PHYS 305 - INTRODUCTION TO MODERN PHYSICS Kepler's laws Question Bank Set 3 | PHYS 305

PHYS 305 - INTRODUCTION TO

MODERN PHYSICS - Kepler’s laws

Question Bank - Set 3

Liberty University

Question 1

Question

Consider a planet in a circular orbit around a star in a distant solar system.

The period of the planet’s orbit is 150 days and the distance between the planet

and the star is 0.7 AU. Determine the mass of the star using Kepler’s third law.

Solution

To find the mass of the star, we will use Kepler’s third law which relates the

period of an orbit (T), the semi-major axis of the orbit (a), and the masses of

the objects involved. The formula for Kepler’s third law is given by:

T2=(4π2

G(M+m))a3

Where: - Tis the orbital period - Gis the gravitational constant (6.674 ×

10−11 m3kg−1s−2) - Mis the mass of the star - ais the semi-major axis of the

orbit - mis the mass of the planet

Given that T= 150 days, a= 0.7AU, and m≪M(the mass of the planet

is negligible compared to the mass of the star), we can solve for M.

Step 1: Convert units First, we need to convert the period Tand semi-

major axis ainto SI units. - 1 day = 86400 seconds - 1 AU = 1.496 ×1011

meters

Converting, we have T= 150 ×86400 seconds and a= 0.7×1.496 ×1011

meters.

Step 2: Substitute values into Kepler’s third law equation Plugging

in the known values, we get:

(150 ×86400)2=(4π2

G(M))(0.7×1.496 ×1011)3

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

M=4π2(0.7×1.496 ×1011)3

G(150 ×86400

2π)2

Calculating this expression will give us the mass of the star in kilograms.

Question 2

Question

A planet is in an elliptical orbit around the Sun. The closest approach of the

planet to the Sun (perihelion) is 28 million miles and the farthest distance from

the Sun (aphelion) is 40 million miles. If the planet takes 440 days to complete

one full orbit, determine the semi-major axis of the planet’s orbit.

Solution

Step 1: Recall 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:

T2=ka3

where Tis the period of the planet’s orbit and ais the semi-major axis of

the planet’s orbit.

Step 2: To determine the semi-major axis of the planet’s orbit, we need to

first calculate the period of the planet’s orbit. Given that the planet takes 440

days to complete one full orbit, we have:

T= 440 days

Step 3: Now, we need to convert the period to years, since the standard unit

for time in Kepler’s third law is years. There are 365 days in a year, so:

T=440 days

365 days/year = 1.205 years

Step 4: Substitute the period into Kepler’s third law equation and solve for

the semi-major axis:

(1.205 years)2=ka3

Step 5: Next, we need to determine the value of the constant k. This can be

done by using the known period and semi-major axis of a known planet, such as

Earth. For Earth, T= 1 year and a= 1 astronomical unit (AU). Substituting

these values into Kepler’s third law, we can solve for k:

(1 year)2=k(1 AU)3

k= 1

Step 6: Substitute the value of kand the period Tinto Kepler’s third law

equation to solve for the semi-major axis a:

(1.205)2=a3

Step 7: Solve for a:

a=3

√(1.205)2= 1.236 AU

Step 8: Therefore, the semi-major axis of the planet’s orbit is approximately

1.236 astronomical units.

Question 3

Question

A comet discovered in a distant solar system has an orbital period of 6 years

and an average distance from the sun of 4 AU. Calculate the mass of the star

around which the comet orbits.

Given: Gravitational constant, G= 6.67 ×10−11 N m2/kg2

Solution

Step 1: Convert the average distance from Astronomical Units (AU) to meters.

Since 1 AU is approximately 1.496 ×1011 meters,

Average distance = 4 AU = 4 ×1.496 ×1011 m

Step 2: Use Kepler’s third law to find the star’s mass. Kepler’s third law

states:

T2=(4π2

G(M1+M2))a3

where: - Tis the orbital period in seconds - Gis the gravitational constant -

M1and M2are the masses of the bodies - ais the semi-major axis of the orbit

Step 3: Rearrange Kepler’s third law to solve for the star’s mass, M1.

M1=4π2

G(a3

T2)−M2

Step 4: Substitute the known values into the equation.

M1=4π2

6.67 ×10−11 ((4 ×1.496 ×1011)3

(6 ×365.25 ×24 ×3600)2)

Step 5: Calculate the mass of the star.

M1=4π2

6.67 ×10−11 ((5.984 ×1011)3

1.894 ×108×102)

Question 4

Question

The planet Mercury has an average orbital speed of 47.87 km/s and an average

orbital radius of 5.79 ×1010 m. Using Kepler’s third law, determine the period

of Mercury’s orbit around the Sun.

Solution

Step 1: Recall Kepler’s third law, which states that the square of the orbital

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

matically, this can be expressed as:

T2=4π2

G(M1+M2)a3

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

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

orbit.

Step 2: Rearrange the equation to solve for the period T:

T=√4π2

G(M1+M2)a3

Step 3: We can assume the mass of Mercury is negligible compared to the

mass of the Sun, so M1≫M2. Also, we know that the mass of the Sun is

approximately 1.989 ×1030 kg and G= 6.67430 ×10−11 m3kg−1s−2.

