 0.194

2(1.52)(0.094) = 0.194

0.28544 = 0.194

a=0.194

2·0.094

a=0.194

0.188

a≈1.03 AU

Step 4: Now we can calculate the closest and farthest distances from Mars

to the Sun. The closest distance occurs when cos(θ) = −1, and the farthest

distance occurs when cos(θ) = 1. For the closest distance:

rmin =1.03(1 −0.0942)

1−0.094(−1)

rmin =1.03(1 −0.008836)

1+0.094

rmin =1.03(0.991164)

1.094

rmin ≈0.900 AU

For the farthest distance:

rmax =1.03(1 −0.0942)

1+0.094

rmax =1.03(1 −0.008836)

1+0.094

rmax =1.03(0.991164)

1.094

rmax ≈1.160 AU

Therefore, the closest distance between Mars and the Sun during its orbit is

approximately 0.900 AU, and the farthest distance is approximately 1.160 AU.

Question 17

Question

Two planets, A and B, are in orbit around a star. Planet A has a semi-major

axis of 2 AU and an orbital period of 3 years. Planet B has a semi-major axis

of 3 AU. Determine the orbital period of planet B.

Solution

Step 1: Recall Kepler’s Third Law, which 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. Mathematically, this can be written as:

A∝a3

where TAis the period of planet A, aAis the semi-major axis of planet A, and

the subscript Adenotes planet A.

Step 2: We are given that planet A has a semi-major axis aA= 2 AU and an

orbital period TA= 3 years. Substituting these values into the proportionality

equation, we get:

A=k·a3

(3)2=k·(2)3

9 = 8k

k=9

Step 3: Now, we can determine the orbital period of planet B using Kepler’s

Third Law:

B=(9

8)·(3)3

B=9

8·27

TB=√243

√8

TB=3√27

2√2

TB=3√9·3

2√2

TB=3·3√3

2√2

TB=9√3

2√2

TB=9

2·

√3

√2

TB=9

2·

√6

TB=9

4√6

Therefore, the orbital period of planet B is 9

4√6years.

Question 18

Question

The period of a planet orbiting the Sun is 1.0 year and its semimajor axis is 2.0

astronomical units (AU). Determine the mass of the Sun. Assume the orbit of

the planet is nearly circular.

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

T2=ka3

where Tis the orbital period of the planet, ais the semimajor axis of the planet’s

orbit, and kis a constant that depends on the system under consideration.

Step 2: Substituting the given values into the equation, we have:

(1.0year)2=k(2.0AU)3

Step 3: Solve for the constant k:

1.0year =k×8.0AU3

k=1.0year

8.0AU3= 0.125 year/AU3

Step 4: Now, to find the mass of the Sun, we can use Newton’s version of

Kepler’s third law:

4π2

G(m1+m2)=a3

where m1is the mass of the Sun, m2is the mass of the planet (which we’ll

ignore), ais the semimajor axis of the planet’s orbit, Tis the orbital period of

the planet, and Gis the gravitational constant.

Step 5: Rearranging the equation to solve for the mass of the Sun, we get:

m2=4π2a3

GT 2

Step 6: Substituting the given values into the equation:

m2=4π2(2.0AU)3

G×(1.0year)2

Step 7: Calculate the mass of the Sun using the value of G(gravitational

constant) which is 6.67430 ×10−11 m3/kg ·s2.

m2=4π2×8.0AU3

6.67430 ×10−11 m3/kg ·s2=Answer

Question 19

Question

Given that a planet has an orbital period of 4.6 years around the Sun, what is

the semi-major axis of its orbit in astronomical units (AU)? Assume the planet

follows a circular orbit and use Kepler’s third law.

Solution

To find the semi-major axis of the planet’s orbit in AU, we can use Kepler’s

third law, which states that the square of a planet’s orbital period (T) is directly

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

Step 1: Convert the period of the planet’s orbit to seconds. The period of

the planet’s orbit is 4.6 years. We need to convert this to seconds:

1 year = 3.15 ×107seconds

So, the period of the planet’s orbit in seconds is:

4.6years ×3.15 ×107seconds/year = 1.449 ×108seconds

Step 2: Use Kepler’s third law to find the semi-major axis in AU. Kepler’s

third law is given by:

T2=k×a3

Where Tis the orbital period in seconds, ais the semi-major axis in AU, and k

is a constant.

Solving for a:

a=(T2

k)1/3

Since we are looking for the semi-major axis in AU, kwill be in units that

will make the result in AU.

k= 4π2AU3/s2

Plugging in the values:

a=((1.449 ×108s)2

4π2)1/3

a=(2.1036 ×1016 s2/39.478 AU3)1/3

a= 2.684 AU

Therefore, the semi-major axis of the planet’s orbit is approximately 2.684

AU.

