PHYS 305 - INTRODUCTION TO MODERN PHYSICS Kepler's laws Question Bank Set 8 | PHYS 305

PHYS 305 - INTRODUCTION TO

MODERN PHYSICS - Kepler’s laws

Question Bank - Set 8

Liberty University

Question 1

Question

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

planet and the star is 40 million kilometers (perihelion) and the farthest distance

is 100 million kilometers (aphelion). If the period of the planet’s orbit is 3 years,

calculate the mass of the star in kilograms. Assume the system consists of only

the star and planet, and consider the gravitational force between the two bodies.

Solution

Step 1: Find the average distance of the planet from the star. The average

distance rof the planet from the star is given by the semi-major axis of the

orbit, which is the average of the perihelion and aphelion distances:

r=rperihelion +raphelion

2=40 ×106+ 100 ×106

2= 70 ×106km

Step 2: Calculate the average orbital speed. Using Kepler’s third law, we

can relate the period of the orbit T, the semi-major axis a, and the gravitational

constant G:

T2=4π2a3

where Mis the mass of the star.

The average orbital speed of the planet is given by:

v=2πa

Step 3: Calculate the gravitational force. The gravitational force between

the star and planet is given by Newton’s law of gravitation:

F=GMm

where mis the mass of the planet.

Step 4: Equate the gravitational force to the centripetal force. The gravita-

tional force provides the centripetal force needed to keep the planet in orbit:

GMm

r2=mv2

Step 5: Solve for the mass of the star. Substitute the expressions for vand

rinto the equation above, then solve for M:

GM2

70 ×106=(2π·70 ×106/3)2

70 ×106

M=(2π·70 ×106/3)2

G·70 ×106

After plugging in the known values and solving for M, we can find the mass of

the star in kilograms.

Question 2

Question

A comet has an elliptical orbit around the Sun with an eccentricity of 0.7. The

closest point of the comet’s orbit to the Sun (perihelion) is at a distance of 0.3

AU (Astronomical Units). Determine the maximum distance of the comet from

the Sun (aphelion) in AU.

(Hint: Use Kepler’s laws of planetary motion to solve this problem.)

Solution

Step 1: Recall Kepler’s laws of planetary motion:

1. The orbit of a planet/comet is an ellipse with the Sun at one of the two

foci.

2. A line segment joining a planet and the Sun sweeps out equal areas in

equal intervals of time.

3. The square of the orbital period of a planet is directly proportional to the

cube of the semi-major axis of its orbit.

Step 2: The eccentricity of the orbit, e, is defined as the ratio of the distance

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

formula for the distance from the center to a focus is c=ae, where ais the

semi-major axis.

Step 3: Given that the perihelion distance is 0.3 AU and the eccentricity

of the orbit is 0.7, we can find the semi-major axis (a) using the equation

a=1

1−e2·rperihelion, where rperihelion is the perihelion distance.

Step 4: Substitute the given values into the equation:

a=1

1−0.72·0.3 = 1

0.51 ·0.3≈0.5882 AU

Step 5: To find the aphelion distance, we need to consider the fact that the

distance from the center to either the aphelion or perihelion is given by aand

the distance from the center to a focus is ae. Therefore, the distance from the

center to the aphelion (a+c) is a+ae.

Step 6: Plug in the values to find the aphelion distance:

a+c=a+ae = 0.5882 + 0.5882 ·0.7 = 0.5882 + 0.4117 ≈0.9999 AU

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

approximately 0.9999 AU.

Question 3

Question

A planet orbits a star in a highly elliptical orbit, such that at its closest approach

to the star (perihelion) it is 4×1010 meters away, and at its furthest point from

the star (aphelion) it is 8×1010 meters away. If the period of the planet’s orbit

is 3 years, determine the eccentricity of the orbit.

Solution

Step 1: Recall Kepler’s third law of planetary motion:

T2=4π2

G(M+m)a3

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

mass of the star, mis the mass of the planet, and ais the semi-major axis of

the orbit.

Step 2: Calculate the semi-major axis aof the orbit using the given values

for perihelion and aphelion:

a=rmin +rmax

a=4×1010 + 8 ×1010

a= 6 ×1010 m

Step 3: Substitute the given period T= 3 years and semi-major axis a=

6×1010 m into Kepler’s third law to solve for M+m:

(3 yrs)2=4π2

G(M+m)(6 ×1010)3

9 = 4π2

G(1)(6 ×1010 )3

9 = 4π2

G(6 ×1010)3

9 = 4π2

6.67 ×10−11 (6 ×1010)3

9 = (36 ×1020)×4π2

6.67 ×10−11

Step 4: Calculate the value of M+mfrom the above equation.

