PHSC 121 - INTRODUCTION TO
ASTRONOMY - Kepler’s Laws of
Planetary Motion
Question Bank - Set 8
Liberty University
Question 1
Question
Given that a planet has an elliptical orbit around the Sun, with the Sun located
at one of the foci, determine whether the following statements regarding the
planet’s motion are true or false:
1. The planet moves fastest when it is closest to the Sun.
2. The line connecting the planet to the Sun sweeps out equal areas in equal
times.
Solution
To determine whether the statements are true or false, we can refer to Kepler’s
Laws of Planetary Motion.
Statement 1: The planet moves fastest when it is closest to the Sun.
Step 1: According to Kepler’s 2nd Law, a planet moves fastest when it is
closest to the Sun and slowest when it is farthest from the Sun. Therefore, this
statement is true.
Statement 2: The line connecting the planet to the Sun sweeps out equal
areas in equal times.
Step 1: According to Kepler’s 2nd Law, the line connecting the planet to
the Sun sweeps out equal areas in equal times. This is because a planet moves
faster when it is closer to the Sun and slower when it is farther away, allowing
it to sweep out equal areas in equal times. Therefore, this statement is true.
In conclusion, both statements are true based on Kepler’s Laws of Planetary
Motion.
Question 2
Question
A planet follows an elliptical orbit around the Sun with a semi-major axis of
2.5 AU. If the planet’s closest approach to the Sun (perihelion) is at 0.5 AU,
determine the distance of the planet from the Sun when it is at its farthest point
(aphelion).
Solution
Step 1: Recall that the semi-major axis (a) of an ellipse is the average of the
closest and farthest distances from the center to the edge. In this case, the semi-
major axis is given as 2.5 AU, and the closest approach to the Sun (perihelion)
is at 0.5 AU. Therefore, the farthest distance (aphelion) is given by 2.5 AU -
0.5 AU = 2 AU.
Therefore, the distance of the planet from the Sun when it is at its farthest
point (aphelion) is 2 AU.
Question 3
Question
According to Kepler’s laws 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 new planet, Planet X, has an orbital period of 10 years and an
average distance from the sun of 5 AU (astronomical units). If another planet,
Planet Y, has an orbital period of 5 years, what is its average distance from the
sun?
Solution
Step 1: Let’s denote the orbital period of Planet Y as TYand its average distance
from the sun as dY. According to Kepler’s third law:
T2
X=k×d3
X
T2
Y=k×d3
Y
where TX= 10 years, dX= 5 AU, and kis the proportionality constant.
Step 2: Plug in the values for Planet X to find the proportionality constant:
102=k×53
100 = k×125
k=100
125 = 0.8
2
Step 3: Now, use the proportionality constant to find the average distance
from the sun of Planet Y:
52= 0.8×d3
Y
25 = 0.8×d3
Y
d3
Y=25
0.8= 31.25
dY=3
√31.25
dY≈3.32 AU
Therefore, the average distance from the sun of Planet Y is approximately
3.32 astronomical units.
Question 4
Question
A planet has an elliptical orbit with an eccentricity of 0.3. If the distance be-
tween the planet and the sun at its closest approach (perihelion) is 0.65 AU,
what is the distance between the planet and the sun at its farthest point (aphe-
lion)? Use Kepler’s laws of planetary motion to solve this problem.
Solution
Step 1: Recall that Kepler’s first law states that planets move in elliptical orbits
with the sun at one of the foci of the ellipse.
Step 2: Given that the eccentricity of the planet’s orbit is 0.3 and the distance
at perihelion is 0.65 AU, we can calculate the distance at aphelion using the
formula:
distance at aphelion =distance at perihelion
1−eccentricity
Step 3: Substitute the given values into the formula to find the distance at
aphelion:
distance at aphelion =0.65
1−0.3
Step 4: Calculate the distance at aphelion:
distance at aphelion =0.65
0.7= 0.9286 AU
Step 5: Therefore, the distance between the planet and the sun at its farthest
point (aphelion) is 0.9286 AU.
3
Question 5
Question
Given that a planet has an orbital period of 11.86 years, calculate the semi-
major axis of its orbit in astronomical units (AU). (Assume the planet’s orbit
is nearly circular.)
Solution
Step 1: Recall Kepler’s Third Law, which relates the orbital period of a planet
to the semi-major axis of its orbit:
T2=(4π2
G(M+m))a3
where: T= orbital period of the planet (in years), G= gravitational constant
(6.67430 ×10−11 m3kg−1s−2), M= mass of the Sun (1.989 ×1030 kg), m=
mass of the planet (assumed to be much smaller than the Sun’s mass), a=
semi-major axis of the planet’s orbit (in meters).
Step 2: The semi-major axis acan be calculated from the given orbital
period Tusing the formula:
a=(T2·G(M+m)
4π2)1/3
Step 3: Given that the orbital period T= 11.86 years, we can substitute
this value into the formula to find the semi-major axis a. Let’s calculate:
a=((11.86 years)2·6.67430 ×10−11 m3kg−1s−2·1.989 ×1030 kg
4π2)1/3
Step 4: Perform the calculations to find the value of the semi-major axis a
in meters.
Step 5: Convert the semi-major axis afrom meters to astronomical units (1
AU = 1.496 ×1011 meters) to express the answer in a more convenient unit.
Step 6: The calculated value will give the semi-major axis of the planet’s
orbit in astronomical units.
Question 6
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is directly proportional to the cube of the semi-major axis of
its orbit. Suppose a newly discovered planet has an orbital period of 12 years
and a semi-major axis of 3 AU (astronomical units). If another planet has a
semi-major axis of 6 AU, what is its orbital period in years?
4
Solution
Step 1: Let T1be the orbital period of the first planet in years and a1be the
semi-major axis of its orbit in AU. Let T2be the orbital period of the second
planet in years and a2be the semi-major axis of its orbit in AU.
According to Kepler’s third law:
(T1
T2)2
=(a1
a2)3
Step 2: Given that T1= 12 years, a1= 3 AU, and a2= 6 AU, we can
substitute these values into the equation:
(12
T2)2
=(3
6)3
Step 3: Simplifying the equation gives:
(12
T2)2
=(1
2)3
(12
T2)2
=1
8
Step 4: Taking the square root of both sides gives:
12
T2
=1
√8
12
T2
=1
2√2
Step 5: Solving for T2:
T2=12 ×2√2
1
T2= 24√2
Therefore, the orbital period of the second planet is 24√2years.
Question 7
Question
Kepler’s third law of planetary motion states that the square of the orbital pe-
riod of a planet is proportional to the cube of its semi-major axis. Suppose
a newly discovered exoplanet has an orbital period of 300 days. If the orbital
period of Earth is 365.25 days and its average distance from the Sun is 1 as-
tronomical unit (AU), what is the semi-major axis of the exoplanet’s orbit in
AU?
5
Solution
Step 1: Understand the relationship described by Kepler’s third law: Kepler’s
third law states that the ratio of the square of the orbital period of a planet
(T) to the cube of its semi-major axis (a) is constant for all planets in the solar
system. Mathematically, we can express this as:
T2
1
a3
1
=T2
2
a3
2
where T1and T2are the orbital periods of two planets, and a1and a2are their
semi-major axes.
Step 2: Calculate the ratio for Earth: Given that the orbital period of Earth
(T1) is 365.25 days and its semi-major axis (a1) is 1 AU, we have:
365.252
13=133225.06
1= 133225.06
Step 3: Find the semi-major axis of the exoplanet: We can now use the ratio
calculated for Earth to solve for the semi-major axis of the exoplanet (a2) when
its orbital period (T2) is 300 days:
3002
a3
2
= 133225.06
Step 4: Solve for a2:
3002= 133225.06 ×a3
2
90000 = 133225.06 ×a3
2
a3
2=90000
133225.06 ≈0.675
a2≈3
√0.675 ≈0.87 AU
Therefore, the semi-major axis of the exoplanet’s orbit is approximately 0.87
astronomical units (AU).
Question 8
Question
Given an exoplanet with a semi-major axis of 1.5 AU, determine the exoplanet’s
orbital period in Earth years. Assume the exoplanet’s orbit is circular.
6
Solution
Step 1: Recall Kepler’s Third Law: T2=4π2
G(M1+M2)a3, where: - T= orbital
period (in seconds) - G= gravitational constant, 6.67430 ×10−11 m3kg−1s−2
-M1and M2= masses of the two objects (in this case, we are assuming the
planet is much smaller than the star so we can neglect the planet’s mass) - a=
semi-major axis of the orbit
Step 2: Convert the semi-major axis from Astronomical Units (AU) to me-
ters. 1 AU is approximately 1.496 ×1011 meters.
a= 1.5AU ×1.496 ×1011 meters/AU = 2.244 ×1011 meters
Step 3: Plug in the numbers and solve for the orbital period:
T2=4π2
6.67430 ×10−11 ·M⊙
(2.244 ×1011)3
Step 4: Solve for T:
T=√4π2
6.67430 ×10−11 ·M⊙
(2.244 ×1011)3
Step 5: Convert the orbital period from seconds to Earth years. 1 Earth
year is approximately 3.154 ×107seconds.
Tyears =Tseconds
3.154 ×107
Question 9
Question
Consider a planet in a circular orbit around the Sun with a radius of 1 AU.