Question 20

Question

Consider a star with a mass five times that of the Sun. If a planet of mass equal

to that of the Earth orbits this star in a circular orbit with a radius of 1 AU,

determine the period of the planet’s orbit around the star according to Kepler’s

Third Law of Planetary Motion.

Solution

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

square of the period of an orbiting planet is proportional to the cube of the

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

=T2

where T1and T2are the periods of the orbits, and R1and R2are the semi-major

axes of the orbits.

Step 2: Given that the planet’s mass is equal to that of the Earth, the mass

of the star is five times that of the Sun, and the planet orbits at a radius of

1 AU, we can use Kepler’s Third Law to find the period of the planet’s orbit

around the star.

Step 3: Since the mass of the star does not affect the period of the orbit in

Kepler’s Third Law, we can use the values for Earth’s orbit around the Sun to

find the period for this planet:

Earth

(1 AU)3=T2

planet

(1 AU)3

Step 4: Solving for the period of the planet’s orbit around the star:

Tplanet =TEarth ×√(1 AU)3

(1 AU)3

Tplanet =TEarth ×1

Step 5: The period of the Earth’s orbit around the Sun is approximately 1

year, which is equivalent to 365.25 days. Therefore, the period of the planet’s

orbit around the star would also be 1 year, or 365.25 days.

Question 21

Question

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

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

axis (a) of its orbit.

Given that Earth’s orbital period is approximately 365.25 days and its av-

erage distance from the Sun is 1 astronomical unit (AU), calculate the orbital

period of a hypothetical planet with an average distance from the Sun of 2 AU.

Solution

Step 1: Write down Kepler’s third law in mathematical form:

T2∝a3

Step 2: Use the data given for Earth:

Earth =k×a3

Earth

365.252=k×13

k=365.252

k= 133,225.06

Step 3: Use the calculated value of kto find the orbital period of the hypo-

thetical planet:

planet = 133,225.06 ×23

planet = 133,225.06 ×8

planet = 1,065,800.48

Step 4: Solve for the orbital period of the hypothetical planet:

Tplanet =√1,065,800.48

Tplanet ≈1031.91 days

Therefore, the orbital period of the hypothetical planet with an average

distance of 2 AU from the Sun would be approximately 1031.91 days.

Question 22

Question

The planet Neptune has an average distance from the Sun of about 4.50 billion

km. Calculate the orbital period of Neptune in years, given that the average

distance from the Sun to Earth is approximately 1.50 billion km.

(Note: You may assume that Neptune’s orbit is nearly circular.)

Solution

Step 1: Use Kepler’s Third Law to derive an expression relating the orbital

periods of two planets to their average distances from the Sun. This law states

that the square of the orbital period (T) of a planet is proportional to the cube

of its average distance from the Sun (r):

Neptune

Earth

=r3

Neptune

Earth

Step 2: Substitute the given values: rNeptune = 4.50 ×109km and rEarth =

1.50 ×109km. Remember that the period of Earth’s orbit is about 1 year:

Neptune

12=(4.50 ×109)3

(1.50 ×109)3

Step 3: Simplify the equation by squaring the right side:

Neptune =(4.50)3

(1.50)3×12

Step 4: Calculate the value on the right side:

Neptune =91.125

Step 5: Find the square root to get the orbital period of Neptune:

TNeptune =√91.125

Step 6: Calculate the value:

TNeptune ≈√30.375 ≈5.51 years

Therefore, the orbital period of Neptune is approximately 5.51 years.

Question 23

Question

The eccentricity of a planet’s orbit is a measure of how much the orbit devi-

ates from a perfect circle. Given that the eccentricity of Earth’s orbit is ap-

proximately 0.017, determine the shape of Earth’s orbit and explain what this

eccentricity value implies about the planet’s motion around the Sun.

Solution

Step 1: The eccentricity of an orbit determines its shape. Step 2: The shape of

an orbit is determined by its eccentricity as follows: - If eccentricity, e= 0, the

orbit is a perfect circle. - If eccentricity, 0< e < 1, the orbit is an ellipse. - If

eccentricity, e= 1, the orbit is a parabola. - If eccentricity, e > 1, the orbit is a

hyperbola. Step 3: Since the eccentricity of Earth’s orbit is approximately 0.017,

the orbit of Earth is an ellipse. Step 4: When the eccentricity of an orbit is close

to 0, as in the case of Earth’s orbit, the shape of the ellipse is very close to a

circle. Step 5: Therefore, the value of Earth’s eccentricity being approximately

0.017 implies that Earth’s orbit around the Sun is almost circular. Step 6: In

summary, the eccentricity value of 0.017 indicates that Earth’s orbit is very

close to being a perfect circle, with only a slight deviation.

Question 24

Question

The planet Mars has an average orbital radius of 1.52 astronomical units (AU)

from the Sun. If Mars takes approximately 687 Earth days to complete one

orbit, calculate the average orbital speed of Mars in kilometers per second.