Step 5: Recall the equation for eccentricity ein terms of semi-major axis a,

and semi-minor axis b:

e=√1−(b

a)2

Step 6: Calculate the semi-minor axis bin terms of the provided values for

perihelion and aphelion:

b=2rminrmax

rmin +rmax

b=2×4×1010 ×8×1010

4×1010 + 8 ×1010

b=64 ×1020

12 ×1010

b= 5.33 ×1010 m

Step 7: Substitute the semi-major axis a= 6 ×1010 m and semi-minor axis

b= 5.33 ×1010 m into the formula for eccentricity to determine the eccentricity

Question 4

Question

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

the foci. The planet’s closest distance to the sun (perihelion) is 0.3 AU, while its

farthest distance from the sun (aphelion) is 0.7 AU. Calculate the eccentricity

of the planet’s orbit.

Solution

Step 1: The eccentricity, e, of an elliptical orbit can be calculated using the

formula:

e=rmax −rmin

rmax +rmin

where rmax is the aphelion distance and rmin is the perihelion distance.

Step 2: Substituting the given values, we have:

e=0.7−0.3

0.7+0.3=0.4

1= 0.4

Therefore, the eccentricity of the planet’s orbit is 0.4.

Question 5

Question

A planet has an orbital period of 10 years around a star. If the planet’s average

distance from the star is 7×1011 m, determine the mass of the star in kilograms.

Assume the planet’s orbit is circular.

Solution

Step 1: Determine the star’s mass using Kepler’s third law.

T2=4π2r3

G(M1+M2)

M1=4π2r3

GT 2−M2

Given that the orbital period T= 10 years, the average distance from the star

r= 7 ×1011 m, and M2is negligible compared to M1.

Step 2: Calculate the star’s mass.

M1=4π2(7 ×1011)3

G(10 years)2

=4π2×73×1033

G×100

≈61544 ×1033

6.674 ×10−11

≈9.2234 ×1044 kg

Therefore, the mass of the star is approximately 9.2234 ×1044 kg.

Question 6

Question

A planet follows an elliptical orbit around the sun. The planet is at its closest

point to the sun (perihelion) at a distance of 0.3 AU and at its farthest point

from the sun (aphelion) at a distance of 0.7 AU. If the planet takes 200 days to

complete one orbit, determine the average orbital speed of the planet.

Solution

Let’s denote the distance of the planet from the sun at any given time as r, and

the speed of the planet at the same time as v. According to Kepler’s second

law, a planet sweeps out equal areas in equal times, so we can use conservation

of angular momentum to find the average orbital speed of the planet.

Step 1: Calculate the angular momentum of the planet. The angular mo-

mentum of the planet is given by L=mvr, where mis the mass of the planet.

Since we are interested in the average speed, we can assume the force is radial

and that angular momentum is conserved. Therefore, at perihelion and aphelion

the angular momentum is the same.

At perihelion (r= 0.3AU):

L1=m·v1·r1

At aphelion (r= 0.7AU):

L2=m·v2·r2

Since L1=L2, we have:

m·v1·r1=m·v2·r2

v1=v2·r2

Step 2: Calculate the time taken to travel from perihelion to aphelion.

Since the total time taken for one orbit is 200 days, the time taken to travel

from perihelion to aphelion is half of that, which is 100 days.

Step 3: Calculate the average orbital speed of the planet. The average

orbital speed is given by:

vavg =Total distance traveled

Total time taken

The total distance traveled is 0.4AU (difference between aphelion and per-

ihelion distances), and the total time taken is 100 days. Converting AU to km

(1 AU = 1.496 ×108km) and days to seconds, we can find the average orbital

speed.

Now,

v1=v2·r2

vavg =0.4×1.496 ×108

100 ×24 ×3600

vavg ≈0.847 km/s

Therefore, the average orbital speed of the planet is approximately 0.847

km/s.

Question 7

Question

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

the foci of the ellipse. The planet’s closest approach to the star is 0.3 AU and

its farthest distance is 0.7 AU. Determine the eccentricity of the planet’s orbit.

Solution

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

of the distance between the foci and the length of the major axis. In this case,

the distance between the foci is 0.7−0.3 = 0.4AU.

Step 2: The length of the major axis of the ellipse is equal to the sum of the

closest approach distance and the farthest distance, which is 0.3+0.7 = 1 AU.

Step 3: The eccentricity, denoted by e, is given by the formula:

e=distance between foci

length of major axis

Step 4: Substituting the values we found earlier:

e=0.4

Step 5: Simplifying the expression:

e= 0.4

Step 6: Therefore, the eccentricity of the planet’s orbit is 0.4.