Step 4: Substitute the given values for the semi-major axis aand solve for

the period T:

T=√4π2

6.67430 ×10−11(1.989 ×1030 )(5.79 ×1010 )3

Step 5: Calculate the period Tusing the average orbital speed of Mercury:

T=√4π2

1.327124 ×1020 (1.2087591 ×1031)

Step 6: Simplify the equation and solve for the period Tto find the answer.

Question 5

Question

Consider a planet in a circular orbit around a star. The planet has a mass of

3.21 ×1024 kg and the star has a mass of 1.99 ×1030 kg. The radius of the orbit

is 1.50 ×1011 m. Calculate the period of the planet’s orbit in seconds.

(Given: G= 6.67 ×10−11 m3kg−1s−2)

Solution

Step 1: Calculate the gravitational force between the planet and the star using

Newton’s law of universal gravitation:

F=G·M1·M2

where Fis the gravitational force, Gis the gravitational constant, M1and M2

are the masses of the objects, and ris the distance between the centers of the

objects.

Step 2: Substitute the given values into the equation to find the gravitational

force.

Step 3: Use the gravitational force to calculate the centripetal force required

to keep the planet in circular motion:

Fcentripetal =M·v2

where Mis the mass of the planet, vis the velocity of the planet, and ris the

radius of the orbit.

Step 4: Equate the centripetal force to the gravitational force and solve for

the velocity vof the planet.

Step 5: Use the formula for the period of circular motion to find the period

T=2πr

where ris the radius of the orbit and vis the velocity of the planet found in

the previous step.

Step 6: Substitute the known values into the period formula to calculate the

period of the planet’s orbit.

Question 6

Question

Kepler’s third law states that the square of the period of revolution of a planet

around the Sun is directly proportional to the cube of its average distance from

the Sun. For a hypothetical planet in our solar system with an average distance

from the Sun of 2 AU, its period of revolution is 5 years. Calculate the average

distance from the Sun of a different planet in our solar system with a period of

revolution of 10 years.

Solution

Step 1: Let T1be the period of revolution of the first planet and R1be its

average distance from the Sun. According to Kepler’s third law:

1∝R3

Step 2: Given that the first planet has T1= 5 years and R1= 2 AU, we

have:

52∝23

25 ∝8

=25

Step 3: Let T2be the period of revolution of the second planet and R2be

its average distance from the Sun. According to Kepler’s third law:

2∝R3

Step 4: Since the second planet has T2= 10 years, we can set up a proportion

using the ratio obtained in Step 2:

=T2

8=102

Step 5: Solve for R2:

2=8×100

2=800

2= 32

R2=3

√32

R2= 2 AU

Therefore, the average distance from the Sun of the different planet in our

solar system with a period of revolution of 10 years is 2 AU.

Question 7

Question

Consider a binary star system consisting of two stars with masses M1= 2.0×

1030 kg and M2= 1.5×1030 kg, separated by a distance of 1.0×1012 m. If the

stars are in circular orbit around their center of mass at a speed of 2.5×104m/s,

determine the period of the stars’ motion and the gravitational force between

them.

Solution

Step 1: Firstly, we need to find the total mass of the system.

M=M1+M2= 2.0×1030 kg + 1.5×1030 kg = 3.5×1030 kg

Step 2: Next, we find the reduced mass of the system.

µ=M1·M2

M1+M2

=(2.0×1030 kg)(1.5×1030 kg)

3.5×1030 kg = 0.857 ×1030 kg

Step 3: Using the formula for the period of an object in circular motion,

T=2πr

v, we can find the period of the stars’ motion.

T=2π×1.0×1012 m

2.5×104m/s = 251.33 days

Step 4: Lastly, we can calculate the gravitational force between the stars

using the formula for gravitational force, F=G·M1·M2

r2.

F=6.67 ×10−11 Nm2/kg2·2.0×1030 kg ·1.5×1030 kg

(1.0×1012 m)2= 2.01 ×1020 N

Therefore, the period of the stars’ motion is 251.33 days and the gravitational

force between them is 2.01 ×1020 N.

Question 8

Question

An asteroid orbits the Sun in a highly elliptical orbit. At its closest approach,

the asteroid is 0.2 AU from the Sun and has a speed of 30 km/s. At its farthest

distance, the asteroid is 3.0 AU from the Sun. Calculate the speed of the asteroid

at this point.

Solution

Step 1: We can use the conservation of angular momentum to solve this problem.

The angular momentum of the asteroid is given by:

L=mrv

where mis the mass of the asteroid, ris the distance of the asteroid from the

Sun, and vis the speed of the asteroid.

Step 2: At the closest approach, the asteroid is 0.2 AU from the Sun and

has a speed of 30 km/s. We can write the angular momentum at this point as:

L1=m×0.2×1.496 ×1011 ×3.0×105

Step 3: At the farthest distance, the asteroid is 3.0 AU from the Sun, and

we need to find the speed of the asteroid. This can be written as:

L2=m×3.0×1.496 ×1011 ×v

Step 4: Since angular momentum is conserved, we can set L1=L2:

m×0.2×1.496 ×1011 ×3.0×105=m×3.0×1.496 ×1011 ×v

Step 5: Solving for v, we find:

v=0.2×3.0×105

3.0≈2.0×104m/s

Step 6: Therefore, the speed of the asteroid at its farthest distance from the

Sun is approximately 2.0×104m/s.

Question 9

Question

Explain the physical significance of Kepler’s laws of planetary motion and how

they are related to Newton’s laws of motion and universal gravitation.

