PHSC 121 - INTRODUCTION TO
ASTRONOMY - Kepler’s Laws of
Planetary Motion
Question Bank - Set 7
Liberty University
Question 1
Question
Kepler’s third law of planetary motion 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.
Given that the period of Jupiter’s orbit around the Sun is 11.86 years, determine
the semi-major axis of Jupiter’s orbit in astronomical units (AU), where 1 AU is
the average distance between the Earth and the Sun (approximately 1.496×1011
meters).
Solution
Step 1: Let’s denote the period of Jupiter’s orbit as T(in years) and the semi-
major axis of its orbit as a(in AU). According to Kepler’s third law, we have
the following proportion:
(T
1year )2
=(a
1AU )3
Step 2: We are given that the period of Jupiter’s orbit, T, is 11.86 years.
Plugging this value into the equation, we get:
(11.86
1)2
=(a
1)3
11.862=a3
Step 3: Calculate the left-hand side of the equation:
11.862= 140.8996
Step 4: Solve for a:
a3= 140.8996
a=3
√140.8996
a≈5.208 AU
Step 5: Therefore, the semi-major axis of Jupiter’s orbit is approximately
5.208 AU.
Question 2
Question
In the study of planetary motion, Kepler’s third law states that the square of
the period of revolution of a planet around the Sun is proportional to the cube
of the planet’s average distance from the Sun. If the period of revolution for
Jupiter is 11.86 years and its average distance from the Sun is 778 million km,
calculate the period of revolution for Saturn given that its average distance from
the Sun is 1.43 billion km.
Solution
Step 1: Find the ratio of the cube of Jupiter’s average distance to the cube
of Saturn’s average distance. Let r1= 778 million km be Jupiter’s average
distance from the Sun and r2= 1.43 billion km be Saturn’s average distance
from the Sun. The ratio of the cubes of their average distances is given by:
(r1
r2)3
=(778
1430)3
=(778
1430)3
Step 2: Use Kepler’s third law to find the period of revolution for Saturn.
Kepler’s third law states that the square of the period of revolution is propor-
tional to the cube of the average distance from the Sun. So, T2
1
T2
2
=(r1
r2)3
where
T1and T2are the periods of revolution for Jupiter and Saturn, respectively.
Since we know T1= 11.86 years, we can solve for T2:
T2
1
T2
2
=(r1
r2)3
11.862
T2
2
=(778
1430)3
T2
2=11.862
(778
1430 )3
T2=√11.862
(778
1430 )3
2
Step 3: Calculate the period of revolution for Saturn. Now, we can substitute
the values and calculate T2:
T2=√11.862
(778
1430 )3≈√140.6596
0.9716
T2≈√144.8864 ≈12.04 years
Therefore, the period of revolution for Saturn is approximately 12.04 years.
Question 3
Question
Given the elliptical orbit of a planet around the Sun, the aphelion distance
(furthest point from the Sun) is 2.5 AU and the perihelion distance (closest
point to the Sun) is 1.5 AU. Calculate the eccentricity of the planet’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 to the length
of the major axis. In this case, the foci are at the center of the ellipse (the Sun)
and either the aphelion or perihelion.
Step 2: The major axis of an ellipse is the longest line that passes through
the center and both foci. It is equal to the sum of the distances to the aphelion
and perihelion, which gives us the equation 2a= 2.5+1.5 = 4 AU.
Step 3: Next, we find the distance between the foci, which is 2c= 2.5−1.5 =
1AU.
Step 4: We can now calculate the eccentricity eusing the formula e=c
a,
where ais the semimajor axis equal to 1
2of the major axis. Substituting in the
known values, we get e=1
4= 0.25.
Step 5: Therefore, the eccentricity of the planet’s orbit is 0.25.
Question 4
Question
The period of a planet orbiting a star is related to the semi-major axis of its orbit
by Kepler’s third law, which states that the square of the period of the orbit is
proportional to the cube of the semi-major axis of the orbit. For a particular
planet in our solar system, its period is 11.86 years and its semi-major axis is 5.2
astronomical units (AU). Determine the period of another planet in the same
star system with a semi-major axis of 2.8 AU.
3
Solution
Let T1be the period of the first planet and a1be its semi-major axis. Similarly,
let T2be the period of the second planet with semi-major axis a2. According to
Kepler’s third law, we have:
T2
1
a3
1
=T2
2
a3
2
Given that T1= 11.86 years and a1= 5.2AU, we want to find T2when a2= 2.8
AU.
Step 1: Substitute the known values into the equation and solve for T2.
(11.86)2
(5.2)3=T2
2
(2.8)3
Step 2: Simplify the equation.
139.96
140.608 =T2
2
21.952
Step 3: Calculate the value of T2.
T2
2=139.96 ×21.952
140.608
T2
2= 21.89
T2=√21.89
T2≈4.68 years
Therefore, the period of the second planet with a semi-major axis of 2.8 AU
is approximately 4.68 years.
Question 5
Question
The period of a planet’s orbit around the sun is 15 years. If the average distance
between the planet and the sun is 3.5 astronomical units (AU), determine the
mass of the sun. (Assume the planet’s orbit is nearly circular.)
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion, which states:
T2
1
a3
1
=T2
2
a3
2
where Tis the period of the orbit (in years) and ais the average distance
between the planet and the sun (in AU).
4
Step 2: Let’s denote the period of the unknown planet as T2and its distance
from the sun as a2. We can rewrite the equation as:
152
3.53=T2
2
a3
2
Step 3: We can solve for a2by rearranging the equation:
a2= 3.5(15
T2)2/3
Step 4: Now, let’s use Newton’s version of Kepler’s Third Law to find the
mass of the sun. Newton’s form is:
GMsun
4π2=a3
2(T2
2π)2
Step 5: Substitute the expression for a2into the equation:
GMsun
4π2=(3.5(15
T2)2/3)3
(T2
2π)2
Step 6: Simplify the equation and solve for Msun to find the mass of the sun.
Question 6
Question
The asteroid belt is a region located between the orbits of Mars and Jupiter
where many small objects orbit the Sun. Consider an asteroid with a semi-
major axis of 2.5 AU (astronomical units) and an orbital period of 4.0 years.
Determine the mass of the Sun using Kepler’s third law.
Solution
Step 1: Identify the given values and the unknown. Let abe the semi-major
axis of the asteroid’s orbit, Tbe the orbital period, and M⊙be the mass of the
Sun. The given values are a= 2.5AU and T= 4.0years. The unknown is M⊙.
Step 2: Convert the given units. Since we are using astronomical units (AU)
for a, we need to convert the semi-major axis to meters to be consistent with SI
units. Given that 1AU = 1.496 ×1011 meters, we have:
a= 2.5AU ×1.496 ×1011 m/AU = 3.74 ×1011 m
Step 3: Apply Kepler’s third law. Kepler’s third law is given by the equation:
T2=(4π2a3
G(M⊙+m))
5
where mis the mass of the asteroid, Gis the gravitational constant, and all
other variables are defined as before.
Step 4: Solve for the mass of the Sun. Substitute the given values into
Kepler’s third law and solve for M⊙:
4.02=(4π2(3.74 ×1011)3
G(M⊙+m))
16 = 4π2(5.59 ×1034)
G(M⊙+m)
16G(M⊙+m) = 4π2(5.59 ×1034)
2G(M⊙+m) = π2(5.59 ×1034)
2GM⊙=π2(5.59 ×1034)−2Gm
Step 5: Calculate the mass of the Sun. Unfortunately, without information
about the mass of the asteroid m, we cannot determine the exact mass of the
Sun using this method. The mass of the Sun can only be found if mis known.
Question 8
Question
Consider a hypothetical planet with a semi-major axis of 2 AU. Determine the
period of this planet’s orbit using Kepler’s Third Law of Planetary Motion.
Solution
To find the period of the planet’s orbit using Kepler’s Third Law of Planetary
Motion, we can use the formula:
T2=(4π2
G(M+m))a3
where: - Tis the period of the planet’s orbit, - Gis the gravitational constant
(6.67430 ×10−11 m3kg−1s−2), - Mis the mass of the sun (1.989 ×1030 kg), -
mis the mass of the planet (assumed to be negligible compared to the sun), -
ais the semi-major axis of the planet’s orbit.
Step 1: Substitute the given values into the formula:
T2=(4π2
G(M+m))a3
T2=(4π2
6.67430 ×10−11 ×(1.989 ×1030))(23)
6
Step 2: Calculate the period of the planet’s orbit:
T2=(4π2
1.329 ×1020 )(8)
T2= (2.972 ×107)s2
Step 3: Find the period of the planet’s orbit by taking the square root:
T=√2.972 ×107s
T≈5455 s
Therefore, the period of the planet’s orbit is approximately 5455 seconds.
Question 9
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is proportional to the cube of its average distance from the
sun. Suppose a newly discovered planet has an orbital period of 4years and an
average distance from the Sun of 6AU (astronomical units). What would be
the orbital period of a different planet located at an average distance of 15 AU
from the Sun, according to Kepler’s third law?
Solution
Step 1: Begin by setting up the proportion based on Kepler’s third law: Let T
be the orbital period of the second planet with an average distance of 15 AU
from the Sun. According to Kepler’s third law,
T2
1
d3
1
=T2
2
d3
2
where T1= 4 years, d1= 6 AU, and d2= 15 AU.
Step 2: Plug in the given values and solve for T2:
42
63=T2
2
153
16
216 =T2
2
3375
Step 3: Simplify the equation:
2
27 =T2
2
3375
7
Step 4: Cross multiply to solve for T2:
2×3375 = 27 ×T2
2
6750 = 27 ×T2
2
T2
2=6750
27
T2
2= 250
T2=√250 = 5√10
Step 5: Therefore, the orbital period of the second planet, located at an
average distance of 15 AU from the Sun, would be approximately 5√10 years
according to Kepler’s third law.
Question 10
Question
An exoplanet orbits a star with a period of 400 days and an average distance of
2 AU. If the star has a mass of 2×1030 kg, determine the mass of the exoplanet
using Kepler’s Third Law of Planetary Motion.