Solution

Step 1: Calculate the circumference of Mars’s orbit using the formula

C= 2πr,

where ris the average orbital radius of Mars in astronomical units (AU). Note

that 1 AU is equal to approximately 149,597,870.7 kilometers. Step 2: Convert

the circumference of Mars’s orbit from astronomical units to kilometers. Step

3: Calculate the time it takes for Mars to complete one orbit in seconds. Step

4: Use the formula

v=C

where vis the average orbital speed of Mars in kilometers per second and Tis

the time it takes for Mars to complete one orbit in seconds.

Step 1: The circumference of Mars’s orbit is given by

C= 2π×1.52 AU.

Step 2: Converting the AU to kilometers, we have

C= 2π×1.52 ×149,597,870.7km.

Step 3: The time it takes for Mars to complete one orbit in seconds is

T= 687 ×24 ×60 ×60 seconds.

Step 4: Finally, the average orbital speed of Mars is

v=2π×1.52 ×149,597,870.7

687 ×24 ×60 ×60 km/s.

Question 25

Question

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

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

sun. Suppose a newly discovered planet, Planet X, has an orbital period of

24 years and an average distance from the sun of 12 astronomical units (AU).

What is the orbital period of another planet, Planet Y, if its average distance

from the sun is 20 AU?

Solution

To solve this problem, we can use Kepler’s third law of planetary motion, which

states that T2

=r3

, where T1and T2are the orbital periods of Planet X and

Planet Y, respectively, and r1and r2are the average distances from the sun for

Planet X and Planet Y, respectively.

Step 1: Substitute the given values into the equation. We are given that

T1= 24 years, r1= 12 AU, and r2= 20 AU. Let’s solve for T2.

242

=123

203

Step 2: Simplify the equation.

576

=1728

8000

Multiplying both sides by T2

2to get T2out of the denominator:

576 = 1728

8000T2

Step 3: Solve for T2.

2=576 ×8000

1728 = 2666.67

T2=√2666.67 ≈51.64 years

Thus, the orbital period of Planet Y, if its average distance from the sun is

20 AU, is approximately 51.64 years.

where: - bis the semi-minor axis of the orbit.

From the definition of eccentricity: eccentricity =√1+b2/a2

Let’s first determine busing the formula:

b=a√1−e2

b= 2√1−e2

We know that the area enclosed by the ellipse is the same as a circle with a

radius of 2 AU:

πab =π(2)2

π×2√1−e2×2 = π×4

4√1−e2= 4

√1−e2= 1

1−e2= 1

e2= 0

e= 0

Therefore, the eccentricity of the planet’s orbit is 0, indicating a circular

orbit.

Question 2

Question

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

planet is proportional to the cube of its semi-major axis. Consider two planets,

Planet A and Planet B, orbiting a star. Planet A has a semi-major axis of 2

AU and a period of 4 years, while Planet B has a semi-major axis of 4 AU.

Calculate the period of Planet B.

Solution

Step 1: We can express Kepler’s Third Law mathematically as:

(TA

TB)2

=(aA

aB)3

where: - TAand TBare the periods of Planet A and Planet B, respectively. -

aAand aBare the semi-major axes of Planet A and Planet B, respectively.

Step 2: Substituting the known values into the equation, we get:

TB)2

=(2

4)3

Step 3: Simplifying the equation gives:

TB)2

=(1

2)3

Step 4: Solving for TBby taking the square root of both sides, we have:

=√(1

2)3

Step 5: Simplifying further yields:

√2

Step 6: Multiplying both sides by TBgives:

TB= 4√2

Therefore, the period of Planet B is 4√2years.

Question 3

Question

An asteroid is orbiting the Sun in an elliptical orbit. The semi-major axis of its

orbit is 2.5 AU and the eccentricity of the orbit is 0.6. Determine the periapsis

and apoapsis distances of the asteroid from the Sun.

Solution

Let’s denote the periapsis distance as rpand the apoapsis distance as ra. The

periapsis distance occurs at the point where the asteroid is closest to the Sun,

while the apoapsis distance occurs at the point where the asteroid is farthest

from the Sun.

Step 1: Calculate the periapsis distance using the formula:

rp=a(1 −e)

where ais the semi-major axis and eis the eccentricity.

Substitute a= 2.5AU and e= 0.6into the formula:

rp= 2.5(1 −0.6) = 2.5×0.4 = 1 AU

Therefore, the periapsis distance of the asteroid from the Sun is 1 AU.

Step 2: Calculate the apoapsis distance using the formula:

ra=a(1 + e)

Substitute a= 2.5AU and e= 0.6into the formula:

ra= 2.5(1 + 0.6) = 2.5×1.6 = 4 AU

Therefore, the apoapsis distance of the asteroid from the Sun is 4 AU.

