PHYS 305 - INTRODUCTION TO MODERN PHYSICS Kepler's laws Question Bank Set 1 | PHYS 305

PHYS 305 - INTRODUCTION TO

MODERN PHYSICS - Kepler’s laws

Question Bank - Set 1

Liberty University

Question 1

Question

A small planet orbits a star in a circular orbit with a period of 20 years. The

radius of the orbit is 4 AU (astronomical units). Calculate the mass of the star

in solar masses, assuming the gravitational constant is 6.67 ×10−11 N m2/kg2

and the mass of the Sun is 2×1030 kg.

Solution

Step 1: First, we need to find the orbital speed of the planet using Kepler’s

Third Law, which states T2∝R3. In this case, the period T= 20 years and

the radius R= 4 AU. Converting AU to meters, 1AU = 1.496 ×1011 m. So,

R= 4 ×1.496 ×1011 m= 5.984 ×1011 m

Thus, T2∝(5.984 ×1011)3, giving us the proportionality constant.

Step 2: Substituting the given period T= 20 years,

202yrs2=k×(5.984 ×1011)3m3

Solving for k, we get

k=202yrs2

(5.984 ×1011)3m3

Step 3: Next, we can find the orbital speed vusing the formula v=2πR

Substituting R= 5.984 ×1011 m and T= 20 years,

v=2π×5.984 ×1011 m

20 ×365 ×24 ×60 ×60 s

Step 4: Now, we can find the mass of the star using the formula F=GMm

R2=

mv2

Rand substituting vfrom the previous step along with Rand G. This gives

M=v2R

Step 5: Substituting the calculated values for v,R, and G, we find

M=(value of vfrom step 3)2×value of R

6.67 ×10−11 N m2/kg2

Step 6: Finally, the mass of the star in solar masses can be calculated by

dividing the obtained mass by the mass of the Sun, 2×1030 kg and converting

it to solar masses.

Question 2

Question

In a distant solar system, a planet orbits its star in an elliptical path with a

semi-major axis of 2.5×1011 meters. The planet’s speed when it is closest to

the star is 3.0×104m/s. Determine the planet’s speed when it is farthest from

the star. Assume the planet moves in accordance with Kepler’s laws.

Solution

Step 1: Recall that according to Kepler’s second law, a planet sweeps out equal

areas in equal times. This implies that the planet moves fastest when it is closest

to the star and slowest when it is farthest from the star.

Step 2: To find the planet’s speed when it is farthest from the star, we

first need to determine its angular momentum. Angular momentum is given by

L=mvr, where mis the planet’s mass, vis its velocity, and ris its distance

from the star.

Step 3: The planet’s angular momentum is constant, so we can set it equal

to the value when it is closest to the star:

m·3.0×104m/s ·2.5×1011 m=m·vf·rf,

where vfis the final speed (when the planet is farthest from the star) and rfis

the distance of the planet from the star when it is farthest.

Step 4: Solving for vf, we get:

3.0×104m/s ·2.5×1011 m=vf·2×2.5×1011 m.

Step 5: Simplifying further, we find:

7.5×1015 = 5 ×1011 ·vf.

Step 6: Therefore, the planet’s speed when it is farthest from the star is:

vf=7.5×1015

5×1011 = 1.5×104m/s .

Question 3

Question

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

and the star at the closest approach (perihelion) is 50 million kilometers, and

at the farthest point (aphelion) it is 100 million kilometers. If the period of the

planet’s orbit is 1 year, determine the semi-major axis of the orbit.

Solution

To determine the semi-major axis of the planet’s orbit, we can use Kepler’s third

law which states that the square of the period of an orbit is proportional to the

cube of the semi-major axis of the orbit.

Step 1: Determine the semi-major axis using the given information.

We know that the sum of the planet’s distances from the star at perihelion

and aphelion is equal to twice the semi-major axis (a) of the orbit. Therefore,

we have:

2a=perihelion distance +aphelion distance

Given that the perihelion distance is 50 million kilometers and the aphelion

distance is 100 million kilometers, we have:

2a= 50 + 100 = 150 million kilometers

a=150

2= 75 million kilometers

So, the semi-major axis of the planet’s orbit is 75 million kilometers.

Question 4

Question

Suppose a planet orbits a star in an elliptical path, with the eccentricity of the

orbit equal to 0.3. The planet’s closest distance to the star (perihelion) is 0.6

AU. What is the planet’s farthest distance from the star (aphelion) in AU?

Solution

Step 1: Recall that the formula relating the perihelion distance (minimum dis-

tance) and the aphelion distance (maximum distance) of an elliptical orbit is

given by the following equation:

a=rmin

1−e

where: a= semi-major axis of the elliptical orbit, rmin = perihelion distance, e

= eccentricity of the orbit.

Step 2: Substitute the given values into the formula:

a=0.6

1−0.3

Step 3: Calculate the semi-major axis a:

a=0.6

0.7= 0.8571 AU

Step 4: Remember that the aphelion distance (rmax ) in the elliptical orbit

is given by:

rmax =a(1 + e)

Step 5: Substitute the values of aand einto the formula:

rmax = 0.8571(1 + 0.3)

Step 6: Calculate the aphelion distance rmax :

rmax = 0.8571(1.3) = 1.1147 AU

Therefore, the planet’s farthest distance from the star (aphelion) is 1.1147

AU.

Question 5

Question

In a binary star system, two stars of equal mass orbit their center of mass. One

star is observed to have a period of 75 days and an average distance from the

center of mass of 1.5 astronomical units (AU). Determine the mass of each star

in the system.

Solution

Step 1: Recall Kepler’s Third Law for binary star systems, which relates the

orbital period and the average distance from the center of mass to the total

mass of the system:

P2=4π2

G(M1+M2)a3

where: - Pis the period of the orbit, - Gis the gravitational constant, - M1and

M2are the masses of the two stars, - ais the average distance from the center

of mass.

Step 2: First, convert the period from days to seconds:

75 days = 75 ×24 ×60 ×60 seconds = 6,480,000 sec

Step 3: Substitute the given values into Kepler’s Third Law formula:

(6,480,000)2=4π2

G(M1+M2)(1.5)3

Step 4: Simplify the equation and solve for the sum of the masses M1+M2.

Step 5: Once you have the sum of the masses, divide by 2 to find the

individual mass of each star, since they are equal in this system.

Step 6: Calculate the mass of each star using the total mass and write the

final answer with appropriate units.

Question 6

Question

A planet is in an elliptical orbit around the sun with a semi-major axis of 3 AU.

If the planet has an orbital period of 5 years, calculate the eccentricity of the

planet’s orbit.

Solution

Step 1: First, we can use Kepler’s third law, 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:

T2=ka3

where Tis the orbital period, ais the semi-major axis, and kis a constant of

proportionality.

Step 2: Given that the semi-major axis is 3 AU and the orbital period is 5

years, we have:

52=k×33

25 = 27k

k=25

Step 3: Next, we can use the definition of eccentricity (e) for an elliptical

orbit, which is given by:

e=√1−b2

where ais the semi-major axis and bis the semi-minor axis.

