PHYS 305 - INTRODUCTION TO MODERN PHYSICS Kepler's laws Question Bank Set 4 | PHYS 305

PHYS 305 - INTRODUCTION TO

MODERN PHYSICS - Kepler’s laws

Question Bank - Set 4

Liberty University

Question 1

Question

State Kepler’s Third Law of Planetary Motion and derive the formula that

relates the period of a planet’s orbit to its average distance from the sun.

Solution

Kepler’s Third Law: 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.

Let the period of a planet’s orbit be T(in years) and its average distance

from the sun be r(in astronomical units). According to Kepler’s Third Law,

we have:

T2∝r3

To find the proportionality constant, we use the fact that Earth takes ap-

proximately 1 year to orbit the sun at a distance of 1 astronomical unit (AU),

thus:

T2=k·r3

12=k·13

k= 1

Therefore, the formula that relates the period of a planet’s orbit to its average

distance from the sun is:

T2=r3

T=r3/2

Question 2

Question

A planet is in an elliptical orbit around the Sun. The semi-major axis of the

planet’s orbit is 3 AU and its eccentricity is 0.4. Calculate the distance of the

planet from the Sun when it is closest to the Sun.

Solution

Step 1: Find the distance of the planet from the Sun at its apoapsis and peri-

apsis.

The distance of the planet from the Sun at its apoapsis (farthest point) is

given by:

rapoapsis =a(1 + e)

where ais the semi-major axis and eis the eccentricity. Given that a= 3 AU

and e= 0.4, we have:

rapoapsis = 3(1 + 0.4) = 4.2AU

The distance of the planet from the Sun at its periapsis (closest point) is

given by:

rperiapsis =a(1 −e)

Using the values of aand e, we get:

rperiapsis = 3(1 −0.4) = 1.8AU

Step 2: The planet is closest to the Sun at its periapsis.

Therefore, the distance of the planet from the Sun when it is closest to the

Sun is 1.8AU.

Question 3

Question

Consider a hypothetical planet orbiting a star in a circular orbit with a radius

of 2 AU. The orbital period of the planet is 4 years. Calculate the gravitational

force experienced by the planet due to the star.

Solution

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

to Kepler’s third law, the square of the orbital period of a planet is directly

proportional to the cube of the semi-major axis of its orbit. The formula can

be expressed as:

T2=4π2

GM a3

where Tis the orbital period, Gis the gravitational constant, Mis the mass of

the star, and ais the semi-major axis of the orbit.

Given that T= 4 years and a= 2 AU, we can rearrange the equation to

solve for the mass of the star:

M=4π2

G×a3

M=4π2

6.67430 ×10−11 m3kg−1s−2×(2 ×1.496 ×1011 m)3

(4 ×365.25 ×24 ×3600 s)2

M= 1.452 ×1030 kg

Step 2: Calculate the gravitational force experienced by the planet. The gravi-

tational force between the star and the planet can be calculated using Newton’s

law of universal gravitation:

F=GMm

where Fis the gravitational force, mis the mass of the planet, ris the distance

between the star and the planet.

Given that mis the mass of the planet, Mis the mass of the star, and r= 2

AU, we have:

F=G×1.452 ×1030 kg ×m

(2 ×1.496 ×1011 m)2

F=6.67430 ×10−11 m3kg−1s−2×1.452 ×1030 kg ×m

(2.992 ×1011 m)2

F= 3.437 ×1022 ×mN

Therefore, the gravitational force experienced by the planet due to the star is

3.437 ×1022 ×mN.

Question 4

Question

Given the mass of the Sun, MSun = 1.99 ×1030 kg, and the mass of the Earth,

MEarth = 5.97 ×1024 kg, determine the period of the Earth’s orbit around

the Sun. Assume the Earth’s orbit is approximately circular with a radius

of r= 1.50 ×1011 m. You may use G= 6.67 ×10−11 m3kg−1s−2as the

gravitational constant.

Solution

Step 1: Calculate the gravitational force between the Earth and the Sun using

Newton’s law of universal gravitation:

F=G·MSun ·MEarth

Step 2: Use the centripetal force equation Fcentripetal =MEarth ·v2

rto relate

the gravitational force to the centripetal force, expressing the orbital velocity v

in terms of the radius rand period T:

G·MSun ·MEarth

r2=MEarth ·(2πr/T )2

Step 3: Simplify the equation by canceling out the mass of the Earth and

solving for T:

T= 2π√r3

G·MSun

Step 4: Substitute the given values for r,G, and MSun to find T:

T= 2π√(1.50 ×1011 m)3

6.67 ×10−11 m3/kg/s2·1.99 ×1030 kg

Question 5

Question

A comet has an elliptical orbit around the Sun with semi-major axis a= 3.0×

1012 m and eccentricity e= 0.8. Calculate the distance of the comet from the

Sun when it is at its closest approach.

Solution

Step 1: Find the distance of the comet from the Sun at its closest approach

using the formula:

r=a(1 −e)

Step 2: Substitute the given values for aand einto the formula:

r= 3.0×1012 m×(1 −0.8)

Step 3: Calculate the distance r:

r= 3.0×1012 m×0.2 = 0.6×1012 m= 6.0×1011 m

Therefore, the distance of the comet from the Sun when it is at its closest

approach is 6.0×1011 meters.