Solution

Step 1: Kepler’s Laws of Planetary Motion Kepler’s laws describe the

motion of planets around the Sun. 1. The Law of Ellipses: Each planet follows

an elliptical orbit with the Sun at one of the two foci. 2. The Law of Equal

Areas: A line segment joining a planet and the Sun sweeps out equal areas

during equal intervals of time. 3. The Law of Harmonies: The square of the

period of revolution of a planet is proportional to the cube of the semimajor

axis of its orbit.

Step 2: Physical Significance of Kepler’s Laws - The first law implies

that planets move in elliptical orbits, not perfect circles as previously thought.

- The second law demonstrates that a planet moves faster when it is closer to

the Sun, leading to equal areas being swept out in the same time. - The third

law shows how the period of revolution of a planet increases with the size of its

orbit.

Step 3: Relationship to Newton’s Laws and Universal Gravitation

- Kepler’s laws were later explained by Newton’s laws of motion and universal

gravitation. - Newton’s first law explains that an object will continue in a state

of rest or motion unless acted upon by an external force, which explains why

planets move in orbits. - Newton’s second law shows how the gravitational

force between two objects causes acceleration and changes in velocity, which

keeps planets in their elliptical paths. - Newton’s law of universal gravitation

provides the mathematical explanation for the observed motion described by

Kepler’s laws.

Therefore, Kepler’s laws of planetary motion are foundational to understand-

ing the motion of celestial bodies, and they find a comprehensive explanation

within Newton’s laws of motion and universal gravitation.

Question 10

Question

A comet, with an orbital period of 35 years, has an average distance from the

Sun of 4.5 AU. Determine the semi-major axis of the comet’s elliptical orbit.

Assume a circular orbit for the planets.

Solution

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

an orbiting object 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.

Step 2: We are given that the comet’s orbital period is 35 years and its aver-

age distance from the Sun is 4.5 AU. Let’s plug these values into the equation:

(35)2=k(4.5)3

Step 3: Solve for the constant k:

k=(35)2

(4.5)3

Step 4: Calculate the value of k:

k=1225

91.125 ≈13.46

Step 5: Now that we have the value of k, we can find the semi-major axis of

the comet’s orbit. Let T′be the period of a circular orbit around the Sun at a

distance of 4.5 AU:

T′2= 4π2(4.5)3

Step 6: Solve for T′:

T′=√4π2(4.5)3

Step 7: Calculate the value of T′:

T′≈√4·9.87 ·91.125 ≈25.16 years

Step 8: The semi-major axis of a circular orbit with a period of 25.16 years

is 4.5 AU. Therefore, the semi-major axis of the comet’s elliptical orbit, with a

period of 35 years, can be found using Kepler’s third law:

(35)2= 13.46a3

Step 9: Solve for the semi-major axis a:

a=√(35)2

13.46

Step 10: Calculate the value of the semi-major axis a:

a≈√919.69/13.46 ≈√68.28 ≈8.26 AU

Therefore, the semi-major axis of the comet’s elliptical orbit is approximately

8.26 AU.

Question 11

Question

A comet travels along an elliptical orbit around the sun. The closest distance

from the center of the ellipse to the sun is 0.1 AU and the farthest distance is

1 AU. If the comet’s orbital period is 3 years, determine the total energy of the

comet’s motion.

Solution

Step 1: Calculate the semi-major axis of the comet’s orbit Given that the closest

distance from the sun is 0.1 AU and the farthest distance is 1 AU, we can find

the semi-major axis (a) using the formula for an ellipse:

a=rmin +rmax

a=0.1+1

a= 0.55 AU

Step 2: Use Kepler’s third law to find the orbital period in terms of the

semi-major axis Kepler’s third law states that the square of the orbital period

(T) is proportional to the cube of the semi-major axis (a):

T2=k·a3

where kis a constant.

Given that the orbital period is 3 years, we can substitute this into the

equation:

32=k·0.553

9 = k·0.166375

k=9

0.166375

k≈54.018

Step 3: Calculate the total energy of the comet’s motion The total energy

Eof the comet’s motion can be expressed as:

E=−GMm

where Gis the gravitational constant, Mis the mass of the sun, mis the mass

of the comet, and ais the semi-major axis.

Substitute the values and constants into the equation:

E=−6.674 ×10−11 N m2/kg2×1.989 ×1030 kg ×m

2×0.55 ×1.496 ×1011 m

E=−6.674 ×1.989 ×m

2×0.55 ×1.496

E=−13.282 ×m

1.646

E≈ −8.07 ×mJ

Therefore, the total energy of the comet’s motion is approximately −8.07

times the mass of the comet in joules.

Question 12

Question

A planet orbits a star in an elliptical orbit. The planet’s farthest distance from

the star is 3 AU, and its closest approach to the star is 1 AU. If the planet takes

one year to complete one orbit, determine the eccentricity of the planet’s orbit.

Solution

Step 1: Recall Kepler’s second law, which states that a line joining a planet and

its star sweeps out equal areas in equal times.