Step 4: The relationship between the semi-major axis a, semi-minor axis b,

and eccentricity eis given by:

a=1

1−e2

Step 5: Since we already know a= 3 AU and have found k, we can rearrange

the equation from Step 2 to solve for bas follows:

b=√a2(1 −e2)

Step 6: Substituting the values we know into the equation from Step 5, we

get:

b=√32(1 −e2)

b= 3√1−e2

Step 7: Since we know that the area of an ellipse is given by A=πab, we

can also express bin terms of Tand k:

b=√k

πT 2

Step 8: Substituting the values we know into the equation from Step 7, we

get:

3√1−e2=√25

27π

9(1 −e2) = 25

27π

9−9e2=25

27π

Step 9: Solving for eccentricity e, we get:

e=√1−25

27π×9

e=√1−25

243π

Question 7

Question

The period of a planet orbiting a star is found to be 15 years. If the distance

between the planet and the star is known to be 2.5×1012 meters, determine

the mass of the star in solar masses. Assume the orbit is nearly circular.

Solution

Step 1: Recall that Kepler’s third law states that the square of the period of

revolution of a planet is directly proportional to the cube of its average distance

from the Sun.

Step 2: The formula for Kepler’s third law is given by:

T2=(4π2

GM )R3

where: T= period of revolution of the planet, G= gravitational constant, M

= mass of the star, and R= distance between the planet and the star.

Step 3: Given that T= 15 years and R= 2.5×1012 meters, we need to

solve for Min solar masses.

Step 4: First, convert the period of revolution Tfrom years to seconds:

Tsec = 15 ×365 ×24 ×3600 = 473040000 sec

Step 5: Now, substitute the known values into Kepler’s third law formula:

(473040000)2=(4π2

G)(2.5×1012)3

Step 6: Solve for M:

M=4π2(2.5×1012)3

G=4π2(2.5×1012)3

6.674 ×10−11

Step 7: Calculate Min kilograms, then convert it to solar masses by dividing

by the mass of the Sun.

Step 8: The mass of the Sun is approximately 1.989 ×1030 kg.

Step 9: Calculate the mass of the star in solar masses:

Mass (solar masses) =M

1.989 ×1030

Step 10: Perform the final calculations to determine the mass of the star in

solar masses.

Question 8

Question

A planet orbits a star in a highly elliptical orbit such that the ratio of the lengths

of the major and minor axes is 3:1. The planet takes 300 days to complete the

orbit. Determine the eccentricity of the planet’s orbit.

Solution

Step 1: Recall that the period of an orbiting body Tis related to the semi-major

axis aby Kepler’s third law:

T2=4π2a3

where Gis the gravitational constant and Mis the mass of the star.

Step 2: For an ellipse, the relationship between the period T, the semi-major

axis a, and the eccentricity eis given by:

T= 2π√a3

GM(1 −e2)

Step 3: Given that the ratio of the major axis ato the minor axis bis 3:1,

we have a= 3b. Since the major axis is twice the semi-major axis, a= 2a0,

where a0is the semi-major axis.

Step 4: Substituting a= 2a0into the relationship in Step 2, we have:

T= 2π√(2a0)3

GM(1 −e2)

Step 5: Substituting T= 300 days into the equation above and simplifying,

we get:

3002= 4π2(2a0)3

GM(1 −e2)

Step 6: Substituting a= 3binto Kepler’s third law, we have:

3002=4π2(3b)3

Step 7: Solving for eby setting the two expressions for 3002equal to each

other gives:

4π2(6b)3

GM(1 −e2)=4π2(3b)3

Step 8: Simplifying the equation above and solving for e, we obtain:

e=√1−(3

6)2/3

Step 9: Therefore, the eccentricity of the planet’s orbit is approximately

0.598.

Question 9

Question

Given the following information about a planet orbiting a star: - The planet has

an orbital radius of 2 AU. - The star has a mass of 2×1030 kg. - The period of

the planet’s orbit is 1 year.

Calculate the eccentricity of the planet’s orbit.

Solution

Step 1: First, we need to find the gravitational force acting on the planet. The

gravitational force between the planet and the star is given by Newton’s law of

gravitation:

F=G·Mm·Ms

where: F= gravitational force between the planet and the star, G= 6.67×10−11

N m2/kg2(gravitational constant), Mm= 6 ×1024 kg (mass of the planet),

Ms= 2 ×1030 kg (mass of the star), r= 2 AU = 2 ×1.496 ×1011 m (orbital

radius of the planet in meters).

Step 2: Calculate the gravitational force between the planet and the star.

F=(6.67 ×10−11 N m2/kg2)·(6 ×1024 kg)·(2 ×1030 kg)

(2 ×1.496 ×1011 m)2

Step 3: Simplify the expression for the gravitational force.

F=8.004 ×105N

1.496 ×1022 m2= 5.35 ×10−17 N

Step 4: Next, we can use Kepler’s Third Law to calculate the eccentricity of

the planet’s orbit. Kepler’s Third Law states: T2=4π2

G(Mm+Ms)r3, where: T=

1year (period of the planet’s orbit), r= 2×1.496×1011 m (orbital radius of the

planet), G= 6.67 ×10−11 N m2/kg2(gravitational constant), Mm= 6 ×1024 kg

(mass of the planet), Ms= 2 ×1030 kg (mass of the star).

Step 5: Rearrange Kepler’s Third Law to solve for the eccentricity eof the

planet’s orbit.

e=√1−(r

a)2

Step 6: Substitute the given values for rand ainto the eccentricity formula

and calculate.

e=√1−(2×1.496 ×1011 m

a)2

Step 7: Solve for aby simplifying the expression.

a=3

√T2·G·(Mm+Ms)

4π2

Step 8: Substitute the given values for T,G,Mm, and Msinto the formula

for aand calculate.

a=3

√(1 year)2·(6.67 ×10−11 N m2/kg2)·(6 ×1024 + 2 ×1030 kg)

4π2

Step 9: Calculate aand substitute it back into the formula for the eccen-

tricity eto find the eccentricity of the planet’s orbit.

Question 10

Question

An asteroid in our solar system orbits the Sun in an elliptical path. The semi-

major axis of its orbit is 2.5 AU and its eccentricity is 0.4. Calculate the period

of revolution of the asteroid around the Sun.

Solution

Let’s denote the semi-major axis of the asteroid’s orbit as aand its eccentricity

as e. The period of revolution Tof an object in an elliptical orbit is related to

the semi-major axis by Kepler’s third law, given by the equation:

T2=(4π2

GM )a3

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

semi-major axis.

Step 1: Calculate the average distance of the asteroid from the Sun. The

semi-major axis ais related to the average distance rav of the asteroid from the

Sun by the equation:

rav =a(1 −e)

Plugging in the values a= 2.5AU and e= 0.4:

rav = 2.5AU ×(1 −0.4) = 2.5×0.6 = 1.5AU

Step 2: Using Newton’s form of Kepler’s third law, substitute a=rav into

the period formula:

T2=(4π2

GM )(1.5)3

Step 3: Calculate the period of revolution (T):

T2=(4π2

GM )(1.5)3

T=√(4π2

GM )(1.5)3

Thus, the period of revolution of the asteroid around the Sun is T=

√(4π2

GM )(1.5)3.

Question 11

Question

Consider a planet in an elliptical orbit around the Sun with a semi-major axis

of 4.0×1011 m. If the planet is closest to the Sun (perihelion) at a distance

of 3.0×1011 m, determine the maximum distance of the planet from the Sun

(aphelion) in meters.

Solution

Step 1: Recall Kepler’s second law which states that a planet sweeps out equal

areas in equal times. This implies that the speed of the planet in its elliptical

orbit is not constant.

Step 2: The semi-major axis of the orbit is given by a=rmin +rmax

2, where

rmin and rmax are the minimum and maximum distances of the planet from the

Sun (perihelion and aphelion).

