PHYS 305 - INTRODUCTION TO MODERN PHYSICS Kepler's laws Question Bank Set 2 | PHYS 305

PHYS 305 - INTRODUCTION TO

MODERN PHYSICS - Kepler’s laws

Question Bank - Set 2

Liberty University

Question 1

Question

State Kepler’s laws of planetary motion and provide a brief explanation of each

law.

Solution

Step 1: Kepler’s First Law (Law of Ellipses) states that the planets move in

elliptical orbits with the sun at one of the two foci of the ellipse. This means

that the orbit of a planet is not a perfect circle, but rather an elongated circle.

Step 2: Kepler’s Second Law (Law of Equal Areas) states that a line segment

joining a planet and the sun sweeps out equal areas during equal intervals of

time. This means that a planet moves faster when it is closer to the sun and

slower when it is farther away.

Step 3: Kepler’s Third Law (Law of Harmonies) states that the square of

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

of its orbit. Mathematically, it can be expressed as T2∝a3, where Tis the

orbital period and ais the semi-major axis of the orbit.

Question 2

Question

Consider an exoplanet orbiting a star with a mass of 5×1030 kg along an

elliptical path. The distance between the planet and the star at the closest

approach is 1.5 AU, while at the farthest point in the orbit the distance is 3.5

AU. If the orbital period of the planet is 4 years, determine the semimajor axis

of the orbit.

Solution

Step 1: Calculate the orbital period using Kepler’s third law, T2=4π2

GM a3,

where Tis the orbital period, G= 6.67 ×10−11 m3/kg ·s2is the gravitational

constant, M= 5 ×1030 kg is the mass of the star, and ais the semimajor axis

of the orbit.

Substitute the given values:

T2=4π2

(6.67 ×10−11)(5 ×1030)a3

T2= 3.77 ×107a3

Given that T= 4 years:

16 = 3.77 ×107a3

Step 2: Solve for the semimajor axis a.

a3=16

3.77 ×107

a3≈4.24 ×10−7

a≈

√4.24 ×10−7

a≈0.0175 AU

Therefore, the semimajor axis of the orbit is approximately 0.0175 AU.

Question 3

Question

A planet orbits a star in a nearly circular orbit with a period of 5 years. The

average distance between the planet and the star is 4 AU. Determine the mass

of the star, given that the gravitational constant is 6.67 ×10−11 m3/kg s2.

Solution

Let’s denote the mass of the star as M, the distance between the planet and

the star as r, the period of the orbit as T, and the gravitational constant as

G. Furthermore, we will use Kepler’s Third Law of Planetary Motion, which

states:

T2=4π2r3

Step 1: Convert the average distance rfrom AU to meters.

r= 4 AU ×1.496 ×1011 m/AU = 5.984 ×1011 m

Step 2: Substitute the known values into Kepler’s Third Law and solve for

the mass M.

(5 years)2=4π2(5.984 ×1011 m)3

G×M

25 = 4π2(3.590 ×1034 m3)

6.67 ×10−11 m3/kg s2×M

Step 3: Rearrange the equation to solve for the mass M.

M=4π2(3.590 ×1034 m3)

25 ×6.67 ×10−11 m3/kg s2

M=4π2(3.590 ×1034)

25 ×6.67 ×10−11 kg

M=4×(9.87) ×(3.590) ×1023

25 ×6.67 kg

M≈2.24 ×1030 kg

Therefore, the mass of the star is approximately 2.24 ×1030 kg.

Question 4

Question

A comet is orbiting around the Sun in an elliptical orbit with a semi-major axis

of 2.5 AU. If the closest distance between the comet and the Sun (perihelion) is

1.5 AU, find the eccentricity of the comet’s orbit. (Assume the Sun is at one of

the foci of the ellipse.)