Step 2: The eccentricity of an elliptical orbit can be defined in terms of the

semi-major axis (a) and the semi-minor axis (b) of the ellipse. The formula for

eccentricity (e) is given by:

e=√1−b2

Step 3: The semi-major axis aof the orbit is the average of the farthest and

closest distances of the planet from the star. Given that the farthest distance

is 3 AU and the closest distance is 1 AU:

a=1

2(3 AU + 1 AU) = 2 AU

Step 4: The semi-minor axis bof the orbit is half the length of the major axis

of the ellipse. The length of the major axis is the total distance of the planet’s

orbit:

Major Axis = 3 AU + 1 AU = 4 AU

b=1

2×Major Axis =1

2×4AU = 2 AU

Step 5: Now, substitute the values of aand binto the eccentricity formula:

e=√1−(2 AU)2

(2 AU)2=√1−4AU2

4AU2=√1−1 = 0

Step 6: Therefore, the eccentricity of the planet’s orbit is 0. This means

that the orbit is a perfect circle.

Question 13

Question

According to Kepler’s laws of planetary motion, the time it takes a planet to

travel around the sun is related to its average distance from the sun. Suppose a

planet has an average distance of 5 AU (astronomical units) from the sun and

takes 16 years to complete one orbit. Calculate the average speed of this planet

in km/s.

Solution

Step 1: Convert the average distance from AU to kilometers. We know that 1

AU is approximately equal to 1.496 ×108km. Therefore, the planet’s average

distance from the sun in kilometers is:

5AU ×1.496 ×108km/AU = 7.48 ×108km

Step 2: Convert the time taken to complete one orbit from years to seconds.

Since 1 year is equal to 365.25 days, and 1 day is equal to 24 hours, and 1 hour

is equal to 3600 seconds, we have:

16 years ×365.25 days/year ×24 hours/day ×3600 s/hour = 5.04 ×108s

Step 3: Calculate the circumference of the orbit using the formula C= 2πr.

Plugging in the average distance, we get:

C= 2π×7.48 ×108km

Step 4: Use the formula v=C

Tto find the average speed. Substitute the

values for Cand T, we have:

v=2π×7.48 ×108km

5.04 ×108s= 2.35 km/s

Therefore, the average speed of the planet is 2.35 km/s.

Question 14

Question

Consider a star with a mass of 2×1030 kg and a planet with a mass of 4×1024 kg

orbiting around it. If the planet is in a circular orbit at a distance of 1.5×1011 m

from the star, calculate the period of the planet’s orbit around the star. Assume

the gravitational constant is 6.67 ×10−11 Nm2/kg2.

Solution

Step 1: Calculate the gravitational force between the planet and the star. The

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

universal gravitation:

F=G·m1·m2

where - Fis the gravitational force, - Gis the gravitational constant (6.67 ×

10−11 Nm2/kg2), - m1and m2are the masses of the planet and the star, respec-

tively, - ris the distance between the planet and the star.

Substitute the given values to find the gravitational force.

Question 15

Question

Consider a planet in an elliptical orbit around a star. The semi-major axis of

the orbit is aand the eccentricity is e. If the planet is closest to the star at

perihelion and farthest from the star at aphelion, prove that the distance from

the star to the planet at any point in its orbit is given by the equation:

r=a(1 −e2)

1 + ecos(θ)

where ris the distance, eis the eccentricity, ais the semi-major axis, and θis

the angle between the planet and the line from the star to the perihelion point.

Solution

Step 1: We start by writing the general equation for an ellipse in polar coordi-

nates:

r=a(1 −e2)

1 + ecos(θ′)

where ris the distance from the focus (star) to the planet, ais the semi-major

axis, eis the eccentricity, and θ′is the angle measured from the focus to the

intersecting line between the ellipse and the circle centered at the star.

Step 2: Next, we relate this general equation to our specific case of interest.

Let θbe the angle between the planet and the line from the star to the perihelion

point. The angle θ′is related to θby θ′=θ−ω, where ωis the argument of

perihelion.

Step 3: Since the planet is closest to the star at perihelion and farthest at

aphelion, we have θ= 0 at perihelion and θ=πat aphelion.

Step 4: By substituting θ′=θ−ωand using the specific values of θ= 0 and

θ=π, we find that θ′=−ωat perihelion and θ′=π−ωat aphelion.

Step 5: Plugging these values into the general equation for an ellipse, we

find the distance rat perihelion and aphelion, and then express the equation r

as a function of θ.

At perihelion: r=a(1 −e2)

1−ecos(ω)

At aphelion: r=a(1 −e2)

1 + ecos(π−ω)

Step 6: Simplifying the expressions, we can rewrite the equation for rin

terms of the angle θbetween the star and the planet. We find the equation:

r=a(1 −e2)

1 + ecos(θ)

This completes the proof that the distance from the star to the planet at any

point in its orbit is given by the provided equation.

Question 16

Question

A planet has an orbital period of 5 years around a star with a mass of 2×1030 kg.

If the average distance between the planet and the star is 3×1011 m, determine

the speed of the planet in its orbit.

Solution

Step 1: First, we need to calculate the gravitational force that the star exerts

on the planet by using Newton’s law of gravitation:

F=G·M·m

where G= 6.67 ×10−11 Nm2/kg2is the gravitational constant, M= 2 ×1030

kg is the mass of the star, mis the mass of the planet (which we can assume to

be negligible compared to the star’s mass), and r= 3 ×1011 m is the average

distance between the planet and the star.

Step 2: Substitute the given values into the formula:

F=(6.67 ×10−11 Nm2/kg2)·(2 ×1030 kg)·m

(3 ×1011 m)2

Step 3: The gravitational force provides the centripetal force to keep the

planet in its orbit. Thus, we can equate the two forces and solve for the planet’s

speed using the formula for centripetal force:

F=m·v2

where vis the speed of the planet in its orbit.