Step 3: We are given a= 4.0×1011 m and rmin = 3.0×1011 m. Plugging

these values into the formula in Step 2 gives us 4.0×1011 =3.0×1011 +rmax

Step 4: Solving for rmax , we find rmax = 5.0×1011 m.

Therefore, the maximum distance of the planet from the Sun (aphelion) is

5.0×1011 meters.

Question 12

Question

A planet, with a mass of 3.20 ×1024 kg, orbits a star in a circular orbit with

a radius of 1.50 ×1011 m. The planet takes 5.64 Earth years to complete one

orbit around the star.

a) Calculate the gravitational force between the planet and the star.

b) Determine the speed of the planet in its orbit.

c) Find the period of the orbit in seconds.

Solution

a) To calculate the gravitational force between the planet and the star, we can

use Newton’s law of universal gravitation given by the formula:

F=G·m1·m2

where: - Fis the gravitational force, - G= 6.674 ×10−11 m3kg−1s−2is the

gravitational constant, - m1= 3.20 ×1024 kg is the mass of the planet, - m2

is the mass of the star (assumed to be much larger than the planet so often is

neglected), - r= 1.50 ×1011 m is the radius of the orbit.

Substitute the values into the formula to find the gravitational force.

Step 1: Calculate the gravitational force:

F=(6.674 ×10−11 m3kg−1s−2)·(3.20 ×1024 kg)

(1.50 ×1011 m)2

F=(2.134 ×1014)

(2.25 ×1022)

F= 9.482 ×106N

Thus, the gravitational force between the planet and the star is 9.482 ×106

b) The speed of the planet in its orbit can be found using the formula for

the centripetal force:

F=m·v2

where: - Fis the gravitational force acting as the centripetal force, - m= 3.20×

1024 kg is the mass of the planet, - vis the speed of the planet, - r= 1.50×1011 m

is the radius of the orbit.

Step 2: Calculate the speed of the planet:

9.482 ×106=(3.20 ×1024)·v2

1.50 ×1011

v2=9.482 ×106·1.50 ×1011

3.20 ×1024

v=√4.457 ×106

v= 2111 m/s

Therefore, the speed of the planet in its orbit is 2111 m/s.

c) The period of the orbit can be found using the formula:

T=2πr

where: - Tis the period of orbit we want to find, - πis the mathematical

constant π, and - r= 1.50 ×1011 m is the radius of the orbit, - v= 2111 m/s is

the speed of the planet.

Step 3: Calculate the period of the orbit:

T=2π·1.50 ×1011

2111

T=3×1011π

2111

T≈4.72 ×107

Question 13

Question

A small comet is orbiting the Sun in a circular orbit with a radius of 3.0 AU.

Calculate the period of the comet’s orbit in years. The mass of the Sun is

2.0×1030 kg, and the universal gravitational constant is 6.67×10−11 N m2/kg2.

Solution

Step 1: Write down Kepler’s third law which relates the period of an orbiting

body to its distance from the center of mass it is orbiting:

T2=(4π2

GM )r3

where: - Tis the period in seconds, - Gis the universal gravitational constant,

-Mis the mass of the object being orbited, - ris the average distance between

the two masses.

Step 2: Convert the given radius from astronomical units (AU) to meters:

3.0AU = 3.0×1.496 ×1011 m= 4.488 ×1011 m

Step 3: Substitute the known values into Kepler’s third law equation:

T2=(4π2

(6.67 ×10−11 N m2/kg2)(2.0×1030 kg))(4.488 ×1011 m)3

Step 4: Solve for Tby taking the square root of both sides of the equation:

T=√(4π2

(6.67 ×10−11 N m2/kg2)(2.0×1030 kg))(4.488 ×1011 m)3

Step 5: Calculate the period of the comet’s orbit:

T=√(4π2

(6.67 ×10−11)(2.0×1030))(4.488 ×1011)3

T=√(4π2

1.334 ×1020 )(9.102 ×1034)

T=√(9.422 ×10−13) (9.102 ×1034)

T≈√8.575 ×1022

T≈9.26 ×1011 seconds

Step 6: Convert the period from seconds to years:

T=9.26 ×1011 seconds

60 ×60 ×24 ×365.25 seconds/year ≈29.35 years

Therefore, the period of the comet’s orbit is approximately 29.35 years.

Question 14

Question

Consider a hypothetical planetary system where Planet X orbits a star in an

elliptical orbit. The semi-major axis of Planet X’s orbit is 2.5 AU and its

eccentricity is 0.4. Determine the closest and farthest distances of Planet X

from the star.

Solution

Let’s denote the semi-major axis of the elliptical orbit as aand the eccentricity

as e. The closest distance of the planet from the star (perihelion) occurs when

the planet is at one of the foci of the ellipse, and the farthest distance (aphelion)

occurs when the planet is at the farthest point from the star.

Step 1: Calculate the closest distance (perihelion): The closest distance

from the star occurs at the perihelion when the planet is at its nearest point.

This distance is given by:

rmin =a(1 −e)

Given that the semi-major axis a= 2.5AU and the eccentricity e= 0.4,

substitute these values into the formula:

rmin = 2.5×(1 −0.4) = 2.5×0.6 = 1.5AU

So, the closest distance of Planet X from the star is 1.5 AU.

Step 2: Calculate the farthest distance (aphelion): The farthest distance

from the star occurs at the aphelion when the planet is at its farthest point.

This distance is given by:

rmax =a(1 + e)

Given the same values of aand e, substitute into the formula:

rmax = 2.5×(1 + 0.4) = 2.5×1.4 = 3.5AU

Therefore, the farthest distance of Planet X from the star is 3.5 AU.

Question 15

Question

An asteroid has an elliptical orbit around the Sun with a semi-major axis of

3.5 AU and an eccentricity of 0.6. Determine the distance from the asteroid to

the Sun when the asteroid is at its closest point in its orbit (perihelion). The

average distance from the Earth to the Sun is 1 AU.

Solution

Step 1: First, we need to recall the formula for the distance from a point in an

elliptical orbit to the center (focus) of the ellipse. The formula is:

r=a(1 −e2)

1 + ecos(θ)

where: - ris the distance from the Sun to the asteroid, - ais the semi-major

axis of the orbit (given as 3.5 AU), - eis the eccentricity of the orbit (given as

0.6), - θis the angle between the asteroid and the point where it is closest to

the Sun.

Step 2: At the perihelion point, the angle θ= 0◦because it is the closest

point to the Sun. Substituting θ= 0◦into the formula, we get:

r=3.5(1 −0.62)

1+0.6 cos(0◦)

Step 3: We know that cos(0◦) = 1, so the equation simplifies to:

r=3.5(1 −0.62)

1+0.6

Step 4: Calculate the expression in the numerator:

r=3.5(1 −0.36)

1+0.6=3.5(0.64)

1.6=2.24

1.6= 1.4AU

Step 5: Therefore, the distance from the asteroid to the Sun when it is at

its closest point in its orbit (perihelion) is 1.4 AU.

Question 16

Question

Consider a planet in a circular orbit around a star. The period of this planet

is 20.0 years. If the planet’s distance from the star is tripled, calculate the new

period of revolution of the planet.

Solution

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

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

major axis of its elliptical orbit. Letting T1be the initial period, a1be the initial

semi-major axis, T2be the new period, and a2be the new semi-major axis, we

have: T2

=T2

Step 2: Given that T1= 20.0years, and the distance from the star is tripled,

we have a2= 3a1.