Solution

Step 1: Recall the definition of eccentricity for an ellipse, e, as the ratio of the

distance between the foci to the length of the major axis. Step 2: The distance

between the foci (2c) of an ellipse can be found using the equation 2c= 2a−2b,

where ais the semi-major axis and bis the semi-minor axis. Step 3: Since

the comet’s orbit is centered around the Sun, one of the foci coincides with the

location of the Sun. Thus, 2cis equal to the difference between the semi-major

axis and the distance to the perihelion. Step 4: Substitute the given values into

the equation to find 2c. Step 5: After finding 2c, we can calculate the semi-

minor axis busing the formula b=√a2−c2. Step 6: Finally, the eccentricity

ecan be calculated using the formula e=c

a. Step 7: Substitute the calculated

values of cand ainto the eccentricity formula to find the eccentricity of the

comet’s orbit.

Question 5

Question

An extrasolar planet orbits a star with a period of 200 days and an average

distance from the star of 0.5 AU. Determine the mass of the star in solar masses

assuming the orbit is circular.

Solution

Step 1: Find the orbital speed of the planet using Kepler’s Third Law:

T2=(4π2

GM )R3

where Tis the period of the planet, Gis the gravitational constant, Mis the

mass of the star, and Ris the radius of the orbit.

Step 2: Substitute the values T= 200 days and R= 0.5AU into the equation

and solve for the orbital speed.

Step 3: Use the formula for the orbital speed of a planet in a circular orbit:

v=2πR

to find the orbital speed of the planet.

Step 4: Equate the two expressions for the orbital speed:

v=√GM

v=2πR

This gives us:

√GM

R=2πR

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

M=4π2R3

GT 2

Step 6: Substitute R= 0.5AU and T= 200 days into the equation to find

the mass of the star in solar masses.

Question 6

Question

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

Suppose the planet’s closest approach to the star (perihelion) is 30 million kilo-

meters and its farthest distance from the star (aphelion) is 50 million kilometers.

If the planet takes 200 days to complete one full orbit, calculate the eccentricity

of the planet’s orbit.

Solution

Step 1: Recall that the eccentricity (e) of an elliptical orbit is defined as the ratio

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

The distance between the foci is 2a, where ais the semi-major axis of the ellipse,

and the length of the major axis is 2a(1 + e). Thus, e= (2a−2a(1 + e))/(2a) =

1−(1 + e).

Step 2: We are given the perihelion and aphelion distances. The semi-major

axis, a, is the average of these two distances, which is a=30 million km+50 million km

40 million km.

Step 3: Using the period of the orbit, we can calculate the average orbital

speed of the planet using Kepler’s third law, which relates the orbital period

(T), the semi-major axis (a), and the gravitational constant (G) and the mass

of the star (M) as T2=4π2

GM a3.

Step 4: The average orbital speed, v, is given by v=2πa

T. Substituting the

values of Tand ainto this equation will give us the average orbital speed of the

planet.

Step 5: Finally, using the formula for the eccentricity of an elliptical orbit,

e= 1 −average orbital speed2·a

GM , we can calculate the eccentricity of the planet’s

orbit. Substituting the known values into this equation will give us the final

answer.

Question 7

Question

A planet is in a circular orbit around a star with a period of 50 years. Determine

the radius of the planet’s orbit in astronomical units (AU), given that the star

has a mass of 2×1030 kg.

Solution

Step 1: We can use Kepler’s Third Law to relate the period of an orbiting body

to its average distance from the central body:

T2=(4π2r3

GM )

where: - Tis the period of the orbit, - ris the average distance between the

planet and the star, - Gis the gravitational constant (approximately 6.67 ×

10−11 m3/kg s2), - Mis the mass of the central body (the star).

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

(50 yr)2=4π2·r3

(6.67 ×10−11 m3/kg s2)·(2 ×1030 kg)

Step 3: Simplifying, we find:

2500 yr2=4π2·r3

1.334 ×1020

Step 4: Rearranging the equation to solve for r, we get:

r3=2500 ·1.334 ×1020

4π2

Step 5: Calculating r, we find:

r≈(2500 ·1.334 ×1020

4π2)1/3

Step 6: Finally, converting the radius into astronomical units (AU) where 1

AU is the average Earth-Sun distance of 149.6 million km:

r≈(2500 ·1.334 ×1020

4π2)1/3

×1

149.6×106

This will give us the radius of the planet’s orbit in astronomical units.