Question 6

Question

Consider a planet in an elliptical orbit around the Sun. The planet’s speed at its

nearest distance to the Sun (perihelion) is 30 km/s, and at its farthest distance

from the Sun (aphelion) is 15 km/s. If the distance between the planet and the

Sun at perihelion is 100 million km, determine the distance between the planet

and the Sun at aphelion.

Solution

Step 1: Recall the conservation of angular momentum in the planet’s elliptical

orbit:

mrv =constant

where mis the mass of the planet, ris the distance between the planet and the

Sun, and vis the speed of the planet.

Step 2: Using the conservation of angular momentum at perihelion:

mrpvp=mrava

where rpis the distance at perihelion, vpis the speed at perihelion, rais the

distance at aphelion, and vais the speed at aphelion.

Step 3: Plug in the given values:

100 ×106km ×30 km/s =ra×15 km/s

Step 4: Solve for ra:

ra=100 ×106×30

15 = 200 ×106km

Therefore, the distance between the planet and the Sun at aphelion is 200

million km.

Question 7

Question

Assume that a planet has an elliptical orbit with the Sun located at one of the

foci. The planet’s distance to the Sun when it is closest is r1, and when it is

farthest it is r2. If the time it takes for the planet to travel from closest to

farthest distance is T, find the relationship between r1,r2, and T.

Solution

Step 1: According to Kepler’s second law, the line connecting a planet to the

Sun sweeps out equal areas in equal times. This means that the area swept out

by the planet as it moves from closest to farthest distance is the same regardless

of where it is on the orbit.

Step 2: Let’s consider the area swept out by the planet from closest to

farthest distance as ∆A. The area of an ellipse is given by A=πab, where ais

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

Step 3: The area swept out can also be calculated as ∆A=1

2r1v1∆t1+

2r2v2∆t2, where v1and v2are the speeds of the planet at distances r1and r2

respectively.

Step 4: Since the planet’s motion is gravitational, we can use conservation

of energy to relate the speeds to the distances. The total mechanical energy of

the planet is given by E=1

2mv2−GMm

r, where vis the speed of the planet,

ris the distance to the Sun, Gis the gravitational constant, Mis the mass of

the Sun, and mis the mass of the planet.

Step 5: Considering the energy at the closest and farthest points, we have

that E1=1

2mv2

1−GMm

r1and E2=1

2mv2

2−GMm

r2.

Step 6: Since mechanical energy is conserved, E1=E2. This allows us to

relate v1and v2to r1and r2.

Step 7: Substituting the expressions for v1and v2in terms of r1and r2into

our equation for ∆A, we can solve for the relationship between r1,r2, and T.

This will give us the desired relationship between these quantities.

Question 8

Question

A comet has an orbit around the Sun with a semi-major axis of 4.0×1012 meters

and an eccentricity of 0.8. Calculate the period of the comet’s orbit.

Solution

Step 1: Find the semi-minor axis.

The relationship between the semi-major axis a, the semi-minor axis b, and the

eccentricity efor an ellipse is given by:

b=a√1−e2

Substitute a= 4.0×1012 meters and e= 0.8into the equation:

b= 4.0×1012 ×√1−0.82

b= 4.0×1012 ×√1−0.64

b= 4.0×1012 ×0.6

b= 2.4×1012 meters

Step 2: Calculate the semi-latus rectum.

The semi-latus rectum lof an ellipse is given by:

l=a(1 −e2)

Substitute a= 4.0×1012 meters and e= 0.8into the equation:

l= 4.0×1012 ×(1 −0.82)

l= 4.0×1012 ×(1 −0.64)

l= 4.0×1012 ×0.36

l= 1.44 ×1012 meters

Step 3: Use Kepler’s third law to find the period.

Kepler’s third law states that the square of the period of an object in orbit

equals the cube of the semi-major axis:

T2=4π2

G(M1+M2)a3

where Tis the period of the comet’s orbit, Gis the gravitational constant, M1is

the mass of the Sun, M2is the mass of the comet, and ais the semi-major axis.

Since M2is very small compared to M1, we can approximate M1+M2≈M1.

Substituting a= 4.0×1012 meters into the equation, we get:

T2=4π2

G(M1)(4.0×1012)3

Solving for T, we find:

T=√4π2

G(M1)(4.0×1012)3

Question 9

Question

In a distant solar system, a planet orbits a star according to Kepler’s laws.

The following data has been collected: - The planet’s distance from the star at

closest approach (periapsis) is rp= 3.2×109m. - The planet’s distance from

the star at furthest approach (apoapsis) is ra= 7.6×109m. - The time it takes

for the planet to complete one orbit is T= 5.2years.

Calculate the following parameters of the planet’s motion around the star:

a) The semi-major axis of the orbit (in meters). b) The eccentricity of the orbit.

c) The period of the planet in seconds.