Step 3: Substitute T1= 20.0years and a2= 3a1into the proportionality

equation:

(20.0)2

=T2

(3a1)3

Step 4: Simplify the equation:

400

=T2

27a3

Step 5: Cancel out a3

1terms:

27 ·400 = T2

Step 6: Solve for T2:

T2=√27 ·400

Step 7: Calculate the new period of revolution of the planet:

T2=√10800 ≈104.0years

Therefore, the new period of revolution of the planet is approximately 104.0

years.

Question 17

Question

Consider a hypothetical solar system where there are two planets orbiting a star.

Planet A has a semi-major axis of 2 AU and a period of 3 years, while Planet B

has a semi-major axis of 3 AU and a period of 5 years. If Planet A and Planet

B are both orbiting the same star, determine which planet has a higher orbital

speed and explain your reasoning using Kepler’s third law.

Solution

To determine which planet has a higher orbital speed, we can use Kepler’s third

law, which relates the period of an orbiting body to the semi-major axis of its

orbit. Kepler’s third law can be written as:

=T2

where Tis the period of the planet and ais the semi-major axis of the planet.

Step 1: Substitute the given values into the equation.

For Planet A: 32

23=9

For Planet B: 52

33=25

Step 2: Determine which planet has a higher orbital speed.

Since 9

8<25

27 , Planet B has a higher orbital speed than Planet A. This means

that Planet B orbits the star at a higher speed compared to Planet A.

Question 18

Question

A planet is orbiting the Sun in an elliptical orbit with semi-major axis 2 AU. If

the planet takes 600 days to complete one full orbit, determine the eccentricity

of the planet’s orbit.

Solution

Given: Semi-major axis, a= 2 AU Orbital period, T= 600 days

We can relate the orbital period of a planet (T), the semi-major axis of its

orbit (a), and the eccentricity of its orbit (e) using Kepler’s third law:

T2=(4π2

G(M+m))a3

where Gis the gravitational constant, Mis the mass of the Sun, and mis

the mass of the planet. However, for simplicity in this problem, we will assume

that the mass of the planet is negligible compared to the mass of the Sun.

Step 1: Convert the orbital period Tfrom days to seconds. Given that 1

day = 86400 seconds, we have T= 600 ×86400 seconds.

Step 2: Substitute the given values into Kepler’s third law equation.

(600 ×86400)2=(4π2

G·M)(2)3

Step 3: Simplify the equation to solve for the eccentricity e.

e=√1−(b2

a2)

where bis the semi-minor axis of the ellipse, related to aand eby b=

a√1−e2in this case. Knowing this relationship, we can now solve for eusing

a= 2 AU.

Step 4: Calculate the eccentricity of the planet’s orbit. Plugging the given

semi-major axis aand the calculated orbital period Tinto the equations, we

can solve for the eccentricity of the planet’s orbit.

Question 19

Question

The semi-major axis of the orbit of a planet is 1.5 AU. If the planet takes 1.2

years to complete one orbit around the Sun, determine the mass of the Sun

using Kepler’s third law.

Solution

Step 1: Convert the semi-major axis to meters

Given that 1 AU (Astronomical Unit) is equal to 1.496 ×1011 meters, we can

convert the semi-major axis from AU to meters as follows:

a= 1.5AU ×1.496 ×1011 m/AU = 2.244 ×1011 m

Step 2: Convert the orbital period to seconds

Given that 1 year is equal to 365.25 days and 1 day is equal to 24 hours, we can

convert the orbital period from years to seconds as follows:

T= 1.2years ×365.25 days/year ×24 hours/day ×3600 s/hour = 3.785 ×107s

Step 3: Calculate the mass of the Sun using Kepler’s third law

Kepler’s third law states:

T2=4π2

G(M1+M2)a3

where: - Tis the orbital period, - Gis the gravitational constant, - M1is the

mass of the Sun, - M2is the mass of the planet, - ais the semi-major axis of

the orbit.

Since the mass of the planet is negligible compared to the mass of the Sun,

we can consider M2to be negligible. Thus, the equation simplifies to:

T2=4π2

GM1

Solving for M1:

M1=4π2

G(a3

T2)

Substitute the known values:

M1=4π2

6.67 ×10−11 m3/kg s2((2.244 ×1011 m)3

(3.785 ×107s)2)

Calculating this expression gives the mass of the Sun.

Question 20

Question

A planet orbits a star in an elliptical orbit, with the star located at one of the

foci of the ellipse. The semi-major axis of the orbit is 3.2 AU and the eccentricity

of the orbit is 0.6. Determine the semi-minor axis of the orbit in AU.

Solution

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

b, and eccentricity eof an ellipse:

a2=b2+ (a·e)2

Step 2: Substitute the given values a= 3.2AU and e= 0.6into the equation

to solve for b:

(3.2)2=b2+ (3.2·0.6)2

10.24 = b2+ (1.92)2

b2= 10.24 −3.6864

b2= 6.5536

Step 3: Take the square root of both sides to find the semi-minor axis b:

b=√6.5536

b= 2.56 AU

Therefore, the semi-minor axis of the orbit is 2.56 AU.

Question 21

Question

Suppose a planet is in an elliptical orbit around the Sun. The aphelion distance

(farthest distance from the Sun) is 1.5 AU, and the eccentricity of the orbit is

0.4. Find the perihelion distance (closest distance to the Sun) of the planet.

(Hint: Use Kepler’s second law)

Solution

Step 1: Recall Kepler’s second law which states that a line segment joining a

planet and the Sun sweeps out equal areas in equal times. This means that the

area of the elliptical sector between the planet and the Sun is constant.

Step 2: Let’s denote the aphelion distance as a(1.5 AU) and the eccentricity

of the orbit as e(0.4). The perihelion distance can be represented as rpand the

distance from the planet to the Sun at any given point as r.

Step 3: According to the conservation of angular momentum, we can relate

the aphelion distance to the perihelion distance using the formula a·(1 + e) =

rp·(1 + 0).

Step 4: Substituting the given values into the formula, we have 1.5×(1 +

0.4) = rp×(1).

Step 5: Solving for rp, we get 1.5×1.4 = rpwhich simplifies to rp= 2.1AU.

Thus, the perihelion distance of the planet is 2.1 AU.

Question 22

Question

Describe Kepler’s laws of planetary motion and explain how they are related to

the motion of planets around the Sun.

Solution

Kepler’s Laws of Planetary Motion:

First Law (Law of Ellipses): Each planet moves in an elliptical orbit,

with the Sun located at one of the two foci of the ellipse.

Second Law (Law of Equal Areas): A line segment joining a planet and

the Sun sweeps out equal areas during equal intervals of time. This implies that

a planet moves fastest when closest to the Sun (perihelion) and slowest when

farthest from the Sun (aphelion).

Third Law (Harmonic Law): 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.

Mathematically, this is expressed as T2∝a3.

Relationship to Planetary Motion:

Step 1: Kepler’s laws describe the motion of planets around the Sun in a

way that is consistent with Newton’s laws of motion and universal gravitation.

Step 2: The first law explains the shape of planetary orbits, which are not

perfect circles but rather elliptical. This understanding was a departure from

the previously held view of perfectly circular orbits.

Step 3: The second law provides insight into the speed at which a planet

moves at different points in its orbit, correlating orbital speed with distance

from the Sun.

Step 4: The third law establishes a relationship between the orbital periods

and distances of planets from the Sun, enabling predictions of the behavior of

planets in our solar system and beyond.

Step 5: Collectively, Kepler’s laws help explain the observed motion of

celestial bodies and laid the foundation for our understanding of gravitational

forces in the universe.