Question 8

Question

Consider a planet that orbits a star in an elliptical path. According to Kepler’s

Second Law of Planetary Motion, the planet moves faster when it is closer to

the star and slower when it is farther away. If the planet takes 60 days to travel

from its closest point to the star (perihelion) to its farthest point (aphelion),

and the distance between perihelion and aphelion is 0.8 astronomical units (AU),

determine the average speed of the planet in the elliptical orbit.

Given:

• Time taken from perihelion to aphelion (∆t) = 60 days

• Distance between perihelion and aphelion (∆r) = 0.8 AU

Solution

Step 1: Calculate the average speed of the planet by using the formula for

average speed, Average speed =Total distance

Total time .

Step 2: First, calculate the total distance traveled by the planet in one

complete elliptical orbit. This distance is equal to the 2 times the semi-major

axis of the elliptical orbit.

The semi-major axis (a) of the elliptical orbit can be found by dividing the

sum of perihelion distance and aphelion distance by 2:

a=rperihelion +raphelion

2=0+0.8

2= 0.4AU

The total distance traveled in one complete orbit is:

Total distance = 2a= 2(0.4) = 0.8AU

Step 3: Now, substitute the values of total distance and total time into the

formula for average speed:

Average speed =0.8AU

60 days =0.8×AU

60 ×days =0.8

60 AU/day

Step 4: Therefore, the average speed of the planet in its elliptical orbit is

0.8

60 AU/day.

Question 9

Question

State Kepler’s Third Law of planetary motion and describe how it can be used

to determine the period of a planet orbiting the Sun.

Solution

Kepler’s Third Law states that the square of the period of revolution (T) of a

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

(r) from the Sun. Mathematically, this can be expressed as:

T2=kr3

where kis a constant that depends on the masses of the planet and the Sun, as

well as the gravitational constant.

Step 1: Consider a planet with a semi-major axis of aorbiting the Sun.

The average distance rfrom the Sun is half of the major axis, so r=1

2a.

Step 2: The period of revolution Tof the planet is related to the average

distance rby Kepler’s Third Law as T2=kr3. Substituting r=1

2a, we get:

T2=k(1

2a)3

8a3

Step 3: Now, consider the period of Earth’s orbit around the Sun, denoted

as TEarth, and the average distance of Earth from the Sun, denoted as rEarth.

We can write:

Earth =k

8r3

Earth

Step 4: Let’s introduce the period of the planet we want to find the period

for, denoted as Tplanet, and the semi-major axis of its orbit around the Sun,

denoted as aplanet. We can then write:

planet =k

8a3

planet

Step 5: To determine the period Tplanet of the planet orbiting the Sun, we

need to know its semi-major axis aplanet. If this information is given, we can

use the relationship T2

planet =k

8a3

planet to calculate the period of the planet.

Question 10

Question

An asteroid is moving in a perfectly circular orbit around the sun. The radius

of the orbit is 3 AU (astronomical units). If the asteroid’s orbital speed is 30

km/s, determine the mass of the asteroid in terms of the mass of the sun (M⊙).

Solution

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

of an orbit (in years) is proportional to the cube of the semi-major axis (in

astronomical units) of the orbit.

1year )2

=(a

1AU )3

Where Tis the orbital period and ais the semi-major axis.

Step 2: We know that the orbital speed of the asteroid can be related to the

orbit’s radius and the gravitational force. The gravitational force is given by:

F=GM⊙m

Where Gis the gravitational constant, M⊙is the mass of the sun, mis the mass

of the asteroid, and ris the radius of the orbit.

Step 3: The centripetal force required to keep the asteroid in a circular orbit

is given by:

F=mv2

Where vis the orbital speed of the asteroid.

Step 4: Setting the gravitational force equal to the centripetal force, we have:

GM⊙m

r2=mv2

Step 5: Substitute r= 3 AU and v= 30 km/s into the equation above and

solve for min terms of M⊙.

GM⊙m

(3 AU)2=m(30 km/s)2

3AU

Step 6: Simplify and solve for m. Remember that 1AU = 1.496 ×108km.

m=(30 ×103)2×32

G×1.496 ×108M⊙

Step 7: Calculate the numerical value for m.

Question 11

Question

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

this planet is T. Determine the expression for the radius of the planet’s orbit

in terms of G,M, and T.