Solution
Step 1: We first convert the average distance of the exoplanet from the star to
meters:
Average distance = 2 AU = 2 ×1.496 ×1011 m/AU = 2.992 ×1011 m
Step 2: We then convert the period of the exoplanet’s orbit to seconds:
Period = 400 days×24 hours/day×60 minutes/hour×60 seconds/minute = 3.456×107seconds
Step 3: Using Kepler’s Third Law of Planetary Motion, we have:
T2
r3=4π2
G(M1+M2)
Step 4: We can rearrange the formula to solve for the mass of the exoplanet
M2:
M2=4π2r3
GT 2−M1
Step 5: Substituting the known values into the formula, we get:
M2=4π2×(2.992 ×1011)3
6.67 ×10−11 ×(3.456 ×107)2−2×1030
Step 6: Calculating the mass of the exoplanet, we find:
M2≈1.171 ×1028 kg
Therefore, the mass of the exoplanet is approximately 1.171 ×1028 kg.
8
Question 11
Question
The planet Mercury has an average distance from the Sun of approximately
0.39 AU. Calculate the period of Mercury’s orbit around the Sun in Earth
years. (Hint: Kepler’s Third Law states that the square of the orbital period of
a planet is proportional to the cube of its average distance from the Sun.)
Solution
Step 1: Given that Mercury’s average distance from the Sun is 0.39 AU, we can
use Kepler’s Third Law to find the period of Mercury’s orbit in Earth years.
Let Tbe the period of Mercury’s orbit.
Step 2: Kepler’s Third Law states that T2
1
T2
2
=a3
1
a3
2
, where T1and T2are the
periods of two planets, and a1and a2are their average distances from the Sun.
Step 3: Substituting the values for Mercury (T=T1,a=a1) and Earth
(T2= 1 year, a2= 1 AU) into the equation, we get:
T2
12=0.393
13
Step 4: Solving for T:
T2= 0.393
T=√0.393≈0.241
Step 5: Therefore, the period of Mercury’s orbit around the Sun is approxi-
mately 0.241 years.
Question 12
Question
During their orbits, planets move faster when they are closer to the Sun and
slower when they are farther away. Explain how Kepler’s laws of planetary
motion account for this observation.
Solution
To explain why planets move faster when they are closer to the Sun and slower
when they are farther away, we will use Kepler’s laws of planetary motion.
Step 1: Kepler’s First Law states that planets move in elliptical orbits with
the Sun at one of the foci. This means that the distance between a planet and
the Sun changes as the planet orbits.
Step 2: When a planet is closer to the Sun, it is at the perihelion point
(closest point to the Sun) of its orbit. At this point, the gravitational force
9
between the planet and the Sun is stronger, causing the planet to move faster
in order to maintain its orbit.
Step 3: Conversely, when a planet is farther away from the Sun, it is at
the aphelion point (farthest point from the Sun) of its orbit. At this point, the
gravitational force between the planet and the Sun is weaker, resulting in the
planet moving slower to stay in orbit.
Step 4: This variation in orbital speed is also explained by Kepler’s Second
Law, which states that a planet sweeps out equal areas in equal times. When a
planet is closer to the Sun and moving faster, it covers a larger area in a given
amount of time. Conversely, when it is farther from the Sun and moving slower,
it covers a smaller area in the same amount of time.
Therefore, Kepler’s laws of planetary motion account for the observation
that planets move faster when they are closer to the Sun and slower when they
are farther away.
Question 13
Question
Kepler’s third law states that the square of the period of revolution of a planet
around the sun is proportional to the cube of its semi-major axis. If a planet has
a period of revolution of 10 years and a semi-major axis of 3 AU (astronomical
units), what is the period of revolution of another planet with a semi-major axis
of 6 AU?
Solution
Step 1: Write down Kepler’s third law in mathematical form:
(T1
T2)2
=(a1
a2)3
where T1and T2are the periods of revolution of the two planets, and a1and a2
are their semi-major axes.
Step 2: Substitute the given values into the equation:
(10
T2)2
=(3
6)3
Step 3: Simplify the equation by squaring and cubing the fractions:
(10
T2)2
=(1
2)3
(102
T2
2)=(1
8)
10
Step 4: Cross multiply to solve for T2:
102=T2
2
8
100 ×8 = T2
2
800 = T2
2
Step 5: Take the square root of both sides to find T2:
T2=√800
T2= 28.28 years
Therefore, the period of revolution of the planet with a semi-major axis of
6 AU is 28.28 years.
Question 14
Question
In a hypothetical solar system, a planet takes 40 years to complete one full orbit
around its star. If the planet’s average distance from the star is 2.5 astronomical
units (AU), determine the mass of the star in terms of solar masses (M⊙).
Solution
Step 1: Recall Kepler’s third law of planetary motion, which relates the orbital
period of a planet (T) to its average distance from the star (r) using the formula:
T2=(4π2
G(m1+m2))r3
where T= orbital period of the planet, G= gravitational constant, m1= mass
of the star, r= average distance between the star and the planet, and m2=
mass of the planet (assumed to be negligible).
Step 2: Given that the planet takes 40 years to complete one full orbit, we
have T= 40 years = 40 ×365 days = 14,600 days.
Step 3: Convert the average distance from astronomical units (AU) to meters
using the conversion factor: 1 AU = 1.496 ×1011 meters. Given the average
distance r= 2.5AU, we have r= 2.5×1.496×1011 meters = 3.74×1011 meters.
Step 4: Substitute the values of Tand rinto Kepler’s third law formula:
(14600)2=(4π2
G(m⊙))(3.74 ×1011)3
Step 5: Solve for m⊙, the mass of the star, in terms of solar masses (M⊙).
Step 6: After finding the mass of the star, express the result in terms of solar
masses by dividing the mass by the mass of the Sun (M⊙= 1.989 ×1030 kg).
11
Question 15
Question
State Kepler’s Third Law of Planetary Motion and provide an example calcula-
tion showing how the law can be applied to determine the period of a planet’s
orbit around the Sun.
Solution
Kepler’s Third Law of Planetary Motion states that the square of the orbital
period of a planet is proportional to the cube of the semi-major axis of its orbit.
Step 1: Let’s denote the orbital period of the planet as T(in years) and
the semi-major axis of its orbit as a(in astronomical units, AU). The equation
for Kepler’s Third Law is given as:
T2=k·a3
where kis a constant.
Step 2: We can use the data for Earth’s orbit around the Sun to find the
value of the constant k. The orbital period of Earth is approximately 1 year
and the semi-major axis of its orbit is about 1 AU. Substituting these values
into the equation:
(1)2=k·(1)3
1 = k
Step 3: Now, let’s apply Kepler’s Third Law to determine the period of
Jupiter’s orbit around the Sun. The semi-major axis of Jupiter’s orbit is ap-
proximately 5.2 AU. Substituting this value into the equation, we can solve for
the orbital period T:
T2= (1) ·(5.2)3
T2= 5.23
T2= 140.608
Step 4: Taking the square root of both sides to find T:
T=√140.608
T≈11.85 years
Therefore, the orbital period of Jupiter around the Sun is approximately
11.85 years according to Kepler’s Third Law of Planetary Motion.
Question 16
Question
A planet is in elliptical orbit around the Sun. The semi-major axis of the planet’s
orbit is 3.5 AU and its eccentricity is 0.2. Determine the semi-minor axis of the
orbit.
12
Solution
Step 1: Recall the formula to calculate the semi-minor axis of an elliptical orbit:
b=a√1−e2
where ais the semi-major axis and eis the eccentricity of the orbit.
Step 2: Substitute the given values into the formula:
b= 3.5AU ·√1−0.22
Step 3: Calculate the value inside the square root:
1−0.22= 1 −0.04 = 0.96
Step 4: Substitute the calculated value back into the formula:
b= 3.5AU ·√0.96
Step 5: Calculate the square root:
√0.96 = 0.960.5= 0.9798
Step 6: Multiply the semi-major axis by the square root:
b= 3.5AU ·0.9798 = 3.43 AU
Therefore, the semi-minor axis of the orbit is 3.43 AU.
Question 17
Question
Kepler’s second law states that a line segment joining a planet and the Sun
sweeps out equal areas during equal intervals of time. Consider a planet orbiting
a Sun in a circular orbit as shown in the figure below. If the planet sweeps out
an area of 10 square units in the first 3 days of its orbit, what area will it sweep
out in the next 5 days?
Sun Planet
r
θ
13
Solution
Let the area swept out by the planet in the next 5 days be A. Since the planet
moves in a circular orbit, the area swept out in a certain time period is given
by:
A=1
2r2∆θ
where ris the radius of the orbit and ∆θis the change in angle subtended
at the center of the circle.
Given that the planet sweeps out an area of 10 square units in the first 3
days of its orbit, we have:
10 = 1
2r2θ1
where θ1is the angle swept out in the first 3 days.
Now, we know that the time period for the next 5 days is twice that of the
first 3 days. Therefore, the angle swept out in the next 5 days, θ2, is twice that
of θ1.
Hence, 2θ1=θ2.
Substitute θ1in terms of rfrom the given information:
10 = 1
2r2θ1
θ1=20
r2
Substitute in θ2:
θ2= 2 ×20
r2=40
r2
So, the area swept out by the planet in the next 5 days is:
A=1
2r2θ2=1
2r2×40
r2= 20 square units
Question 18
Question
Consider a planet with a semi-major axis of 2.5 AU that orbits a star with a
period of 5 years. Calculate the period of a hypothetical moon that orbits this
planet at a distance of 0.8 AU.
14
Solution
To solve this problem, we can use Kepler’s third law of planetary motion, which
states that the square of the period of an orbiting body is proportional to the
cube of the semi-major axis of its orbit. Mathematically, this can be expressed
as: T2
1
a3
1
=T2
2
a3
2
where: - T1and a1are the period and semi-major axis of the planet, - T2
and a2are the period and semi-major axis of the moon.
Given that the period of the planet (T1) is 5 years, the semi-major axis of
the planet (a1) is 2.5 AU, and the semi-major axis of the moon (a2) is 0.8 AU,
we can substitute these values into the formula to find the period of the moon
(T2).
Step 1: Substitute the known values into the formula.
52
2.53=T2
2
0.83
Step 2: Simplify the equation.
25
15.625 =T2
2
0.512
1.6 = T2
2
0.512
Step 3: Solve for T2.
T2
2= 1.6×0.512
T2
2= 0.8192
Step 4: Find T2by taking the square root.