Question 8

Question

Consider a planet with a semi-major axis of 1 AU (astronomical unit) orbiting

a star with a mass of 2×1030 kg. Determine the orbital period of this planet

in years. Assume the star is stationary.

Solution

To determine the orbital period of the planet, we can use Kepler’s third law

of planetary motion. Kepler’s third law states that the square of the orbital

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

its orbit.

Step 1: Calculate the gravitational constant G.

G= 6.67 ×10−11 m3kg−1s−2

Step 2: Calculate the orbital period Tusing Kepler’s third law.

T= 2π√a3

where: - ais the semi-major axis of the orbit, - Gis the gravitational constant,

-Mis the mass of the star.

Plugging in the values:

T= 2π√(1 AU)3

6.67 ×10−11 m3kg−1s−2×2×1030 kg

Step 3: Convert the orbital period Tfrom seconds to years.

1year = 3.15 ×107seconds

Now, calculate Tin years.

Question 9

Question

In a distant solar system, a planet has an orbital period of 150 days and an

average distance from its star of 0.3 AU. Calculate the mass of the star in terms

of the mass of the sun (M⊙), given that the mass of the planet is 0.2M⊕(M⊕

is the mass of the Earth).

Solution

Step 1: Calculate the orbital speed of the planet. The orbital speed of the

planet can be calculated using Kepler’s Third Law of planetary motion:

r3=4π2

G(M1+M2)

where: T= orbital period of the planet (in seconds), r= average distance of

the planet from its star, G= gravitational constant, M1= mass of the star, M2

= mass of the planet.

Given that T= 150 days and r= 0.3AU, we convert Tto seconds:

T= 150 ×24 ×3600 = 12,960,000 seconds

Plugging in the values and solving for the orbital speed v:

v=2πr

v=2π×0.3AU

12,960,000 s= 7.3×10−5AU/s

Step 2: Calculate the mass of the star. The gravitational force between the

star and the planet provides the centripetal force causing the planet to orbit.

The gravitational force can be calculated using Newton’s law of gravitation:

F=GM1M2

r2=M2

Substitute the known values:

GM1M2

r2=M2

GM1=v2r3

M1=v2r3

M1=(7.3×10−5AU/s)2×(0.3AU)3

Given G= 6.674 ×10−11 N m2/kg2:

M1=(7.3×10−5)2×(0.3)3

6.674 ×10−11 ≈0.55M⊙

Therefore, the mass of the star is approximately 0.55 times the mass of the

sun.

Question 10

Question

A distant planet has an orbital period of 3.5 Earth years. If the average distance

between the planet and the Sun is 4 AU (astronomical units), determine the

semi-major axis of the planet’s elliptical orbit.

Solution

Step 1: Recall Kepler’s third law, which relates the orbital period (T) of a planet

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

T2=k·a3

Step 2: We are given that the orbital period (T) of the planet is 3.5 Earth

years and the average distance (r) between the planet and the Sun is 4 AU. We

need to determine the semi-major axis (a).

Step 3: Since we are given the average distance, we first need to find the

semi-major axis using the relationship between semi-major axis and average

distance in an elliptical orbit:

a=r

1−e2

where eis the eccentricity of the orbit. Since we are not given the eccentricity,

we can assume it to be zero for a circular orbit. So, e= 0.

Step 4: Plugging in the given value of the average distance r= 4 AU and

e= 0 into the formula, we find:

a=4

1−0= 4 AU

Step 5: Now, we can use Kepler’s third law to find the period Tfor the

semi-major axis awe just found:

T2=k·a3

Step 6: We can solve for the constant kusing the known period of Earth’s

orbit around the Sun: TEarth = 1 year and aEarth = 1 AU.

k=T2

Earth

= 12/13= 1

Step 7: Substituting the values of T= 3.5years and a= 4 AU into Kepler’s

third law equation, we find:

(3.5)2= 43

Step 8: Simplifying the equation gives us the semi-major axis a:

3.52= 16

12.25 = 16

12.25 = 43

Step 9: Therefore, the semi-major axis of the planet’s elliptical orbit is 4

AU.

Question 11

Question

Assuming a circular orbit for the Earth around the Sun, determine the ratio of

the speed of the Earth at perihelion (closest point to the Sun) to the speed at

aphelion (farthest point from the Sun). The distance between the Sun and the

Earth at perihelion is 147.1 million km, and at aphelion is 152.1 million km.