Solution

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

period of a planet is proportional to the cube of the semimajor axis of its orbit.

Mathematically, this is represented as:

T2=4π2

GM a3

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

semimajor axis of the orbit.

Step 2: Since the orbit is circular, the semimajor axis is equal to the radius

of the orbit. Therefore, we can rewrite the equation as:

T2=4π2

GM r3

where ris the radius of the planet’s orbit.

Step 3: Solve for the radius of the planet’s orbit by isolating rin the equation.

Taking the cube root of both sides, we get:

r=(GMT 2

4π2)1/3

Step 4: Simplifying the expression further, we find:

r=(GM

4π2)1/3

T2/3

Therefore, the radius of the planet’s orbit in terms of G,M, and Tis

(GM

4π2)1/3T2/3.

Question 12

Question

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

axis of the planet’s orbit is 3.5×108km. If the period of the planet’s orbit is

400 days, calculate the following: a) The semi-minor axis of the planet’s orbit.

b) The speed of the planet when it is at its closest approach to the star. c) The

speed of the planet when it is at its farthest distance from the star.

Solution

a) To find the semi-minor axis of the planet’s orbit, we use the relationship

between the semi-major axis aand the semi-minor axis bin an elliptical orbit:

a2=b2+c2

where cis the distance from the center to a focus of the ellipse.

Step 1: Calculate cSince the eccentricity of the planet’s orbit is 0.2, we

have:

c= 0.2×3.5×108km

c= 7 ×107km

Step 2: Find bSubstitute a= 3.5×108km and c= 7 ×107km into the

equation:

b=√a2−c2

b=√(3.5×108)2−(7 ×107)2

b) To find the speed of the planet when it is at its closest approach to the

star, we use the formula for the speed of an object in orbit:

v=√G(M+m)

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

of the planet (assumed negligible), and ris the distance between the planet and

the star.

Step 1: Find rat closest approach At closest approach, the distance between

the planet and the star is a−c.

Step 2: Calculate the speed vat closest approach

c) To find the speed of the planet when it is at its farthest distance from the

star, we repeat the process as in part (b) but using r=a+c. Calculate the

speed of the planet when it is at its farthest distance.

Question 13

Question

Suppose a planet orbits a star in an elliptical path. The closest and farthest

distances of the planet from the star are 30 million kilometers and 60 million

kilometers, respectively. Calculate the eccentricity of the planet’s orbit.

Solution

Step 1: Recall the definition of eccentricity for an elliptical orbit. The eccen-

tricity (e) of an orbit is defined as the ratio of the distance between the foci of

the ellipse to the length of the major axis. The formula for eccentricity is given

by:

e=c

where cis the distance between the center of the ellipse and one of its foci, and

ais the length of the semi-major axis.

Step 2: Identify the center of the ellipse, and the foci based on the informa-

tion given. In this case, the center of the ellipse is the star around which the

planet orbits, and the two foci lie along the major axis of the ellipse.

Step 3: Calculate the length of the semi-major axis (a) using the given closest

and farthest distances of the planet from the star. The length of the semi-major

axis (a) is equal to half of the sum of the closest and farthest distances:

a=30 million km + 60 million km

2= 45 million km

Step 4: Calculate the distance between the center and one of the foci (c).

The distance between the center and one of the foci is equal to half of the

distance between the closest and farthest distances:

c=60 million km −30 million km

2= 15 million km

Step 5: Substitute the values of aand cinto the formula for eccentricity to

find the eccentricity of the planet’s orbit:

e=15 million km

45 million km =1

Therefore, the eccentricity of the planet’s orbit is 1

3or approximately 0.333.

Question 14

Question

Assume a hypothetical planetary system in which a planet follows an elliptical

orbit around a star. The semi-major axis of the orbit is 2 AU, and the eccen-

tricity of the orbit is 0.4. If the distance between the planet and the star at

perihelion is 1 AU, calculate the distance between the planet and the star at

aphelion.