Given that the period of the planet’s orbit is 2 years, calculate the mass of the
Sun. (Assume the gravitational constant G= 6.67 ×10−11 m3/kg ·s2.)
Solution
Step 1: Recall Kepler’s Third Law, which states that the square of the period
of any planet is proportional to the cube of the semi-major axis of its orbit. Let
Tbe the period of the planet’s orbit, abe the semi-major axis of the planet’s
orbit, and Mbe the mass of the Sun. Then, the formula from Kepler’s Third
Law can be expressed as:
T2=4π2
GM a3
Step 2: Given that T= 2 years and a= 1 AU, we can rewrite the equation
using these values:
(2 years)2=4π2
G·M·(1 AU)3
7
Step 3: Convert the units to SI units. Since 1 year is equal to 3.15 ×107
seconds, and 1 AU is equal to 1.496 ×1011 meters, we substitute these values
into the equation:
(2 ×3.15 ×107s)2=4π2
G·M·(1.496 ×1011 m)3
Step 4: Solve for the mass of the Sun, M, by isolating it in the equation:
M=4π2
G·(1.496 ×1011 m
2×3.15 ×107s2)3
Step 5: Now, calculate the mass of the Sun using the given values for G,T,
and a:
M=4π2
6.67 ×10−11 m3/kg ·s2·(1.496 ×1011 m
2×3.15 ×107s2)3
Step 6: Calculate the mass of the Sun using the above expression and ensure
that the final answer is in kilograms.
Question 10
Question
An exoplanet orbits a star in an elliptical orbit. The semi-major axis of the
orbit is 3 AU and the eccentricity of the orbit is 0.5. Determine the distance of
the exoplanet from the star when it is at its closest point (perihelion) and when
it is at its farthest point (aphelion).
Solution
Step 1: Recall the formula for the distance from a focus to a point on an ellipse
given its semi-major axis aand eccentricity e:
r=a(1 −e)
where ris the distance from the focus to the point on the ellipse.
Step 2: Substitute the given values into the formula to find the distance at
perihelion:
rperihelion = 3 AU(1 −0.5)
rperihelion = 3 AU(0.5)
rperihelion = 1.5AU
Therefore, the exoplanet is 1.5 AU away from the star at perihelion.
Step 3: Now, calculate the distance at aphelion using the same formula:
raphelion = 3 AU(1 + 0.5)
raphelion = 3 AU(1.5)
raphelion = 4.5AU
Hence, the exoplanet is 4.5 AU away from the star at aphelion.
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Question 11
Question
Suppose a planet has an orbital period of 8 years and an average distance from
the sun of 2.5 AU. Determine the mass of the sun in terms of its mass compared
to Earth’s mass (M⊙).
Solution
Step 1: We can use Kepler’s Third Law to relate the orbital period (T) of a
planet to its average distance from the sun (a). Kepler’s Third Law is given by
the equation:
T2=4π2
G(M⊙+Mplanet)a3
where: T= orbital period of the planet (in years), G= gravitational constant
(6.67 ×10−11 N m2/kg2), M⊙= mass of the sun (in kg), Mplanet = mass of the
planet (we will neglect this compared to the sun’s mass), a= average distance
from the sun (in astronomical units, AU).
Step 2: We are given that the orbital period of the planet is 8 years and the
average distance from the sun is 2.5 AU. Substituting these values into Kepler’s
Third Law, we have:
(8)2=4π2
G(M⊙)(2.5)3
Step 3: Simplifying the equation, we get:
64 = 4π2
G(M⊙)(15.625)
Step 4: Next, we isolate M⊙by rearranging the equation:
M⊙=4π2×15.625
64 ×G
Step 5: Now, we substitute the values G= 6.67 ×10−11 N m2/kg2and solve
for M⊙:
M⊙=4π2×15.625
64 ×6.67 ×10−11
Step 6: Calculating the expression, we find:
M⊙≈1.03 ×1030 kg
Therefore, the mass of the sun is approximately 1.03 ×1030 kg, or about
333,000 times the mass of Earth.
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Question 12
Question
Consider a hypothetical solar system consisting of a star with a mass of 2×1030
kg and a planet with a mass of 6×1024 kg. The planet orbits the star at a
distance of 3×1011 meters. Calculate the orbital period of the planet in this
system according to Kepler’s third law of planetary motion.
Solution
To find the orbital period of the planet in the given system, we can use Kepler’s
third law of planetary motion, which 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: Find the gravitational constant The universal gravitational
constant, G, is 6.67430 ×10−11 m3kg−1s−2.
Step 2: Calculate the combined mass of the system The combined
mass of the star and the planet is M=Mstar +Mplanet = 2 ×1030 kg + 6 ×
1024 kg = 2.00006 ×1030 kg.
Step 3: Calculate the semi-major axis of the planet’s orbit The
semi-major axis of the planet’s orbit, a, is given as 3×1011 m.
Step 4: Use Kepler’s third law to find the orbital period According
to Kepler’s third law, we have:
T2=(4π2
G(M))a3
T2=(4π2
6.67430 ×10−11 )(2.00006 ×1030)(3 ×1011)3
Now, calculate the orbital period Tby taking the square root of the right-
hand side of the equation.
Step 5: Calculate the orbital period
T=√(4π2
6.67430 ×10−11 )(2.00006 ×1030) (3 ×1011)3
After this calculation, you will find the orbital period of the planet in this
system.
Question 13
Question
Consider a planet with a semimajor axis of 2 AU orbiting a star with a mass
of 2×1030 kg. If the eccentricity of the planet’s orbit is 0.5, determine the
following: (a) The distance of closest approach of the planet to the star. (b)
The distance of farthest separation of the planet from the star.
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Solution
(a) To find the distance of closest approach of the planet to the star, we need
to determine the perihelion distance. The perihelion distance can be calculated
using the formula:
rperihelion =a(1 −ε)
where ais the semimajor axis and εis the eccentricity.
Step 1: Calculate the perihelion distance:
rperihelion = 2 AU ×(1 −0.5) = 2 AU ×0.5 = 1 AU
Therefore, the distance of closest approach of the planet to the star is 1 AU.
(b) To find the distance of farthest separation of the planet from the star,
we need to determine the aphelion distance. The aphelion distance can be
calculated using the formula:
raphelion =a(1 + ε)
Step 2: Calculate the aphelion distance:
raphelion = 2 AU ×(1 + 0.5) = 2 AU ×1.5 = 3 AU
Therefore, the distance of farthest separation of the planet from the star is
3 AU.
Question 14
Question
Kepler’s Third Law of Planetary Motion states that the square of the period of
revolution of a planet is directly proportional to the cube of its average distance
from the Sun. Given that the period of Venus is 0.62 Earth years and its average
distance from the Sun is 0.723 AU (Astronomical Units), determine the period
of revolution of Mars, whose average distance from the Sun is 1.524 AU.
Solution
Step 1: Let’s denote the period of revolution of Mars as T(in Earth years) and
its average distance from the Sun as r(in AU). According to Kepler’s Third
Law, we have the following proportion:
T2
r3=T2
Venus
r3
Venus
Step 2: Substituting the given values, we have:
T2
1.5243=0.622
0.7233
11
Step 3: Solving for T, we get:
T2=0.622×1.5243
0.7233
Step 4: Simplifying the expression, we find:
T2≈0.3844 ×3.7221
0.3732 ≈1.4328
0.3732 ≈3.8393
Step 5: Taking the square root of both sides, we get:
T≈√3.8393 ≈1.959 Earth years
Step 6: Therefore, the period of revolution of Mars is approximately 1.959
Earth years.
Question 15
Question
Consider a hypothetical planetary system where Planet X orbits a star in an
elliptical orbit. The distance between the star and Planet X at its closest ap-
proach is 0.3 Astronomical Units (AU) and at its farthest point is 0.7 AU. If
the time taken for Planet X to complete one full orbit is 2.5 years, determine
the eccentricity of the orbit.
Solution
Step 1: Recall Kepler’s Laws of Planetary Motion, specifically Kepler’s First
Law. It states that all planets move in elliptical orbits with the Sun at one of
the foci.
Step 2: In an elliptical orbit, the distance from the center of the ellipse to
its farthest point (major axis) is the semi-major axis, a, and the distance from
the center to the closest point (minor axis) is the semi-minor axis, b.
Step 3: The eccentricity of an ellipse, denoted by e, is given by the formula:
e=distance between the foci
length of the major axis
Step 4: In this case, the distance between the foci is equal to the difference
between the closest and farthest distances from the star to Planet X, which is
0.7 AU - 0.3 AU = 0.4 AU.
Step 5: The length of the major axis is the sum of the closest and farthest
distances, divided by 2. So, the major axis 2a= 0.7AU + 0.3AU = 1 AU.
Step 6: Therefore, the eccentricity eis:
e=0.4AU
1AU = 0.4
Step 7: Thus, the eccentricity of the orbit of Planet X is 0.4.
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Question 16
Question
An exoplanet orbits a star with a semi-major axis of 0.7 AU. If the period of
the exoplanet’s orbit is 260 days, determine the approximate mass of the star
in solar masses (assuming a circular orbit).
Solution
To solve this problem, we can use Kepler’s Third Law, which states:
T2
a3=4π2
G(M1+M2)
Where: - T= period of the orbit (in seconds), - a= semi-major axis of the
orbit (in meters), - G= gravitational constant (6.67 ×10−11 m3/kg ·s2), - M1
= mass of the star (in kg), - M2= mass of the exoplanet (negligible compared
to the star).