Solution

Step 1: Let’s denote the distance of Earth from the Sun at perihelion as r1=

147.1million km and at aphelion as r2= 152.1million km.

Step 2: We first compute the speed of the Earth at perihelion using Kepler’s

third law: T2

1/r3

1=T2

2/r3

2where T1and T2are the time periods at perihelion

and aphelion respectively. Since the orbit is circular, speed is constant.

Step 3: The time period of the Earth’s orbit can be calculated using T=2πr

where vis the speed of the Earth.

Step 4: Let v1be the speed of the Earth at perihelion. We have T1=2πr1

v1.

Step 5: Now we can rearrange the Kepler’s third law equation to solve for

v2, the speed of the Earth at aphelion.

Step 6: Finally, we calculate the ratio v1

v2to find the desired ratio of speeds

at perihelion and aphelion.

Question 12

Question

A planet follows an elliptical orbit around the Sun. The semi-major axis of its

orbit is 2.5 AU and the semi-minor axis is 1.5 AU. Determine the eccentricity

of the planet’s orbit.

Solution

To find the eccentricity of the planet’s orbit, we first need to recall the definition

of eccentricity for an ellipse. The eccentricity (e) of an ellipse is a measure of

how much the ellipse deviates from being a perfect circle. It is defined as the

ratio of the distance between the foci of the ellipse (2c) to the length of the

major axis (2a). The relationship is given by

e=c

where ais the length of the semi-major axis and cis the distance from the

center of the ellipse to one of the foci.

Step 1: Find the value of aand cusing the given semi-major and

semi-minor axes. Given that the semi-major axis (a) is 2.5 AU and the

semi-minor axis (b) is 1.5 AU, we can find the value of cusing the relationship

a2=b2+c2for an ellipse. Solving for c:

c=√a2−b2

c=√2.52−1.52

c=√6.25 −2.25

c=√4

c= 2

Step 2: Calculate the eccentricity of the planet’s orbit. Now that

we have found the values of aand c, we can calculate the eccentricity (e) using

the formula:

e=c

a=2

2.5= 0.8

Therefore, the eccentricity of the planet’s orbit is 0.8.

Question 13

Question

An asteroid is in an elliptical orbit around the Sun. The semimajor axis of the

orbit is 3.0 AU and the eccentricity is 0.4. Determine the closest distance and

the farthest distance of the asteroid from the Sun.

Solution

Step 1: First, we need to find the distance from the focus of the ellipse to the

center of the ellipse, which is the Sun. We know that c=ae, where ais the

semimajor axis and eis the eccentricity.

Given: a= 3.0AU e= 0.4

Calculating c:c= 3.0×0.4 = 1.2AU

Step 2: The closest distance of the asteroid from the Sun is when the asteroid

is at the perihelion. The perihelion distance (rmin) is given by: rmin =a−c

Substitute a= 3.0AU and c= 1.2AU into the formula: rmin = 3.0−1.2 =

1.8AU

Therefore, the closest distance of the asteroid from the Sun is 1.8 AU.

Step 3: The farthest distance of the asteroid from the Sun is when the

asteroid is at the aphelion. The aphelion distance (rmax) is given by: rmax =a+c

Substitute a= 3.0AU and c= 1.2AU into the formula: rmax = 3.0 + 1.2 =

4.2AU

Therefore, the farthest distance of the asteroid from the Sun is 4.2 AU.

Question 14

Question

Consider a planet in a circular orbit around a star with a radius of 3.2×1011

m. If the planet’s orbital period is 500 days, determine the mass of the star.

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

Solution

Step 1: Recall Kepler’s Third Law, which relates the orbital period Tof a planet

to the radius rof its orbit:

T2=(4π2

GM )r3

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

radius of the planet’s orbit.

Step 2: Rearranging the formula gives:

M=4π2

GT 2r3

Step 3: Now plug in the given values. The radius of the orbit is r= 3.2×1011

m and the period T= 500 days. First, we need to convert the period to seconds

since the standard SI unit of time is seconds. 1day is 86400 seconds.

T= 500 days ×86400 s/day = 43200000 s

Step 4: Now substitute the values into the formula:

M=4π2

G(43200000)2(3.2×1011)3

Step 5: Next, plug in the values for G= 6.67×10−11 N m2/kg2and calculate

the mass of the star.

Step 6: After performing the calculations, you should find the mass of the

star to be approximately 1.99 ×1030 kg.

Question 15

Question

A comet has a highly elliptical orbit around the Sun with an eccentricity of 0.9.

The comet’s closest distance to the Sun (perihelion) is 0.2 AU. Determine the

comet’s farthest distance from the Sun (aphelion) in AU.