Question 4

Question

In a distant planetary system, the semi-major axis of a certain planet’s orbit is

3.5 AU. The period of the planet’s orbit around its star is 9.2 years. Determine

the mass of the star (in solar masses) given that the gravitational constant

G= 6.67 ×10−11 m3kg−1s−2.

Solution

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

of an orbit (T) and the semi-major axis of the orbit (a), as follows:

T2=(4π2

G·M)a3,

where Mis the mass of the star.

Step 2: We are given a= 3.5AU and T= 9.2years. We will convert the

semi-major axis from astronomical units (AU) to meters using the conversion

factor 1AU = 1.496 ×1011 m.

a= 3.5×1.496 ×1011 m= 5.236 ×1011 m.

Step 3: Substitute aand Tinto Kepler’s Third Law to solve for the mass of

the star M:

(9.2years)2=(4π2

6.67 ×10−11 m3kg−1s−2·M)(5.236 ×1011 m)3.

Step 4: Simplify and solve the equation for M:

9.22=4π2

6.67 ×10−11 ·M·(5.236 ×1011)3.

84.64 = 4π2

6.67 ×10−11 ·M·1.190 ×1035.

M=4π2

6.67 ×10−11 ·1.190 ×1035 .

Step 5: Calculate the mass of the star M:

M=4π2

6.67 ×10−11 ·1.190 ×1035 ≈39.48

7.95 ×1024 ≈4.96 ×1023 kg.

Step 6: Finally, convert the mass of the star from kilograms to solar masses

using the conversion factor 1solar mass = 1.989 ×1030 kg:

Mstar =4.96 ×1023 kg

1.989 ×1030 kg/solar mass ≈2.49 ×10−7solar masses.

Therefore, the mass of the star is approximately 2.49 ×10−7solar masses.

Question 5

Question

According to Kepler’s laws 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 3.5 astronomical units (AU). If another planet has a period of revolution

around the Sun that is twice as long as the first planet, what is the average

distance of the second planet from the Sun?

Solution

Step 1: Let Tbe the period of revolution of the first planet, dbe the average

distance from the Sun of the first planet, and T′and d′be the corresponding

quantities for the second planet. We know that according to Kepler’s Third

Law: T2∝d3. Therefore, for the first planet: T2=k·d3, where kis the

constant of proportionality.

Step 2: Since the second planet has a period of revolution that is twice as

long as the first planet, we have T′= 2T, which implies that T′= 2√k·d3.

Step 3: We now solve for the average distance of the second planet from

the Sun by appealing to Kepler’s Third Law. For the second planet, we have:

T′2=k·(d′)3. Substituting T′= 2√k·d3, we get: (2√k·d3)2=k·(d′)3.

Step 4: Simplifying, we have: 4·k·d3=k·(d′)3. Dividing both sides by k,

we have: 4·d3= (d′)3.

Step 5: Taking the cube root of both sides, we find: d′=3

√4·d. Given

that the average distance of the first planet from the Sun is 3.5 AU, we have:

d′=3

√4·3.5AU.

Step 6: Calculating the result, we find: d′= 3.5·3

√4≈3.5·1.587 =

5.557 AU. Therefore, the average distance of the second planet from the Sun is

approximately 5.557 astronomical units.

Question 6

Question

At perihelion, a planet is at its closest point to the Sun in its elliptical orbit.

Consider a planet with a semi-major axis of 2.5 AU and an eccentricity of 0.3:

1. Calculate the distance of the planet from the Sun at perihelion.

2. Calculate the distance of the planet from the Sun at aphelion.

Solution

1. To calculate the distance of the planet from the Sun at perihelion, we can

use the following formula based on the properties of an elliptical orbit:

rperi =a(1 −e)

where: - rperi is the distance at perihelion, - ais the semi-major axis, - eis the

eccentricity.

Substitute a= 2.5AU and e= 0.3into the formula:

rperi = 2.5×(1 −0.3)

rperi = 2.5×0.7 = 1.75 AU

Therefore, the planet’s distance from the Sun at perihelion is 1.75 AU.

2. To calculate the distance of the planet from the Sun at aphelion, we can

use a similar formula:

rapo =a(1 + e)

where: - rapo is the distance at aphelion, - ais the semi-major axis, - eis the

eccentricity.

Substitute a= 2.5AU and e= 0.3into the formula:

rapo = 2.5×(1 + 0.3)

rapo = 2.5×1.3 = 3.25 AU

Therefore, the planet’s distance from the Sun at aphelion is 3.25 AU.

Question 7

Question

Consider a planet orbiting a star in an elliptical orbit with the star located at

one of the foci. If the semi-major axis of the orbit is 3 AU and the eccentricity

of the orbit is 0.4, calculate the semi-minor axis of the orbit.