Step 4: Equate the gravitational force to the centripetal force:

(6.67 ×10−11 Nm2/kg2)·(2 ×1030 kg)·m

(3 ×1011 m)2=m·v2

3×1011 m

Step 5: Solve for the speed v:

v=√(6.67 ×10−11 Nm2/kg2)·(2 ×1030 kg)

3×1011 m

Step 6: Now, calculate the speed of the planet:

v=√(6.67 ×10−11 Nm2/kg2)·(2 ×1030 kg)

3×1011 m

v=√1.334 ×1020 Nm2/kg

3×1011 m

v=√4.447 ×108m/s

v≈2.11 ×104m/s

Therefore, the speed of the planet in its orbit is approximately 2.11 ×104

m/s.

Question 17

Question

A planet orbits a star in an elliptical orbit with the star at one focus. The planet

is closest to the star at a distance of 0.3 astronomical units (AU) and farthest

from the star at a distance of 0.7 AU. If the period of the planet’s orbit is 0.5

years, determine the semi-major axis of the orbit in AU.

Solution

Step 1: Recall Kepler’s Third Law, which relates the period of an orbit (T) to

the semi-major axis of the orbit (a):

T2=(4π2

GM )a3

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

Step 2: We are given that the period Tis 0.5 years. We also know the

distances of closest approach and furthest distance, which correspond to the

semi-major axis aalong with the eccentricity of the orbit e:

a=rmin +rmax

e=rmax −rmin

rmax +rmin

Step 3: Calculate the semi-major axis ausing the given distances:

a=0.3AU + 0.7AU

2= 0.5AU

Step 4: Next, calculate the eccentricity e:

e=0.7AU −0.3AU

0.7AU + 0.3AU =0.4AU

1AU = 0.4

Step 5: Substitute the known values into Kepler’s Third Law equation:

0.52=(4π2

GM )(0.5)3

Step 6: Solve for the mass Mof the star:

M=4π2a3

GT 2=4π2(0.5)3

G(0.5)2

Step 7: Substitute the values of a,T, and Ginto the equation and calculate

Mto find the semi-major axis of the orbit.

Question 18

Question

Assume a planet is orbiting a star in an elliptical orbit such that the closest

approach to the star (perihelion) is 0.25 AU and the farthest distance from the

star (aphelion) is 0.75 AU. If the period of the planet’s orbit is 1 year, determine

the eccentricity of the orbit.

Solution

Step 1: Recall the formula relating period, semi-major axis, and eccentricity for

Keplerian orbits:

T2=(4π2

G(M+m))a3(1−e2)

where T= period of the orbit, G= gravitational constant, M= mass of

the star, m= mass of the planet, a= semi-major axis, and e= eccentricity.

Step 2: We first need to find the semi-major axis aof the orbit. The semi-

major axis is simply the average of the perihelion and aphelion distances:

a=rperihelion +raphelion

2=0.25 + 0.75

2AU = 0.5AU

Step 3: Substituting the known values into the formula, we have:

(1 year)2=(4π2

G(M+m))(0.5AU)3(1−e2)

Step 4: Recall that the period of the Earth’s orbit around the Sun is 1

year, so Mis the mass of the Sun and mis the mass of the Earth. We can

consider mas negligible compared to M. The gravitational constant Gis 6.674

× 10−11Nm2/kg2.

Step 5: Solving for the eccentricity e:

e=√1−T2G(M+m)

4π2a3=√1−(1 year)2·6.674 ×10−11 N m2/kg2·(1.989 ×1030 kg)

4π2·(0.5AU)3

Step 6: Calculate the eccentricity using the formula and the given values.

Question 19

Question

Consider a planet with an eccentricity of 0.2 orbiting around a star. The semi-

major axis of the planet’s orbit is 3 AU. Calculate the closest distance of the

planet to the star and the farthest distance of the planet from the star.

Solution

Step 1: Find the eccentricity of the elliptical orbit, e, which is given as 0.2.

Step 2: Find the semi-minor axis of the orbit, b, using the relationship

between the semi-major axis (a) and eccentricity (e) of an ellipse:

b=a√1−e2

Since a= 3 AU and e= 0.2:

b= 3√1−0.22= 3√1−0.04 = 3√0.96

Step 3: Calculate the closest distance of the planet to the star (perihelion),

which occurs at the distance of a−b:

Perihelion distance =a−b= 3 −3√0.96

Step 4: Calculate the farthest distance of the planet from the star (aphelion),

which occurs at the distance of a+b:

Aphelion distance =a+b= 3 + 3√0.96

Therefore, the closest distance of the planet to the star is 3−3√0.96 AU

and the farthest distance is 3+3√0.96 AU.

Question 20

Question

Consider a planet in a circular orbit around a star in a distant solar system.

The radius of the planet’s orbit is 5.0×108meters and its period is 5.0 Earth

years. Calculate the mass of the star using Kepler’s third law. Assume the mass

of the planet is negligible compared to the mass of the star.

Solution

Step 1: Write down Kepler’s third law in terms of the period and radius of the

planet’s orbit.

=T2

where T1and R1are the period and radius of the Earth’s orbit (known values),

and T2and R2are the period and radius of the planet’s orbit.

Step 2: Convert the period of the planet’s orbit to seconds.