Converting the period to seconds and the semi-major axis to meters, we
have:
T= 260 days×24 hours/day×60 minutes/hour×60 seconds/minute = 22,464,000 seconds
a= 0.7AU ×1.496 ×1011 meters/AU = 1.0442 ×1011 meters
Now we can substitute the values into Kepler’s Third Law:
(22,464,000 s)2
(1.0442 ×1011 m)3=4π2
6.67 ×10−11 m3/kg ·s2(M1)
5.034 ×1014 s2
1.237 ×1033 m3=4π2
6.67 ×10−11 (M1)
4.07 ×10−19 = 7.444 ×1010 ×M1
Solving for M1, we get:
M1≈4.07 ×10−19
7.444 ×1010 ≈5.46 ×10−30 kg
Finally, we can convert the mass of the star from kilograms to solar masses
by dividing by the mass of the Sun:
M1≈5.46 ×10−30 kg
1.989 ×1030 kg/solar mass ≈2.75 ×10−60 solar masses
So, the approximate mass of the star in solar masses is 2.75 ×10−60.
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Question 17
Question
Kepler’s third law of planetary motion states that the square of the orbital
period of a planet is proportional to the cube of its semi-major axis. Suppose a
planet has an orbital period of 4 years and a semi-major axis of 2 astronomical
units (AU). If another planet has an orbital period of 16 years, what is the
semi-major axis of its orbit in AU?
Solution
Step 1: Let’s denote the orbital period of the second planet as T2and the semi-
major axis of its orbit as a2. According to Kepler’s third law, we have the
following relation:
(T1
T2)2
=(a1
a2)3
Step 2: Substituting the given values for the first planet (T1= 4 years,
a1= 2 AU) into the equation, we get:
(4
16)2
=(2
a2)3
Step 3: Simplifying the left side of the equation gives us:
(1
4)2
=(2
a2)3
1
16 =(2
a2)3
Step 4: Taking the cube root of both sides to solve for a2, we have:
3
√1
16 =2
a2
Step 5: Simplifying the cube root gives:
1
2=2
a2
Step 6: Cross multiplying yields:
a2= 4 AU
Therefore, the semi-major axis of the second planet’s orbit is 4 AU.
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Question 18
Question
A planet orbits a star in an elliptical orbit. The planet’s closest approach to the
star is 0.3 AU and its furthest distance from the star is 0.7 AU. If the period of
the planet’s orbit is 1 year, what is the eccentricity of the planet’s orbit?
Solution
Step 1: Recall the formula for the eccentricity (e) of an elliptical orbit:
e=ra−rp
ra+rp
where rais the aphelion distance (furthest distance from the star) and rpis the
perihelion distance (closest approach to the star).
Step 2: Substitute ra= 0.7AU and rp= 0.3AU into the formula:
e=0.7−0.3
0.7+0.3
Step 3: Simplify the expression:
e=0.4
1= 0.4
Step 4: Therefore, the eccentricity of the planet’s orbit is 0.4.
Question 19
Question
According to Kepler’s Third Law of Planetary Motion, the square of the period
of revolution of a planet (T) is proportional to the cube of the semimajor axis
of its orbit (a). Suppose a new planet is discovered with a semimajor axis of 2.5
AU (astronomical units). If the period of revolution of this planet is 7.5 years,
find the period of revolution of a different planet with a semimajor axis of 5 AU.
Solution
Step 1: Let T1and a1represent the period and semimajor axis of the newly dis-
covered planet, and let T2and a2represent the unknown period and semimajor
axis of the different planet, respectively. According to Kepler’s Third Law, we
have the equation:
T2
1
a3
1
=T2
2
a3
2
15
Step 2: Substituting the values T1= 7.5years and a1= 2.5AU into the
equation, we get:
7.52
2.53=T2
2
53
Step 3: Solving for T2
2, we have:
7.52
2.53·53=T2
2
T2
2= 225
Step 4: Taking the square root of both sides, we find the period of revolution
of the different planet:
T2=√225
T2= 15
Therefore, the period of revolution of the different planet with a semimajor
axis of 5 AU is 15 years.
Question 20
Question
An asteroid orbits around the Sun in an elliptical path such that its closest
distance to the Sun (perihelion) is 0.8 AU and its farthest distance from the
Sun (aphelion) is 2.4 AU. Calculate the eccentricity of the asteroid’s orbit.
Solution
Step 1: Recall the definition of eccentricity for an elliptical orbit. The eccen-
tricity of an ellipse is a measure of how ”stretched out” the ellipse is, with a
value between 0 and 1 given by the formula:
e=rmax −rmin
rmax +rmin
where eis the eccentricity, rmax is the distance from the center of the ellipse
to the farthest point along the major axis (aphelion), and rmin is the distance
from the center to the closest point along the major axis (perihelion).
Step 2: Substitute the given values into the eccentricity formula. In this
case, rmin = 0.8AU and rmax = 2.4AU:
e=2.4−0.8
2.4+0.8
e=1.6
3.2
e= 0.5
Step 3: Therefore, the eccentricity of the asteroid’s orbit is 0.5.
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Question 21
Question
On which law of planetary motion does the following statement align: ”The
square of the orbital period of a planet is directly proportional to the cube of
the semi-major axis of its orbit”?
Solution
Step 1: Kepler’s third law of planetary motion 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.
Therefore, the given statement aligns with Kepler’s third law of planetary
motion.
Question 22
Question
In a distant solar system, a planet has an orbital period of 230 days. If the
average distance from the planet to the star is 0.8 AU, determine the mass of
the star in terms of solar masses. (1 AU is the average distance between the
Earth and the Sun, which is approximately 1.496 ×1011 meters.)
Solution
Step 1: First, let’s find the orbital speed of the planet using Kepler’s third law,
which states: T2
r3=4π2
GM . Here, Tis the orbital period, ris the average distance,
Gis the gravitational constant, and Mis the mass of the star.
Given: Orbital period, T= 230 days Average distance, r= 0.8AU =
0.8×1.496 ×1011 meters
Step 2: Let’s convert the orbital period to seconds: T= 230 days = 230 ×
24 ×60 ×60 seconds = 19,872,000 seconds.
Step 3: Now, let’s plug in the values into the equation and solve for the
orbital speed: (19,872,000)2
(0.8×1.496×1011 )3=4π2
GM .
Step 4: Calculate the orbital speed: (19,872,000)2= 394,906,590,000,000
(0.8×1.496 ×1011)3= 2.860672
The equation becomes: 394,906,590,000,000
2,860672 =4π2
GM .
Step 5: Solve for Mto find the mass of the star: M=4π2×2.860672
394,906,590,000,000 ≈
0.003003 solar masses.
Therefore, the mass of the star in terms of solar masses is approximately
0.003003.
17
Question 23
Question
According to Kepler’s third law of planetary motion, the ratio of the square of
a planet’s orbital period (T) to the cube of its average distance from the sun
(r) is constant.
If the average distance from the sun to Earth is approximately 1 astronom-
ical unit (AU) and the orbital period of Earth is 1 year, determine the average
distance from the sun to Mars given that Mars has an orbital period of approx-
imately 1.88 years.
Solution
Step 1: Let the average distance from the sun to Mars be represented by rMars.
According to Kepler’s third law, we have:
(TEarth
r3
Earth )=(TMars
r3
Mars )
Step 2: Substitute the values for Earth’s orbital period TEarth, Earth’s av-
erage distance from the sun rEarth, and Mars’s orbital period TMars into the
equation:
(1year
1AU3)=(1.88 years
r3
Mars )
Step 3: Simplify the equation by squaring the values on the left side:
(1
1)2
=(1.88
r3
Mars )
1 = 1.88
r3
Mars
Step 4: Solve for rMars:
r3
Mars = 1.88
rMars =3
√1.88 ≈1.24 AU
Therefore, the average distance from the sun to Mars is approximately 1.24
astronomical units.
Question 24
Question
According to Kepler’s Third Law of Planetary Motion, the square of the period
of a planet’s orbit is proportional to the cube of its average distance from the
Sun. Consider two planets, Planet A and Planet B, with orbital periods of 5
18
years and 10 years, respectively. If Planet A is located 1 AU from the Sun, how
far is Planet B from the Sun? (1 AU is the average distance between the Earth
and the Sun.)
Solution
To find the average distance of Planet B from the Sun, we can use Kepler’s
Third Law of Planetary Motion:
T2
A
r3
A
=T2
B
r3
B
where TAand rAare the period and average distance of Planet A, and TB
and rBare the period and average distance of Planet B, respectively.
Given that TA= 5 years, rA= 1 AU, and TB= 10 years, we can substitute
these values into the equation and solve for rB:
52
13=102
r3
B
25 = 100
r3
B
25r3
B= 100
r3
B=100
25 = 4
rB=3
√4 = 2
The average distance of Planet B from the Sun is 2 AU.
Question 25
Question
An extrasolar planet, Kepler-186f, has an orbital period of approximately 129.9
Earth days. If Kepler-186f is located at a distance of 0.36 AU from its host star,
determine the approximate mass of the star in terms of solar masses. (Note: 1
AU is the average distance from the Earth to the Sun, approximately 1.5×1011
meters, and the mass of the Sun is approximately 2×1030 kg.)