Solution

Step 1: Recall the relationship between the semi-major axis a, semi-minor axis

b, and eccentricity efor an elliptical orbit:

b=a√1−e2

Step 2: Given that the semi-major axis a= 3 AU and the eccentricity

e= 0.4, we can now substitute these values into the equation:

b= 3√1−0.42

Step 3: Calculate 1−0.42:

1−0.42= 0.84

Step 4: Substitute the result back into the equation:

b= 3√0.84

Step 5: Calculate √0.84:

√0.84 ≈0.917

Step 6: Finally, multiply the semi-major axis by the calculated value:

b= 3 ×0.917 = 2.751 AU

Therefore, the semi-minor axis of the orbit is approximately 2.751 AU.

Question 8

Question

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

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

has an orbital period of 5 years and an average distance from the sun of 3

astronomical units (AU), what would be the orbital period of another planet

with an average distance from the sun of 12 AU?

Solution

To solve this problem, we will first express Kepler’s third law mathematically

and then use the given information to find the orbital period of the second

planet.

Step 1: Write Kepler’s third law mathematically: Kepler’s third law states

that:

T2∝a3

where Tis the orbital period (in years) and ais the semi-major axis (in AU).

Step 2: Use the given values for the first planet: For the first planet, we

have:

T1= 5 years

a1= 3 AU

This gives us:

1∝a3

52∝33

25 ∝27

Step 3: Set up the proportion to find the orbital period of the second planet:

We can now set up a proportion using the information from the first planet and

the average distance from the sun of the second planet:

=T2

Step 4: Substitute the values and solve for the orbital period of the second

planet: Substitute the values we know:

27 =T2

123

27 =T2

1728

Now, solve for T2:

2=25

27 ×1728

2= 1600

T2=√1600

T2= 40 years

Therefore, the orbital period of the second planet with an average distance

from the sun of 12 AU would be 40 years.

Question 9

Question

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

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

Given that Earth has an orbital period of approximately 365.25 days and a

semi-major axis of approximately 1 astronomical unit (AU), calculate the orbital

period of a planet with a semi-major axis of 4 AU.

Solution

Step 1: Let T1be the orbital period of Earth and a1be the semi-major axis of

Earth’s orbit. Let T2be the orbital period of the unknown planet and a2be

the semi-major axis of the unknown planet’s orbit. According to Kepler’s third

law:

=T2

Step 2: Substitute the values for Earth’s orbital period (T1= 365.25 days)

and semi-major axis (a1= 1 AU) into the equation:

(365.25)2

13=T2

Step 3: Simplify the equation to solve for T2.

365.252=T2

2×43

2=365.252

2= 365.252×1

2=365.252

T2=√365.252

T2=365.25

8= 45.65625 days

Step 4: Therefore, the orbital period of a planet with a semi-major axis of 4

AU is approximately 45.66 days.

Question 10

Question

An exoplanet with a semi-major axis of 0.4 AU orbits a star with a mass of

1.5×1030 kg. If the exoplanet takes 100 Earth days to complete one full orbit

around the star, calculate the period of the exoplanet’s orbit (in Earth days)

according to Kepler’s third law.

Solution

Step 1: Convert the given values to SI units: The semi-major axis of the exo-

planet’s orbit a= 0.4AU = 0.4×1.496 ×1011 m= 5.984 ×1010 m. The mass

of the star M= 1.5×1030 kg. The period of the orbit T= 100 Earth days =

100 ×24 ×3600 seconds = 8.64 ×106seconds.

Step 2: Use Kepler’s third law to calculate the period of the exoplanet’s

orbit: Kepler’s third law states that T2=4π2

G(M+m)a3, where Tis the period of

the orbit, Gis the gravitational constant, Mis the mass of the star, mis the

mass of the exoplanet, and ais the semi-major axis of the exoplanet’s orbit.

Since the mass of the exoplanet is much smaller compared to the star’s mass,

we can neglect it in the equation.

Substitute the given values into the equation:

T2=4π2

6.67430×10−11 ×(1.5×1030 )×(5.984 ×1010)3

Step 3: Solve for the period of the exoplanet’s orbit: T2=4π2

1.001145×1020 ×

2.149052 ×1032

T2= 8.587107 ×104

T=√8.587107 ×104

T≈293 Earth days

Therefore, the period of the exoplanet’s orbit around the star is approxi-

mately 293 Earth days.

Question 11

Question

According to Kepler’s third law 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 2.5 astronomical

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

axis of 4.2 AU.

Solution

Step 1: 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 of planetary motion, the following

relationship holds:

=T2

Step 3: Substituting the given values for the first planet (orbital period = 8

years, semi-major axis = 2.5 AU) and the unknown values for the second planet,

we have: 82

2.53=T2

4.23

Step 4: Solving for T2, we get:

T2= 82×4.23

2.53

T2= 64 ×74.088

15.625

T2≈304.56

Step 5: Taking the square root of both sides to find the orbital period T, we

get:

T≈√304.56

T≈17.45 years

Step 6: Therefore, the orbital period of the second planet with a semi-major

axis of 4.2 AU is approximately 17.45 years.