T2=√0.8192
T2≈0.9041
Therefore, the period of the hypothetical moon orbiting the planet at a
distance of 0.8 AU is approximately 0.9041 years.
Question 19
Question
In the context of Kepler’s Laws of Planetary Motion, explain the significance of
Kepler’s Third Law and how it relates to the period and semi-major axis of a
planet’s orbit.
15
Solution
Step 1: Kepler’s Third Law states that the square of the period of revolution of
a planet around the Sun is directly proportional to the cube of the semi-major
axis of its orbit.
Step 2: Mathematically, Kepler’s Third Law can be represented as:
T2
1
a3
1
=T2
2
a3
2
where: T1and T2are the periods of revolution of the planet in orbits with
semi-major axes a1and a2, respectively.
Step 3: This law essentially implies that planets farther away from the Sun
(larger semi-major axis) take longer to complete one orbit (larger period) com-
pared to planets closer to the Sun (smaller semi-major axis).
Step 4: For example, if we compare the Earth and Mars, with respective
semi-major axes of approximately 1 AU and 1.52 AU, we can apply Kepler’s
Third Law to find the ratio of their periods:
T2
Earth
13=T2
Mars
1.523
Step 5: Solving for the ratio of the periods gives:
TMars
TEarth
=√1.523
1≈1.88
This indicates that Mars takes about 1.88 times longer than Earth to com-
plete one orbit around the Sun, which matches observations.
Question 20
Question
Suppose a planet orbits a star in an elliptical orbit, with the star located at one of
the foci of the ellipse. If the planet is at its closest point to the star (perihelion),
which of Kepler’s laws of planetary motion can be used to determine its velocity
at that point? Justify your answer.
Solution
Step 1: Kepler’s laws of planetary motion state: 1. The orbit of a planet around
the Sun is an ellipse with the Sun at one of the two foci. 2. A line joining a
planet and the Sun sweeps out equal areas during 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: When the planet is at its closest point to the star (perihelion), it is
moving fastest in its orbit due to the conservation of angular momentum.
16
Step 3: Therefore, at perihelion, we can use Kepler’s Second Law to deter-
mine the planet’s velocity. This law states that a line connecting the planet to
the star sweeps out equal areas in equal times.
Step 4: Since the planet moves fastest at perihelion, the area swept out in
a given time interval is greatest at this point. This implies that the planet
covers the arc from perihelion to the next point within the elliptical orbit in the
shortest amount of time, resulting in the highest velocity at perihelion.
Step 5: In conclusion, at perihelion, the velocity of a planet can be deter-
mined using Kepler’s Second Law of planetary motion.
Question 21
Question
Consider a hypothetical planetary system where a planet orbits a star at a
distance of 2.5AU with an orbital period of 3.5years. If another planet is
located at a distance of 5AU from the same star, what is the orbital period of
the second planet?
Solution
Step 1: Apply Kepler’s Third Law of Planetary Motion, which 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. Let T1be the orbital period of the first planet
and a1be the semi-major axis of the first planet’s orbit. Similarly, let T2be the
orbital period of the second planet and a2be the semi-major axis of the second
planet’s orbit.
Step 2: Use Kepler’s Third Law formula:
T2
1
T2
2
=a3
1
a3
2
Step 3: Substitute the given values into the formula:
3.52
T2
2
=2.53
53
Step 4: Solve for T2:
3.52
T2
2
=2.53
53
T2
2=53·3.52
2.53
T2
2=875
15.625
T2
2= 56
17
T2=√56
T2≈7.48 years
Therefore, the orbital period of the second planet is approximately 7.48
years.
Question 22
Question
Kepler’s third law of planetary motion relates the orbital period of a planet to
its mean distance from the sun. For a certain planet, the mean distance from
the sun is 2.5 AU (astronomical units). If the orbital period of this planet is 4.5
years, determine the mass of the sun. (Use G= 6.67 ×10−11 m3kg−1s−2)
Solution
Step 1: Recall Kepler’s third law of planetary motion, which states:
T2
a3=4π2
GM
where: T= orbital period of the planet, a= mean distance from the sun, G=
universal gravitational constant, M= mass of the sun.
Step 2: Substitute the given values into the formula:
(4.5years)2
(2.5AU)3=4π2
G·M
Step 3: Convert the orbital period from years to seconds (since Gis given
in SI units):
T= 4.5years ×(365.25 days/year)×(24 hours/day)×(3600 s/hour)
Step 4: Convert the mean distance from AU to meters:
1AU = 1.496 ×1011 m
a= 2.5AU ×1.496 ×1011 m/AU
Step 5: Substitute the converted values into the formula and solve for the
mass of the sun, M.
Question 23
Question
One star has a mass of 2×1030 kg and another star has a mass of 5×1030 kg.
If the first star is 10 times as bright as the second star, what is the ratio of their
luminosities?
18
Solution
Step 1: Use the definition of luminosity. The luminosity of a star is directly
proportional to its mass. We can express the luminosity Lof a star in terms of
its mass Mas L∝M. Therefore, the ratio of the luminosities of the two stars
can be expressed as:
L1
L2
=M1
M2
Step 2: Plug in the given values. Given that the mass of the first star (M1)
is 2×1030 kg, the mass of the second star (M2) is 5×1030 kg, and the first star
is 10 times as bright as the second star, we have:
L1
L2
=2×1030 kg
5×1030 kg
Step 3: Simplify the ratio.
L1
L2
=2
5= 0.4
Therefore, the ratio of the luminosities of the two stars is 0.4.
Question 24
Question
The semi-major axis of Mercury’s orbit around the Sun is 0.387 AU. If the semi-
major axis of Venus’s orbit around the Sun is 0.723 AU, what is the ratio of the
squares of their orbital periods?
Solution
Step 1: Recall Kepler’s Third Law for planetary motion, which 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. Mathematically, we can express this as:
T2
1
T2
2
=a3
1
a3
2
where T1and T2are the orbital periods of the two planets, and a1and a2
are the semi-major axes of their orbits.
Step 2: We are given that the semi-major axis of Mercury’s orbit (a1) is 0.387
AU and the semi-major axis of Venus’s orbit (a2) is 0.723 AU. Let’s substitute
these values into the formula:
T2
1
T2
2
=0.3873
0.7233
Step 3: Calculate the ratio of the squares of the orbital periods:
19
T2
1
T2
2
=0.056187
0.373243 ≈0.15044
Therefore, the ratio of the squares of the orbital periods of Mercury and
Venus is approximately 0.15044.
Question 25
Question
Kepler’s Third Law states that the square of the period of revolution of a planet
around the Sun is directly proportional to the cube of its average distance from
the Sun. Suppose a newly discovered planet has an average distance from the
Sun that is 4 times farther than Earth’s average distance. If Earth takes 365
days to orbit the Sun, how long does it take for this new planet to orbit the
Sun?
Solution
Let Tbe the period of revolution of the new planet around the Sun in days and
rbe its average distance from the Sun in astronomical units (AU). We are given
that r= 4 AU and the period of revolution for Earth is TEarth = 365 days.
According to Kepler’s Third Law, we have the following equation:
(T
TEarth )2
=(r
1)3
Step 1: Plug in the given values into the equation.
(T
365)2
=(4
1)3
Step 2: Square both sides of the equation.
(T
365)2
= 64
Step 3: Take the square root of both sides to solve for T.
T
365 =√64 = 8
Step 4: Multiply both sides by 365 to solve for T.
T= 8 ×365 = 2920
Therefore, the period of revolution for the new planet is 2920 days, which is
8 times longer than Earth’s period.
20
Step 4: Solve for a:
a3= 140.8996
a=3
√140.8996
a≈5.208 AU
Step 5: Therefore, the semi-major axis of Jupiter’s orbit is approximately
5.208 AU.
Question 2
Question
In the study of planetary motion, Kepler’s third law states that the square of
the period of revolution of a planet around the Sun is proportional to the cube
of the planet’s average distance from the Sun. If the period of revolution for
Jupiter is 11.86 years and its average distance from the Sun is 778 million km,
calculate the period of revolution for Saturn given that its average distance from
the Sun is 1.43 billion km.
Solution
Step 1: Find the ratio of the cube of Jupiter’s average distance to the cube
of Saturn’s average distance. Let r1= 778 million km be Jupiter’s average
distance from the Sun and r2= 1.43 billion km be Saturn’s average distance
from the Sun. The ratio of the cubes of their average distances is given by:
(r1
r2)3
=(778
1430)3
=(778
1430)3
Step 2: Use Kepler’s third law to find the period of revolution for Saturn.
Kepler’s third law states that the square of the period of revolution is propor-
tional to the cube of the average distance from the Sun. So, T2
1
T2
2
=(r1
r2)3
where
T1and T2are the periods of revolution for Jupiter and Saturn, respectively.
Since we know T1= 11.86 years, we can solve for T2:
T2
1
T2
2
=(r1
r2)3
11.862
T2
2
=(778
1430)3
T2
2=11.862
(778
1430 )3
T2=√11.862
(778
1430 )3
2
Step 3: Calculate the period of revolution for Saturn. Now, we can substitute
the values and calculate T2:
T2=√11.862
(778
1430 )3≈√140.6596
0.9716
T2≈√144.8864 ≈12.04 years
Therefore, the period of revolution for Saturn is approximately 12.04 years.
Question 3
Question
Given the elliptical orbit of a planet around the Sun, the aphelion distance
(furthest point from the Sun) is 2.5 AU and the perihelion distance (closest
point to the Sun) is 1.5 AU. Calculate the eccentricity of the planet’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 to the length
of the major axis. In this case, the foci are at the center of the ellipse (the Sun)
and either the aphelion or perihelion.
Step 2: The major axis of an ellipse is the longest line that passes through
the center and both foci. It is equal to the sum of the distances to the aphelion
and perihelion, which gives us the equation 2a= 2.5+1.5 = 4 AU.
Step 3: Next, we find the distance between the foci, which is 2c= 2.5−1.5 =
1AU.
Step 4: We can now calculate the eccentricity eusing the formula e=c
a,
where ais the semimajor axis equal to 1
2of the major axis. Substituting in the
known values, we get e=1
4= 0.25.
Step 5: Therefore, the eccentricity of the planet’s orbit is 0.25.