19
Solution
Step 1: Calculate the orbital radius of Kepler-186f in meters. Given that the
distance of Kepler-186f from its host star is 0.36 AU, we can calculate the
distance in meters:
(0.36 AU)×(1.5×1011 m/AU) = 5.4×1010 m
Step 2: Use Kepler’s Third Law to determine the star’s mass. Kepler’s Third
Law states that for any planet, the square of its orbital period is proportional
to the cube of its semi-major axis (orbital radius). This can be represented as:
T2=4π2
G(M⋆+Mp)a3
where: - Tis the orbital period of the planet, - Gis the gravitational constant,
-M⋆is the mass of the host star, - Mpis the mass of the planet, - ais the
semi-major axis of the planet’s orbit.
Since the mass of the planet is much smaller than the mass of the star, we
can consider M⋆≫Mp.
Substitute T= 129.9Earth days (converted to seconds) and a= 5.4×
1010 m into Kepler’s Third Law and solve for M⋆:
(129.9×24 ×3600)2=4π2
G
a3
M⋆
(129.9×24 ×3600)2×M⋆=4π2
G×a3
M⋆=4π2
G×a3∇ · (129.9×24 ×3600)2
Step 3: Calculate the mass of the star in terms of solar masses. Substitute
G= 6.674 ×10−11 m3kg−1s−2into the equation and convert the mass into
solar masses:
M⋆=4π2
6.674 ×10−11 ×(5.4×1010)3∇ · (129.9×24 ×3600)2
M⋆≈0.48 solar masses
20
Question 2
Question
A planet follows an elliptical orbit around the Sun with a semi-major axis of
2.5 AU. If the planet’s closest approach to the Sun (perihelion) is at 0.5 AU,
determine the distance of the planet from the Sun when it is at its farthest point
(aphelion).
Solution
Step 1: Recall that the semi-major axis (a) of an ellipse is the average of the
closest and farthest distances from the center to the edge. In this case, the semi-
major axis is given as 2.5 AU, and the closest approach to the Sun (perihelion)
is at 0.5 AU. Therefore, the farthest distance (aphelion) is given by 2.5 AU -
0.5 AU = 2 AU.
Therefore, the distance of the planet from the Sun when it is at its farthest
point (aphelion) is 2 AU.
Question 3
Question
According to Kepler’s laws 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 new planet, Planet X, has an orbital period of 10 years and an
average distance from the sun of 5 AU (astronomical units). If another planet,
Planet Y, has an orbital period of 5 years, what is its average distance from the
sun?
Solution
Step 1: Let’s denote the orbital period of Planet Y as TYand its average distance
from the sun as dY. According to Kepler’s third law:
T2
X=k×d3
X
T2
Y=k×d3
Y
where TX= 10 years, dX= 5 AU, and kis the proportionality constant.
Step 2: Plug in the values for Planet X to find the proportionality constant:
102=k×53
100 = k×125
k=100
125 = 0.8
2
Step 3: Now, use the proportionality constant to find the average distance
from the sun of Planet Y:
52= 0.8×d3
Y
25 = 0.8×d3
Y
d3
Y=25
0.8= 31.25
dY=3
√31.25
dY≈3.32 AU
Therefore, the average distance from the sun of Planet Y is approximately
3.32 astronomical units.
Question 4
Question
A planet has an elliptical orbit with an eccentricity of 0.3. If the distance be-
tween the planet and the sun at its closest approach (perihelion) is 0.65 AU,
what is the distance between the planet and the sun at its farthest point (aphe-
lion)? Use Kepler’s laws of planetary motion to solve this problem.
Solution
Step 1: Recall that Kepler’s first law states that planets move in elliptical orbits
with the sun at one of the foci of the ellipse.
Step 2: Given that the eccentricity of the planet’s orbit is 0.3 and the distance
at perihelion is 0.65 AU, we can calculate the distance at aphelion using the
formula:
distance at aphelion =distance at perihelion
1−eccentricity
Step 3: Substitute the given values into the formula to find the distance at
aphelion:
distance at aphelion =0.65
1−0.3
Step 4: Calculate the distance at aphelion:
distance at aphelion =0.65
0.7= 0.9286 AU
Step 5: Therefore, the distance between the planet and the sun at its farthest
point (aphelion) is 0.9286 AU.
3
Question 5
Question
Given that a planet has an orbital period of 11.86 years, calculate the semi-
major axis of its orbit in astronomical units (AU). (Assume the planet’s orbit
is nearly circular.)
Solution
Step 1: Recall Kepler’s Third Law, which relates the orbital period of a planet
to the semi-major axis of its orbit:
T2=(4π2
G(M+m))a3
where: T= orbital period of the planet (in years), G= gravitational constant
(6.67430 ×10−11 m3kg−1s−2), M= mass of the Sun (1.989 ×1030 kg), m=
mass of the planet (assumed to be much smaller than the Sun’s mass), a=
semi-major axis of the planet’s orbit (in meters).
Step 2: The semi-major axis acan be calculated from the given orbital
period Tusing the formula:
a=(T2·G(M+m)
4π2)1/3
Step 3: Given that the orbital period T= 11.86 years, we can substitute
this value into the formula to find the semi-major axis a. Let’s calculate:
a=((11.86 years)2·6.67430 ×10−11 m3kg−1s−2·1.989 ×1030 kg
4π2)1/3
Step 4: Perform the calculations to find the value of the semi-major axis a
in meters.
Step 5: Convert the semi-major axis afrom meters to astronomical units (1
AU = 1.496 ×1011 meters) to express the answer in a more convenient unit.
Step 6: The calculated value will give the semi-major axis of the planet’s
orbit in astronomical units.
Question 6
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is directly proportional to the cube of the semi-major axis of
its orbit. Suppose a newly discovered planet has an orbital period of 12 years
and a semi-major axis of 3 AU (astronomical units). If another planet has a
semi-major axis of 6 AU, what is its orbital period in years?
4
Solution
Step 1: Let T1be the orbital period of the first planet in years and a1be the
semi-major axis of its orbit in AU. Let T2be the orbital period of the second
planet in years and a2be the semi-major axis of its orbit in AU.
According to Kepler’s third law:
(T1
T2)2
=(a1
a2)3
Step 2: Given that T1= 12 years, a1= 3 AU, and a2= 6 AU, we can
substitute these values into the equation:
(12
T2)2
=(3
6)3
Step 3: Simplifying the equation gives:
(12
T2)2
=(1
2)3
(12
T2)2
=1
8
Step 4: Taking the square root of both sides gives:
12
T2
=1
√8
12
T2
=1
2√2
Step 5: Solving for T2:
T2=12 ×2√2
1
T2= 24√2
Therefore, the orbital period of the second planet is 24√2years.
Question 7
Question
Kepler’s third law of planetary motion states that the square of the orbital pe-
riod of a planet is proportional to the cube of its semi-major axis. Suppose
a newly discovered exoplanet has an orbital period of 300 days. If the orbital
period of Earth is 365.25 days and its average distance from the Sun is 1 as-
tronomical unit (AU), what is the semi-major axis of the exoplanet’s orbit in
AU?
5
Solution
Step 1: Understand the relationship described by Kepler’s third law: Kepler’s
third law states that the ratio of the square of the orbital period of a planet
(T) to the cube of its semi-major axis (a) is constant for all planets in the solar
system. Mathematically, we can express this as:
T2
1
a3
1
=T2
2
a3
2
where T1and T2are the orbital periods of two planets, and a1and a2are their
semi-major axes.
Step 2: Calculate the ratio for Earth: Given that the orbital period of Earth
(T1) is 365.25 days and its semi-major axis (a1) is 1 AU, we have:
365.252
13=133225.06
1= 133225.06
Step 3: Find the semi-major axis of the exoplanet: We can now use the ratio
calculated for Earth to solve for the semi-major axis of the exoplanet (a2) when
its orbital period (T2) is 300 days:
3002
a3
2
= 133225.06
Step 4: Solve for a2:
3002= 133225.06 ×a3
2
90000 = 133225.06 ×a3
2
a3
2=90000
133225.06 ≈0.675
a2≈3
√0.675 ≈0.87 AU
Therefore, the semi-major axis of the exoplanet’s orbit is approximately 0.87
astronomical units (AU).
Question 8
Question
Given an exoplanet with a semi-major axis of 1.5 AU, determine the exoplanet’s
orbital period in Earth years. Assume the exoplanet’s orbit is circular.
6
Solution
Step 1: Recall Kepler’s Third Law: T2=4π2
G(M1+M2)a3, where: - T= orbital
period (in seconds) - G= gravitational constant, 6.67430 ×10−11 m3kg−1s−2
-M1and M2= masses of the two objects (in this case, we are assuming the
planet is much smaller than the star so we can neglect the planet’s mass) - a=
semi-major axis of the orbit
Step 2: Convert the semi-major axis from Astronomical Units (AU) to me-
ters. 1 AU is approximately 1.496 ×1011 meters.
a= 1.5AU ×1.496 ×1011 meters/AU = 2.244 ×1011 meters
Step 3: Plug in the numbers and solve for the orbital period:
T2=4π2
6.67430 ×10−11 ·M⊙
(2.244 ×1011)3
Step 4: Solve for T:
T=√4π2
6.67430 ×10−11 ·M⊙
(2.244 ×1011)3
Step 5: Convert the orbital period from seconds to Earth years. 1 Earth
year is approximately 3.154 ×107seconds.