Question 12

Question

A planet orbits around the sun in an elliptical path according to Kepler’s Laws

of Planetary Motion. The planet’s closest approach to the sun (perihelion) is

0.3 astronomical units (AU) and its farthest point from the sun (aphelion) is

0.7 AU. If the planet takes 250 days to complete one full orbit, find the average

speed of the planet in its orbit.

Solution

Step 1: First, we find the semi-major axis of the planet’s elliptical orbit, denoted

by a, which is the average of the perihelion and aphelion distances:

a=0.3+0.7

2= 0.5AU

Step 2: We calculate the eccentricity of the orbit, denoted by e, as the ratio

of the distance between the foci of the ellipse to the length of the major axis.

e=0.7−0.5

0.5= 0.4

Step 3: To find the average speed of the planet in its orbit, we can use

Kepler’s Third Law which relates the period of orbit (T) and the semi-major

axis (a) of the orbit: T2= 4π2a3/(G(M+m)), where Gis the gravitational

constant, Mis the mass of the sun, and mis the mass of the planet.

Step 4: Since we are interested in the average speed of the planet in its orbit,

we can find the total distance traveled by the planet in one complete orbit. This

distance is equal to the circumference of the ellipse, which can be approximated

by 2πa.

Step 5: Using the total distance traveled and the period of orbit, we can

calculate the average speed of the planet:

Average speed =Total distance

Total time

Step 6: Substituting the relevant values into the equation, we find:

Average speed =2π×0.5AU

250 days

Step 7: Converting the average speed from AU/day to km/s (1 AU = 1.496×

108km):

Average speed =2π×0.5×1.496 ×108

250 ×24 ×3600

Step 8: Calculating the final value gives us the average speed of the planet

in its orbit.

Question 13

Question

Consider a planet orbiting the Sun in an elliptical orbit. The planet’s closest

distance to the Sun is 0.3 AU and its farthest distance is 0.6 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 relates the period

of an orbit (T) to the semi-major axis of the orbit (a) through the equation:

T2=k·a3

where kis a constant specific for the system of measurements being used.

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

and farthest distances from the central body (in this case, the Sun). Therefore,

the semi-major axis ais given by:

a=rclosest +rfarthest

Plugging in the values, we have:

a=0.3+0.6

a= 0.45 AU

Step 3: We are given that the period Tof the orbit is 0.5 years. Substituting

T= 0.5years and a= 0.45 AU into Kepler’s Third Law, we get:

(0.5)2=k·(0.45)3

0.25 = k·0.091125

k≈2.743]

Step 4: Now, we can calculate the eccentricity (e) of the planet’s orbit using

the formula relating eccentricity to the closest and farthest distances:

e=rfarthest −rclosest

rfarthest +rclosest

Substitute rclosest = 0.3AU and rfarthest = 0.6AU into the formula:

e=0.6−0.3

0.6+0.3

e=0.3

0.9

e= 0.333]

Therefore, the eccentricity of the planet’s orbit is 0.333.

Question 14

Question

Let’s consider a hypothetical binary star system with two stars, A and B, orbit-

ing their common center of mass. Star A has a mass of 2.0×1030 kg, while Star

B has a mass of 3.0×1030 kg. If the distance between the two stars is 1.5×1011

m, determine the position of the center of mass of the system from Star A.

Solution

Step 1: Calculate the total mass of the system.

Given: Mass of Star A, ma= 2.0×1030 kg

Mass of Star B, mb= 3.0×1030 kg

The total mass of the system, mtotal, is the sum of the masses of both stars:

mtotal =ma+mb= 2.0×1030 kg + 3.0×1030 kg

mtotal = 5.0×1030 kg

Step 2: Calculate the position of the center of mass from Star A.

Let the position of the center of mass from Star A be x. To find this position,

we use the formula for the center of mass of a two-body system:

x=mb

mtotal

where: mbis the mass of Star B, mtotal is the total mass of the system, and d

is the distance between the two stars.

Substitute the given values into the formula:

x=3.0×1030 kg

5.0×1030 kg ×1.5×1011 m

x= 0.6×1.5×1011 m

x= 0.9×1011 m

Therefore, the position of the center of mass of the system from Star A is

9.0×1010 m.

Question 15

Question

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

of a planet’s orbit (T) is proportional to the cube of the semimajor axis of its

orbit (a). Given that Mars has an average orbital period of 687 days and an

average semimajor axis of 1.524 astronomical units (AU), calculate the orbital

period of Jupiter, whose average semimajor axis is 5.204 AU.

Solution

Step 1: First, let’s set up the proportion using Kepler’s third law:

Jupiter

=T2

Mars

where TJupiter is the orbital period of Jupiter, aJupiter is the semimajor axis of

Jupiter, TMars is the orbital period of Mars, and aMars is the semimajor axis of

Mars.