Question 23

Question

An asteroid orbits the Sun in an elliptical path with semi-major axis a=

3.5×1011 m and eccentricity e= 0.7. Find the distance of closest approach

(perihelion distance) and the distance of farthest retreat (aphelion distance) of

the asteroid from the Sun.

Solution

1. To find the perihelion distance (rmin ), we use the relationship between the

semi-major axis (a), eccentricity (e), and perihelion distance:

rmin =a(1 −e)

2. Substitute the given values of aand einto the formula:

rmin = 3.5×1011 m·(1 −0.7)

3. Calculate the perihelion distance:

rmin = 3.5×1011 m·0.3 = 1.05 ×1011 m

Therefore, the perihelion distance of the asteroid from the Sun is 1.05 ×1011

meters.

4. To find the aphelion distance (rmax ), we use the relationship:

rmax =a(1 + e)

5. Substitute the given values of aand einto the formula:

rmax = 3.5×1011 m·(1 + 0.7)

6. Calculate the aphelion distance:

rmax = 3.5×1011 m·1.7 = 5.95 ×1011 m

Therefore, the aphelion distance of the asteroid from the Sun is 5.95 ×1011

meters.

Question 24

Question

In a distant solar system, a planet follows an elliptical orbit around its star.

The semi-major axis of the orbit is 5.2 AU, and the eccentricity of the orbit is

0.4. Determine the distance between the planet and its star when the planet is

at its closest approach to the star.

Solution

Step 1: Recall the equation for the distance between a planet and its star in an

elliptical orbit:

r=a(1 −e)

1 + ecos θ

where: - ris the distance between the planet and its star, - ais the semi-major

axis of the orbit, - eis the eccentricity of the orbit, and - θis the true anomaly

of the planet.

Step 2: At the closest approach to the star, the planet is at its perihelion,

where θ= 0. Thus, we have:

r=a(1 −e)

1 + ecos 0 =a(1 −e)

1 + e

Step 3: Substitute the given values into the formula:

r=5.2AU ×(1 −0.4)

1+0.4

Step 4: Simplify the expression:

r=5.2×0.6

1.4=3.12

1.4= 2.23 AU

Step 5: Therefore, the distance between the planet and its star when the

planet is at its closest approach is 2.23 AU.

Question 25

Question

A planet orbits a star following Kepler’s laws of planetary motion. The semi-

major axis of the planet’s orbit is 2.5 AU. Calculate the period of the planet’s

orbit in years.

Given: G= 6.67 ×10−11 m3kg−1s−2M⋆= 2.0×1030 kg

Solution

Step 1: Calculate the mass of the planet using the third law of Kepler.

=T2

M⋆+m=4π2a3

GT 2

m=4π2a3

GT 2−M⋆

Given a= 2.5AU, M⋆= 2.0×1030 kg, and G= 6.67 ×10−11 m3kg−1s−2:

m=4π2(2.5×1.496 ×1011)3

6.67 ×10−11T2−2.0×1030

Step 2: Calculate the period of the planet using the second law of Kepler.

2πm =T

Given that the angular momentum, L=mv⊥rwhere v⊥is the speed of the

planet perpendicular to the line connecting the planet to the star:

mv⊥r

2πm =T

Step 3: Calculate the speed of the planet in its orbit.

v⊥=2πa

Given that a= 2.5AU, we can substitute this into the equation above:

v⊥=2π×2.5×1.496 ×1011

Step 4: Substitute the expression for v⊥into the equation for the period to

find T.

m(2π×2.5×1.496×1011

T)×a

2πm =T

Solving this equation will give us the period of the planet’s orbit in years.

Step 4: Now, we can find the mass of the star using the formula F=GMm

R2=

mv2

Rand substituting vfrom the previous step along with Rand G. This gives

M=v2R

Step 5: Substituting the calculated values for v,R, and G, we find

M=(value of vfrom step 3)2×value of R

6.67 ×10−11 N m2/kg2

Step 6: Finally, the mass of the star in solar masses can be calculated by

dividing the obtained mass by the mass of the Sun, 2×1030 kg and converting

it to solar masses.

Question 2

Question

In a distant solar system, a planet orbits its star in an elliptical path with a

semi-major axis of 2.5×1011 meters. The planet’s speed when it is closest to

the star is 3.0×104m/s. Determine the planet’s speed when it is farthest from

the star. Assume the planet moves in accordance with Kepler’s laws.

Solution

Step 1: Recall that according to Kepler’s second law, a planet sweeps out equal

areas in equal times. This implies that the planet moves fastest when it is closest

to the star and slowest when it is farthest from the star.

Step 2: To find the planet’s speed when it is farthest from the star, we

first need to determine its angular momentum. Angular momentum is given by

L=mvr, where mis the planet’s mass, vis its velocity, and ris its distance

from the star.

Step 3: The planet’s angular momentum is constant, so we can set it equal

to the value when it is closest to the star:

m·3.0×104m/s ·2.5×1011 m=m·vf·rf,

where vfis the final speed (when the planet is farthest from the star) and rfis

the distance of the planet from the star when it is farthest.

Step 4: Solving for vf, we get:

3.0×104m/s ·2.5×1011 m=vf·2×2.5×1011 m.

Step 5: Simplifying further, we find:

7.5×1015 = 5 ×1011 ·vf.

Step 6: Therefore, the planet’s speed when it is farthest from the star is:

vf=7.5×1015

5×1011 = 1.5×104m/s .

Question 3

Question

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

and the star at the closest approach (perihelion) is 50 million kilometers, and

at the farthest point (aphelion) it is 100 million kilometers. If the period of the

planet’s orbit is 1 year, determine the semi-major axis of the orbit.

Solution

To determine the semi-major axis of the planet’s orbit, we can use Kepler’s third

law which states that the square of the period of an orbit is proportional to the

cube of the semi-major axis of the orbit.

Step 1: Determine the semi-major axis using the given information.

We know that the sum of the planet’s distances from the star at perihelion

and aphelion is equal to twice the semi-major axis (a) of the orbit. Therefore,

we have:

2a=perihelion distance +aphelion distance

Given that the perihelion distance is 50 million kilometers and the aphelion

distance is 100 million kilometers, we have:

2a= 50 + 100 = 150 million kilometers

a=150

2= 75 million kilometers

So, the semi-major axis of the planet’s orbit is 75 million kilometers.

Question 4

Question

Suppose a planet orbits a star in an elliptical path, with the eccentricity of the

orbit equal to 0.3. The planet’s closest distance to the star (perihelion) is 0.6

AU. What is the planet’s farthest distance from the star (aphelion) in AU?

Solution

Step 1: Recall that the formula relating the perihelion distance (minimum dis-

tance) and the aphelion distance (maximum distance) of an elliptical orbit is

given by the following equation:

a=rmin

1−e

where: a= semi-major axis of the elliptical orbit, rmin = perihelion distance, e

= eccentricity of the orbit.

Step 2: Substitute the given values into the formula:

a=0.6

1−0.3

Step 3: Calculate the semi-major axis a:

a=0.6

0.7= 0.8571 AU

Step 4: Remember that the aphelion distance (rmax ) in the elliptical orbit

is given by:

rmax =a(1 + e)

Step 5: Substitute the values of aand einto the formula:

rmax = 0.8571(1 + 0.3)

Step 6: Calculate the aphelion distance rmax :

rmax = 0.8571(1.3) = 1.1147 AU

Therefore, the planet’s farthest distance from the star (aphelion) is 1.1147

AU.