Tyears =Tseconds
3.154 ×107
Question 9
Question
Consider a planet in a circular orbit around the Sun with a radius of 1 AU.
Given that the period of the planet’s orbit is 2 years, calculate the mass of the
Sun. (Assume the gravitational constant G= 6.67 ×10−11 m3/kg ·s2.)
Solution
Step 1: Recall Kepler’s Third Law, which states that the square of the period
of any planet is proportional to the cube of the semi-major axis of its orbit. Let
Tbe the period of the planet’s orbit, abe the semi-major axis of the planet’s
orbit, and Mbe the mass of the Sun. Then, the formula from Kepler’s Third
Law can be expressed as:
T2=4π2
GM a3
Step 2: Given that T= 2 years and a= 1 AU, we can rewrite the equation
using these values:
(2 years)2=4π2
G·M·(1 AU)3
7
Step 3: Convert the units to SI units. Since 1 year is equal to 3.15 ×107
seconds, and 1 AU is equal to 1.496 ×1011 meters, we substitute these values
into the equation:
(2 ×3.15 ×107s)2=4π2
G·M·(1.496 ×1011 m)3
Step 4: Solve for the mass of the Sun, M, by isolating it in the equation:
M=4π2
G·(1.496 ×1011 m
2×3.15 ×107s2)3
Step 5: Now, calculate the mass of the Sun using the given values for G,T,
and a:
M=4π2
6.67 ×10−11 m3/kg ·s2·(1.496 ×1011 m
2×3.15 ×107s2)3
Step 6: Calculate the mass of the Sun using the above expression and ensure
that the final answer is in kilograms.
Question 10
Question
An exoplanet orbits a star in an elliptical orbit. The semi-major axis of the
orbit is 3 AU and the eccentricity of the orbit is 0.5. Determine the distance of
the exoplanet from the star when it is at its closest point (perihelion) and when
it is at its farthest point (aphelion).
Solution
Step 1: Recall the formula for the distance from a focus to a point on an ellipse
given its semi-major axis aand eccentricity e:
r=a(1 −e)
where ris the distance from the focus to the point on the ellipse.
Step 2: Substitute the given values into the formula to find the distance at
perihelion:
rperihelion = 3 AU(1 −0.5)
rperihelion = 3 AU(0.5)
rperihelion = 1.5AU
Therefore, the exoplanet is 1.5 AU away from the star at perihelion.
Step 3: Now, calculate the distance at aphelion using the same formula:
raphelion = 3 AU(1 + 0.5)
raphelion = 3 AU(1.5)
raphelion = 4.5AU
Hence, the exoplanet is 4.5 AU away from the star at aphelion.
8
Question 11
Question
Suppose a planet has an orbital period of 8 years and an average distance from
the sun of 2.5 AU. Determine the mass of the sun in terms of its mass compared
to Earth’s mass (M⊙).
Solution
Step 1: We can use Kepler’s Third Law to relate the orbital period (T) of a
planet to its average distance from the sun (a). Kepler’s Third Law is given by
the equation:
T2=4π2
G(M⊙+Mplanet)a3
where: T= orbital period of the planet (in years), G= gravitational constant
(6.67 ×10−11 N m2/kg2), M⊙= mass of the sun (in kg), Mplanet = mass of the
planet (we will neglect this compared to the sun’s mass), a= average distance
from the sun (in astronomical units, AU).
Step 2: We are given that the orbital period of the planet is 8 years and the
average distance from the sun is 2.5 AU. Substituting these values into Kepler’s
Third Law, we have:
(8)2=4π2
G(M⊙)(2.5)3
Step 3: Simplifying the equation, we get:
64 = 4π2
G(M⊙)(15.625)
Step 4: Next, we isolate M⊙by rearranging the equation:
M⊙=4π2×15.625
64 ×G
Step 5: Now, we substitute the values G= 6.67 ×10−11 N m2/kg2and solve
for M⊙:
M⊙=4π2×15.625
64 ×6.67 ×10−11
Step 6: Calculating the expression, we find:
M⊙≈1.03 ×1030 kg
Therefore, the mass of the sun is approximately 1.03 ×1030 kg, or about
333,000 times the mass of Earth.
9
Question 12
Question
Consider a hypothetical solar system consisting of a star with a mass of 2×1030
kg and a planet with a mass of 6×1024 kg. The planet orbits the star at a
distance of 3×1011 meters. Calculate the orbital period of the planet in this
system according to Kepler’s third law of planetary motion.
Solution
To find the orbital period of the planet in the given system, we can use Kepler’s
third law of planetary motion, which 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: Find the gravitational constant The universal gravitational
constant, G, is 6.67430 ×10−11 m3kg−1s−2.
Step 2: Calculate the combined mass of the system The combined
mass of the star and the planet is M=Mstar +Mplanet = 2 ×1030 kg + 6 ×
1024 kg = 2.00006 ×1030 kg.
Step 3: Calculate the semi-major axis of the planet’s orbit The
semi-major axis of the planet’s orbit, a, is given as 3×1011 m.
Step 4: Use Kepler’s third law to find the orbital period According
to Kepler’s third law, we have:
T2=(4π2
G(M))a3
T2=(4π2
6.67430 ×10−11 )(2.00006 ×1030)(3 ×1011)3
Now, calculate the orbital period Tby taking the square root of the right-
hand side of the equation.
Step 5: Calculate the orbital period
T=√(4π2
6.67430 ×10−11 )(2.00006 ×1030) (3 ×1011)3
After this calculation, you will find the orbital period of the planet in this
system.
Question 13
Question
Consider a planet with a semimajor axis of 2 AU orbiting a star with a mass
of 2×1030 kg. If the eccentricity of the planet’s orbit is 0.5, determine the
following: (a) The distance of closest approach of the planet to the star. (b)
The distance of farthest separation of the planet from the star.
10
Solution
(a) To find the distance of closest approach of the planet to the star, we need
to determine the perihelion distance. The perihelion distance can be calculated
using the formula:
rperihelion =a(1 −ε)
where ais the semimajor axis and εis the eccentricity.
Step 1: Calculate the perihelion distance:
rperihelion = 2 AU ×(1 −0.5) = 2 AU ×0.5 = 1 AU
Therefore, the distance of closest approach of the planet to the star is 1 AU.
(b) To find the distance of farthest separation of the planet from the star,
we need to determine the aphelion distance. The aphelion distance can be
calculated using the formula:
raphelion =a(1 + ε)
Step 2: Calculate the aphelion distance:
raphelion = 2 AU ×(1 + 0.5) = 2 AU ×1.5 = 3 AU
Therefore, the distance of farthest separation of the planet from the star is
3 AU.
Question 14
Question
Kepler’s Third Law of Planetary Motion states that the square of the period of
revolution of a planet is directly proportional to the cube of its average distance
from the Sun. Given that the period of Venus is 0.62 Earth years and its average
distance from the Sun is 0.723 AU (Astronomical Units), determine the period
of revolution of Mars, whose average distance from the Sun is 1.524 AU.
Solution
Step 1: Let’s denote the period of revolution of Mars as T(in Earth years) and
its average distance from the Sun as r(in AU). According to Kepler’s Third
Law, we have the following proportion:
T2
r3=T2
Venus
r3
Venus
Step 2: Substituting the given values, we have:
T2
1.5243=0.622
0.7233
11
Step 3: Solving for T, we get:
T2=0.622×1.5243
0.7233
Step 4: Simplifying the expression, we find:
T2≈0.3844 ×3.7221
0.3732 ≈1.4328
0.3732 ≈3.8393
Step 5: Taking the square root of both sides, we get:
T≈√3.8393 ≈1.959 Earth years
Step 6: Therefore, the period of revolution of Mars is approximately 1.959
Earth years.
Question 15
Question
Consider a hypothetical planetary system where Planet X orbits a star in an
elliptical orbit. The distance between the star and Planet X at its closest ap-
proach is 0.3 Astronomical Units (AU) and at its farthest point is 0.7 AU. If
the time taken for Planet X to complete one full orbit is 2.5 years, determine
the eccentricity of the orbit.
Solution
Step 1: Recall Kepler’s Laws of Planetary Motion, specifically Kepler’s First
Law. It states that all planets move in elliptical orbits with the Sun at one of
the foci.
Step 2: In an elliptical orbit, the distance from the center of the ellipse to
its farthest point (major axis) is the semi-major axis, a, and the distance from
the center to the closest point (minor axis) is the semi-minor axis, b.
Step 3: The eccentricity of an ellipse, denoted by e, is given by the formula:
e=distance between the foci
length of the major axis
Step 4: In this case, the distance between the foci is equal to the difference
between the closest and farthest distances from the star to Planet X, which is
0.7 AU - 0.3 AU = 0.4 AU.
Step 5: The length of the major axis is the sum of the closest and farthest
distances, divided by 2. So, the major axis 2a= 0.7AU + 0.3AU = 1 AU.
Step 6: Therefore, the eccentricity eis:
e=0.4AU
1AU = 0.4
Step 7: Thus, the eccentricity of the orbit of Planet X is 0.4.
12
Question 16
Question
An exoplanet orbits a star with a semi-major axis of 0.7 AU. If the period of
the exoplanet’s orbit is 260 days, determine the approximate mass of the star
in solar masses (assuming a circular orbit).