Question 4
Question
The period of a planet orbiting a star is related to the semi-major axis of its orbit
by Kepler’s third law, which states that the square of the period of the orbit is
proportional to the cube of the semi-major axis of the orbit. For a particular
planet in our solar system, its period is 11.86 years and its semi-major axis is 5.2
astronomical units (AU). Determine the period of another planet in the same
star system with a semi-major axis of 2.8 AU.
3
Solution
Let T1be the period of the first planet and a1be its semi-major axis. Similarly,
let T2be the period of the second planet with semi-major axis a2. According to
Kepler’s third law, we have:
T2
1
a3
1
=T2
2
a3
2
Given that T1= 11.86 years and a1= 5.2AU, we want to find T2when a2= 2.8
AU.
Step 1: Substitute the known values into the equation and solve for T2.
(11.86)2
(5.2)3=T2
2
(2.8)3
Step 2: Simplify the equation.
139.96
140.608 =T2
2
21.952
Step 3: Calculate the value of T2.
T2
2=139.96 ×21.952
140.608
T2
2= 21.89
T2=√21.89
T2≈4.68 years
Therefore, the period of the second planet with a semi-major axis of 2.8 AU
is approximately 4.68 years.
Question 5
Question
The period of a planet’s orbit around the sun is 15 years. If the average distance
between the planet and the sun is 3.5 astronomical units (AU), determine the
mass of the sun. (Assume the planet’s orbit is nearly circular.)
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion, which states:
T2
1
a3
1
=T2
2
a3
2
where Tis the period of the orbit (in years) and ais the average distance
between the planet and the sun (in AU).
4
Step 2: Let’s denote the period of the unknown planet as T2and its distance
from the sun as a2. We can rewrite the equation as:
152
3.53=T2
2
a3
2
Step 3: We can solve for a2by rearranging the equation:
a2= 3.5(15
T2)2/3
Step 4: Now, let’s use Newton’s version of Kepler’s Third Law to find the
mass of the sun. Newton’s form is:
GMsun
4π2=a3
2(T2
2π)2
Step 5: Substitute the expression for a2into the equation:
GMsun
4π2=(3.5(15
T2)2/3)3
(T2
2π)2
Step 6: Simplify the equation and solve for Msun to find the mass of the sun.
Question 6
Question
The asteroid belt is a region located between the orbits of Mars and Jupiter
where many small objects orbit the Sun. Consider an asteroid with a semi-
major axis of 2.5 AU (astronomical units) and an orbital period of 4.0 years.
Determine the mass of the Sun using Kepler’s third law.
Solution
Step 1: Identify the given values and the unknown. Let abe the semi-major
axis of the asteroid’s orbit, Tbe the orbital period, and M⊙be the mass of the
Sun. The given values are a= 2.5AU and T= 4.0years. The unknown is M⊙.
Step 2: Convert the given units. Since we are using astronomical units (AU)
for a, we need to convert the semi-major axis to meters to be consistent with SI
units. Given that 1AU = 1.496 ×1011 meters, we have:
a= 2.5AU ×1.496 ×1011 m/AU = 3.74 ×1011 m
Step 3: Apply Kepler’s third law. Kepler’s third law is given by the equation:
T2=(4π2a3
G(M⊙+m))
5
where mis the mass of the asteroid, Gis the gravitational constant, and all
other variables are defined as before.
Step 4: Solve for the mass of the Sun. Substitute the given values into
Kepler’s third law and solve for M⊙:
4.02=(4π2(3.74 ×1011)3
G(M⊙+m))
16 = 4π2(5.59 ×1034)
G(M⊙+m)
16G(M⊙+m) = 4π2(5.59 ×1034)
2G(M⊙+m) = π2(5.59 ×1034)
2GM⊙=π2(5.59 ×1034)−2Gm
Step 5: Calculate the mass of the Sun. Unfortunately, without information
about the mass of the asteroid m, we cannot determine the exact mass of the
Sun using this method. The mass of the Sun can only be found if mis known.
Question 8
Question
Consider a hypothetical planet with a semi-major axis of 2 AU. Determine the
period of this planet’s orbit using Kepler’s Third Law of Planetary Motion.
Solution
To find the period of the planet’s orbit using Kepler’s Third Law of Planetary
Motion, we can use the formula:
T2=(4π2
G(M+m))a3
where: - Tis the period of the planet’s orbit, - Gis the gravitational constant
(6.67430 ×10−11 m3kg−1s−2), - Mis the mass of the sun (1.989 ×1030 kg), -
mis the mass of the planet (assumed to be negligible compared to the sun), -
ais the semi-major axis of the planet’s orbit.
Step 1: Substitute the given values into the formula:
T2=(4π2
G(M+m))a3
T2=(4π2
6.67430 ×10−11 ×(1.989 ×1030))(23)
6
Step 2: Calculate the period of the planet’s orbit:
T2=(4π2
1.329 ×1020 )(8)
T2= (2.972 ×107)s2
Step 3: Find the period of the planet’s orbit by taking the square root:
T=√2.972 ×107s
T≈5455 s
Therefore, the period of the planet’s orbit is approximately 5455 seconds.
Question 9
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is proportional to the cube of its average distance from the
sun. Suppose a newly discovered planet has an orbital period of 4years and an
average distance from the Sun of 6AU (astronomical units). What would be
the orbital period of a different planet located at an average distance of 15 AU
from the Sun, according to Kepler’s third law?
Solution
Step 1: Begin by setting up the proportion based on Kepler’s third law: Let T
be the orbital period of the second planet with an average distance of 15 AU
from the Sun. According to Kepler’s third law,
T2
1
d3
1
=T2
2
d3
2
where T1= 4 years, d1= 6 AU, and d2= 15 AU.
Step 2: Plug in the given values and solve for T2:
42
63=T2
2
153
16
216 =T2
2
3375
Step 3: Simplify the equation:
2
27 =T2
2
3375
7
Step 4: Cross multiply to solve for T2:
2×3375 = 27 ×T2
2
6750 = 27 ×T2
2
T2
2=6750
27
T2
2= 250
T2=√250 = 5√10
Step 5: Therefore, the orbital period of the second planet, located at an
average distance of 15 AU from the Sun, would be approximately 5√10 years
according to Kepler’s third law.
Question 10
Question
An exoplanet orbits a star with a period of 400 days and an average distance of
2 AU. If the star has a mass of 2×1030 kg, determine the mass of the exoplanet
using Kepler’s Third Law of Planetary Motion.
Solution
Step 1: We first convert the average distance of the exoplanet from the star to
meters:
Average distance = 2 AU = 2 ×1.496 ×1011 m/AU = 2.992 ×1011 m
Step 2: We then convert the period of the exoplanet’s orbit to seconds:
Period = 400 days×24 hours/day×60 minutes/hour×60 seconds/minute = 3.456×107seconds
Step 3: Using Kepler’s Third Law of Planetary Motion, we have:
T2
r3=4π2
G(M1+M2)
Step 4: We can rearrange the formula to solve for the mass of the exoplanet
M2:
M2=4π2r3
GT 2−M1
Step 5: Substituting the known values into the formula, we get:
M2=4π2×(2.992 ×1011)3
6.67 ×10−11 ×(3.456 ×107)2−2×1030
Step 6: Calculating the mass of the exoplanet, we find:
M2≈1.171 ×1028 kg
Therefore, the mass of the exoplanet is approximately 1.171 ×1028 kg.
8
Question 11
Question
The planet Mercury has an average distance from the Sun of approximately
0.39 AU. Calculate the period of Mercury’s orbit around the Sun in Earth
years. (Hint: Kepler’s Third Law states that the square of the orbital period of
a planet is proportional to the cube of its average distance from the Sun.)
Solution
Step 1: Given that Mercury’s average distance from the Sun is 0.39 AU, we can
use Kepler’s Third Law to find the period of Mercury’s orbit in Earth years.
Let Tbe the period of Mercury’s orbit.
Step 2: Kepler’s Third Law states that T2
1
T2
2
=a3
1
a3
2
, where T1and T2are the
periods of two planets, and a1and a2are their average distances from the Sun.
Step 3: Substituting the values for Mercury (T=T1,a=a1) and Earth
(T2= 1 year, a2= 1 AU) into the equation, we get:
T2
12=0.393
13
Step 4: Solving for T:
T2= 0.393
T=√0.393≈0.241
Step 5: Therefore, the period of Mercury’s orbit around the Sun is approxi-
mately 0.241 years.
Question 12
Question
During their orbits, planets move faster when they are closer to the Sun and
slower when they are farther away. Explain how Kepler’s laws of planetary
motion account for this observation.
Solution
To explain why planets move faster when they are closer to the Sun and slower
when they are farther away, we will use Kepler’s laws of planetary motion.
Step 1: Kepler’s First Law states that planets move in elliptical orbits with
the Sun at one of the foci. This means that the distance between a planet and
the Sun changes as the planet orbits.
Step 2: When a planet is closer to the Sun, it is at the perihelion point
(closest point to the Sun) of its orbit. At this point, the gravitational force
9
between the planet and the Sun is stronger, causing the planet to move faster
in order to maintain its orbit.
Step 3: Conversely, when a planet is farther away from the Sun, it is at
the aphelion point (farthest point from the Sun) of its orbit. At this point, the
gravitational force between the planet and the Sun is weaker, resulting in the
planet moving slower to stay in orbit.
Step 4: This variation in orbital speed is also explained by Kepler’s Second
Law, which states that a planet sweeps out equal areas in equal times. When a
planet is closer to the Sun and moving faster, it covers a larger area in a given
amount of time. Conversely, when it is farther from the Sun and moving slower,
it covers a smaller area in the same amount of time.
Therefore, Kepler’s laws of planetary motion account for the observation
that planets move faster when they are closer to the Sun and slower when they
are farther away.
Question 13
Question
Kepler’s third law states that the square of the period of revolution of a planet
around the sun is proportional to the cube of its semi-major axis. If a planet has
a period of revolution of 10 years and a semi-major axis of 3 AU (astronomical
units), what is the period of revolution of another planet with a semi-major axis
of 6 AU?
Solution
Step 1: Write down Kepler’s third law in mathematical form:
(T1
T2)2
=(a1
a2)3
where T1and T2are the periods of revolution of the two planets, and a1and a2
are their semi-major axes.