T2= 5.0Earth years×365 days/year×24 hours/day×60 minutes/hour×60 seconds/minute

T2= 1.58 ×108seconds

Step 3: Substitute the known values into Kepler’s third law and solve for

the mass of the star. (1 year)2

(1 AU)3=(T2)2

(R2)3

(1.5×1011)3=(1.58 ×108)2

(5.0×108)3

1 = (1.58 ×108)2×(1.5×1011)3

(5.0×108)3

Step 4: Solve for the mass of the star.

M=R3

=(5.0×108)3

(1.58 ×108)2

M≈1.18 ×1030 kg

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

1.18 ×1030 kg.

Question 21

Question

A spaceship is in a circular orbit around a planet of mass M. If the spaceship

increases its speed, it will move to a larger orbit. Explain this in terms of

Kepler’s laws of planetary motion.

Solution

Step 1: According to Kepler’s laws of planetary motion, the orbit of a planet

(or spaceship) around a star (or planet) is an ellipse with the star (or planet)

at one focus.

Step 2: If the spaceship is in a circular orbit and it increases its speed, its

orbit will change from a circle to an ellipse. The ellipse will have the same

center as the original circle, but one of the foci (the planet) will coincide with

the center.

Step 3: As the spaceship increases its speed, it moves further away from the

planet along the major axis of the ellipse. This is because the speed of an object

in orbit is related to its distance from the central body.

Step 4: The increase in speed causes the spaceship to move into a larger

orbit where the planet is located at one of the foci of the ellipse. This new orbit

will be elongated in the direction of the spaceship’s motion.

Step 5: In summary, by increasing its speed, the spaceship transitions from

a circular orbit to an elliptical orbit with the planet at one focus, in accordance

with Kepler’s laws of planetary motion.

Question 22

Question

A planet in a certain solar system has an elliptical orbit with an eccentricity of

0.6. If the planet is closest to the sun at a distance of 0.4 AU and has a speed

of 30 km/s at its closest approach, determine the farthest distance of the planet

from the sun.

Solution

Step 1: Recall Kepler’s Second Law, which states that a planet sweeps out equal

areas in equal times. This means that the planet moves faster when it is closer

to the sun and slower when it is farther away.

Step 2: The planet’s speed at its farthest distance from the sun can be found

using the conservation of angular momentum. The formula for conservation of

angular momentum is given by:

r1·v1=r2·v2

where: - r1= 0.4AU (closest distance to the sun) - v1= 30 km/s (speed at

closest approach) - r2is the farthest distance we are trying to find - v2is the

speed at the farthest distance

Step 3: Substituting the given values into the conservation of angular mo-

mentum equation:

0.4·30 = r2·v2

Step 4: Solving for v2:

12 = r2·v2

Step 5: When the planet is at its farthest distance from the sun, it is moving

at its slowest speed. At this point, the kinetic energy is equal to the gravitational

potential energy. The formula for kinetic energy is KE =1

2mv2and the formula

for gravitational potential energy is P E =−GM m

r, where: - KE is the kinetic

energy - mis the mass of the planet (cancel out in the equations) - Gis the

gravitational constant - Mis the mass of the sun - ris the distance between

the planet and the sun

Step 6: Setting the kinetic energy equal to the potential energy:

2v2

2=−GM

Step 7: Substituting the known values:

2·122=−GM

Step 8: Solving for r2:

72 = GM

Step 9: Since the eccentricity e=rmax −rmin

rmax +rmin , in this case, e= 0.6:

0.6 = r2−0.4

r2+ 0.4

Step 10: Solving for r2:

0.6(r2+ 0.4) = r2−0.4

0.6r2+ 0.24 = r2−0.4

0.4r2=−0.64

r2=−1.6AU

Step 11: The farthest distance of the planet from the sun is 1.6AU.

Question 23

Question

Consider a planet in a highly eccentric orbit around a star. At the point of

closest approach, the planet is moving at a speed of 50 km/s. At the point of

farthest distance, the planet is moving at a speed of 20 km/s. Determine the

semi-major axis of the orbit.

Solution

Step 1: We start by applying Kepler’s second law, which states that the line

joining the planet to the sun sweeps out equal areas in equal times. This implies

that the angular momentum of the planet with respect to the sun is constant.

Mathematically, we can express this as:

r1·v1=r2·v2

where r1and r2are the distances at the closest and farthest points from the

sun respectively, and v1and v2are the speeds at those points.

Step 2: Given that r1< r2, the planet moves faster when it is closer to the

sun (conservation of angular momentum). Therefore, we have:

r1·50 km/s =r2·20 km/s

Step 3: Let’s denote the semi-major axis of the orbit as a, the distance from

the star to the closest point as r1=a(1 −e), and the distance to the farthest

point as r2=a(1 + e), where eis the eccentricity of the orbit.

Step 4: Substituting r1and r2into the angular momentum equation, we get:

a(1 −e)·50 = a(1 + e)·20

Step 5: Simplifying the equation above, we find:

50a−50ae = 20a+ 20ae

Step 6: Rearranging terms, we get:

30a= 70ae

Step 7: Solving for the semi-major axis a, we find:

a=7

Therefore, the semi-major axis of the orbit is 7

3e.

Question 24

Question

Given that a planet in a circular orbit around a star has a period of 10 years,

determine the ratio of its average orbital radius to the radius of Earth’s orbit

(i.e., to the semimajor axis of Earth’s elliptical orbit around the Sun).