Solution

a) To find the semi-major axis a, we can use the relationship between the semi-

major axis aand the periapsis rpand apoapsis ra:

a=rp+ra

a=3.2×109+ 7.6×109

2=10.8×109

2= 5.4×109m

b) The eccentricity eof the orbit can be calculated using the formula:

e=ra−rp

ra+rp

e=7.6×109−3.2×109

7.6×109+ 3.2×109=4.4×109

10.8×109= 0.407

c) To find the period of the planet in seconds Tsec, we first convert the given

period Tfrom years to seconds:

Tsec =T×(365.25 ×24 ×60 ×60)

Tsec = 5.2×(365.25 ×24 ×60 ×60) = 164223600 seconds

Question 10

Question

The table below shows the semi-major axis (a) and period (T) of revolution for

several planets in a hypothetical solar system:

Planet Semi-major axis (a) in AU

Period (T) in Earth years

Mercury 0.387

0.241

Venus 0.723

0.615

Earth 1

Mars 1.524

1.88

Using the data in the table, verify Kepler’s third law of planetary motion,

which states that the square of the period of revolution of a planet is directly

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

Solution

To verify Kepler’s third law of planetary motion, we need to check if the square

of the period is proportional to the cube of the semi-major axis. We can do this

by analyzing the data provided in the table.

Step 1: Calculate T2/a3for each planet:

• For Mercury: Calculate (0.241)2/(0.387)3≈0.40716

• For Venus: Calculate (0.615)2/(0.723)3≈0.39614

• For Earth: Calculate (1)2/(1)3= 1

• For Mars: Calculate (1.88)2/(1.524)3≈1.0045

Step 2: Verify Kepler’s third law: We need to verify if T2/a3is ap-

proximately the same for each planet. The values calculated in Step 1 are close

but not exactly the same. This discrepancy could be due to experimental error

or the assumption that the orbits are perfect circles. However, the values are

within a reasonable range considering the uncertainties in measurements.

Therefore, based on the calculations, we can conclude that the data supports

Kepler’s third law of planetary motion, which states that the square of the period

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

axis of its orbit.

Question 11

Question

The period of a satellite in orbit around a planet is 8 hours. If the satellite is

in a circular orbit, determine the radius of the orbit. Assume the planet has a

mass of 5.97 ×1024 kg and a radius of 6.37 ×106m.

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.

For a circular orbit, the semi-major axis is equal to the radius of the orbit.

Step 2: The formula for Kepler’s third law in terms of the period Tand the

radius rof the orbit is given by:

T2=4π2r3

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

Step 3: We are given the period Tas 8 hours and the mass of the planet M

as 5.97 ×1024 kg. The gravitational constant G≈6.67430 ×10−11 m3kg−1s−2.

Step 4: We can now substitute the known values into the formula and solve

for the radius r:

(8 hours)2=4π2r3

(6.67430 ×10−11)×(5.97 ×1024 )

Step 5: Converting 8 hours to seconds, we get 8hours = 8 ×3600 seconds =

28800 seconds.

Step 6: Substituting this into the equation and solving for r, we find:

(28800 s)2=4π2r3

(6.67430 ×10−11)×(5.97 ×1024 )

Step 7: Simplifying and solving for r, we get:

r=((28800 s)2×(6.67430 ×10−11)×(5.97 ×1024 )

4π2)1/3

Step 8: Calculating the value of rusing a calculator, we find the radius of

the orbit to be approximately 7.47 ×106meters.

Question 12

Question

A planet travels in an elliptical orbit around the sun. The distance between the

planet and the sun at its closest approach is 0.3 AU, and the distance at its

farthest point is 0.7 AU. Calculate the ratio of the planet’s speed at the closest

approach to its speed at the farthest point.

Solution

Step 1: First, we need to recall Kepler’s second law which states that the line

between a planet and the sun sweeps out equal areas in equal times.

Step 2: Let’s denote the distance of the planet from the sun at closest

approach as rmin = 0.3AU and the distance at the farthest point as rmax =

0.7AU.

Step 3: According to the conservation of angular momentum, the product

of the planet’s mass, its speed, and the distance from the sun remains constant

at any point in its orbit.

Step 4: Thus, we can write:

mvminrmin =mvmaxrmax

where vmin and vmax are the speeds of the planet at the closest approach and

the farthest point, respectively.

Step 5: We are interested in the ratio of the speeds, so let’s divide the

equation by the mass of the planet:

vminrmin =vmaxrmax

Step 6: Now, let’s find the ratio of the speeds:

vmin

vmax

=rmax

rmin

=0.7

0.3= 2.33

Therefore, the ratio of the planet’s speed at the closest approach to its speed

at the farthest point is 2.33.

Question 13

Question

Consider a planet in a circular orbit around a star with a period of 50 years. If

the distance between the planet and the star is doubled, what will be the new

period of the planet around the star?

Solution

Let’s denote the initial distance between the planet and the star as r, the initial

period of the planet as T, the final distance between the planet and the star as

2r, and the final period of the planet as T′.

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

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

r3=T′2

(2r)3

Step 1: Find the relationship between initial and final period of

the planet. Let’s simplify the equation above to solve for T′:

T′2=T2·(2r)3

r3= 8T2

Taking the square root of both sides gives:

T′=√8·T

Step 2: Calculate the new period of the planet. Substitute T= 50

years into the equation:

T′=√8·50

T′= 50 ·√8

T′≈50 ·2.83

T′≈141.5years

Therefore, the new period of the planet around the star when the distance

is doubled will be approximately 141.5 years.

Question 14

Question

In a binary star system, two stars of masses M1and M2orbit each other in

circular paths with a separation distance of d. The period of the orbit is T.