Step 2: Substitute the given values into the equation:
(10
T2)2
=(3
6)3
Step 3: Simplify the equation by squaring and cubing the fractions:
(10
T2)2
=(1
2)3
(102
T2
2)=(1
8)
10
Step 4: Cross multiply to solve for T2:
102=T2
2
8
100 ×8 = T2
2
800 = T2
2
Step 5: Take the square root of both sides to find T2:
T2=√800
T2= 28.28 years
Therefore, the period of revolution of the planet with a semi-major axis of
6 AU is 28.28 years.
Question 14
Question
In a hypothetical solar system, a planet takes 40 years to complete one full orbit
around its star. If the planet’s average distance from the star is 2.5 astronomical
units (AU), determine the mass of the star in terms of solar masses (M⊙).
Solution
Step 1: Recall Kepler’s third law of planetary motion, which relates the orbital
period of a planet (T) to its average distance from the star (r) using the formula:
T2=(4π2
G(m1+m2))r3
where T= orbital period of the planet, G= gravitational constant, m1= mass
of the star, r= average distance between the star and the planet, and m2=
mass of the planet (assumed to be negligible).
Step 2: Given that the planet takes 40 years to complete one full orbit, we
have T= 40 years = 40 ×365 days = 14,600 days.
Step 3: Convert the average distance from astronomical units (AU) to meters
using the conversion factor: 1 AU = 1.496 ×1011 meters. Given the average
distance r= 2.5AU, we have r= 2.5×1.496×1011 meters = 3.74×1011 meters.
Step 4: Substitute the values of Tand rinto Kepler’s third law formula:
(14600)2=(4π2
G(m⊙))(3.74 ×1011)3
Step 5: Solve for m⊙, the mass of the star, in terms of solar masses (M⊙).
Step 6: After finding the mass of the star, express the result in terms of solar
masses by dividing the mass by the mass of the Sun (M⊙= 1.989 ×1030 kg).
11
Question 15
Question
State Kepler’s Third Law of Planetary Motion and provide an example calcula-
tion showing how the law can be applied to determine the period of a planet’s
orbit around the Sun.
Solution
Kepler’s Third Law of Planetary Motion states that the square of the orbital
period of a planet is proportional to the cube of the semi-major axis of its orbit.
Step 1: Let’s denote the orbital period of the planet as T(in years) and
the semi-major axis of its orbit as a(in astronomical units, AU). The equation
for Kepler’s Third Law is given as:
T2=k·a3
where kis a constant.
Step 2: We can use the data for Earth’s orbit around the Sun to find the
value of the constant k. The orbital period of Earth is approximately 1 year
and the semi-major axis of its orbit is about 1 AU. Substituting these values
into the equation:
(1)2=k·(1)3
1 = k
Step 3: Now, let’s apply Kepler’s Third Law to determine the period of
Jupiter’s orbit around the Sun. The semi-major axis of Jupiter’s orbit is ap-
proximately 5.2 AU. Substituting this value into the equation, we can solve for
the orbital period T:
T2= (1) ·(5.2)3
T2= 5.23
T2= 140.608
Step 4: Taking the square root of both sides to find T:
T=√140.608
T≈11.85 years
Therefore, the orbital period of Jupiter around the Sun is approximately
11.85 years according to Kepler’s Third Law of Planetary Motion.
Question 16
Question
A planet is in elliptical orbit around the Sun. The semi-major axis of the planet’s
orbit is 3.5 AU and its eccentricity is 0.2. Determine the semi-minor axis of the
orbit.
12
Solution
Step 1: Recall the formula to calculate the semi-minor axis of an elliptical orbit:
b=a√1−e2
where ais the semi-major axis and eis the eccentricity of the orbit.
Step 2: Substitute the given values into the formula:
b= 3.5AU ·√1−0.22
Step 3: Calculate the value inside the square root:
1−0.22= 1 −0.04 = 0.96
Step 4: Substitute the calculated value back into the formula:
b= 3.5AU ·√0.96
Step 5: Calculate the square root:
√0.96 = 0.960.5= 0.9798
Step 6: Multiply the semi-major axis by the square root:
b= 3.5AU ·0.9798 = 3.43 AU
Therefore, the semi-minor axis of the orbit is 3.43 AU.
Question 17
Question
Kepler’s second law states that a line segment joining a planet and the Sun
sweeps out equal areas during equal intervals of time. Consider a planet orbiting
a Sun in a circular orbit as shown in the figure below. If the planet sweeps out
an area of 10 square units in the first 3 days of its orbit, what area will it sweep
out in the next 5 days?
Sun Planet
r
θ
13
Solution
Let the area swept out by the planet in the next 5 days be A. Since the planet
moves in a circular orbit, the area swept out in a certain time period is given
by:
A=1
2r2∆θ
where ris the radius of the orbit and ∆θis the change in angle subtended
at the center of the circle.
Given that the planet sweeps out an area of 10 square units in the first 3
days of its orbit, we have:
10 = 1
2r2θ1
where θ1is the angle swept out in the first 3 days.
Now, we know that the time period for the next 5 days is twice that of the
first 3 days. Therefore, the angle swept out in the next 5 days, θ2, is twice that
of θ1.
Hence, 2θ1=θ2.
Substitute θ1in terms of rfrom the given information:
10 = 1
2r2θ1
θ1=20
r2
Substitute in θ2:
θ2= 2 ×20
r2=40
r2
So, the area swept out by the planet in the next 5 days is:
A=1
2r2θ2=1
2r2×40
r2= 20 square units
Question 18
Question
Consider a planet with a semi-major axis of 2.5 AU that orbits a star with a
period of 5 years. Calculate the period of a hypothetical moon that orbits this
planet at a distance of 0.8 AU.
14
Solution
To solve this problem, we can use Kepler’s third law of planetary motion, which
states that the square of the period of an orbiting body is proportional to the
cube of the semi-major axis of its orbit. Mathematically, this can be expressed
as: T2
1
a3
1
=T2
2
a3
2
where: - T1and a1are the period and semi-major axis of the planet, - T2
and a2are the period and semi-major axis of the moon.
Given that the period of the planet (T1) is 5 years, the semi-major axis of
the planet (a1) is 2.5 AU, and the semi-major axis of the moon (a2) is 0.8 AU,
we can substitute these values into the formula to find the period of the moon
(T2).
Step 1: Substitute the known values into the formula.
52
2.53=T2
2
0.83
Step 2: Simplify the equation.
25
15.625 =T2
2
0.512
1.6 = T2
2
0.512
Step 3: Solve for T2.
T2
2= 1.6×0.512
T2
2= 0.8192
Step 4: Find T2by taking the square root.
T2=√0.8192
T2≈0.9041
Therefore, the period of the hypothetical moon orbiting the planet at a
distance of 0.8 AU is approximately 0.9041 years.
Question 19
Question
In the context of Kepler’s Laws of Planetary Motion, explain the significance of
Kepler’s Third Law and how it relates to the period and semi-major axis of a
planet’s orbit.
15
Solution
Step 1: Kepler’s Third Law states that the square of the period of revolution of
a planet around the Sun is directly proportional to the cube of the semi-major
axis of its orbit.
Step 2: Mathematically, Kepler’s Third Law can be represented as:
T2
1
a3
1
=T2
2
a3
2
where: T1and T2are the periods of revolution of the planet in orbits with
semi-major axes a1and a2, respectively.
Step 3: This law essentially implies that planets farther away from the Sun
(larger semi-major axis) take longer to complete one orbit (larger period) com-
pared to planets closer to the Sun (smaller semi-major axis).
Step 4: For example, if we compare the Earth and Mars, with respective
semi-major axes of approximately 1 AU and 1.52 AU, we can apply Kepler’s
Third Law to find the ratio of their periods:
T2
Earth
13=T2
Mars
1.523
Step 5: Solving for the ratio of the periods gives:
TMars
TEarth
=√1.523
1≈1.88
This indicates that Mars takes about 1.88 times longer than Earth to com-
plete one orbit around the Sun, which matches observations.
Question 20
Question
Suppose a planet orbits a star in an elliptical orbit, with the star located at one of
the foci of the ellipse. If the planet is at its closest point to the star (perihelion),
which of Kepler’s laws of planetary motion can be used to determine its velocity
at that point? Justify your answer.
Solution
Step 1: Kepler’s laws of planetary motion state: 1. The orbit of a planet around
the Sun is an ellipse with the Sun at one of the two foci. 2. A line joining a
planet and the Sun sweeps out equal areas during 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: When the planet is at its closest point to the star (perihelion), it is
moving fastest in its orbit due to the conservation of angular momentum.
16
Step 3: Therefore, at perihelion, we can use Kepler’s Second Law to deter-
mine the planet’s velocity. This law states that a line connecting the planet to
the star sweeps out equal areas in equal times.
Step 4: Since the planet moves fastest at perihelion, the area swept out in
a given time interval is greatest at this point. This implies that the planet
covers the arc from perihelion to the next point within the elliptical orbit in the
shortest amount of time, resulting in the highest velocity at perihelion.
Step 5: In conclusion, at perihelion, the velocity of a planet can be deter-
mined using Kepler’s Second Law of planetary motion.
Question 21
Question
Consider a hypothetical planetary system where a planet orbits a star at a
distance of 2.5AU with an orbital period of 3.5years. If another planet is
located at a distance of 5AU from the same star, what is the orbital period of
the second planet?
Solution
Step 1: Apply Kepler’s Third Law of Planetary Motion, which 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. Let T1be the orbital period of the first planet
and a1be the semi-major axis of the first planet’s orbit. Similarly, let T2be the
orbital period of the second planet and a2be the semi-major axis of the second
planet’s orbit.
Step 2: Use Kepler’s Third Law formula:
T2
1
T2
2
=a3
1
a3
2
Step 3: Substitute the given values into the formula:
3.52
T2
2
=2.53
53
Step 4: Solve for T2:
3.52
T2
2
=2.53
53
T2
2=53·3.52
2.53
T2
2=875
15.625
T2
2= 56
17
T2=√56
T2≈7.48 years
Therefore, the orbital period of the second planet is approximately 7.48
years.