Question 5

Question

In a binary star system, two stars of equal mass orbit their center of mass. One

star is observed to have a period of 75 days and an average distance from the

center of mass of 1.5 astronomical units (AU). Determine the mass of each star

in the system.

Solution

Step 1: Recall Kepler’s Third Law for binary star systems, which relates the

orbital period and the average distance from the center of mass to the total

mass of the system:

P2=4π2

G(M1+M2)a3

where: - Pis the period of the orbit, - Gis the gravitational constant, - M1and

M2are the masses of the two stars, - ais the average distance from the center

of mass.

Step 2: First, convert the period from days to seconds:

75 days = 75 ×24 ×60 ×60 seconds = 6,480,000 sec

Step 3: Substitute the given values into Kepler’s Third Law formula:

(6,480,000)2=4π2

G(M1+M2)(1.5)3

Step 4: Simplify the equation and solve for the sum of the masses M1+M2.

Step 5: Once you have the sum of the masses, divide by 2 to find the

individual mass of each star, since they are equal in this system.

Step 6: Calculate the mass of each star using the total mass and write the

final answer with appropriate units.

Question 6

Question

A planet is in an elliptical orbit around the sun with a semi-major axis of 3 AU.

If the planet has an orbital period of 5 years, calculate the eccentricity of the

planet’s orbit.

Solution

Step 1: First, we can use Kepler’s third law, 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:

T2=ka3

where Tis the orbital period, ais the semi-major axis, and kis a constant of

proportionality.

Step 2: Given that the semi-major axis is 3 AU and the orbital period is 5

years, we have:

52=k×33

25 = 27k

k=25

Step 3: Next, we can use the definition of eccentricity (e) for an elliptical

orbit, which is given by:

e=√1−b2

where ais the semi-major axis and bis the semi-minor axis.

Step 4: The relationship between the semi-major axis a, semi-minor axis b,

and eccentricity eis given by:

a=1

1−e2

Step 5: Since we already know a= 3 AU and have found k, we can rearrange

the equation from Step 2 to solve for bas follows:

b=√a2(1 −e2)

Step 6: Substituting the values we know into the equation from Step 5, we

get:

b=√32(1 −e2)

b= 3√1−e2

Step 7: Since we know that the area of an ellipse is given by A=πab, we

can also express bin terms of Tand k:

b=√k

πT 2

Step 8: Substituting the values we know into the equation from Step 7, we

get:

3√1−e2=√25

27π

9(1 −e2) = 25

27π

9−9e2=25

27π

Step 9: Solving for eccentricity e, we get:

e=√1−25

27π×9

e=√1−25

243π

Question 7

Question

The period of a planet orbiting a star is found to be 15 years. If the distance

between the planet and the star is known to be 2.5×1012 meters, determine

the mass of the star in solar masses. Assume the orbit is nearly circular.

Solution

Step 1: Recall that Kepler’s third law states that the square of the period of

revolution of a planet is directly proportional to the cube of its average distance

from the Sun.

Step 2: The formula for Kepler’s third law is given by:

T2=(4π2

GM )R3

where: T= period of revolution of the planet, G= gravitational constant, M

= mass of the star, and R= distance between the planet and the star.

Step 3: Given that T= 15 years and R= 2.5×1012 meters, we need to

solve for Min solar masses.

Step 4: First, convert the period of revolution Tfrom years to seconds:

Tsec = 15 ×365 ×24 ×3600 = 473040000 sec

Step 5: Now, substitute the known values into Kepler’s third law formula:

(473040000)2=(4π2

G)(2.5×1012)3

Step 6: Solve for M:

M=4π2(2.5×1012)3

G=4π2(2.5×1012)3

6.674 ×10−11

Step 7: Calculate Min kilograms, then convert it to solar masses by dividing

by the mass of the Sun.

Step 8: The mass of the Sun is approximately 1.989 ×1030 kg.

Step 9: Calculate the mass of the star in solar masses:

Mass (solar masses) =M

1.989 ×1030

Step 10: Perform the final calculations to determine the mass of the star in

solar masses.

Question 8

Question

A planet orbits a star in a highly elliptical orbit such that the ratio of the lengths

of the major and minor axes is 3:1. The planet takes 300 days to complete the

orbit. Determine the eccentricity of the planet’s orbit.

Solution

Step 1: Recall that the period of an orbiting body Tis related to the semi-major

axis aby Kepler’s third law:

T2=4π2a3

where Gis the gravitational constant and Mis the mass of the star.

Step 2: For an ellipse, the relationship between the period T, the semi-major

axis a, and the eccentricity eis given by:

T= 2π√a3

GM(1 −e2)

Step 3: Given that the ratio of the major axis ato the minor axis bis 3:1,

we have a= 3b. Since the major axis is twice the semi-major axis, a= 2a0,

where a0is the semi-major axis.

Step 4: Substituting a= 2a0into the relationship in Step 2, we have:

T= 2π√(2a0)3

GM(1 −e2)

Step 5: Substituting T= 300 days into the equation above and simplifying,

we get:

3002= 4π2(2a0)3

GM(1 −e2)

Step 6: Substituting a= 3binto Kepler’s third law, we have:

3002=4π2(3b)3

Step 7: Solving for eby setting the two expressions for 3002equal to each

other gives:

4π2(6b)3

GM(1 −e2)=4π2(3b)3

Step 8: Simplifying the equation above and solving for e, we obtain:

e=√1−(3

6)2/3

Step 9: Therefore, the eccentricity of the planet’s orbit is approximately

0.598.

Question 9

Question

Given the following information about a planet orbiting a star: - The planet has

an orbital radius of 2 AU. - The star has a mass of 2×1030 kg. - The period of

the planet’s orbit is 1 year.

Calculate the eccentricity of the planet’s orbit.

Solution

Step 1: First, we need to find the gravitational force acting on the planet. The

gravitational force between the planet and the star is given by Newton’s law of

gravitation:

F=G·Mm·Ms

where: F= gravitational force between the planet and the star, G= 6.67×10−11

N m2/kg2(gravitational constant), Mm= 6 ×1024 kg (mass of the planet),

Ms= 2 ×1030 kg (mass of the star), r= 2 AU = 2 ×1.496 ×1011 m (orbital

radius of the planet in meters).

Step 2: Calculate the gravitational force between the planet and the star.

F=(6.67 ×10−11 N m2/kg2)·(6 ×1024 kg)·(2 ×1030 kg)

(2 ×1.496 ×1011 m)2

Step 3: Simplify the expression for the gravitational force.

F=8.004 ×105N

1.496 ×1022 m2= 5.35 ×10−17 N

Step 4: Next, we can use Kepler’s Third Law to calculate the eccentricity of

the planet’s orbit. Kepler’s Third Law states: T2=4π2

G(Mm+Ms)r3, where: T=

1year (period of the planet’s orbit), r= 2 ×1.496 ×1011 m (orbital radius of the

planet), G= 6.67 ×10−11 N m2/kg2(gravitational constant), Mm= 6 ×1024 kg

(mass of the planet), Ms= 2 ×1030 kg (mass of the star).

Step 5: Rearrange Kepler’s Third Law to solve for the eccentricity eof the

planet’s orbit.

e=√1−(r

a)2

Step 6: Substitute the given values for rand ainto the eccentricity formula

and calculate.

e=√1−(2×1.496 ×1011 m

a)2

Step 7: Solve for aby simplifying the expression.

a=3

√T2·G·(Mm+Ms)

4π2

Step 8: Substitute the given values for T,G,Mm, and Msinto the formula

for aand calculate.

a=3

√(1 year)2·(6.67 ×10−11 N m2/kg2)·(6 ×1024 + 2 ×1030 kg)

4π2

Step 9: Calculate aand substitute it back into the formula for the eccen-

tricity eto find the eccentricity of the planet’s orbit.