Solution

Step 1: Recall the formula for the distance from the focus of an ellipse to a

point on its boundary:

r=a(1 −e2)

1 + ecos(θ)

where: r= distance between the planet and the star a= semi-major axis of the

orbit e= eccentricity of the orbit θ= true anomaly

Step 2: Identify the known values: For perihelion: rperihelion = 1 AU, a= 2

AU, e= 0.4

Step 3: Use the formula to determine the true anomaly at perihelion:

1 = 2(1 −0.42)

1+0.4 cos(θperihelion)

1 = 2(0.84)

1+0.4 cos(θperihelion)

1 = 1.68

1+0.4 cos(θperihelion)

1+0.4 cos(θperihelion) = 1.68

0.4 cos(θperihelion) = 0.68

cos(θperihelion) = 0.68

0.4= 1.7

Step 4: Find the angle θperihelion: Since cos−1(1.7) is not a valid angle, we

made a mistake in our calculation. Let’s find the correct angle.

cos(θperihelion) = −1.7

θperihelion = cos−1(−1.7) = 2π−cos−1(1.7)

Step 5: Calculate the distance from the star at aphelion:

raphelion =2(1 −0.42)

1+0.4 cos(θaphelion)

Substitute the correct value of θperihelion to find raphelion.

Question 15

Question

During an experiment to study the motion of planets, a team of scientists ob-

served a specific planet with an elliptical orbit around the Sun. The planet’s

closest distance to the Sun (perihelion) is 0.3 AU and its farthest distance (aphe-

lion) is 0.7 AU. Calculate the eccentricity of the planet’s orbit.

Solution

Step 1: Recall the formula for eccentricity (e) of an ellipse:

e=distance between foci

length of major axis

Step 2: In a planet’s elliptical orbit, the distance between the foci is twice

the semi-major axis, so

distance between foci = 2a

Step 3: The major axis (2a) is the sum of the perihelion and aphelion dis-

tances:

2a=perihelion +aphelion

Step 4: Substitute the given values to find the length of the major axis:

2a= 0.3+0.7 = 1.0AU

Step 5: Substitute into the eccentricity formula:

e=2a

2a=1.0

1.0= 1.0

Step 6: Therefore, the eccentricity of the planet’s orbit is 1.0.

Question 16

Question

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

the foci. The planet’s closest approach to the star is 28 million kilometers and

its farthest distance is 40 million kilometers. Calculate the eccentricity of the

planet’s orbit.

Solution

Step 1: Recall the definition of eccentricity, e, for an ellipse:

e=distance from the center to a focus

distance from the center to a point on the ellipse

Step 2: In this case, the closest approach of the planet to the star is the

distance from the center to the point on the ellipse, so rmin = 28 million km.

Step 3: The farthest distance of the planet from the star is the sum of the

distance from the center to a focus and the distance to a point on the ellipse.

Therefore, we have rmax =rmin + 2f= 40 million km, where fis the distance

from the center to a focus.

Step 4: Substituting the given values into the equation for e:

e=f

rmin

=rmax −rmin

2rmin

Step 5: Plugging in the values, we get:

e=40 −28

2×28 =12

56 =3

14 ≈0.2143

Step 6: Therefore, the eccentricity of the planet’s orbit is approximately

0.2143.

Question 17

Question

In a distant solar system, a planet has an orbital period of 2.5 Earth years. The

planet’s average distance from its star is 3×108km. Determine the mass of the

star using Kepler’s third law.

Solution

Step 1: Recall Kepler’s third law, which relates the square of the orbital period

of a planet (T) to the cube of its average distance from the star (r) and the

mass of the star (M):

T2=4π2

GM r3

where: - T= 2.5Earth years = 2.5 Earth years ×365.25 days

1Earth year ×24 hours

1day ×

3600 s

1hour = 7.89 ×107s. - r= 3 ×108km = 3×108km ×103m

1km = 3 ×1011 m. -

G= 6.67 ×10−11 N m2/ kg2(gravitational constant).

Step 2: Plug in the known values into the formula:

(7.89 ×107s)2=4π2

6.67 ×10−11 N m2/kg2·M·(3 ×1011 m)3

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

M=4π2

6.67 ×10−11 ·(3 ×1011)3

(7.89 ×107)2

Step 4: Calculate the mass of the star using a calculator:

M≈5.33 ×1030 kg

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

5.33 ×1030 kg.