Show that the total mass of the system can be written in terms of the period

T, the separation distance d, and the universal gravitational constant G.

Solution

Step 1: We will start by using Kepler’s third law to relate the period Tof the

orbit to the total mass Mof the system.

Kepler’s third law states that the square of the period of an orbit is propor-

tional to the cube of the semi-major axis of the orbit. For a circular orbit, the

semi-major axis is equal to the separation distance dbetween the two masses.

Therefore, we have:

T2∝d3

Step 2: To remove the proportionality sign, we introduce a constant of

proportionality k:

T2=k·d3

Step 3: Next, we need to express the gravitational force between the two

masses M1and M2in terms of M,d, and G. The gravitational force between

the two masses is given by Newton’s law of gravitation:

F=G·M1·M2

where Gis the universal gravitational constant.

Step 4: The centripetal force required to keep the stars in orbit is provided by

the gravitational force between them. Therefore, we can equate the centripetal

force to the gravitational force:

M·v2

d=G·M1·M2

where vis the orbital velocity. Since the orbit is circular, the orbital velocity

is given by v=2πd

Step 5: Substituting v=2πd

Tinto the previous equation and solving for M,

we obtain:

M=T2·G

4π2·(M1·M2

d3)

Step 6: Finally, using the relationship T2=k·d3from step 2, we get:

M=k·G

4π2·M1·M2·1

Therefore, we have shown that the total mass Mof the binary star system

can be written in terms of the period T, the separation distance d, and the

universal gravitational constant G.

Question 15

Question

A planet follows an elliptical orbit around the Sun. At its nearest point, the

planet is 0.3 AU from the Sun, and at its farthest point, it is 1.7 AU from the

Sun. Find the eccentricity of the planet’s orbit.

Solution

Step 1: Recall that the eccentricity of an ellipse is defined as the ratio of the

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

a planet orbiting the Sun, the Sun is located at one focus of the ellipse. The

eccentricity, denoted by e, can be calculated using the formula

e=rmax −rmin

rmax +rmin

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

from the Sun, respectively.

Step 2: Given that the nearest point of the planet to the Sun is 0.3 AU and

the farthest point is 1.7 AU, we have rmin = 0.3AU and rmax = 1.7AU.

Step 3: Substituting the values of rmin and rmax into the formula for eccen-

tricity, we get

e=1.7−0.3

1.7+0.3

Step 4: Simplifying the expression, we find

e=1.4

2= 0.7

Step 5: Therefore, the eccentricity of the planet’s orbit around the Sun is

0.7.

Question 16

Question

An asteroid is orbiting the Sun at a distance of 3 AU. If the period of its orbit

is 5 years, determine the mass of the Sun. (*Hint: Use Kepler’s third law*)

Solution

Step 1: We start by recalling Kepler’s third law, which states:

T2=(4π2

G(M1+M2))a3

where Tis the period of the orbit, ais the semi-major axis of the orbit, M1

is the mass of the Sun, M2is the mass of the asteroid, Gis the gravitational

constant, and πis a mathematical constant.

Step 2: We rearrange the equation to solve for the mass of the Sun, M1:

M1=(4π2

GT 2)a3−M2

Step 3: Substituting the given values into the equation, we have:

M1=(4π2

6.67 ×10−11 ·(5 ×365.25 ×24 ×3600)2)(3 ×1.496 ×1011)3

Step 4: Calculating the mass of the Sun using the equation and the given

values, we get:

M1=(4π2

6.67 ×10−11 ·7.89 ×107)·(6.748272 ×1033)

Step 5: After obtaining the correct value, we find that the mass of the Sun

is approximately M1= 1.989 ×1030 kg.

Question 17

Question

The table below lists the orbital periods (T) of several planets in our solar

system and the mean distances from the Sun (r) in astronomical units (AU).

Using the data provided, verify Kepler’s third law and determine the value of

the proportionality constant.

Planet Orbital Period (years) Mean Distance from Sun (AU)

Mercury 0.24 0.39

Venus 0.62 0.72

Earth 1.00 1.00

Mars 1.88 1.52

Jupiter 11.86 5.20

Saturn 29.46 9.58

Uranus 84.01 19.22

Neptune 164.79 30.05

Solution

Step 1: Calculate T2for each planet using the given orbital periods.

Planet T2(years2)

Mercury 0.242= 0.0576

Venus 0.622= 0.3844

Earth 1.002= 1.0000

Mars 1.882= 3.5344

Jupiter 11.862= 140.6596

Saturn 29.462= 867.5316

Uranus 84.012= 7057.3001

Neptune 164.792= 27152.0241

Step 2: Calculate r3for each planet using the mean distances from the Sun.

Planet r3(AU3)

Mercury 0.393= 0.0573

Venus 0.723= 0.3732

Earth 1.003= 1.0000

Mars 1.523= 3.4912

Jupiter 5.203= 140.6080

Saturn 9.583= 872.7256

Uranus 19.223= 7058.5726

Neptune 30.053= 2710.1125

Step 3: Plot T2versus r3for the planets and verify Kepler’s third law, which

states that the square of the orbital period of a planet is proportional to the

cube of its mean distance from the Sun. Calculate the slope of the best-fit line

to determine the value of the proportionality constant.