Solution
To solve this problem, we can use Kepler’s Third Law, which states:
T2
a3=4π2
G(M1+M2)
Where: - T= period of the orbit (in seconds), - a= semi-major axis of the
orbit (in meters), - G= gravitational constant (6.67 ×10−11 m3/kg ·s2), - M1
= mass of the star (in kg), - M2= mass of the exoplanet (negligible compared
to the star).
Converting the period to seconds and the semi-major axis to meters, we
have:
T= 260 days×24 hours/day×60 minutes/hour×60 seconds/minute = 22,464,000 seconds
a= 0.7AU ×1.496 ×1011 meters/AU = 1.0442 ×1011 meters
Now we can substitute the values into Kepler’s Third Law:
(22,464,000 s)2
(1.0442 ×1011 m)3=4π2
6.67 ×10−11 m3/kg ·s2(M1)
5.034 ×1014 s2
1.237 ×1033 m3=4π2
6.67 ×10−11 (M1)
4.07 ×10−19 = 7.444 ×1010 ×M1
Solving for M1, we get:
M1≈4.07 ×10−19
7.444 ×1010 ≈5.46 ×10−30 kg
Finally, we can convert the mass of the star from kilograms to solar masses
by dividing by the mass of the Sun:
M1≈5.46 ×10−30 kg
1.989 ×1030 kg/solar mass ≈2.75 ×10−60 solar masses
So, the approximate mass of the star in solar masses is 2.75 ×10−60.
13
Question 17
Question
Kepler’s third law of planetary motion states that the square of the orbital
period of a planet is proportional to the cube of its semi-major axis. Suppose a
planet has an orbital period of 4 years and a semi-major axis of 2 astronomical
units (AU). If another planet has an orbital period of 16 years, what is the
semi-major axis of its orbit in AU?
Solution
Step 1: Let’s denote the orbital period of the second planet as T2and the semi-
major axis of its orbit as a2. According to Kepler’s third law, we have the
following relation:
(T1
T2)2
=(a1
a2)3
Step 2: Substituting the given values for the first planet (T1= 4 years,
a1= 2 AU) into the equation, we get:
(4
16)2
=(2
a2)3
Step 3: Simplifying the left side of the equation gives us:
(1
4)2
=(2
a2)3
1
16 =(2
a2)3
Step 4: Taking the cube root of both sides to solve for a2, we have:
3
√1
16 =2
a2
Step 5: Simplifying the cube root gives:
1
2=2
a2
Step 6: Cross multiplying yields:
a2= 4 AU
Therefore, the semi-major axis of the second planet’s orbit is 4 AU.
14
Question 18
Question
A planet orbits a star in an elliptical orbit. The planet’s closest approach to the
star is 0.3 AU and its furthest distance from the star is 0.7 AU. If the period of
the planet’s orbit is 1 year, what is the eccentricity of the planet’s orbit?
Solution
Step 1: Recall the formula for the eccentricity (e) of an elliptical orbit:
e=ra−rp
ra+rp
where rais the aphelion distance (furthest distance from the star) and rpis the
perihelion distance (closest approach to the star).
Step 2: Substitute ra= 0.7AU and rp= 0.3AU into the formula:
e=0.7−0.3
0.7+0.3
Step 3: Simplify the expression:
e=0.4
1= 0.4
Step 4: Therefore, the eccentricity of the planet’s orbit is 0.4.
Question 19
Question
According to Kepler’s Third Law of Planetary Motion, the square of the period
of revolution of a planet (T) is proportional to the cube of the semimajor axis
of its orbit (a). Suppose a new planet is discovered with a semimajor axis of 2.5
AU (astronomical units). If the period of revolution of this planet is 7.5 years,
find the period of revolution of a different planet with a semimajor axis of 5 AU.
Solution
Step 1: Let T1and a1represent the period and semimajor axis of the newly dis-
covered planet, and let T2and a2represent the unknown period and semimajor
axis of the different planet, respectively. According to Kepler’s Third Law, we
have the equation:
T2
1
a3
1
=T2
2
a3
2
15
Step 2: Substituting the values T1= 7.5years and a1= 2.5AU into the
equation, we get:
7.52
2.53=T2
2
53
Step 3: Solving for T2
2, we have:
7.52
2.53·53=T2
2
T2
2= 225
Step 4: Taking the square root of both sides, we find the period of revolution
of the different planet:
T2=√225
T2= 15
Therefore, the period of revolution of the different planet with a semimajor
axis of 5 AU is 15 years.
Question 20
Question
An asteroid orbits around the Sun in an elliptical path such that its closest
distance to the Sun (perihelion) is 0.8 AU and its farthest distance from the
Sun (aphelion) is 2.4 AU. Calculate the eccentricity of the asteroid’s orbit.
Solution
Step 1: Recall the definition of eccentricity for an elliptical orbit. The eccen-
tricity of an ellipse is a measure of how ”stretched out” the ellipse is, with a
value between 0 and 1 given by the formula:
e=rmax −rmin
rmax +rmin
where eis the eccentricity, rmax is the distance from the center of the ellipse
to the farthest point along the major axis (aphelion), and rmin is the distance
from the center to the closest point along the major axis (perihelion).
Step 2: Substitute the given values into the eccentricity formula. In this
case, rmin = 0.8AU and rmax = 2.4AU:
e=2.4−0.8
2.4+0.8
e=1.6
3.2
e= 0.5
Step 3: Therefore, the eccentricity of the asteroid’s orbit is 0.5.
16
Question 21
Question
On which law of planetary motion does the following statement align: ”The
square of the orbital period of a planet is directly proportional to the cube of
the semi-major axis of its orbit”?
Solution
Step 1: Kepler’s third law of planetary motion 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.
Therefore, the given statement aligns with Kepler’s third law of planetary
motion.
Question 22
Question
In a distant solar system, a planet has an orbital period of 230 days. If the
average distance from the planet to the star is 0.8 AU, determine the mass of
the star in terms of solar masses. (1 AU is the average distance between the
Earth and the Sun, which is approximately 1.496 ×1011 meters.)
Solution
Step 1: First, let’s find the orbital speed of the planet using Kepler’s third law,
which states: T2
r3=4π2
GM . Here, Tis the orbital period, ris the average distance,
Gis the gravitational constant, and Mis the mass of the star.
Given: Orbital period, T= 230 days Average distance, r= 0.8AU =
0.8×1.496 ×1011 meters
Step 2: Let’s convert the orbital period to seconds: T= 230 days = 230 ×
24 ×60 ×60 seconds = 19,872,000 seconds.
Step 3: Now, let’s plug in the values into the equation and solve for the
orbital speed: (19,872,000)2
(0.8×1.496×1011 )3=4π2
GM .
Step 4: Calculate the orbital speed: (19,872,000)2= 394,906,590,000,000
(0.8×1.496 ×1011)3= 2.860672
The equation becomes: 394,906,590,000,000
2,860672 =4π2
GM .
Step 5: Solve for Mto find the mass of the star: M=4π2×2.860672
394,906,590,000,000 ≈
0.003003 solar masses.
Therefore, the mass of the star in terms of solar masses is approximately
0.003003.
17
Question 23
Question
According to Kepler’s third law of planetary motion, the ratio of the square of
a planet’s orbital period (T) to the cube of its average distance from the sun
(r) is constant.
If the average distance from the sun to Earth is approximately 1 astronom-
ical unit (AU) and the orbital period of Earth is 1 year, determine the average
distance from the sun to Mars given that Mars has an orbital period of approx-
imately 1.88 years.
Solution
Step 1: Let the average distance from the sun to Mars be represented by rMars.
According to Kepler’s third law, we have:
(TEarth
r3
Earth )=(TMars
r3
Mars )
Step 2: Substitute the values for Earth’s orbital period TEarth, Earth’s av-
erage distance from the sun rEarth, and Mars’s orbital period TMars into the
equation:
(1year
1AU3)=(1.88 years
r3
Mars )
Step 3: Simplify the equation by squaring the values on the left side:
(1
1)2
=(1.88
r3
Mars )
1 = 1.88
r3
Mars
Step 4: Solve for rMars:
r3
Mars = 1.88
rMars =3
√1.88 ≈1.24 AU
Therefore, the average distance from the sun to Mars is approximately 1.24
astronomical units.
Question 24
Question
According to Kepler’s Third Law of Planetary Motion, the square of the period
of a planet’s orbit is proportional to the cube of its average distance from the
Sun. Consider two planets, Planet A and Planet B, with orbital periods of 5
18
years and 10 years, respectively. If Planet A is located 1 AU from the Sun, how
far is Planet B from the Sun? (1 AU is the average distance between the Earth
and the Sun.)
Solution
To find the average distance of Planet B from the Sun, we can use Kepler’s
Third Law of Planetary Motion:
T2
A
r3
A
=T2
B
r3
B
where TAand rAare the period and average distance of Planet A, and TB
and rBare the period and average distance of Planet B, respectively.
Given that TA= 5 years, rA= 1 AU, and TB= 10 years, we can substitute
these values into the equation and solve for rB:
52
13=102
r3
B
25 = 100
r3
B
25r3
B= 100
r3
B=100
25 = 4
rB=3
√4 = 2
The average distance of Planet B from the Sun is 2 AU.