Solution

Step 1: Recall Kepler’s Third Law for circular orbits, which states that the

square of the period of an orbit is proportional to the cube of the average

orbital radius:

T2=k·a3

where Tis the period of the orbit, ais the average orbital radius, and kis a

constant.

Step 2: Let’s denote the average orbital radius of the planet as ap, and

the radius of Earth’s orbit (semimajor axis of Earth’s elliptical orbit) as ae.

Therefore, we are looking for the ratio ap/ae.

Step 3: We can set up a ratio of Kepler’s Third Law for the planet and

Earth: T2

=T2

where Tpis the period of the planet’s orbit (given as 10 years) and Teis the

period of Earth’s orbit (1 year).

Step 4: Substitute the given values and solve for the ratio ap/ae:

102

=12

p= 100a3

ap=3

√100 ·ae

Step 5: Therefore, the ratio of the average orbital radius of the planet to the

radius of Earth’s orbit is:

√100 ·ae

√100

Step 6: Thus, the ratio of the average orbital radius of the planet to the

radius of Earth’s orbit is 3

√100.

Question 25

Question

A planet orbits a star in a nearly circular orbit with a radius of 2 AU. The

period of this orbit is 4 years. Calculate the mass of the central star in solar

masses given that the gravitational constant is 6.674 ×10−11 m3kg−1s−2.

Solution

Step 1: Write down Kepler’s Third Law which relates the orbital period of a

planet around a star to the semi-major axis of the orbit:

T2=4π2

G(M+m)a3

where: - T= 4 years is the period of the orbit, - G= 6.674 ×10−11 m3kg−1s−2

is the gravitational constant, - Mis the mass of the central star in solar masses

(unknown), - a= 2 AU = 2 ×1.496 ×1011 m is the semi-major axis of the orbit.

Step 2: Substitute the given values into Kepler’s Third Law equation:

(4 years)2=4π2

6.674 ×10−11 m3kg−1s−2

M⊙

M(2 ×1.496 ×1011 m)3

Step 3: Simplify the equation and solve for the mass of the central star M

in solar masses:

16 = 4π2

6.674 ×10−11

M⊙

M(8.976 ×1033)

M=4π2

6.674 ×10−11 ×16 ×8.976 ×1033 M⊙

M≈0.93M⊙

Therefore, the mass of the central star is approximately 0.93 solar masses.

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

M=4π2(0.7×1.496 ×1011)3

G(150 ×86400

2π)2

Calculating this expression will give us the mass of the star in kilograms.

Question 2

Question

A planet is in an elliptical orbit around the Sun. The closest approach of the

planet to the Sun (perihelion) is 28 million miles and the farthest distance from

the Sun (aphelion) is 40 million miles. If the planet takes 440 days to complete

one full orbit, determine the semi-major axis of the planet’s orbit.

Solution

Step 1: Recall 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:

T2=ka3

where Tis the period of the planet’s orbit and ais the semi-major axis of

the planet’s orbit.

Step 2: To determine the semi-major axis of the planet’s orbit, we need to

first calculate the period of the planet’s orbit. Given that the planet takes 440

days to complete one full orbit, we have:

T= 440 days

Step 3: Now, we need to convert the period to years, since the standard unit

for time in Kepler’s third law is years. There are 365 days in a year, so:

T=440 days

365 days/year = 1.205 years

Step 4: Substitute the period into Kepler’s third law equation and solve for

the semi-major axis:

(1.205 years)2=ka3

Step 5: Next, we need to determine the value of the constant k. This can be

done by using the known period and semi-major axis of a known planet, such as

Earth. For Earth, T= 1 year and a= 1 astronomical unit (AU). Substituting

these values into Kepler’s third law, we can solve for k:

(1 year)2=k(1 AU)3

k= 1

Step 6: Substitute the value of kand the period Tinto Kepler’s third law

equation to solve for the semi-major axis a:

(1.205)2=a3

Step 7: Solve for a:

a=3

√(1.205)2= 1.236 AU

Step 8: Therefore, the semi-major axis of the planet’s orbit is approximately

1.236 astronomical units.

Question 3

Question

A comet discovered in a distant solar system has an orbital period of 6 years

and an average distance from the sun of 4 AU. Calculate the mass of the star

around which the comet orbits.

Given: Gravitational constant, G= 6.67 ×10−11 N m2/kg2

Solution

Step 1: Convert the average distance from Astronomical Units (AU) to meters.

Since 1 AU is approximately 1.496 ×1011 meters,

Average distance = 4 AU = 4 ×1.496 ×1011 m

Step 2: Use Kepler’s third law to find the star’s mass. Kepler’s third law

states:

T2=(4π2

G(M1+M2))a3

where: - Tis the orbital period in seconds - Gis the gravitational constant -

M1and M2are the masses of the bodies - ais the semi-major axis of the orbit

Step 3: Rearrange Kepler’s third law to solve for the star’s mass, M1.

M1=4π2

G(a3

T2)−M2

Step 4: Substitute the known values into the equation.

M1=4π2

6.67 ×10−11 ((4 ×1.496 ×1011)3

(6 ×365.25 ×24 ×3600)2)

Step 5: Calculate the mass of the star.

M1=4π2

6.67 ×10−11 ((5.984 ×1011)3

1.894 ×108×102)

Question 4

Question

The planet Mercury has an average orbital speed of 47.87 km/s and an average

orbital radius of 5.79 ×1010 m. Using Kepler’s third law, determine the period

of Mercury’s orbit around the Sun.