Question 18

Question

Given an elliptical orbit of a planet around the Sun, find the relationship between

the orbital period Tand the semi-major axis a. Kepler’s third law states that

the square of the orbital period is proportional to the cube of the semi-major

axis: T2∝a3. Show that this relationship holds for an elliptical orbit.

Solution

To show the relationship T2∝a3for an elliptical orbit, we first need to express

the orbital period Tand the semi-major axis ain terms of the physical properties

of the orbit.

Step 1: Expressing Tin terms of afor an elliptical orbit

The orbital period Tcan be related to the semi-major axis ausing Kepler’s

second law, which states that the area swept by the line connecting the planet

to the Sun is constant. The area swept by this line in a small time interval dt

is given by 1

2r2dθ, where ris the distance from the planet to the Sun and dθ is

the change in angle.

Since the orbit is elliptical, rcan be expressed in terms of the semi-major

axis aand the eccentricity eof the ellipse: r=a(1−e2)

1+ecos(θ), where θis the true

anomaly.

Integrating dA

dt =1

2r2dθ

dt over one full orbit gives:

T=∫2π

2(a(1 −e2)

1 + ecos(θ))2dθ

dt dθ

Step 2: Simplifying the expression for T

The integral for Tmay not have a simple closed-form solution. However,

using numerical methods, we can compute the numerical value of Tfor a given

a. This demonstrates that Tdepends on afor an elliptical orbit.

Step 3: Showing the relationship T2∝a3

From Step 2, we have shown that Tis dependent on afor an elliptical orbit.

By representing Tin terms of a, we can then square Tand cube ato compare

them. This comparison will confirm that T2∝a3holds for an elliptical orbit,

consistent with Kepler’s third law.

Question 19

Question

A planet has an elliptical orbit around the Sun, with the Sun located at one of

the foci. The closest and farthest distances between the planet and the Sun are

0.3 AU and 0.7 AU, respectively. What is the eccentricity of the planet’s orbit?

Solution

Step 1: The eccentricity of an elliptical orbit is defined as the ratio of the

distance between the foci to the length of the major axis. Since the Sun is

located at one of the foci, the distance between the foci is twice the distance

from the Sun to the farthest point on the orbit: 2×0.7AU = 1.4AU.

Step 2: The major axis of an ellipse is the sum of the distances from the

center of the ellipse to the two points on the ellipse farthest apart. In this

case, the major axis is the sum of the closest and farthest distances between the

planet and the Sun: 0.3AU + 0.7AU = 1.0AU.

Step 3: The eccentricity, denoted by e, is given by the formula e=distance between foci

length of major axis .

Thus, in this case, the eccentricity of the planet’s orbit is:

e=1.4AU

1.0AU = 1.4

Step 4: Therefore, the eccentricity of the planet’s orbit is 1.4.

Question 20

Question

Suppose an asteroid is orbiting the Sun in an elliptical orbit. The asteroid’s

closest approach to the Sun (perihelion) is 0.25 AU and its furthest distance

from the Sun (aphelion) is 0.75 AU. If the time it takes for the asteroid to

complete one full orbit is 800 days, determine the average orbital speed of the

asteroid in km/s.

Solution

Step 1: Find the semi-major axis of the asteroid’s orbit. The semi-major axis

of an elliptical orbit is given by the average of the perihelion and aphelion

distances:

a=rperihelion +raphelion

Given that rperihelion = 0.25 AU and raphelion = 0.75 AU, we have:

a=0.25 + 0.75

2=1

2AU

Step 2: Convert the semi-major axis to kilometers. Since 1 AU is equivalent

to 1.496 ×108km, we can convert the semi-major axis to kilometers:

a=1

2AU ×1.496 ×108km/AU = 7.48 ×107km

Step 3: Find the orbital period in seconds. Given that the orbital period is

800 days, we first convert this to seconds:

Orbital period = 800 days×24 hours/day×60 minutes/hour×60 seconds/minute

= 6.912 ×107seconds

Step 4: Calculate the average orbital speed. The average orbital speed of an

object in an elliptical orbit is given by:

vavg =2πa

Orbital period

Substitute the values we found above:

vavg =2π×7.48 ×107

6.912 ×107

vavg =2π×7.48

6.912

vavg ≈4.08 km/s

Therefore, the average orbital speed of the asteroid is approximately 4.08

km/s.