Question 22
Question
Kepler’s third law of planetary motion relates the orbital period of a planet to
its mean distance from the sun. For a certain planet, the mean distance from
the sun is 2.5 AU (astronomical units). If the orbital period of this planet is 4.5
years, determine the mass of the sun. (Use G= 6.67 ×10−11 m3kg−1s−2)
Solution
Step 1: Recall Kepler’s third law of planetary motion, which states:
T2
a3=4π2
GM
where: T= orbital period of the planet, a= mean distance from the sun, G=
universal gravitational constant, M= mass of the sun.
Step 2: Substitute the given values into the formula:
(4.5years)2
(2.5AU)3=4π2
G·M
Step 3: Convert the orbital period from years to seconds (since Gis given
in SI units):
T= 4.5years ×(365.25 days/year)×(24 hours/day)×(3600 s/hour)
Step 4: Convert the mean distance from AU to meters:
1AU = 1.496 ×1011 m
a= 2.5AU ×1.496 ×1011 m/AU
Step 5: Substitute the converted values into the formula and solve for the
mass of the sun, M.
Question 23
Question
One star has a mass of 2×1030 kg and another star has a mass of 5×1030 kg.
If the first star is 10 times as bright as the second star, what is the ratio of their
luminosities?
18
Solution
Step 1: Use the definition of luminosity. The luminosity of a star is directly
proportional to its mass. We can express the luminosity Lof a star in terms of
its mass Mas L∝M. Therefore, the ratio of the luminosities of the two stars
can be expressed as:
L1
L2
=M1
M2
Step 2: Plug in the given values. Given that the mass of the first star (M1)
is 2×1030 kg, the mass of the second star (M2) is 5×1030 kg, and the first star
is 10 times as bright as the second star, we have:
L1
L2
=2×1030 kg
5×1030 kg
Step 3: Simplify the ratio.
L1
L2
=2
5= 0.4
Therefore, the ratio of the luminosities of the two stars is 0.4.
Question 24
Question
The semi-major axis of Mercury’s orbit around the Sun is 0.387 AU. If the semi-
major axis of Venus’s orbit around the Sun is 0.723 AU, what is the ratio of the
squares of their orbital periods?
Solution
Step 1: Recall Kepler’s Third Law for planetary motion, which 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. Mathematically, we can express this as:
T2
1
T2
2
=a3
1
a3
2
where T1and T2are the orbital periods of the two planets, and a1and a2
are the semi-major axes of their orbits.
Step 2: We are given that the semi-major axis of Mercury’s orbit (a1) is 0.387
AU and the semi-major axis of Venus’s orbit (a2) is 0.723 AU. Let’s substitute
these values into the formula:
T2
1
T2
2
=0.3873
0.7233
Step 3: Calculate the ratio of the squares of the orbital periods:
19
T2
1
T2
2
=0.056187
0.373243 ≈0.15044
Therefore, the ratio of the squares of the orbital periods of Mercury and
Venus is approximately 0.15044.
Question 25
Question
Kepler’s Third Law states that the square of the period of revolution of a planet
around the Sun is directly proportional to the cube of its average distance from
the Sun. Suppose a newly discovered planet has an average distance from the
Sun that is 4 times farther than Earth’s average distance. If Earth takes 365
days to orbit the Sun, how long does it take for this new planet to orbit the
Sun?
Solution
Let Tbe the period of revolution of the new planet around the Sun in days and
rbe its average distance from the Sun in astronomical units (AU). We are given
that r= 4 AU and the period of revolution for Earth is TEarth = 365 days.
According to Kepler’s Third Law, we have the following equation:
(T
TEarth )2
=(r
1)3
Step 1: Plug in the given values into the equation.
(T
365)2
=(4
1)3
Step 2: Square both sides of the equation.
(T
365)2
= 64
Step 3: Take the square root of both sides to solve for T.
T
365 =√64 = 8
Step 4: Multiply both sides by 365 to solve for T.
T= 8 ×365 = 2920
Therefore, the period of revolution for the new planet is 2920 days, which is
8 times longer than Earth’s period.
20
Step 4: Solve for a:
a3= 140.8996
a=3
√140.8996
a≈5.208 AU
Step 5: Therefore, the semi-major axis of Jupiter’s orbit is approximately
5.208 AU.
Question 2
Question
In the study of planetary motion, Kepler’s third law states that the square of
the period of revolution of a planet around the Sun is proportional to the cube
of the planet’s average distance from the Sun. If the period of revolution for
Jupiter is 11.86 years and its average distance from the Sun is 778 million km,
calculate the period of revolution for Saturn given that its average distance from
the Sun is 1.43 billion km.
Solution
Step 1: Find the ratio of the cube of Jupiter’s average distance to the cube
of Saturn’s average distance. Let r1= 778 million km be Jupiter’s average
distance from the Sun and r2= 1.43 billion km be Saturn’s average distance
from the Sun. The ratio of the cubes of their average distances is given by:
(r1
r2)3
=(778
1430)3
=(778
1430)3
Step 2: Use Kepler’s third law to find the period of revolution for Saturn.
Kepler’s third law states that the square of the period of revolution is propor-
tional to the cube of the average distance from the Sun. So, T2
1
T2
2
=(r1
r2)3
where
T1and T2are the periods of revolution for Jupiter and Saturn, respectively.
Since we know T1= 11.86 years, we can solve for T2:
T2
1
T2
2
=(r1
r2)3
11.862
T2
2
=(778
1430)3
T2
2=11.862
(778
1430 )3
T2=√11.862
(778
1430 )3
2
Step 3: Calculate the period of revolution for Saturn. Now, we can substitute
the values and calculate T2:
T2=√11.862
(778
1430 )3≈√140.6596
0.9716
T2≈√144.8864 ≈12.04 years
Therefore, the period of revolution for Saturn is approximately 12.04 years.
Question 3
Question
Given the elliptical orbit of a planet around the Sun, the aphelion distance
(furthest point from the Sun) is 2.5 AU and the perihelion distance (closest
point to the Sun) is 1.5 AU. Calculate the eccentricity of the planet’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 to the length
of the major axis. In this case, the foci are at the center of the ellipse (the Sun)
and either the aphelion or perihelion.
Step 2: The major axis of an ellipse is the longest line that passes through
the center and both foci. It is equal to the sum of the distances to the aphelion
and perihelion, which gives us the equation 2a= 2.5+1.5 = 4 AU.
Step 3: Next, we find the distance between the foci, which is 2c= 2.5−1.5 =
1AU.
Step 4: We can now calculate the eccentricity eusing the formula e=c
a,
where ais the semimajor axis equal to 1
2of the major axis. Substituting in the
known values, we get e=1
4= 0.25.
Step 5: Therefore, the eccentricity of the planet’s orbit is 0.25.
Question 4
Question
The period of a planet orbiting a star is related to the semi-major axis of its orbit
by Kepler’s third law, which states that the square of the period of the orbit is
proportional to the cube of the semi-major axis of the orbit. For a particular
planet in our solar system, its period is 11.86 years and its semi-major axis is 5.2
astronomical units (AU). Determine the period of another planet in the same
star system with a semi-major axis of 2.8 AU.
3
Solution
Let T1be the period of the first planet and a1be its semi-major axis. Similarly,
let T2be the period of the second planet with semi-major axis a2. According to
Kepler’s third law, we have:
T2
1
a3
1
=T2
2
a3
2
Given that T1= 11.86 years and a1= 5.2AU, we want to find T2when a2= 2.8
AU.
Step 1: Substitute the known values into the equation and solve for T2.
(11.86)2
(5.2)3=T2
2
(2.8)3
Step 2: Simplify the equation.
139.96
140.608 =T2
2
21.952
Step 3: Calculate the value of T2.
T2
2=139.96 ×21.952
140.608
T2
2= 21.89
T2=√21.89
T2≈4.68 years
Therefore, the period of the second planet with a semi-major axis of 2.8 AU
is approximately 4.68 years.
Question 5
Question
The period of a planet’s orbit around the sun is 15 years. If the average distance
between the planet and the sun is 3.5 astronomical units (AU), determine the
mass of the sun. (Assume the planet’s orbit is nearly circular.)
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion, which states:
T2
1
a3
1
=T2
2
a3
2
where Tis the period of the orbit (in years) and ais the average distance
between the planet and the sun (in AU).
4
Step 2: Let’s denote the period of the unknown planet as T2and its distance
from the sun as a2. We can rewrite the equation as:
152
3.53=T2
2
a3
2
Step 3: We can solve for a2by rearranging the equation:
a2= 3.5(15
T2)2/3
Step 4: Now, let’s use Newton’s version of Kepler’s Third Law to find the
mass of the sun. Newton’s form is:
GMsun
4π2=a3
2(T2
2π)2
Step 5: Substitute the expression for a2into the equation:
GMsun
4π2=(3.5(15
T2)2/3)3
(T2
2π)2
Step 6: Simplify the equation and solve for Msun to find the mass of the sun.
Question 6
Question
The asteroid belt is a region located between the orbits of Mars and Jupiter
where many small objects orbit the Sun. Consider an asteroid with a semi-
major axis of 2.5 AU (astronomical units) and an orbital period of 4.0 years.
Determine the mass of the Sun using Kepler’s third law.
Solution
Step 1: Identify the given values and the unknown. Let abe the semi-major
axis of the asteroid’s orbit, Tbe the orbital period, and M⊙be the mass of the
Sun. The given values are a= 2.5AU and T= 4.0years. The unknown is M⊙.
Step 2: Convert the given units. Since we are using astronomical units (AU)
for a, we need to convert the semi-major axis to meters to be consistent with SI
units. Given that 1AU = 1.496 ×1011 meters, we have:
a= 2.5AU ×1.496 ×1011 m/AU = 3.74 ×1011 m
Step 3: Apply Kepler’s third law. Kepler’s third law is given by the equation:
T2=(4π2a3
G(M⊙+m))
5
where mis the mass of the asteroid, Gis the gravitational constant, and all
other variables are defined as before.
Step 4: Solve for the mass of the Sun. Substitute the given values into
Kepler’s third law and solve for M⊙:
4.02=(4π2(3.74 ×1011)3
G(M⊙+m))
16 = 4π2(5.59 ×1034)
G(M⊙+m)
16G(M⊙+m) = 4π2(5.59 ×1034)
2G(M⊙+m) = π2(5.59 ×1034)
2GM⊙=π2(5.59 ×1034)−2Gm
Step 5: Calculate the mass of the Sun. Unfortunately, without information
about the mass of the asteroid m, we cannot determine the exact mass of the
Sun using this method. The mass of the Sun can only be found if mis known.