Solution

Step 1: Recall Kepler’s third law, which states that the square of the orbital

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

matically, this can be expressed as:

T2=4π2

G(M1+M2)a3

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

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

orbit.

Step 2: Rearrange the equation to solve for the period T:

T=√4π2

G(M1+M2)a3

Step 3: We can assume the mass of Mercury is negligible compared to the

mass of the Sun, so M1≫M2. Also, we know that the mass of the Sun is

approximately 1.989 ×1030 kg and G= 6.67430 ×10−11 m3kg−1s−2.

Step 4: Substitute the given values for the semi-major axis aand solve for

the period T:

T=√4π2

6.67430 ×10−11(1.989 ×1030 )(5.79 ×1010 )3

Step 5: Calculate the period Tusing the average orbital speed of Mercury:

T=√4π2

1.327124 ×1020 (1.2087591 ×1031)

Step 6: Simplify the equation and solve for the period Tto find the answer.

Question 5

Question

Consider a planet in a circular orbit around a star. The planet has a mass of

3.21 ×1024 kg and the star has a mass of 1.99 ×1030 kg. The radius of the orbit

is 1.50 ×1011 m. Calculate the period of the planet’s orbit in seconds.

(Given: G= 6.67 ×10−11 m3kg−1s−2)

Solution

Step 1: Calculate the gravitational force between the planet and the star using

Newton’s law of universal gravitation:

F=G·M1·M2

where Fis the gravitational force, Gis the gravitational constant, M1and M2

are the masses of the objects, and ris the distance between the centers of the

objects.

Step 2: Substitute the given values into the equation to find the gravitational

force.

Step 3: Use the gravitational force to calculate the centripetal force required

to keep the planet in circular motion:

Fcentripetal =M·v2

where Mis the mass of the planet, vis the velocity of the planet, and ris the

radius of the orbit.

Step 4: Equate the centripetal force to the gravitational force and solve for

the velocity vof the planet.

Step 5: Use the formula for the period of circular motion to find the period

T=2πr

where ris the radius of the orbit and vis the velocity of the planet found in

the previous step.

Step 6: Substitute the known values into the period formula to calculate the

period of the planet’s orbit.

Question 6

Question

Kepler’s third law states that the square of the period of revolution of a planet

around the Sun is directly proportional to the cube of its average distance from

the Sun. For a hypothetical planet in our solar system with an average distance

from the Sun of 2 AU, its period of revolution is 5 years. Calculate the average

distance from the Sun of a different planet in our solar system with a period of

revolution of 10 years.

Solution

Step 1: Let T1be the period of revolution of the first planet and R1be its

average distance from the Sun. According to Kepler’s third law:

1∝R3

Step 2: Given that the first planet has T1= 5 years and R1= 2 AU, we

have:

52∝23

25 ∝8

=25

Step 3: Let T2be the period of revolution of the second planet and R2be

its average distance from the Sun. According to Kepler’s third law:

2∝R3

Step 4: Since the second planet has T2= 10 years, we can set up a proportion

using the ratio obtained in Step 2:

=T2

8=102

Step 5: Solve for R2:

2=8×100

2=800

2= 32

R2=3

√32

R2= 2 AU

Therefore, the average distance from the Sun of the different planet in our

solar system with a period of revolution of 10 years is 2 AU.

Question 7

Question

Consider a binary star system consisting of two stars with masses M1= 2.0×

1030 kg and M2= 1.5×1030 kg, separated by a distance of 1.0×1012 m. If the

stars are in circular orbit around their center of mass at a speed of 2.5×104m/s,

determine the period of the stars’ motion and the gravitational force between

them.

Solution

Step 1: Firstly, we need to find the total mass of the system.

M=M1+M2= 2.0×1030 kg + 1.5×1030 kg = 3.5×1030 kg

Step 2: Next, we find the reduced mass of the system.

µ=M1·M2

M1+M2

=(2.0×1030 kg)(1.5×1030 kg)

3.5×1030 kg = 0.857 ×1030 kg

Step 3: Using the formula for the period of an object in circular motion,

T=2πr

v, we can find the period of the stars’ motion.

T=2π×1.0×1012 m

2.5×104m/s = 251.33 days

Step 4: Lastly, we can calculate the gravitational force between the stars

using the formula for gravitational force, F=G·M1·M2

r2.

F=6.67 ×10−11 Nm2/kg2·2.0×1030 kg ·1.5×1030 kg

(1.0×1012 m)2= 2.01 ×1020 N

Therefore, the period of the stars’ motion is 251.33 days and the gravitational

force between them is 2.01 ×1020 N.

Question 8

Question

An asteroid orbits the Sun in a highly elliptical orbit. At its closest approach,

the asteroid is 0.2 AU from the Sun and has a speed of 30 km/s. At its farthest

distance, the asteroid is 3.0 AU from the Sun. Calculate the speed of the asteroid

at this point.

Solution

Step 1: We can use the conservation of angular momentum to solve this problem.

The angular momentum of the asteroid is given by:

L=mrv

where mis the mass of the asteroid, ris the distance of the asteroid from the

Sun, and vis the speed of the asteroid.

Step 2: At the closest approach, the asteroid is 0.2 AU from the Sun and

has a speed of 30 km/s. We can write the angular momentum at this point as:

L1=m×0.2×1.496 ×1011 ×3.0×105