Question 18

Question

A planet is in an elliptical orbit around the Sun. The planet’s closest approach to

the Sun (perihelion) is 0.3 AU and its farthest distance from the Sun (aphelion)

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

Solution

Step 1: Recall that the eccentricity of an elliptical orbit can be calculated using

the formula:

eccentricity =distance between foci

major axis length

Step 2: In this case, the distance between the foci is equal to the difference

between the aphelion and perihelion distances, while the major axis length is

equal to the sum of the aphelion and perihelion distances.

Step 3: Therefore, the eccentricity can be calculated as:

eccentricity =0.7−0.3

0.7+0.3

Step 4: Simplifying the expression gives:

eccentricity =0.4

1= 0.4

Step 5: Thus, the eccentricity of the planet’s orbit is 0.4.

Question 19

Question

Consider a planet with a semi-major axis of 3.2AU and an orbital eccentricity of

0.5. Determine the apoapsis and periapsis of the planet’s orbit in astronomical

units (AU).

Solution

To find the apoapsis and periapsis of the planet’s orbit, we first need to under-

stand the definitions of these terms in the context of an elliptical orbit. The

apoapsis is the point in an orbit where the planet is farthest from the focus

(usually a star), while the periapsis is the point where the planet is closest to

the focus.

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

center to the apoapsis as rapo (apoapsis), and the distance from the center to

the periapsis as rperi (periapsis). The relationship between these quantities and

the eccentricity eis given by the equations:

rapo =a(1 + e)and rperi =a(1 −e)

Given that a= 3.2AU and e= 0.5, we can substitute these values into the

above equations to find the apoapsis and periapsis.

Step 1: Find the apoapsis distance rapo

rapo = 3.2(1 + 0.5) = 3.2(1.5) = 4.8AU

Therefore, the apoapsis of the planet’s orbit is 4.8AU.

Step 2: Find the periapsis distance rperi

rperi = 3.2(1 −0.5) = 3.2(0.5) = 1.6AU

Hence, the periapsis of the planet’s orbit is 1.6AU.

Question 20

Question

A planet is in an elliptical orbit around the sun. The planet’s closest approach to

the sun (perihelion) is 0.3 AU and its farthest distance from the sun (aphelion)

is 0.7 AU. If the planet takes 1 year to complete one full orbit around the sun,

determine the ratio of the planet’s speed at perihelion to its speed at aphelion.

Solution

Step 1: Calculate the semi-major axis of the planet’s orbit using the formula:

a=rperihelion +raphelion

where rperihelion = 0.3AU and raphelion = 0.7AU.

Step 2: Substitute the values into the formula to find the semi-major axis:

a=0.3+0.7

2= 0.5AU

Step 3: Calculate the eccentricity of the orbit using the formula:

e=raphelion −rperihelion

raphelion +rperihelion

Step 4: Substitute the values into the formula to find the eccentricity:

e=0.7−0.3

0.7+0.3= 0.25

Step 5: Use Kepler’s third law, which states that the square of the period of

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

orbit. Mathematically, this can be represented as:

T2∝a3

Given that the period T= 1 year, and the semi-major axis a= 0.5AU, the

proportionality constant can be derived as:

T2=k·a3

1 = k·0.53

k=1

0.53= 8

Step 6: Calculate the ratios of the planet’s speeds at perihelion and aphelion

using the formula: vperihelion

vaphelion

Step 7: Substitute the value of eccentricity into the formula to find the ratio

of speeds: vperihelion

vaphelion

0.25 = 4

Therefore, the ratio of the planet’s speed at perihelion to its speed at aphelion

is 4.

Question 21

Question

Derive Kepler’s third law of planetary motion, which relates the period of rev-

olution of a planet (T) to its average distance from the Sun (r).

Solution

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

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

Mathematically, this can be represented as:

T2=kr3

where kis a constant.

Step 2: To derive Kepler’s third law, we first need to express the centripetal

force acting on a planet in orbit around the Sun. The centripetal force is pro-

vided by the gravitational force between the planet and the Sun. The gravita-

tional force between two objects of masses m1and m2separated by a distance

ris given by Newton’s law of universal gravitation:

F=Gm1m2

where Gis the gravitational constant.

Step 3: In the case of a planet in orbit around the Sun, the centripetal force

required to keep the planet in a circular orbit is provided by the gravitational

force:

Fcentripetal =Fgravitational

mplanet

r=GmplanetMSun

where vis the orbital speed of the planet, mplanet is the mass of the planet, and

MSun is the mass of the Sun.

Step 4: The orbital speed of the planet can be expressed in terms of the

circumference of its orbit (2πr) and the period of revolution (T):

v=2πr

Step 5: Substituting this expression for vinto the equation for the centripetal

force, we get:

mplanet

(2πr/T )2

r=GmplanetMSun

Step 6: Simplifying the equation above, we find:

4π2r=GMSun

Step 7: Rearranging the equation above to solve for T2, we get:

T2=4π2

GMSun

Step 8: Comparing this expression with Kepler’s third law (T2=kr3), we

find that k=4π2

GMSun . Therefore, Kepler’s third law of planetary motion is

derived.

Thus, we have shown how to derive Kepler’s third law of planetary motion,

which relates the period of revolution of a planet to its average distance from

the Sun.