Question 21

Question

A planet has an average distance of 1.5 AU from the Sun. If the planet takes

1.5 years to complete one orbit, determine the mass of the Sun in terms of the

mass of Earth (M⊕).

Solution

Step 1: Find the period of the planet’s orbit in seconds. Given that the planet

takes 1.5 years to complete one orbit, we first convert this to seconds. Since 1

year is approximately 3.15 ×107seconds, we have:

T= 1.5years ×3.15 ×107seconds/year

T= 4.725 ×107seconds

Step 2: Calculate the mass of the Sun using Kepler’s third law. Kepler’s

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

cube of the semi-major axis of the orbit.

T2=4π2

GM a3

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

a= semi-major axis of the orbit.

Given T= 4.725 ×107seconds and a= 1.5AU = 1.5×1.496 ×1011 m

= 2.244 ×1011 m, we can rearrange the equation to solve for M:

M=4π2

G(a3

T2)

Step 3: Substitute the known values and calculate the mass of the Sun. Using

G= 6.674 ×10−11 m3kg−1s−2, we substitute the values into the equation to

find:

M=4π2

6.674 ×10−11 ((2.244 ×1011)3

(4.725 ×107)2)

Calculating this expression will give us the mass of the Sun in terms of the

mass of Earth (M⊕).

Question 22

Question

A planet has an elliptical orbit around a star, with the star located at one of

the foci of the ellipse. The semi-major axis of the planet’s orbit is 2 AU and its

eccentricity is 0.6. If the planet’s speed at the position of closest approach to

the star is v0, find its speed at the position of farthest distance from the star in

terms of v0.

Solution

Step 1: Start by recalling the relation between the semi-major axis, eccentricity,

and the distance of closest approach and farthest distance from the focus in an

elliptical orbit. The distance to the focus from the center of an ellipse is given

by a(1 −e)where ais the semi-major axis and eis the eccentricity.

Step 2: For the distance of closest approach, the distance to the focus is

a(1 −e) = 2(1 −0.6) = 0.8AU. This is the point where the speed is v0.

Step 3: For the distance of farthest distance, the distance to the focus is

a(1 + e) = 2(1 + 0.6) = 3.2AU. This is the point where we want to find the

speed.

Step 4: Using conservation of angular momentum, we know that r1v1=r2v2

where ris the distance from the star and vis the speed at that distance.

Step 5: At the position of closest approach, r1= 0.8AU and v1=v0. At

the position of farthest distance, r2= 3.2AU. Let v2=vbe the speed at this

point.

Step 6: Setting up the conservation of angular momentum equation gives

0.8v0= 3.2v.

Step 7: Solving for vgives v=0.8

3.2v0=1

4v0. Therefore, the speed of the

planet at the position of farthest distance from the star is 1

4v0.

Question 23

Question

A planet orbits a star in a nearly circular orbit with a period of 385 days. If

the planet is located at a distance of 1.2 AU from the star, calculate the mass

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

Solution

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

(T) of a planet’s orbit is proportional to the cube of the semi-major axis (r) of

the orbit. Mathematically, this can be expressed as:

T2∝r3

Step 2: Convert the period of the orbit to seconds since Kepler’s Third Law

typically uses SI units. Given that there are 365.25 days in a year and 24 hours

in a day, we have:

T= 385 days ×24 hours/day ×60 minutes/hour ×60 seconds/minute

T= 33,264,000 seconds

Step 3: Next, convert the distance from AU to meters since 1 AU is approx-

imately 1.496 ×1011 meters. Therefore, the distance rin meters is:

r= 1.2AU ×1.496 ×1011 meters/AU

r= 1.7952 ×1011 meters

Step 4: Substitute the values of Tand rinto Kepler’s Third Law equation:

T2=k×r3

332640002=k×(1.7952 ×1011)3

Step 5: Solve for the constant of proportionality k:

k=332640002

(1.7952 ×1011)3

Step 6: Now, knowing the value of k, we can calculate the mass (M) of the

star using the equation for the period of the orbit:

T2=4π2

G×M×r3

M=4π2

G×k×r3

Step 7: Finally, substitute the known values of kand rinto the equation to

find the mass of the star.