Step 2: Substitute the given values into the equation:

Jupiter

(5.204)3=(687)2

(1.524)3

Step 3: Solve for TJupiter by cross multiplying:

Jupiter =(687)2

(1.524)3×(5.204)3

Step 4: Calculate the square root to find TJupiter:

TJupiter =√(687)2

(1.524)3×(5.204)3

Step 5: Finally, compute the value of TJupiter:

TJupiter =√(472,369)

(3.276) ×(142.715) ≈4,332 days

Therefore, the orbital period of Jupiter is approximately 4,332 days.

Question 16

Question

The planet Mars orbits the Sun in an elliptical path with an eccentricity of

0.094, where the distance between the foci of the ellipse is 0.194 AU. If the

average distance between Mars and the Sun is about 1.52 AU, determine the

closest and farthest distance between Mars and the Sun during its orbit.

Solution

Step 1: Recall the equation for the distance between a planet and the Sun in

an elliptical orbit:

r=a(1 −e2)

1 + ecos(θ)

where ris the distance from the planet to the Sun, ais the semi-major axis, e

is the eccentricity, and θis the true anomaly.

Step 2: Given that the average distance between Mars and the Sun is 1.52

AU, we have:

a= 1.52 AU

Step 3: The distance between the foci of the ellipse is 2ae = 0.194 AU. Since

we know the eccentricity is 0.094, we can solve for a:

2ae = 0.194

2(1.52)(0.094) = 0.194

0.28544 = 0.194

a=0.194

2·0.094

a=0.194

0.188

a≈1.03 AU

Step 4: Now we can calculate the closest and farthest distances from Mars

to the Sun. The closest distance occurs when cos(θ) = −1, and the farthest

distance occurs when cos(θ) = 1. For the closest distance:

rmin =1.03(1 −0.0942)

1−0.094(−1)

rmin =1.03(1 −0.008836)

1+0.094

rmin =1.03(0.991164)

1.094

rmin ≈0.900 AU

For the farthest distance:

rmax =1.03(1 −0.0942)

1+0.094

rmax =1.03(1 −0.008836)

1+0.094

rmax =1.03(0.991164)

1.094

rmax ≈1.160 AU

Therefore, the closest distance between Mars and the Sun during its orbit is

approximately 0.900 AU, and the farthest distance is approximately 1.160 AU.

Question 17

Question

Two planets, A and B, are in orbit around a star. Planet A has a semi-major

axis of 2 AU and an orbital period of 3 years. Planet B has a semi-major axis

of 3 AU. Determine the orbital period of planet B.

Solution

Step 1: Recall Kepler’s Third Law, which 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. Mathematically, this can be written as:

A∝a3

where TAis the period of planet A, aAis the semi-major axis of planet A, and

the subscript Adenotes planet A.

Step 2: We are given that planet A has a semi-major axis aA= 2 AU and an

orbital period TA= 3 years. Substituting these values into the proportionality

equation, we get:

A=k·a3

(3)2=k·(2)3

9 = 8k

k=9

Step 3: Now, we can determine the orbital period of planet B using Kepler’s

Third Law:

B=(9

8)·(3)3

B=9

8·27

TB=√243

√8

TB=3√27

2√2

TB=3√9·3

2√2

TB=3·3√3

2√2

TB=9√3

2√2

TB=9

2·

√3

√2

TB=9

2·

√6

TB=9

4√6

Therefore, the orbital period of planet B is 9

4√6years.

Question 18

Question

The period of a planet orbiting the Sun is 1.0 year and its semimajor axis is 2.0

astronomical units (AU). Determine the mass of the Sun. Assume the orbit of

the planet is nearly circular.

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

T2=ka3

where Tis the orbital period of the planet, ais the semimajor axis of the planet’s

orbit, and kis a constant that depends on the system under consideration.

Step 2: Substituting the given values into the equation, we have:

(1.0year)2=k(2.0AU)3

Step 3: Solve for the constant k:

1.0year =k×8.0AU3

k=1.0year

8.0AU3= 0.125 year/AU3

Step 4: Now, to find the mass of the Sun, we can use Newton’s version of

Kepler’s third law:

4π2

G(m1+m2)=a3

where m1is the mass of the Sun, m2is the mass of the planet (which we’ll

ignore), ais the semimajor axis of the planet’s orbit, Tis the orbital period of

the planet, and Gis the gravitational constant.

Step 5: Rearranging the equation to solve for the mass of the Sun, we get:

m2=4π2a3

GT 2

Step 6: Substituting the given values into the equation:

m2=4π2(2.0AU)3

G×(1.0year)2

Step 7: Calculate the mass of the Sun using the value of G(gravitational

constant) which is 6.67430 ×10−11 m3/kg ·s2.

m2=4π2×8.0AU3

6.67430 ×10−11 m3/kg ·s2=Answer

Question 19

Question

Given that a planet has an orbital period of 4.6 years around the Sun, what is

the semi-major axis of its orbit in astronomical units (AU)? Assume the planet

follows a circular orbit and use Kepler’s third law.