Question 10

Question

An asteroid in our solar system orbits the Sun in an elliptical path. The semi-

major axis of its orbit is 2.5 AU and its eccentricity is 0.4. Calculate the period

of revolution of the asteroid around the Sun.

Solution

Let’s denote the semi-major axis of the asteroid’s orbit as aand its eccentricity

as e. The period of revolution Tof an object in an elliptical orbit is related to

the semi-major axis by Kepler’s third law, given by the equation:

T2=(4π2

GM )a3

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

semi-major axis.

Step 1: Calculate the average distance of the asteroid from the Sun. The

semi-major axis ais related to the average distance rav of the asteroid from the

Sun by the equation:

rav =a(1 −e)

Plugging in the values a= 2.5AU and e= 0.4:

rav = 2.5AU ×(1 −0.4) = 2.5×0.6 = 1.5AU

Step 2: Using Newton’s form of Kepler’s third law, substitute a=rav into

the period formula:

T2=(4π2

GM )(1.5)3

Step 3: Calculate the period of revolution (T):

T2=(4π2

GM )(1.5)3

T=√(4π2

GM )(1.5)3

Thus, the period of revolution of the asteroid around the Sun is T=

√(4π2

GM )(1.5)3.

Question 11

Question

Consider a planet in an elliptical orbit around the Sun with a semi-major axis

of 4.0×1011 m. If the planet is closest to the Sun (perihelion) at a distance

of 3.0×1011 m, determine the maximum distance of the planet from the Sun

(aphelion) in meters.

Solution

Step 1: Recall Kepler’s second law which states that a planet sweeps out equal

areas in equal times. This implies that the speed of the planet in its elliptical

orbit is not constant.

Step 2: The semi-major axis of the orbit is given by a=rmin +rmax

2, where

rmin and rmax are the minimum and maximum distances of the planet from the

Sun (perihelion and aphelion).

Step 3: We are given a= 4.0×1011 m and rmin = 3.0×1011 m. Plugging

these values into the formula in Step 2 gives us 4.0×1011 =3.0×1011 +rmax

Step 4: Solving for rmax , we find rmax = 5.0×1011 m.

Therefore, the maximum distance of the planet from the Sun (aphelion) is

5.0×1011 meters.

Question 12

Question

A planet, with a mass of 3.20 ×1024 kg, orbits a star in a circular orbit with

a radius of 1.50 ×1011 m. The planet takes 5.64 Earth years to complete one

orbit around the star.

a) Calculate the gravitational force between the planet and the star.

b) Determine the speed of the planet in its orbit.

c) Find the period of the orbit in seconds.

Solution

a) To calculate the gravitational force between the planet and the star, we can

use Newton’s law of universal gravitation given by the formula:

F=G·m1·m2

where: - Fis the gravitational force, - G= 6.674 ×10−11 m3kg−1s−2is the

gravitational constant, - m1= 3.20 ×1024 kg is the mass of the planet, - m2

is the mass of the star (assumed to be much larger than the planet so often is

neglected), - r= 1.50 ×1011 m is the radius of the orbit.

Substitute the values into the formula to find the gravitational force.

Step 1: Calculate the gravitational force:

F=(6.674 ×10−11 m3kg−1s−2)·(3.20 ×1024 kg)

(1.50 ×1011 m)2

F=(2.134 ×1014)

(2.25 ×1022)

F= 9.482 ×106N

Thus, the gravitational force between the planet and the star is 9.482 ×106

b) The speed of the planet in its orbit can be found using the formula for

the centripetal force:

F=m·v2

where: - Fis the gravitational force acting as the centripetal force, - m= 3.20×

1024 kg is the mass of the planet, - vis the speed of the planet, - r= 1.50×1011 m

is the radius of the orbit.

Step 2: Calculate the speed of the planet:

9.482 ×106=(3.20 ×1024)·v2

1.50 ×1011

v2=9.482 ×106·1.50 ×1011

3.20 ×1024

v=√4.457 ×106

v= 2111 m/s

Therefore, the speed of the planet in its orbit is 2111 m/s.

c) The period of the orbit can be found using the formula:

T=2πr

where: - Tis the period of orbit we want to find, - πis the mathematical

constant π, and - r= 1.50 ×1011 m is the radius of the orbit, - v= 2111 m/s is

the speed of the planet.

Step 3: Calculate the period of the orbit:

T=2π·1.50 ×1011

2111

T=3×1011π

2111

T≈4.72 ×107

Question 13

Question

A small comet is orbiting the Sun in a circular orbit with a radius of 3.0 AU.

Calculate the period of the comet’s orbit in years. The mass of the Sun is

2.0×1030 kg, and the universal gravitational constant is 6.67×10−11 N m2/kg2.

Solution

Step 1: Write down Kepler’s third law which relates the period of an orbiting

body to its distance from the center of mass it is orbiting:

T2=(4π2

GM )r3

where: - Tis the period in seconds, - Gis the universal gravitational constant,

-Mis the mass of the object being orbited, - ris the average distance between

the two masses.

Step 2: Convert the given radius from astronomical units (AU) to meters:

3.0AU = 3.0×1.496 ×1011 m= 4.488 ×1011 m

Step 3: Substitute the known values into Kepler’s third law equation:

T2=(4π2

(6.67 ×10−11 N m2/kg2)(2.0×1030 kg))(4.488 ×1011 m)3

Step 4: Solve for Tby taking the square root of both sides of the equation:

T=√(4π2

(6.67 ×10−11 N m2/kg2)(2.0×1030 kg))(4.488 ×1011 m)3

Step 5: Calculate the period of the comet’s orbit:

T=√(4π2

(6.67 ×10−11)(2.0×1030))(4.488 ×1011)3

T=√(4π2

1.334 ×1020 )(9.102 ×1034)

T=√(9.422 ×10−13) (9.102 ×1034)

T≈√8.575 ×1022

T≈9.26 ×1011 seconds

Step 6: Convert the period from seconds to years:

T=9.26 ×1011 seconds

60 ×60 ×24 ×365.25 seconds/year ≈29.35 years

Therefore, the period of the comet’s orbit is approximately 29.35 years.

Question 14

Question

Consider a hypothetical planetary system where Planet X orbits a star in an

elliptical orbit. The semi-major axis of Planet X’s orbit is 2.5 AU and its

eccentricity is 0.4. Determine the closest and farthest distances of Planet X

from the star.

Solution

Let’s denote the semi-major axis of the elliptical orbit as aand the eccentricity

as e. The closest distance of the planet from the star (perihelion) occurs when

the planet is at one of the foci of the ellipse, and the farthest distance (aphelion)

occurs when the planet is at the farthest point from the star.

Step 1: Calculate the closest distance (perihelion): The closest distance

from the star occurs at the perihelion when the planet is at its nearest point.

This distance is given by:

rmin =a(1 −e)

Given that the semi-major axis a= 2.5AU and the eccentricity e= 0.4,

substitute these values into the formula:

rmin = 2.5×(1 −0.4) = 2.5×0.6 = 1.5AU

So, the closest distance of Planet X from the star is 1.5 AU.

Step 2: Calculate the farthest distance (aphelion): The farthest distance

from the star occurs at the aphelion when the planet is at its farthest point.