Solution

Step 1: Recall Kepler’s laws of planetary motion. Kepler’s first law states that

each planet’s orbit about the Sun is an ellipse with the Sun at one focus. The

distance from the center of the ellipse to a focus is a constant value called the

semi-major axis, denoted by a. The eccentricity, e, of the orbit describes how

elongated the ellipse is, with e= 0 for a circular orbit and 0<e<1for an

elliptical orbit.

Step 2: The distance between the Sun and a point on the ellipse can be

calculated using the following formula:

r=a(1 −e2)

1 + ecos(θ)

where: r= the distance between the Sun and the point on the ellipse; a= the

semi-major axis of the orbit; e= the eccentricity of the orbit; θ= the angle

between the point on the ellipse, the Sun, and the periapsis (closest point to

the Sun) direction.

Step 3: Given that the eccentricity e= 0.9and the perihelion distance

rperihelion = 0.2AU, we can solve for the semi-major axis a. At perihelion,

cos(θ) = −1:

rperihelion =a(1 −e2)

1 + ecos(θperihelion)

0.2 = a(1 −0.92)

1+0.9(−1)

Step 4: Solve for a:

0.2 = a(1 −0.81)

1−0.9

0.2 = 0.19a

0.1

a=0.2×0.1

0.19 =0.02

0.19 ≈0.1053 AU

Step 5: Now, to find the aphelion distance raphelion, where cos(θ) = 1:

raphelion =a(1 −e2)

1 + e

raphelion =0.1053(1 −0.92)

1+0.9

raphelion =0.1053 ×0.19

1.9=0.01999

1.9≈0.0105 AU

Therefore, the comet’s farthest distance from the Sun (aphelion) is approx-

imately 0.0105 AU.

Question 16

Question

In a distant solar system, a planet orbits a star in an elliptical path such that

when the planet is at its closest distance to the star (perihelion), it is 50 million

kilometers away. When the planet is at its farthest distance from the star

(aphelion), it is 100 million kilometers away. The orbital period of this planet

is 1 year. Calculate the eccentricity of the planet’s orbit.

Solution

Step 1: Recall the formula relating the semi-major axis, a, semi-minor axis, b,

and eccentricity, e, of an ellipse:

e=√1−(b

a)2

Step 2: The semi-major axis, a, of an orbit is the average of the perihelion

and aphelion distances:

a=rperihelion +raphelion

2=50 million km + 100 million km

2= 75 million km

Step 3: The semi-minor axis, b, of an orbit is calculated using the relationship

between a,b, and eccentricity:

b=a√1−e2

Step 4: Since we already know a= 75 million km, we need to find e. But

before that let’s find b:

b= 75√1−e2

Step 5: The orbital period, T, satisfies Kepler’s third law:

T2=4π2a3

G(M+m)

where Mand mare the masses of the star and the planet, respectively.

Step 6: We can simplify the above equation using the given values and the

formula for the semi-major axis:

1year =√4π2(75 million km)3

G(M+m)

Step 7: Since T= 1 year, a= 75 million km, and rperihelion = 50 million km:

rperihelion =a(1 −e)

Step 8: Substitute the given values into the equation and solve for e:

50 million km = 75 million km(1 −e)

Step 9: After solving for e, substitute it into the formula for eccentricity to

get the final answer.

Question 17

Question

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

foci. The farthest distance of the planet from the star is 4.5 AU and the closest

distance is 3.0 AU. Calculate the eccentricity of the planet’s orbit.

Solution

Step 1: Recall that in an ellipse, the sum of the distances from any point on the

ellipse to two fixed points (called foci) is constant, and this constant is equal to

the major axis of the ellipse.

Step 2: The farthest distance of the planet from the star is known as the

semi-major axis a, which is 4.5 AU.

Step 3: The closest distance of the planet from the star is known as the

semi-minor axis b, which is 3.0 AU.

Step 4: The eccentricity eof an ellipse is given by the formula

e=√1−(b

a)2

Step 5: Substitute a= 4.5AU and b= 3.0AU into the formula for eccen-

tricity:

e=√1−(3.0

4.5)2

Step 6: Calculate the eccentricity:

e=√1−(2

3)2

=√1−4

9=√5

Step 7: Therefore, the eccentricity of the planet’s orbit is √5

Question 18

Question

A planet orbits a star with a period of 9.00 years and an average distance from

the star of 1.50 AU. Determine the mass of the star. (Hint: Use Kepler’s third

law)

Solution

Step 1: First, we need to understand Kepler’s third law, which states:

r3=4π2

G(M1+M2)

where: T= orbital period of the planet, r= average distance between the planet

and the star, G= gravitational constant (approximated as 6.67 ×10−11 m3/kg ·

s2), M1= mass of the star, M2= mass of the planet.