Question 25
Question
An extrasolar planet, Kepler-186f, has an orbital period of approximately 129.9
Earth days. If Kepler-186f is located at a distance of 0.36 AU from its host star,
determine the approximate mass of the star in terms of solar masses. (Note: 1
AU is the average distance from the Earth to the Sun, approximately 1.5×1011
meters, and the mass of the Sun is approximately 2×1030 kg.)
19
Solution
Step 1: Calculate the orbital radius of Kepler-186f in meters. Given that the
distance of Kepler-186f from its host star is 0.36 AU, we can calculate the
distance in meters:
(0.36 AU)×(1.5×1011 m/AU) = 5.4×1010 m
Step 2: Use Kepler’s Third Law to determine the star’s mass. Kepler’s Third
Law states that for any planet, the square of its orbital period is proportional
to the cube of its semi-major axis (orbital radius). This can be represented as:
T2=4π2
G(M⋆+Mp)a3
where: - Tis the orbital period of the planet, - Gis the gravitational constant,
-M⋆is the mass of the host star, - Mpis the mass of the planet, - ais the
semi-major axis of the planet’s orbit.
Since the mass of the planet is much smaller than the mass of the star, we
can consider M⋆≫Mp.
Substitute T= 129.9Earth days (converted to seconds) and a= 5.4×
1010 m into Kepler’s Third Law and solve for M⋆:
(129.9×24 ×3600)2=4π2
G
a3
M⋆
(129.9×24 ×3600)2×M⋆=4π2
G×a3
M⋆=4π2
G×a3∇ · (129.9×24 ×3600)2
Step 3: Calculate the mass of the star in terms of solar masses. Substitute
G= 6.674 ×10−11 m3kg−1s−2into the equation and convert the mass into
solar masses:
M⋆=4π2
6.674 ×10−11 ×(5.4×1010)3∇ · (129.9×24 ×3600)2
M⋆≈0.48 solar masses
20
Question 2
Question
A planet follows an elliptical orbit around the Sun with a semi-major axis of
2.5 AU. If the planet’s closest approach to the Sun (perihelion) is at 0.5 AU,
determine the distance of the planet from the Sun when it is at its farthest point
(aphelion).
Solution
Step 1: Recall that the semi-major axis (a) of an ellipse is the average of the
closest and farthest distances from the center to the edge. In this case, the semi-
major axis is given as 2.5 AU, and the closest approach to the Sun (perihelion)
is at 0.5 AU. Therefore, the farthest distance (aphelion) is given by 2.5 AU -
0.5 AU = 2 AU.
Therefore, the distance of the planet from the Sun when it is at its farthest
point (aphelion) is 2 AU.
Question 3
Question
According to Kepler’s laws 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 new planet, Planet X, has an orbital period of 10 years and an
average distance from the sun of 5 AU (astronomical units). If another planet,
Planet Y, has an orbital period of 5 years, what is its average distance from the
sun?
Solution
Step 1: Let’s denote the orbital period of Planet Y as TYand its average distance
from the sun as dY. According to Kepler’s third law:
T2
X=k×d3
X
T2
Y=k×d3
Y
where TX= 10 years, dX= 5 AU, and kis the proportionality constant.
Step 2: Plug in the values for Planet X to find the proportionality constant:
102=k×53
100 = k×125
k=100
125 = 0.8
2
Step 3: Now, use the proportionality constant to find the average distance
from the sun of Planet Y:
52= 0.8×d3
Y
25 = 0.8×d3
Y
d3
Y=25
0.8= 31.25
dY=3
√31.25
dY≈3.32 AU
Therefore, the average distance from the sun of Planet Y is approximately
3.32 astronomical units.
Question 4
Question
A planet has an elliptical orbit with an eccentricity of 0.3. If the distance be-
tween the planet and the sun at its closest approach (perihelion) is 0.65 AU,
what is the distance between the planet and the sun at its farthest point (aphe-
lion)? Use Kepler’s laws of planetary motion to solve this problem.
Solution
Step 1: Recall that Kepler’s first law states that planets move in elliptical orbits
with the sun at one of the foci of the ellipse.
Step 2: Given that the eccentricity of the planet’s orbit is 0.3 and the distance
at perihelion is 0.65 AU, we can calculate the distance at aphelion using the
formula:
distance at aphelion =distance at perihelion
1−eccentricity
Step 3: Substitute the given values into the formula to find the distance at
aphelion:
distance at aphelion =0.65
1−0.3
Step 4: Calculate the distance at aphelion:
distance at aphelion =0.65
0.7= 0.9286 AU
Step 5: Therefore, the distance between the planet and the sun at its farthest
point (aphelion) is 0.9286 AU.
3
Question 5
Question
Given that a planet has an orbital period of 11.86 years, calculate the semi-
major axis of its orbit in astronomical units (AU). (Assume the planet’s orbit
is nearly circular.)
Solution
Step 1: Recall Kepler’s Third Law, which relates the orbital period of a planet
to the semi-major axis of its orbit:
T2=(4π2
G(M+m))a3
where: T= orbital period of the planet (in years), G= gravitational constant
(6.67430 ×10−11 m3kg−1s−2), M= mass of the Sun (1.989 ×1030 kg), m=
mass of the planet (assumed to be much smaller than the Sun’s mass), a=
semi-major axis of the planet’s orbit (in meters).
Step 2: The semi-major axis acan be calculated from the given orbital
period Tusing the formula:
a=(T2·G(M+m)
4π2)1/3
Step 3: Given that the orbital period T= 11.86 years, we can substitute
this value into the formula to find the semi-major axis a. Let’s calculate:
a=((11.86 years)2·6.67430 ×10−11 m3kg−1s−2·1.989 ×1030 kg
4π2)1/3
Step 4: Perform the calculations to find the value of the semi-major axis a
in meters.
Step 5: Convert the semi-major axis afrom meters to astronomical units (1
AU = 1.496 ×1011 meters) to express the answer in a more convenient unit.
Step 6: The calculated value will give the semi-major axis of the planet’s
orbit in astronomical units.
Question 6
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is directly proportional to the cube of the semi-major axis of
its orbit. Suppose a newly discovered planet has an orbital period of 12 years
and a semi-major axis of 3 AU (astronomical units). If another planet has a
semi-major axis of 6 AU, what is its orbital period in years?
4
Solution
Step 1: Let T1be the orbital period of the first planet in years and a1be the
semi-major axis of its orbit in AU. Let T2be the orbital period of the second
planet in years and a2be the semi-major axis of its orbit in AU.
According to Kepler’s third law:
(T1
T2)2
=(a1
a2)3
Step 2: Given that T1= 12 years, a1= 3 AU, and a2= 6 AU, we can
substitute these values into the equation:
(12
T2)2
=(3
6)3
Step 3: Simplifying the equation gives:
(12
T2)2
=(1
2)3
(12
T2)2
=1
8
Step 4: Taking the square root of both sides gives:
12
T2
=1
√8
12
T2
=1
2√2
Step 5: Solving for T2:
T2=12 ×2√2
1
T2= 24√2
Therefore, the orbital period of the second planet is 24√2years.
Question 7
Question
Kepler’s third law of planetary motion states that the square of the orbital pe-
riod of a planet is proportional to the cube of its semi-major axis. Suppose
a newly discovered exoplanet has an orbital period of 300 days. If the orbital
period of Earth is 365.25 days and its average distance from the Sun is 1 as-
tronomical unit (AU), what is the semi-major axis of the exoplanet’s orbit in
AU?
5
Solution
Step 1: Understand the relationship described by Kepler’s third law: Kepler’s
third law states that the ratio of the square of the orbital period of a planet
(T) to the cube of its semi-major axis (a) is constant for all planets in the solar
system. Mathematically, we can express this as:
T2
1
a3
1
=T2
2
a3
2
where T1and T2are the orbital periods of two planets, and a1and a2are their
semi-major axes.
Step 2: Calculate the ratio for Earth: Given that the orbital period of Earth
(T1) is 365.25 days and its semi-major axis (a1) is 1 AU, we have:
365.252
13=133225.06
1= 133225.06
Step 3: Find the semi-major axis of the exoplanet: We can now use the ratio
calculated for Earth to solve for the semi-major axis of the exoplanet (a2) when
its orbital period (T2) is 300 days:
3002
a3
2
= 133225.06
Step 4: Solve for a2:
3002= 133225.06 ×a3
2
90000 = 133225.06 ×a3
2
a3
2=90000
133225.06 ≈0.675
a2≈3
√0.675 ≈0.87 AU
Therefore, the semi-major axis of the exoplanet’s orbit is approximately 0.87
astronomical units (AU).
Question 8
Question
Given an exoplanet with a semi-major axis of 1.5 AU, determine the exoplanet’s
orbital period in Earth years. Assume the exoplanet’s orbit is circular.
6
Solution
Step 1: Recall Kepler’s Third Law: T2=4π2
G(M1+M2)a3, where: - T= orbital
period (in seconds) - G= gravitational constant, 6.67430 ×10−11 m3kg−1s−2
-M1and M2= masses of the two objects (in this case, we are assuming the
planet is much smaller than the star so we can neglect the planet’s mass) - a=
semi-major axis of the orbit
Step 2: Convert the semi-major axis from Astronomical Units (AU) to me-
ters. 1 AU is approximately 1.496 ×1011 meters.
a= 1.5AU ×1.496 ×1011 meters/AU = 2.244 ×1011 meters
Step 3: Plug in the numbers and solve for the orbital period:
T2=4π2
6.67430 ×10−11 ·M⊙
(2.244 ×1011)3
Step 4: Solve for T:
T=√4π2
6.67430 ×10−11 ·M⊙
(2.244 ×1011)3
Step 5: Convert the orbital period from seconds to Earth years. 1 Earth
year is approximately 3.154 ×107seconds.