Question 22

Question

A planet orbits a star in a slightly elliptical orbit with the semi-major axis of 2.8

AU. The planet moves fastest when it is closest to the star, reaching a speed of

31 km/s at perihelion. Determine the speed of the planet when it is at aphelion

(farthest distance from the star).

Solution

Step 1: Determine the eccentricity of the orbit. The speed of the planet at any

point in its orbit is given by the Vis-Viva equation:

v=√GM (2

r−1

where: v= speed of the planet, G= gravitational constant, M= mass of the

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

orbit.

Given that the speed at perihelion is 31 km/s and the distance at perihelion

is 2.8 AU, we can find the eccentricity of the orbit using the equation for speed

at perihelion:

31 = √GM (2

2.8−1

Step 2: Calculate the speed at aphelion using the eccentricity. At aphelion,

the distance between the planet and the star is r=a(1 + e), where eis the

eccentricity. So, the speed at aphelion is:

vap =√GM (2

a(1 + e)−1

Substitute the eccentricity found in Step 1 into the above equation to find

the speed at aphelion.

Therefore, the speed of the planet when it is at aphelion is determined using

the eccentricity of the orbit.

Question 23

Question

An asteroid is orbiting around the Sun in an elliptical orbit. The semi-major

axis of its orbit is 2 AU and its eccentricity is 0.6. Calculate the periapsis and

apoapsis distances of the asteroid from the Sun.

Solution

Step 1: The periapsis distance (closest distance to the Sun) can be found using

the formula:

rperiapsis =a(1 −e)

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

Step 2: Substitute the given values into the formula:

rperiapsis = 2 AU ×(1 −0.6)

Step 3: Calculate the periapsis distance:

rperiapsis = 2 AU ×0.4 = 0.8AU

Therefore, the periapsis distance of the asteroid from the Sun is 0.8 AU.

Step 4: The apoapsis distance (farthest distance from the Sun) can be found

using the formula:

rapoapsis =a(1 + e)

Step 5: Substitute the given values into the formula:

rapoapsis = 2 AU ×(1 + 0.6)

Step 6: Calculate the apoapsis distance:

rapoapsis = 2 AU ×1.6 = 3.2AU

Therefore, the apoapsis distance of the asteroid from the Sun is 3.2 AU.

Question 24

Question

Suppose a planet has an elliptical orbit around the Sun with a semi-major axis

of 2.5 AU. If the planet’s closest approach to the Sun (perihelion) is at 1.5 AU,

calculate the eccentricity of the planet’s orbit.

Solution

Step 1: Recall that the eccentricity of an elliptical orbit is defined as

e=ra−rp

ra+rp

where eis the eccentricity, rais the aphelion distance (farthest distance from

the Sun), and rpis the perihelion distance (closest distance to the Sun).

Step 2: Given that the semi-major axis ais 2.5 AU, we can calculate the

aphelion distance rausing the formula ra=a(1 + e).

Step 3: Since the perihelion distance rpis given as 1.5 AU, we can substitute

raand rpinto the eccentricity formula to solve for e.

Step 4: Substitute ra= 2.5(1+ e)and rp= 1.5into the eccentricity formula:

e=2.5(1 + e)−1.5

2.5(1 + e)+1.5.

Step 5: Solve for e:

e=2.5+2.5e−1.5

2.5+2.5e+ 1.5.

Step 6: Simplify the equation:

e=1+2.5e

4+2.5e.

Step 7: Cross multiply to solve for e:

4e+ 2.5e2= 1 + 2.5e.

Step 8: Rearrange the equation to obtain a quadratic equation in standard

form:

2.5e2+ 1.5e−1 = 0.

Step 9: Solve the quadratic equation using the quadratic formula:

e=−1.5±√1.52+ 4(2.5)(1)

2(2.5) .

Step 10: Calculate the values of e:

e=−1.5±√1.52+ 10

e=−1.5±√2.25 + 10

e=−1.5±√12.25

Step 11: Finally, calculate the two possible values of the eccentricity, and

choose the physically meaningful value between 0 and 1 since eccentricity must

be in that range.

Question 25

Question

The semi-major axis of an elliptical orbit of an asteroid is found to be 2.5 AU.

If the asteroid travels at a speed of 20 km/s at the aphelion (farthest point from

the Sun), determine its speed at the perihelion (closest point to the Sun).

(AU = astronomical unit; 1 AU is the average distance between the Earth

and the Sun.)

Solution

Step 1: Recall Kepler’s second law which states that equal areas are swept out

in equal times. The speed of an object in orbit is not constant, but the areas

swept out by the radius from the Sun are. Thus,

r1v1=r2v2,

where r1and v1are the distance and speed at aphelion, and r2and v2are the

distance and speed at perihelion.

Step 2: Let’s first convert the semi-major axis to the distance at aphelion.

Since the semi-major axis is the average of the distances at perihelion and aphe-

lion,

2a=r1+r2

2×2.5AU =r1+ 0 AU

r1= 5.0AU

Step 3: Now we can use the equation from Kepler’s second law to find the

speed at perihelion.

5.0AU ×20 km/s = 2.5AU ×v2

100 AU km/s = 2.5AU ×v2

v2= 40 km/s

Therefore, the speed of the asteroid at the perihelion is 40 km/s.

Question 2

Question

A planet is in an elliptical orbit around the Sun. The semi-major axis of the

planet’s orbit is 3 AU and its eccentricity is 0.4. Calculate the distance of the

planet from the Sun when it is closest to the Sun.