Step 2: Convert the period of the planet to seconds:

T= 9.00 years ×365.25 days/year ×24 hours/day ×3600 s/hour

T= 2.84 ×108s

Step 3: Convert the distance between the planet and the star to meters:

r= 1.50 ×AU ×1.496 ×1011 m/AU

r= 2.25 ×1011 m

Step 4: Substitute the values into Kepler’s third law equation:

(2.84 ×108)2

(2.25 ×1011)3=4π2

6.67 ×10−11M1

Step 5: Solve for the mass of the star, M1:

M1=4π2(2.25 ×1011)3

6.67 ×10−11(2.84 ×108)2

Step 6: Calculate the mass of the star using a calculator:

M1≈2.25 ×1030 kg

Therefore, the mass of the star is approximately 2.25 ×1030 kg.

Question 19

Question

The semi-major axis of an elliptical orbit is 1.5×1011 meters and the eccentricity

is 0.5. Calculate the distance of the farthest point of the orbit from the focus.

Solution

Step 1: Recall Kepler’s first law which states that each planet follows an elliptical

path with the sun at one focus. The distance from the sun to the farthest point

of the orbit (apoapsis) is given by:

rapoapsis =a(1 + e)

where ais the semi-major axis, and eis the eccentricity of the elliptical orbit.

Step 2: Substitute a= 1.5×1011 meters and e= 0.5into the formula:

rapoapsis = (1.5×1011 )(1 + 0.5)

Step 3: Calculate the distance to the farthest point of the orbit:

rapoapsis = (1.5×1011 )(1.5) = 2.25 ×1011 meters

Therefore, the distance to the farthest point of the orbit from the focus is

2.25 ×1011 meters.

Question 20

Question

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

foci. The minor axis of the elliptical orbit is 0.2 AU in length. The maximum

distance of the planet from the star is 1.5 AU. Find the eccentricity of the orbit.

Solution

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

tricity (e) of an ellipse is a measure of how elongated the ellipse is. It is defined

as the ratio of the distance between the foci of the ellipse and the length of the

major axis. Mathematically, e=c/a, where cis the distance from the center of

the ellipse to a focus and ais the length of the semi-major axis.

Step 2: Determine the length of the semi-major axis: Since the maximum

distance of the planet from the star is 1.5 AU, this distance is equal to the

sum of the semi-major axis (a) and the distance from the center to a focus (c).

Therefore, a+c= 1.5AU.

Step 3: Determine the distance from the center to a focus: The distance

from the center to a focus (c) can be found by using the relationship between

the semi-major axis, the semi-minor axis, and the distance from the center to a

focus in an ellipse: a2=b2+c2, where bis the length of the semi-minor axis.

Since the minor axis is 0.2 AU, b= 0.1AU.

Step 4: Calculate the semi-major axis: Substitute b= 0.1AU into a2=

b2+c2and solve for a. Then, solve for cusing the relationship a+c= 1.5AU.

Step 5: Determine the eccentricity of the orbit: Once you have found the

values of aand c, calculate the eccentricity eusing the formula e=c/a. This

value represents how elongated the orbit is.

Question 21

Question

A planet has an orbital period of 5.2 years and a semi-major axis of 2.8 AU.

Determine the mass of the star it orbits around using Kepler’s third law. (Hint:

Remember Kepler’s third law relates the orbital period of a planet squared to

the semi-major axis of its orbit cubed.)

Solution

Step 1: Recall Kepler’s Third Law which states:

T2=(4π2

G(M+m))a3

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

mass of the star, - mis the mass of the planet, - ais the semi-major axis of the

orbit.

Step 2: We are solving for the mass of the star (M), so we can rewrite the

formula as:

T2=4π2

GM a3

Step 3: Plug in the given values:

5.22=4π2

G×M(2.8)3

Step 4: Simplify the equation:

27.04 = 4π2×(2.8)3

G×M

Step 5: Rearrange the equation to solve for the mass of the star, M:

M=4π2×(2.8)3

27.04 ×G

Step 6: Calculate the mass of the star:

M=4π2×(2.8)3

27.04 ×6.67430 ×10−11

Step 7: Perform the calculations to find the mass of the star M.

Step 8: The mass of the star is approximately 1.98 ×1030 kg.

Question 22

Question

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

farthest distance from the star is 0.3 AU and the closest distance is 0.1 AU. If

the planet travels from its closest distance to its farthest distance in 50 days,

determine the period of the planet’s orbit around the star in years.