Therefore, the mass of the star can be calculated using the above steps.

Question 24

Question

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

approach to the star is 0.1 AU and the farthest point in the orbit is 0.4 AU.

Find the eccentricity of the orbit.

Solution

Step 1: Recall the definition of eccentricity for an elliptical orbit.

The eccentricity, e, of an elliptical orbit is defined as the ratio of the distance

between the foci of the ellipse to the length of the major axis. It can also be

expressed in terms of the major (a) and minor (b) semi-axes of the ellipse:

e=√1−(b

a)2.

Step 2: Determine the major and minor semi-axes of the ellipse.

Given that the planet’s closest approach to the star is 0.1 AU and the farthest

point in the orbit is 0.4 AU, we can calculate the major semi-axis as half of the

sum of these distances (a=0.1+0.4

2) and the minor semi-axis as half of the

difference between these distances (b=0.4−0.1

2).

Step 3: Calculate the values of aand b.

a=0.1+0.4

2=0.5

2= 0.25 AU

b=0.4−0.1

2=0.3

2= 0.15 AU

Step 4: Substitute the values of aand binto the formula for eccentricity.

Now, we substitute a= 0.25 AU and b= 0.15 AU into the formula for

eccentricity:

e=√1−(0.15

0.25 )2=√1−0.36 = √0.64 = 0.8

Therefore, the eccentricity of the orbit is 0.8.

Question 25

Question

Consider a planet in circular orbit around a star. The planet takes 300 days

to complete one full orbit. If the distance between the planet and the star is

4.5×1010 m, determine the mass of the star.

Solution

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

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

Mathematically, this can be expressed as:

T2=k·a3

where: T= period of orbit (in seconds), a= semi-major axis of orbit (in meters),

k= constant of proportionality.

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

1 day is equal to 86400 seconds, we have:

T= 300 days ×86400 seconds/day = 25920000 seconds

Step 3: Plug in the values of Tand ainto Kepler’s Third Law equation:

(25920000)2=k·(4.5×1010)3

Step 4: Solve for the constant of proportionality k:

k=(25920000)2

(4.5×1010)3

Step 5: Now that we have determined k, we can find the mass of the star

(M) using the formula for the mass of the star in the form of:

M=4π2

G·a3/k

where Gis the gravitational constant (6.67430 ×10−11 m3kg−1s−2).

Step 6: Plug in the values of a,k, and Ginto the equation to calculate the

mass of the star. Calculate the numerical value to obtain the final answer.

Solution

Step 1: Calculate the orbital period using Kepler’s third law, T2=4π2

GM a3,

where Tis the orbital period, G= 6.67 ×10−11 m3/kg ·s2is the gravitational

constant, M= 5 ×1030 kg is the mass of the star, and ais the semimajor axis

of the orbit.

Substitute the given values:

T2=4π2

(6.67 ×10−11)(5 ×1030)a3

T2= 3.77 ×107a3

Given that T= 4 years:

16 = 3.77 ×107a3

Step 2: Solve for the semimajor axis a.

a3=16

3.77 ×107

a3≈4.24 ×10−7

a≈

√4.24 ×10−7

a≈0.0175 AU

Therefore, the semimajor axis of the orbit is approximately 0.0175 AU.

Question 3

Question

A planet orbits a star in a nearly circular orbit with a period of 5 years. The

average distance between the planet and the star is 4 AU. Determine the mass

of the star, given that the gravitational constant is 6.67 ×10−11 m3/kg s2.

Solution

Let’s denote the mass of the star as M, the distance between the planet and

the star as r, the period of the orbit as T, and the gravitational constant as

G. Furthermore, we will use Kepler’s Third Law of Planetary Motion, which

states:

T2=4π2r3

Step 1: Convert the average distance rfrom AU to meters.

r= 4 AU ×1.496 ×1011 m/AU = 5.984 ×1011 m