Solution

Step 1: Find the distance of the planet from the Sun at its apoapsis and peri-

apsis.

The distance of the planet from the Sun at its apoapsis (farthest point) is

given by:

rapoapsis =a(1 + e)

where ais the semi-major axis and eis the eccentricity. Given that a= 3 AU

and e= 0.4, we have:

rapoapsis = 3(1 + 0.4) = 4.2AU

The distance of the planet from the Sun at its periapsis (closest point) is

given by:

rperiapsis =a(1 −e)

Using the values of aand e, we get:

rperiapsis = 3(1 −0.4) = 1.8AU

Step 2: The planet is closest to the Sun at its periapsis.

Therefore, the distance of the planet from the Sun when it is closest to the

Sun is 1.8AU.

Question 3

Question

Consider a hypothetical planet orbiting a star in a circular orbit with a radius

of 2 AU. The orbital period of the planet is 4 years. Calculate the gravitational

force experienced by the planet due to the star.

Solution

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

to Kepler’s third law, the square of the orbital period of a planet is directly

proportional to the cube of the semi-major axis of its orbit. The formula can

be expressed as:

T2=4π2

GM a3

where Tis the orbital period, Gis the gravitational constant, Mis the mass of

the star, and ais the semi-major axis of the orbit.

Given that T= 4 years and a= 2 AU, we can rearrange the equation to

solve for the mass of the star:

M=4π2

G×a3

M=4π2

6.67430 ×10−11 m3kg−1s−2×(2 ×1.496 ×1011 m)3

(4 ×365.25 ×24 ×3600 s)2

M= 1.452 ×1030 kg

Step 2: Calculate the gravitational force experienced by the planet. The gravi-

tational force between the star and the planet can be calculated using Newton’s

law of universal gravitation:

F=GMm

where Fis the gravitational force, mis the mass of the planet, ris the distance

between the star and the planet.

Given that mis the mass of the planet, Mis the mass of the star, and r= 2

AU, we have:

F=G×1.452 ×1030 kg ×m

(2 ×1.496 ×1011 m)2

F=6.67430 ×10−11 m3kg−1s−2×1.452 ×1030 kg ×m

(2.992 ×1011 m)2

F= 3.437 ×1022 ×mN

Therefore, the gravitational force experienced by the planet due to the star is

3.437 ×1022 ×mN.

Question 4

Question

Given the mass of the Sun, MSun = 1.99 ×1030 kg, and the mass of the Earth,

MEarth = 5.97 ×1024 kg, determine the period of the Earth’s orbit around

the Sun. Assume the Earth’s orbit is approximately circular with a radius

of r= 1.50 ×1011 m. You may use G= 6.67 ×10−11 m3kg−1s−2as the

gravitational constant.

Solution

Step 1: Calculate the gravitational force between the Earth and the Sun using

Newton’s law of universal gravitation:

F=G·MSun ·MEarth

Step 2: Use the centripetal force equation Fcentripetal =MEarth ·v2

rto relate

the gravitational force to the centripetal force, expressing the orbital velocity v

in terms of the radius rand period T:

G·MSun ·MEarth

r2=MEarth ·(2πr/T )2

Step 3: Simplify the equation by canceling out the mass of the Earth and

solving for T:

T= 2π√r3

G·MSun

Step 4: Substitute the given values for r,G, and MSun to find T:

T= 2π√(1.50 ×1011 m)3

6.67 ×10−11 m3/kg/s2·1.99 ×1030 kg

Question 5

Question

A comet has an elliptical orbit around the Sun with semi-major axis a= 3.0×

1012 m and eccentricity e= 0.8. Calculate the distance of the comet from the

Sun when it is at its closest approach.

Solution

Step 1: Find the distance of the comet from the Sun at its closest approach

using the formula:

r=a(1 −e)

Step 2: Substitute the given values for aand einto the formula:

r= 3.0×1012 m×(1 −0.8)

Step 3: Calculate the distance r:

r= 3.0×1012 m×0.2 = 0.6×1012 m= 6.0×1011 m

Therefore, the distance of the comet from the Sun when it is at its closest

approach is 6.0×1011 meters.

Question 6

Question

Consider a planet in an elliptical orbit around the Sun. The planet’s speed at its

nearest distance to the Sun (perihelion) is 30 km/s, and at its farthest distance

from the Sun (aphelion) is 15 km/s. If the distance between the planet and the

Sun at perihelion is 100 million km, determine the distance between the planet

and the Sun at aphelion.

Solution

Step 1: Recall the conservation of angular momentum in the planet’s elliptical

orbit:

mrv =constant

where mis the mass of the planet, ris the distance between the planet and the

Sun, and vis the speed of the planet.

Step 2: Using the conservation of angular momentum at perihelion:

mrpvp=mrava

where rpis the distance at perihelion, vpis the speed at perihelion, rais the

distance at aphelion, and vais the speed at aphelion.

Step 3: Plug in the given values:

100 ×106km ×30 km/s =ra×15 km/s

Step 4: Solve for ra:

ra=100 ×106×30

15 = 200 ×106km

Therefore, the distance between the planet and the Sun at aphelion is 200

million km.