Solution

Step 1: Determine the semi-major axis of the planet’s elliptical orbit. Given

that the farthest distance from the star is 0.3 AU and the closest distance is

0.1 AU, the semi-major axis of the elliptical orbit, a, is the average of these two

distances.

a=0.3AU + 0.1AU

2= 0.2AU

Step 2: Use Kepler’s third law to find the period of the planet’s orbit.

Kepler’s third law states that the square of the orbit period, T, of a planet

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

T2=k·a3

where kis a constant of proportionality. We can solve for kusing the information

that the planet takes 50 days to travel from its closest distance to its farthest

distance.

502=k·0.23

2500 = k·0.008

k=2500

0.008 = 312500

Step 3: Find the period of the planet’s orbit in years. Now that we have the

value of k, we can find the period of the planet’s orbit in years by substituting

a= 0.2AU back into Kepler’s third law.

T2= 312500 ·0.23

T2= 312500 ·0.008

T2= 2500

T=√2500

T= 50 years

Therefore, the period of the planet’s orbit around the star is 50 years.

Question 23

Question

A planet is in an elliptical orbit around the Sun such that the closest approach

is 0.3 AU and the farthest distance is 0.7 AU. Calculate the eccentricity of the

orbit.

Solution

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

e=ra−rp

ra+rp

where rais the distance of the farthest point and rpis the distance of the closest

point.

Step 2: Substitute the given values into the equation:

e=0.7−0.3

0.7+0.3.

Step 3: Calculate the eccentricity:

e=0.4

1= 0.4.

Therefore, the eccentricity of the orbit is 0.4.

Question 24

Question

A comet has an elliptical orbit around the Sun with an eccentricity of 0.7. If

the distance between the farthest point of its orbit from the Sun and the closest

point of its orbit is 3 AU, determine the semi-major axis of the comet’s orbit.

Solution

Step 1: Recall the definition of eccentricity for an ellipse: The eccentricity eof

an ellipse is defined as the ratio of the distance between the foci of the ellipse

(2c) to the length of the major axis of the ellipse (2a). Mathematically, e=c

Step 2: Use the given eccentricity to find the distance between the foci 2cin

terms of the semi-major axis a: Since the distance between the farthest point

and the closest point of the orbit is 2c= 3 AU, and the eccentricity e= 0.7, we

can write:

2c= 3 AU

e=c

a= 0.7

Step 3: Solve for the distance between the foci 2cin terms of the semi-major

axis a: From the definition of eccentricity, we have:

c=ea = 0.7a

2c= 2(0.7a) = 1.4a= 3 AU

Step 4: Find the semi-major axis ausing the value obtained for 2c: Now,

we can solve for the semi-major axis a:

1.4a= 3 AU

a=3AU

1.4

Step 5: Calculate the value of the semi-major axis a:

a=3

1.4= 2.14 AU

Therefore, the semi-major axis of the comet’s orbit is 2.14 AU.

Question 25

Question

A planet has an elliptical orbit with eccentricity e= 0.6. The closest distance

between the planet and the sun is R1= 0.5AU. Find the farthest distance

between the planet and the sun, R2, in AU.

Solution

Step 1: Recall the formula relating the closest and farthest distances from the

focus in an elliptical orbit with eccentricity e:

R2=R1

1−e

Step 2: Substitute R1= 0.5AU and e= 0.6into the formula:

R2=0.5

1−0.6

Step 3: Calculate the farthest distance R2:

R2=0.5

0.4

R2= 1.25 AU

Step 4: Therefore, the farthest distance between the planet and the sun is

R2= 1.25 AU.

Step 3: Calculate the gravitational force. The gravitational force between

the star and planet is given by Newton’s law of gravitation:

F=GMm

where mis the mass of the planet.

Step 4: Equate the gravitational force to the centripetal force. The gravita-

tional force provides the centripetal force needed to keep the planet in orbit:

GMm

r2=mv2

Step 5: Solve for the mass of the star. Substitute the expressions for vand

rinto the equation above, then solve for M:

GM2

70 ×106=(2π·70 ×106/3)2

70 ×106

M=(2π·70 ×106/3)2

G·70 ×106

After plugging in the known values and solving for M, we can find the mass of

the star in kilograms.

Question 2

Question

A comet has an elliptical orbit around the Sun with an eccentricity of 0.7. The

closest point of the comet’s orbit to the Sun (perihelion) is at a distance of 0.3

AU (Astronomical Units). Determine the maximum distance of the comet from

the Sun (aphelion) in AU.