This distance is given by:

rmax =a(1 + e)

Given the same values of aand e, substitute into the formula:

rmax = 2.5×(1 + 0.4) = 2.5×1.4 = 3.5AU

Therefore, the farthest distance of Planet X from the star is 3.5 AU.

Question 15

Question

An asteroid has an elliptical orbit around the Sun with a semi-major axis of

3.5 AU and an eccentricity of 0.6. Determine the distance from the asteroid to

the Sun when the asteroid is at its closest point in its orbit (perihelion). The

average distance from the Earth to the Sun is 1 AU.

Solution

Step 1: First, we need to recall the formula for the distance from a point in an

elliptical orbit to the center (focus) of the ellipse. The formula is:

r=a(1 −e2)

1 + ecos(θ)

where: - ris the distance from the Sun to the asteroid, - ais the semi-major

axis of the orbit (given as 3.5 AU), - eis the eccentricity of the orbit (given as

0.6), - θis the angle between the asteroid and the point where it is closest to

the Sun.

Step 2: At the perihelion point, the angle θ= 0◦because it is the closest

point to the Sun. Substituting θ= 0◦into the formula, we get:

r=3.5(1 −0.62)

1+0.6 cos(0◦)

Step 3: We know that cos(0◦) = 1, so the equation simplifies to:

r=3.5(1 −0.62)

1+0.6

Step 4: Calculate the expression in the numerator:

r=3.5(1 −0.36)

1+0.6=3.5(0.64)

1.6=2.24

1.6= 1.4AU

Step 5: Therefore, the distance from the asteroid to the Sun when it is at

its closest point in its orbit (perihelion) is 1.4 AU.

Question 16

Question

Consider a planet in a circular orbit around a star. The period of this planet

is 20.0 years. If the planet’s distance from the star is tripled, calculate the new

period of revolution of the planet.

Solution

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

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

major axis of its elliptical orbit. Letting T1be the initial period, a1be the initial

semi-major axis, T2be the new period, and a2be the new semi-major axis, we

have: T2

=T2

Step 2: Given that T1= 20.0years, and the distance from the star is tripled,

we have a2= 3a1.

Step 3: Substitute T1= 20.0years and a2= 3a1into the proportionality

equation:

(20.0)2

=T2

(3a1)3

Step 4: Simplify the equation:

400

=T2

27a3

Step 5: Cancel out a3

1terms:

27 ·400 = T2

Step 6: Solve for T2:

T2=√27 ·400

Step 7: Calculate the new period of revolution of the planet:

T2=√10800 ≈104.0years

Therefore, the new period of revolution of the planet is approximately 104.0

years.

Question 17

Question

Consider a hypothetical solar system where there are two planets orbiting a star.

Planet A has a semi-major axis of 2 AU and a period of 3 years, while Planet B

has a semi-major axis of 3 AU and a period of 5 years. If Planet A and Planet

B are both orbiting the same star, determine which planet has a higher orbital

speed and explain your reasoning using Kepler’s third law.

Solution

To determine which planet has a higher orbital speed, we can use Kepler’s third

law, which relates the period of an orbiting body to the semi-major axis of its

orbit. Kepler’s third law can be written as:

=T2

where Tis the period of the planet and ais the semi-major axis of the planet.

Step 1: Substitute the given values into the equation.

For Planet A: 32

23=9

For Planet B: 52

33=25

Step 2: Determine which planet has a higher orbital speed.

Since 9

8<25

27 , Planet B has a higher orbital speed than Planet A. This means

that Planet B orbits the star at a higher speed compared to Planet A.

Question 18

Question

A planet is orbiting the Sun in an elliptical orbit with semi-major axis 2 AU. If

the planet takes 600 days to complete one full orbit, determine the eccentricity

of the planet’s orbit.

Solution

Given: Semi-major axis, a= 2 AU Orbital period, T= 600 days

We can relate the orbital period of a planet (T), the semi-major axis of its

orbit (a), and the eccentricity of its orbit (e) using Kepler’s third law:

T2=(4π2

G(M+m))a3

where Gis the gravitational constant, Mis the mass of the Sun, and mis

the mass of the planet. However, for simplicity in this problem, we will assume

that the mass of the planet is negligible compared to the mass of the Sun.

Step 1: Convert the orbital period Tfrom days to seconds. Given that 1

day = 86400 seconds, we have T= 600 ×86400 seconds.

Step 2: Substitute the given values into Kepler’s third law equation.

(600 ×86400)2=(4π2

G·M)(2)3

Step 3: Simplify the equation to solve for the eccentricity e.

e=√1−(b2

a2)

where bis the semi-minor axis of the ellipse, related to aand eby b=

a√1−e2in this case. Knowing this relationship, we can now solve for eusing

a= 2 AU.

Step 4: Calculate the eccentricity of the planet’s orbit. Plugging the given

semi-major axis aand the calculated orbital period Tinto the equations, we

can solve for the eccentricity of the planet’s orbit.

Question 19

Question

The semi-major axis of the orbit of a planet is 1.5 AU. If the planet takes 1.2

years to complete one orbit around the Sun, determine the mass of the Sun

using Kepler’s third law.

Solution

Step 1: Convert the semi-major axis to meters

Given that 1 AU (Astronomical Unit) is equal to 1.496 ×1011 meters, we can

convert the semi-major axis from AU to meters as follows:

a= 1.5AU ×1.496 ×1011 m/AU = 2.244 ×1011 m

Step 2: Convert the orbital period to seconds

Given that 1 year is equal to 365.25 days and 1 day is equal to 24 hours, we can

convert the orbital period from years to seconds as follows:

T= 1.2years ×365.25 days/year ×24 hours/day ×3600 s/hour = 3.785 ×107s

Step 3: Calculate the mass of the Sun using Kepler’s third law

Kepler’s third law states:

T2=4π2

G(M1+M2)a3

where: - Tis the orbital period, - Gis the gravitational constant, - M1is the

mass of the Sun, - M2is the mass of the planet, - ais the semi-major axis of

the orbit.

Since the mass of the planet is negligible compared to the mass of the Sun,

we can consider M2to be negligible. Thus, the equation simplifies to:

T2=4π2

GM1

Solving for M1:

M1=4π2

G(a3

T2)

Substitute the known values:

M1=4π2

6.67 ×10−11 m3/kg s2((2.244 ×1011 m)3

(3.785 ×107s)2)

Calculating this expression gives the mass of the Sun.

Question 20

Question

A planet orbits a star in an elliptical orbit, with the star located at one of the

foci of the ellipse. The semi-major axis of the orbit is 3.2 AU and the eccentricity

of the orbit is 0.6. Determine the semi-minor axis of the orbit in AU.

Solution

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

b, and eccentricity eof an ellipse:

a2=b2+ (a·e)2

Step 2: Substitute the given values a= 3.2AU and e= 0.6into the equation

to solve for b:

(3.2)2=b2+ (3.2·0.6)2

10.24 = b2+ (1.92)2

b2= 10.24 −3.6864

b2= 6.5536

Step 3: Take the square root of both sides to find the semi-minor axis b:

b=√6.5536

b= 2.56 AU

Therefore, the semi-minor axis of the orbit is 2.56 AU.

Question 21

Question

Suppose a planet is in an elliptical orbit around the Sun. The aphelion distance

(farthest distance from the Sun) is 1.5 AU, and the eccentricity of the orbit is

0.4. Find the perihelion distance (closest distance to the Sun) of the planet.

(Hint: Use Kepler’s second law)