Solution

To find the semi-major axis of the planet’s orbit in AU, we can use Kepler’s

third law, which states that the square of a planet’s orbital period (T) is directly

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

Step 1: Convert the period of the planet’s orbit to seconds. The period of

the planet’s orbit is 4.6 years. We need to convert this to seconds:

1 year = 3.15 ×107seconds

So, the period of the planet’s orbit in seconds is:

4.6years ×3.15 ×107seconds/year = 1.449 ×108seconds

Step 2: Use Kepler’s third law to find the semi-major axis in AU. Kepler’s

third law is given by:

T2=k×a3

Where Tis the orbital period in seconds, ais the semi-major axis in AU, and k

is a constant.

Solving for a:

a=(T2

k)1/3

Since we are looking for the semi-major axis in AU, kwill be in units that

will make the result in AU.

k= 4π2AU3/s2

Plugging in the values:

a=((1.449 ×108s)2

4π2)1/3

a=(2.1036 ×1016 s2/39.478 AU3)1/3

a= 2.684 AU

Therefore, the semi-major axis of the planet’s orbit is approximately 2.684

AU.

Question 20

Question

Consider a star with a mass five times that of the Sun. If a planet of mass equal

to that of the Earth orbits this star in a circular orbit with a radius of 1 AU,

determine the period of the planet’s orbit around the star according to Kepler’s

Third Law of Planetary Motion.

Solution

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

square of the period of an orbiting planet is proportional to the cube of the

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

=T2

where T1and T2are the periods of the orbits, and R1and R2are the semi-major

axes of the orbits.

Step 2: Given that the planet’s mass is equal to that of the Earth, the mass

of the star is five times that of the Sun, and the planet orbits at a radius of

1 AU, we can use Kepler’s Third Law to find the period of the planet’s orbit

around the star.

Step 3: Since the mass of the star does not affect the period of the orbit in

Kepler’s Third Law, we can use the values for Earth’s orbit around the Sun to

find the period for this planet:

Earth

(1 AU)3=T2

planet

(1 AU)3

Step 4: Solving for the period of the planet’s orbit around the star:

Tplanet =TEarth ×√(1 AU)3

(1 AU)3

Tplanet =TEarth ×1

Step 5: The period of the Earth’s orbit around the Sun is approximately 1

year, which is equivalent to 365.25 days. Therefore, the period of the planet’s

orbit around the star would also be 1 year, or 365.25 days.

Question 21

Question

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

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

axis (a) of its orbit.

Given that Earth’s orbital period is approximately 365.25 days and its av-

erage distance from the Sun is 1 astronomical unit (AU), calculate the orbital

period of a hypothetical planet with an average distance from the Sun of 2 AU.

Solution

Step 1: Write down Kepler’s third law in mathematical form:

T2∝a3

Step 2: Use the data given for Earth:

Earth =k×a3

Earth

365.252=k×13

k=365.252

k= 133,225.06

Step 3: Use the calculated value of kto find the orbital period of the hypo-

thetical planet:

planet = 133,225.06 ×23

planet = 133,225.06 ×8

planet = 1,065,800.48

Step 4: Solve for the orbital period of the hypothetical planet:

Tplanet =√1,065,800.48

Tplanet ≈1031.91 days

Therefore, the orbital period of the hypothetical planet with an average

distance of 2 AU from the Sun would be approximately 1031.91 days.

Question 22

Question

The planet Neptune has an average distance from the Sun of about 4.50 billion

km. Calculate the orbital period of Neptune in years, given that the average

distance from the Sun to Earth is approximately 1.50 billion km.

(Note: You may assume that Neptune’s orbit is nearly circular.)

Solution

Step 1: Use Kepler’s Third Law to derive an expression relating the orbital

periods of two planets to their average distances from the Sun. This law states

that the square of the orbital period (T) of a planet is proportional to the cube

of its average distance from the Sun (r):

Neptune

Earth

=r3

Neptune

Earth

Step 2: Substitute the given values: rNeptune = 4.50 ×109km and rEarth =

1.50 ×109km. Remember that the period of Earth’s orbit is about 1 year:

Neptune

12=(4.50 ×109)3

(1.50 ×109)3

Step 3: Simplify the equation by squaring the right side:

Neptune =(4.50)3

(1.50)3×12

Step 4: Calculate the value on the right side:

Neptune =91.125

Step 5: Find the square root to get the orbital period of Neptune:

TNeptune =√91.125

Step 6: Calculate the value:

TNeptune ≈√30.375 ≈5.51 years

Therefore, the orbital period of Neptune is approximately 5.51 years.