Question 8
Question
Consider a hypothetical planet with a semi-major axis of 2 AU. Determine the
period of this planet’s orbit using Kepler’s Third Law of Planetary Motion.
Solution
To find the period of the planet’s orbit using Kepler’s Third Law of Planetary
Motion, we can use the formula:
T2=(4π2
G(M+m))a3
where: - Tis the period of the planet’s orbit, - Gis the gravitational constant
(6.67430 ×10−11 m3kg−1s−2), - Mis the mass of the sun (1.989 ×1030 kg), -
mis the mass of the planet (assumed to be negligible compared to the sun), -
ais the semi-major axis of the planet’s orbit.
Step 1: Substitute the given values into the formula:
T2=(4π2
G(M+m))a3
T2=(4π2
6.67430 ×10−11 ×(1.989 ×1030))(23)
6
Step 2: Calculate the period of the planet’s orbit:
T2=(4π2
1.329 ×1020 )(8)
T2= (2.972 ×107)s2
Step 3: Find the period of the planet’s orbit by taking the square root:
T=√2.972 ×107s
T≈5455 s
Therefore, the period of the planet’s orbit is approximately 5455 seconds.
Question 9
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is proportional to the cube of its average distance from the
sun. Suppose a newly discovered planet has an orbital period of 4years and an
average distance from the Sun of 6AU (astronomical units). What would be
the orbital period of a different planet located at an average distance of 15 AU
from the Sun, according to Kepler’s third law?
Solution
Step 1: Begin by setting up the proportion based on Kepler’s third law: Let T
be the orbital period of the second planet with an average distance of 15 AU
from the Sun. According to Kepler’s third law,
T2
1
d3
1
=T2
2
d3
2
where T1= 4 years, d1= 6 AU, and d2= 15 AU.
Step 2: Plug in the given values and solve for T2:
42
63=T2
2
153
16
216 =T2
2
3375
Step 3: Simplify the equation:
2
27 =T2
2
3375
7
Step 4: Cross multiply to solve for T2:
2×3375 = 27 ×T2
2
6750 = 27 ×T2
2
T2
2=6750
27
T2
2= 250
T2=√250 = 5√10
Step 5: Therefore, the orbital period of the second planet, located at an
average distance of 15 AU from the Sun, would be approximately 5√10 years
according to Kepler’s third law.
Question 10
Question
An exoplanet orbits a star with a period of 400 days and an average distance of
2 AU. If the star has a mass of 2×1030 kg, determine the mass of the exoplanet
using Kepler’s Third Law of Planetary Motion.
Solution
Step 1: We first convert the average distance of the exoplanet from the star to
meters:
Average distance = 2 AU = 2 ×1.496 ×1011 m/AU = 2.992 ×1011 m
Step 2: We then convert the period of the exoplanet’s orbit to seconds:
Period = 400 days×24 hours/day×60 minutes/hour×60 seconds/minute = 3.456×107seconds
Step 3: Using Kepler’s Third Law of Planetary Motion, we have:
T2
r3=4π2
G(M1+M2)
Step 4: We can rearrange the formula to solve for the mass of the exoplanet
M2:
M2=4π2r3
GT 2−M1
Step 5: Substituting the known values into the formula, we get:
M2=4π2×(2.992 ×1011)3
6.67 ×10−11 ×(3.456 ×107)2−2×1030
Step 6: Calculating the mass of the exoplanet, we find:
M2≈1.171 ×1028 kg
Therefore, the mass of the exoplanet is approximately 1.171 ×1028 kg.
8
Question 11
Question
The planet Mercury has an average distance from the Sun of approximately
0.39 AU. Calculate the period of Mercury’s orbit around the Sun in Earth
years. (Hint: Kepler’s Third Law states that the square of the orbital period of
a planet is proportional to the cube of its average distance from the Sun.)
Solution
Step 1: Given that Mercury’s average distance from the Sun is 0.39 AU, we can
use Kepler’s Third Law to find the period of Mercury’s orbit in Earth years.
Let Tbe the period of Mercury’s orbit.
Step 2: Kepler’s Third Law states that T2
1
T2
2
=a3
1
a3
2
, where T1and T2are the
periods of two planets, and a1and a2are their average distances from the Sun.
Step 3: Substituting the values for Mercury (T=T1,a=a1) and Earth
(T2= 1 year, a2= 1 AU) into the equation, we get:
T2
12=0.393
13
Step 4: Solving for T:
T2= 0.393
T=√0.393≈0.241
Step 5: Therefore, the period of Mercury’s orbit around the Sun is approxi-
mately 0.241 years.
Question 12
Question
During their orbits, planets move faster when they are closer to the Sun and
slower when they are farther away. Explain how Kepler’s laws of planetary
motion account for this observation.
Solution
To explain why planets move faster when they are closer to the Sun and slower
when they are farther away, we will use Kepler’s laws of planetary motion.
Step 1: Kepler’s First Law states that planets move in elliptical orbits with
the Sun at one of the foci. This means that the distance between a planet and
the Sun changes as the planet orbits.
Step 2: When a planet is closer to the Sun, it is at the perihelion point
(closest point to the Sun) of its orbit. At this point, the gravitational force
9
between the planet and the Sun is stronger, causing the planet to move faster
in order to maintain its orbit.
Step 3: Conversely, when a planet is farther away from the Sun, it is at
the aphelion point (farthest point from the Sun) of its orbit. At this point, the
gravitational force between the planet and the Sun is weaker, resulting in the
planet moving slower to stay in orbit.
Step 4: This variation in orbital speed is also explained by Kepler’s Second
Law, which states that a planet sweeps out equal areas in equal times. When a
planet is closer to the Sun and moving faster, it covers a larger area in a given
amount of time. Conversely, when it is farther from the Sun and moving slower,
it covers a smaller area in the same amount of time.
Therefore, Kepler’s laws of planetary motion account for the observation
that planets move faster when they are closer to the Sun and slower when they
are farther away.
Question 13
Question
Kepler’s third law states that the square of the period of revolution of a planet
around the sun is proportional to the cube of its semi-major axis. If a planet has
a period of revolution of 10 years and a semi-major axis of 3 AU (astronomical
units), what is the period of revolution of another planet with a semi-major axis
of 6 AU?
Solution
Step 1: Write down Kepler’s third law in mathematical form:
(T1
T2)2
=(a1
a2)3
where T1and T2are the periods of revolution of the two planets, and a1and a2
are their semi-major axes.
Step 2: Substitute the given values into the equation:
(10
T2)2
=(3
6)3
Step 3: Simplify the equation by squaring and cubing the fractions:
(10
T2)2
=(1
2)3
(102
T2
2)=(1
8)
10
Step 4: Cross multiply to solve for T2:
102=T2
2
8
100 ×8 = T2
2
800 = T2
2
Step 5: Take the square root of both sides to find T2:
T2=√800
T2= 28.28 years
Therefore, the period of revolution of the planet with a semi-major axis of
6 AU is 28.28 years.
Question 14
Question
In a hypothetical solar system, a planet takes 40 years to complete one full orbit
around its star. If the planet’s average distance from the star is 2.5 astronomical
units (AU), determine the mass of the star in terms of solar masses (M⊙).
Solution
Step 1: Recall Kepler’s third law of planetary motion, which relates the orbital
period of a planet (T) to its average distance from the star (r) using the formula:
T2=(4π2
G(m1+m2))r3
where T= orbital period of the planet, G= gravitational constant, m1= mass
of the star, r= average distance between the star and the planet, and m2=
mass of the planet (assumed to be negligible).
Step 2: Given that the planet takes 40 years to complete one full orbit, we
have T= 40 years = 40 ×365 days = 14,600 days.
Step 3: Convert the average distance from astronomical units (AU) to meters
using the conversion factor: 1 AU = 1.496 ×1011 meters. Given the average
distance r= 2.5AU, we have r= 2.5×1.496×1011 meters = 3.74×1011 meters.
Step 4: Substitute the values of Tand rinto Kepler’s third law formula:
(14600)2=(4π2
G(m⊙))(3.74 ×1011)3
Step 5: Solve for m⊙, the mass of the star, in terms of solar masses (M⊙).
Step 6: After finding the mass of the star, express the result in terms of solar
masses by dividing the mass by the mass of the Sun (M⊙= 1.989 ×1030 kg).
11
Question 15
Question
State Kepler’s Third Law of Planetary Motion and provide an example calcula-
tion showing how the law can be applied to determine the period of a planet’s
orbit around the Sun.
Solution
Kepler’s Third Law of Planetary Motion states that the square of the orbital
period of a planet is proportional to the cube of the semi-major axis of its orbit.
Step 1: Let’s denote the orbital period of the planet as T(in years) and
the semi-major axis of its orbit as a(in astronomical units, AU). The equation
for Kepler’s Third Law is given as:
T2=k·a3
where kis a constant.
Step 2: We can use the data for Earth’s orbit around the Sun to find the
value of the constant k. The orbital period of Earth is approximately 1 year
and the semi-major axis of its orbit is about 1 AU. Substituting these values
into the equation:
(1)2=k·(1)3
1 = k
Step 3: Now, let’s apply Kepler’s Third Law to determine the period of
Jupiter’s orbit around the Sun. The semi-major axis of Jupiter’s orbit is ap-
proximately 5.2 AU. Substituting this value into the equation, we can solve for
the orbital period T:
T2= (1) ·(5.2)3
T2= 5.23
T2= 140.608
Step 4: Taking the square root of both sides to find T:
T=√140.608
T≈11.85 years
Therefore, the orbital period of Jupiter around the Sun is approximately
11.85 years according to Kepler’s Third Law of Planetary Motion.
Question 16
Question
A planet is in elliptical orbit around the Sun. The semi-major axis of the planet’s
orbit is 3.5 AU and its eccentricity is 0.2. Determine the semi-minor axis of the
orbit.
12
Solution
Step 1: Recall the formula to calculate the semi-minor axis of an elliptical orbit:
b=a√1−e2
where ais the semi-major axis and eis the eccentricity of the orbit.