(Hint: Use Kepler’s laws of planetary motion to solve this problem.)

Solution

Step 1: Recall Kepler’s laws of planetary motion:

1. The orbit of a planet/comet is an ellipse with the Sun at one of the two

foci.

2. A line segment joining a planet and the Sun sweeps out equal areas in

equal intervals of time.

3. The square of the orbital period of a planet is directly proportional to the

cube of the semi-major axis of its orbit.

Step 2: The eccentricity of the orbit, e, is defined as the ratio of the distance

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

formula for the distance from the center to a focus is c=ae, where ais the

semi-major axis.

Step 3: Given that the perihelion distance is 0.3 AU and the eccentricity

of the orbit is 0.7, we can find the semi-major axis (a) using the equation

a=1

1−e2·rperihelion, where rperihelion is the perihelion distance.

Step 4: Substitute the given values into the equation:

a=1

1−0.72·0.3 = 1

0.51 ·0.3≈0.5882 AU

Step 5: To find the aphelion distance, we need to consider the fact that the

distance from the center to either the aphelion or perihelion is given by aand

the distance from the center to a focus is ae. Therefore, the distance from the

center to the aphelion (a+c) is a+ae.

Step 6: Plug in the values to find the aphelion distance:

a+c=a+ae = 0.5882 + 0.5882 ·0.7 = 0.5882 + 0.4117 ≈0.9999 AU

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

approximately 0.9999 AU.

Question 3

Question

A planet orbits a star in a highly elliptical orbit, such that at its closest approach

to the star (perihelion) it is 4×1010 meters away, and at its furthest point from

the star (aphelion) it is 8×1010 meters away. If the period of the planet’s orbit

is 3 years, determine the eccentricity of the orbit.

Solution

Step 1: Recall Kepler’s third law of planetary motion:

T2=4π2

G(M+m)a3

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

mass of the star, mis the mass of the planet, and ais the semi-major axis of

the orbit.

Step 2: Calculate the semi-major axis aof the orbit using the given values

for perihelion and aphelion:

a=rmin +rmax

a=4×1010 + 8 ×1010

a= 6 ×1010 m

Step 3: Substitute the given period T= 3 years and semi-major axis a=

6×1010 m into Kepler’s third law to solve for M+m:

(3 yrs)2=4π2

G(M+m)(6 ×1010)3

9 = 4π2

G(1)(6 ×1010 )3

9 = 4π2

G(6 ×1010)3

9 = 4π2

6.67 ×10−11 (6 ×1010)3

9 = (36 ×1020)×4π2

6.67 ×10−11

Step 4: Calculate the value of M+mfrom the above equation.

Step 5: Recall the equation for eccentricity ein terms of semi-major axis a,

and semi-minor axis b:

e=√1−(b

a)2

Step 6: Calculate the semi-minor axis bin terms of the provided values for

perihelion and aphelion:

b=2rminrmax

rmin +rmax

b=2×4×1010 ×8×1010

4×1010 + 8 ×1010

b=64 ×1020

12 ×1010

b= 5.33 ×1010 m

Step 7: Substitute the semi-major axis a= 6 ×1010 m and semi-minor axis

b= 5.33 ×1010 m into the formula for eccentricity to determine the eccentricity

Question 4

Question

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

the foci. The planet’s closest distance to the sun (perihelion) is 0.3 AU, while its

farthest distance from the sun (aphelion) is 0.7 AU. Calculate the eccentricity

of the planet’s orbit.

Solution

Step 1: The eccentricity, e, of an elliptical orbit can be calculated using the

formula:

e=rmax −rmin

rmax +rmin

where rmax is the aphelion distance and rmin is the perihelion distance.

Step 2: Substituting the given values, we have:

e=0.7−0.3

0.7+0.3=0.4

1= 0.4

Therefore, the eccentricity of the planet’s orbit is 0.4.

Question 5

Question

A planet has an orbital period of 10 years around a star. If the planet’s average

distance from the star is 7×1011 m, determine the mass of the star in kilograms.

Assume the planet’s orbit is circular.

Solution

Step 1: Determine the star’s mass using Kepler’s third law.

T2=4π2r3

G(M1+M2)

M1=4π2r3

GT 2−M2

Given that the orbital period T= 10 years, the average distance from the star

r= 7 ×1011 m, and M2is negligible compared to M1.

Step 2: Calculate the star’s mass.

M1=4π2(7 ×1011)3

G(10 years)2

=4π2×73×1033

G×100

≈61544 ×1033

6.674 ×10−11

≈9.2234 ×1044 kg

Therefore, the mass of the star is approximately 9.2234 ×1044 kg.