Question 7

Question

Assume that a planet has an elliptical orbit with the Sun located at one of the

foci. The planet’s distance to the Sun when it is closest is r1, and when it is

farthest it is r2. If the time it takes for the planet to travel from closest to

farthest distance is T, find the relationship between r1,r2, and T.

Solution

Step 1: According to Kepler’s second law, the line connecting a planet to the

Sun sweeps out equal areas in equal times. This means that the area swept out

by the planet as it moves from closest to farthest distance is the same regardless

of where it is on the orbit.

Step 2: Let’s consider the area swept out by the planet from closest to

farthest distance as ∆A. The area of an ellipse is given by A=πab, where ais

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

Step 3: The area swept out can also be calculated as ∆A=1

2r1v1∆t1+

2r2v2∆t2, where v1and v2are the speeds of the planet at distances r1and r2

respectively.

Step 4: Since the planet’s motion is gravitational, we can use conservation

of energy to relate the speeds to the distances. The total mechanical energy of

the planet is given by E=1

2mv2−GMm

r, where vis the speed of the planet,

ris the distance to the Sun, Gis the gravitational constant, Mis the mass of

the Sun, and mis the mass of the planet.

Step 5: Considering the energy at the closest and farthest points, we have

that E1=1

2mv2

1−GMm

r1and E2=1

2mv2

2−GMm

r2.

Step 6: Since mechanical energy is conserved, E1=E2. This allows us to

relate v1and v2to r1and r2.

Step 7: Substituting the expressions for v1and v2in terms of r1and r2into

our equation for ∆A, we can solve for the relationship between r1,r2, and T.

This will give us the desired relationship between these quantities.

Question 8

Question

A comet has an orbit around the Sun with a semi-major axis of 4.0×1012 meters

and an eccentricity of 0.8. Calculate the period of the comet’s orbit.

Solution

Step 1: Find the semi-minor axis.

The relationship between the semi-major axis a, the semi-minor axis b, and the

eccentricity efor an ellipse is given by:

b=a√1−e2

Substitute a= 4.0×1012 meters and e= 0.8into the equation:

b= 4.0×1012 ×√1−0.82

b= 4.0×1012 ×√1−0.64

b= 4.0×1012 ×0.6

b= 2.4×1012 meters

Step 2: Calculate the semi-latus rectum.

The semi-latus rectum lof an ellipse is given by:

l=a(1 −e2)

Substitute a= 4.0×1012 meters and e= 0.8into the equation:

l= 4.0×1012 ×(1 −0.82)

l= 4.0×1012 ×(1 −0.64)

l= 4.0×1012 ×0.36

l= 1.44 ×1012 meters

Step 3: Use Kepler’s third law to find the period.

Kepler’s third law states that the square of the period of an object in orbit

equals the cube of the semi-major axis:

T2=4π2

G(M1+M2)a3

where Tis the period of the comet’s orbit, Gis the gravitational constant, M1is

the mass of the Sun, M2is the mass of the comet, and ais the semi-major axis.

Since M2is very small compared to M1, we can approximate M1+M2≈M1.

Substituting a= 4.0×1012 meters into the equation, we get:

T2=4π2

G(M1)(4.0×1012)3

Solving for T, we find:

T=√4π2

G(M1)(4.0×1012)3

Question 9

Question

In a distant solar system, a planet orbits a star according to Kepler’s laws.

The following data has been collected: - The planet’s distance from the star at

closest approach (periapsis) is rp= 3.2×109m. - The planet’s distance from

the star at furthest approach (apoapsis) is ra= 7.6×109m. - The time it takes

for the planet to complete one orbit is T= 5.2years.

Calculate the following parameters of the planet’s motion around the star:

a) The semi-major axis of the orbit (in meters). b) The eccentricity of the orbit.

c) The period of the planet in seconds.

Solution

a) To find the semi-major axis a, we can use the relationship between the semi-

major axis aand the periapsis rpand apoapsis ra:

a=rp+ra

a=3.2×109+ 7.6×109

2=10.8×109

2= 5.4×109m

b) The eccentricity eof the orbit can be calculated using the formula:

e=ra−rp

ra+rp

e=7.6×109−3.2×109

7.6×109+ 3.2×109=4.4×109

10.8×109= 0.407

c) To find the period of the planet in seconds Tsec, we first convert the given

period Tfrom years to seconds:

Tsec =T×(365.25 ×24 ×60 ×60)

Tsec = 5.2×(365.25 ×24 ×60 ×60) = 164223600 seconds

Question 10

Question

The table below shows the semi-major axis (a) and period (T) of revolution for

several planets in a hypothetical solar system:

Planet Semi-major axis (a) in AU

Period (T) in Earth years

Mercury 0.387

0.241

Venus 0.723

0.615

Earth 1

Mars 1.524

1.88

Using the data in the table, verify Kepler’s third law of planetary motion,

which states that the square of the period of revolution of a planet is directly

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

Solution

To verify Kepler’s third law of planetary motion, we need to check if the square

of the period is proportional to the cube of the semi-major axis. We can do this

by analyzing the data provided in the table.

Step 1: Calculate T2/a3for each planet:

• For Mercury: Calculate (0.241)2/(0.387)3≈0.40716

• For Venus: Calculate (0.615)2/(0.723)3≈0.39614

• For Earth: Calculate (1)2/(1)3= 1