Step 2: Substitute the given values into the formula:
b= 3.5AU ·√1−0.22
Step 3: Calculate the value inside the square root:
1−0.22= 1 −0.04 = 0.96
Step 4: Substitute the calculated value back into the formula:
b= 3.5AU ·√0.96
Step 5: Calculate the square root:
√0.96 = 0.960.5= 0.9798
Step 6: Multiply the semi-major axis by the square root:
b= 3.5AU ·0.9798 = 3.43 AU
Therefore, the semi-minor axis of the orbit is 3.43 AU.
Question 17
Question
Kepler’s second law states that a line segment joining a planet and the Sun
sweeps out equal areas during equal intervals of time. Consider a planet orbiting
a Sun in a circular orbit as shown in the figure below. If the planet sweeps out
an area of 10 square units in the first 3 days of its orbit, what area will it sweep
out in the next 5 days?
Sun Planet
r
θ
13
Solution
Let the area swept out by the planet in the next 5 days be A. Since the planet
moves in a circular orbit, the area swept out in a certain time period is given
by:
A=1
2r2∆θ
where ris the radius of the orbit and ∆θis the change in angle subtended
at the center of the circle.
Given that the planet sweeps out an area of 10 square units in the first 3
days of its orbit, we have:
10 = 1
2r2θ1
where θ1is the angle swept out in the first 3 days.
Now, we know that the time period for the next 5 days is twice that of the
first 3 days. Therefore, the angle swept out in the next 5 days, θ2, is twice that
of θ1.
Hence, 2θ1=θ2.
Substitute θ1in terms of rfrom the given information:
10 = 1
2r2θ1
θ1=20
r2
Substitute in θ2:
θ2= 2 ×20
r2=40
r2
So, the area swept out by the planet in the next 5 days is:
A=1
2r2θ2=1
2r2×40
r2= 20 square units
Question 18
Question
Consider a planet with a semi-major axis of 2.5 AU that orbits a star with a
period of 5 years. Calculate the period of a hypothetical moon that orbits this
planet at a distance of 0.8 AU.
14
Solution
To solve this problem, we can use Kepler’s third law of planetary motion, which
states that the square of the period of an orbiting body is proportional to the
cube of the semi-major axis of its orbit. Mathematically, this can be expressed
as: T2
1
a3
1
=T2
2
a3
2
where: - T1and a1are the period and semi-major axis of the planet, - T2
and a2are the period and semi-major axis of the moon.
Given that the period of the planet (T1) is 5 years, the semi-major axis of
the planet (a1) is 2.5 AU, and the semi-major axis of the moon (a2) is 0.8 AU,
we can substitute these values into the formula to find the period of the moon
(T2).
Step 1: Substitute the known values into the formula.
52
2.53=T2
2
0.83
Step 2: Simplify the equation.
25
15.625 =T2
2
0.512
1.6 = T2
2
0.512
Step 3: Solve for T2.
T2
2= 1.6×0.512
T2
2= 0.8192
Step 4: Find T2by taking the square root.
T2=√0.8192
T2≈0.9041
Therefore, the period of the hypothetical moon orbiting the planet at a
distance of 0.8 AU is approximately 0.9041 years.
Question 19
Question
In the context of Kepler’s Laws of Planetary Motion, explain the significance of
Kepler’s Third Law and how it relates to the period and semi-major axis of a
planet’s orbit.
15
Solution
Step 1: Kepler’s Third Law states that the square of the period of revolution of
a planet around the Sun is directly proportional to the cube of the semi-major
axis of its orbit.
Step 2: Mathematically, Kepler’s Third Law can be represented as:
T2
1
a3
1
=T2
2
a3
2
where: T1and T2are the periods of revolution of the planet in orbits with
semi-major axes a1and a2, respectively.
Step 3: This law essentially implies that planets farther away from the Sun
(larger semi-major axis) take longer to complete one orbit (larger period) com-
pared to planets closer to the Sun (smaller semi-major axis).
Step 4: For example, if we compare the Earth and Mars, with respective
semi-major axes of approximately 1 AU and 1.52 AU, we can apply Kepler’s
Third Law to find the ratio of their periods:
T2
Earth
13=T2
Mars
1.523
Step 5: Solving for the ratio of the periods gives:
TMars
TEarth
=√1.523
1≈1.88
This indicates that Mars takes about 1.88 times longer than Earth to com-
plete one orbit around the Sun, which matches observations.
Question 20
Question
Suppose a planet orbits a star in an elliptical orbit, with the star located at one of
the foci of the ellipse. If the planet is at its closest point to the star (perihelion),
which of Kepler’s laws of planetary motion can be used to determine its velocity
at that point? Justify your answer.
Solution
Step 1: Kepler’s laws of planetary motion state: 1. The orbit of a planet around
the Sun is an ellipse with the Sun at one of the two foci. 2. A line joining a
planet and the Sun sweeps out equal areas during 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: When the planet is at its closest point to the star (perihelion), it is
moving fastest in its orbit due to the conservation of angular momentum.
16
Step 3: Therefore, at perihelion, we can use Kepler’s Second Law to deter-
mine the planet’s velocity. This law states that a line connecting the planet to
the star sweeps out equal areas in equal times.
Step 4: Since the planet moves fastest at perihelion, the area swept out in
a given time interval is greatest at this point. This implies that the planet
covers the arc from perihelion to the next point within the elliptical orbit in the
shortest amount of time, resulting in the highest velocity at perihelion.
Step 5: In conclusion, at perihelion, the velocity of a planet can be deter-
mined using Kepler’s Second Law of planetary motion.
Question 21
Question
Consider a hypothetical planetary system where a planet orbits a star at a
distance of 2.5AU with an orbital period of 3.5years. If another planet is
located at a distance of 5AU from the same star, what is the orbital period of
the second planet?
Solution
Step 1: Apply Kepler’s Third Law of Planetary Motion, which 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. Let T1be the orbital period of the first planet
and a1be the semi-major axis of the first planet’s orbit. Similarly, let T2be the
orbital period of the second planet and a2be the semi-major axis of the second
planet’s orbit.
Step 2: Use Kepler’s Third Law formula:
T2
1
T2
2
=a3
1
a3
2
Step 3: Substitute the given values into the formula:
3.52
T2
2
=2.53
53
Step 4: Solve for T2:
3.52
T2
2
=2.53
53
T2
2=53·3.52
2.53
T2
2=875
15.625
T2
2= 56
17
T2=√56
T2≈7.48 years
Therefore, the orbital period of the second planet is approximately 7.48
years.
Question 22
Question
Kepler’s third law of planetary motion relates the orbital period of a planet to
its mean distance from the sun. For a certain planet, the mean distance from
the sun is 2.5 AU (astronomical units). If the orbital period of this planet is 4.5
years, determine the mass of the sun. (Use G= 6.67 ×10−11 m3kg−1s−2)
Solution
Step 1: Recall Kepler’s third law of planetary motion, which states:
T2
a3=4π2
GM
where: T= orbital period of the planet, a= mean distance from the sun, G=
universal gravitational constant, M= mass of the sun.
Step 2: Substitute the given values into the formula:
(4.5years)2
(2.5AU)3=4π2
G·M
Step 3: Convert the orbital period from years to seconds (since Gis given
in SI units):
T= 4.5years ×(365.25 days/year)×(24 hours/day)×(3600 s/hour)
Step 4: Convert the mean distance from AU to meters:
1AU = 1.496 ×1011 m
a= 2.5AU ×1.496 ×1011 m/AU
Step 5: Substitute the converted values into the formula and solve for the
mass of the sun, M.
Question 23
Question
One star has a mass of 2×1030 kg and another star has a mass of 5×1030 kg.
If the first star is 10 times as bright as the second star, what is the ratio of their
luminosities?
18
Solution
Step 1: Use the definition of luminosity. The luminosity of a star is directly
proportional to its mass. We can express the luminosity Lof a star in terms of
its mass Mas L∝M. Therefore, the ratio of the luminosities of the two stars
can be expressed as:
L1
L2
=M1
M2
Step 2: Plug in the given values. Given that the mass of the first star (M1)
is 2×1030 kg, the mass of the second star (M2) is 5×1030 kg, and the first star
is 10 times as bright as the second star, we have:
L1
L2
=2×1030 kg
5×1030 kg
Step 3: Simplify the ratio.
L1
L2
=2
5= 0.4
Therefore, the ratio of the luminosities of the two stars is 0.4.
Question 24
Question
The semi-major axis of Mercury’s orbit around the Sun is 0.387 AU. If the semi-
major axis of Venus’s orbit around the Sun is 0.723 AU, what is the ratio of the
squares of their orbital periods?
Solution
Step 1: Recall Kepler’s Third Law for planetary motion, which 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. Mathematically, we can express this as:
T2
1
T2
2
=a3
1
a3
2
where T1and T2are the orbital periods of the two planets, and a1and a2
are the semi-major axes of their orbits.
Step 2: We are given that the semi-major axis of Mercury’s orbit (a1) is 0.387
AU and the semi-major axis of Venus’s orbit (a2) is 0.723 AU. Let’s substitute
these values into the formula:
T2
1
T2
2
=0.3873
0.7233
Step 3: Calculate the ratio of the squares of the orbital periods:
19
T2
1
T2
2
=0.056187
0.373243 ≈0.15044
Therefore, the ratio of the squares of the orbital periods of Mercury and
Venus is approximately 0.15044.
Question 25
Question
Kepler’s Third Law states that the square of the period of revolution of a planet
around the Sun is directly proportional to the cube of its average distance from
the Sun. Suppose a newly discovered planet has an average distance from the
Sun that is 4 times farther than Earth’s average distance. If Earth takes 365
days to orbit the Sun, how long does it take for this new planet to orbit the
Sun?
Solution
Let Tbe the period of revolution of the new planet around the Sun in days and
rbe its average distance from the Sun in astronomical units (AU). We are given
that r= 4 AU and the period of revolution for Earth is TEarth = 365 days.
According to Kepler’s Third Law, we have the following equation:
(T
TEarth )2
=(r
1)3
Step 1: Plug in the given values into the equation.
(T
365)2
=(4
1)3
Step 2: Square both sides of the equation.
(T
365)2
= 64
Step 3: Take the square root of both sides to solve for T.
T
365 =√64 = 8
Step 4: Multiply both sides by 365 to solve for T.
T= 8 ×365 = 2920
Therefore, the period of revolution for the new planet is 2920 days, which is
8 times longer than Earth’s period.
20