Tyears =Tseconds
3.154 ×107
Question 9
Question
Consider a planet in a circular orbit around the Sun with a radius of 1 AU.
Given that the period of the planet’s orbit is 2 years, calculate the mass of the
Sun. (Assume the gravitational constant G= 6.67 ×10−11 m3/kg ·s2.)
Solution
Step 1: Recall Kepler’s Third Law, which states that the square of the period
of any planet is proportional to the cube of the semi-major axis of its orbit. Let
Tbe the period of the planet’s orbit, abe the semi-major axis of the planet’s
orbit, and Mbe the mass of the Sun. Then, the formula from Kepler’s Third
Law can be expressed as:
T2=4π2
GM a3
Step 2: Given that T= 2 years and a= 1 AU, we can rewrite the equation
using these values:
(2 years)2=4π2
G·M·(1 AU)3
7
Step 3: Convert the units to SI units. Since 1 year is equal to 3.15 ×107
seconds, and 1 AU is equal to 1.496 ×1011 meters, we substitute these values
into the equation:
(2 ×3.15 ×107s)2=4π2
G·M·(1.496 ×1011 m)3
Step 4: Solve for the mass of the Sun, M, by isolating it in the equation:
M=4π2
G·(1.496 ×1011 m
2×3.15 ×107s2)3
Step 5: Now, calculate the mass of the Sun using the given values for G,T,
and a:
M=4π2
6.67 ×10−11 m3/kg ·s2·(1.496 ×1011 m
2×3.15 ×107s2)3
Step 6: Calculate the mass of the Sun using the above expression and ensure
that the final answer is in kilograms.
Question 10
Question
An exoplanet orbits a star in an elliptical orbit. The semi-major axis of the
orbit is 3 AU and the eccentricity of the orbit is 0.5. Determine the distance of
the exoplanet from the star when it is at its closest point (perihelion) and when
it is at its farthest point (aphelion).
Solution
Step 1: Recall the formula for the distance from a focus to a point on an ellipse
given its semi-major axis aand eccentricity e:
r=a(1 −e)
where ris the distance from the focus to the point on the ellipse.
Step 2: Substitute the given values into the formula to find the distance at
perihelion:
rperihelion = 3 AU(1 −0.5)
rperihelion = 3 AU(0.5)
rperihelion = 1.5AU
Therefore, the exoplanet is 1.5 AU away from the star at perihelion.
Step 3: Now, calculate the distance at aphelion using the same formula:
raphelion = 3 AU(1 + 0.5)
raphelion = 3 AU(1.5)
raphelion = 4.5AU
Hence, the exoplanet is 4.5 AU away from the star at aphelion.
8
Question 11
Question
Suppose a planet has an orbital period of 8 years and an average distance from
the sun of 2.5 AU. Determine the mass of the sun in terms of its mass compared
to Earth’s mass (M⊙).
Solution
Step 1: We can use Kepler’s Third Law to relate the orbital period (T) of a
planet to its average distance from the sun (a). Kepler’s Third Law is given by
the equation:
T2=4π2
G(M⊙+Mplanet)a3
where: T= orbital period of the planet (in years), G= gravitational constant
(6.67 ×10−11 N m2/kg2), M⊙= mass of the sun (in kg), Mplanet = mass of the
planet (we will neglect this compared to the sun’s mass), a= average distance
from the sun (in astronomical units, AU).
Step 2: We are given that the orbital period of the planet is 8 years and the
average distance from the sun is 2.5 AU. Substituting these values into Kepler’s
Third Law, we have:
(8)2=4π2
G(M⊙)(2.5)3
Step 3: Simplifying the equation, we get:
64 = 4π2
G(M⊙)(15.625)
Step 4: Next, we isolate M⊙by rearranging the equation:
M⊙=4π2×15.625
64 ×G
Step 5: Now, we substitute the values G= 6.67 ×10−11 N m2/kg2and solve
for M⊙:
M⊙=4π2×15.625
64 ×6.67 ×10−11
Step 6: Calculating the expression, we find:
M⊙≈1.03 ×1030 kg
Therefore, the mass of the sun is approximately 1.03 ×1030 kg, or about
333,000 times the mass of Earth.
9
Question 12
Question
Consider a hypothetical solar system consisting of a star with a mass of 2×1030
kg and a planet with a mass of 6×1024 kg. The planet orbits the star at a
distance of 3×1011 meters. Calculate the orbital period of the planet in this
system according to Kepler’s third law of planetary motion.
Solution
To find the orbital period of the planet in the given system, we can use Kepler’s
third law of planetary motion, which 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: Find the gravitational constant The universal gravitational
constant, G, is 6.67430 ×10−11 m3kg−1s−2.
Step 2: Calculate the combined mass of the system The combined
mass of the star and the planet is M=Mstar +Mplanet = 2 ×1030 kg + 6 ×
1024 kg = 2.00006 ×1030 kg.
Step 3: Calculate the semi-major axis of the planet’s orbit The
semi-major axis of the planet’s orbit, a, is given as 3×1011 m.
Step 4: Use Kepler’s third law to find the orbital period According
to Kepler’s third law, we have:
T2=(4π2
G(M))a3
T2=(4π2
6.67430 ×10−11 )(2.00006 ×1030)(3 ×1011)3
Now, calculate the orbital period Tby taking the square root of the right-
hand side of the equation.
Step 5: Calculate the orbital period
T=√(4π2
6.67430 ×10−11 )(2.00006 ×1030) (3 ×1011)3
After this calculation, you will find the orbital period of the planet in this
system.
Question 13
Question
Consider a planet with a semimajor axis of 2 AU orbiting a star with a mass
of 2×1030 kg. If the eccentricity of the planet’s orbit is 0.5, determine the
following: (a) The distance of closest approach of the planet to the star. (b)
The distance of farthest separation of the planet from the star.
10
Solution
(a) To find the distance of closest approach of the planet to the star, we need
to determine the perihelion distance. The perihelion distance can be calculated
using the formula:
rperihelion =a(1 −ε)
where ais the semimajor axis and εis the eccentricity.
Step 1: Calculate the perihelion distance:
rperihelion = 2 AU ×(1 −0.5) = 2 AU ×0.5 = 1 AU
Therefore, the distance of closest approach of the planet to the star is 1 AU.
(b) To find the distance of farthest separation of the planet from the star,
we need to determine the aphelion distance. The aphelion distance can be
calculated using the formula:
raphelion =a(1 + ε)
Step 2: Calculate the aphelion distance:
raphelion = 2 AU ×(1 + 0.5) = 2 AU ×1.5 = 3 AU
Therefore, the distance of farthest separation of the planet from the star is
3 AU.
Question 14
Question
Kepler’s Third Law of Planetary Motion states that the square of the period of
revolution of a planet is directly proportional to the cube of its average distance
from the Sun. Given that the period of Venus is 0.62 Earth years and its average
distance from the Sun is 0.723 AU (Astronomical Units), determine the period
of revolution of Mars, whose average distance from the Sun is 1.524 AU.
Solution
Step 1: Let’s denote the period of revolution of Mars as T(in Earth years) and
its average distance from the Sun as r(in AU). According to Kepler’s Third
Law, we have the following proportion:
T2
r3=T2
Venus
r3
Venus
Step 2: Substituting the given values, we have:
T2
1.5243=0.622
0.7233
11
Step 3: Solving for T, we get:
T2=0.622×1.5243
0.7233
Step 4: Simplifying the expression, we find:
T2≈0.3844 ×3.7221
0.3732 ≈1.4328
0.3732 ≈3.8393
Step 5: Taking the square root of both sides, we get:
T≈√3.8393 ≈1.959 Earth years
Step 6: Therefore, the period of revolution of Mars is approximately 1.959
Earth years.
Question 15
Question
Consider a hypothetical planetary system where Planet X orbits a star in an
elliptical orbit. The distance between the star and Planet X at its closest ap-
proach is 0.3 Astronomical Units (AU) and at its farthest point is 0.7 AU. If
the time taken for Planet X to complete one full orbit is 2.5 years, determine
the eccentricity of the orbit.
Solution
Step 1: Recall Kepler’s Laws of Planetary Motion, specifically Kepler’s First
Law. It states that all planets move in elliptical orbits with the Sun at one of
the foci.
Step 2: In an elliptical orbit, the distance from the center of the ellipse to
its farthest point (major axis) is the semi-major axis, a, and the distance from
the center to the closest point (minor axis) is the semi-minor axis, b.
Step 3: The eccentricity of an ellipse, denoted by e, is given by the formula:
e=distance between the foci
length of the major axis
Step 4: In this case, the distance between the foci is equal to the difference
between the closest and farthest distances from the star to Planet X, which is
0.7 AU - 0.3 AU = 0.4 AU.
Step 5: The length of the major axis is the sum of the closest and farthest
distances, divided by 2. So, the major axis 2a= 0.7AU + 0.3AU = 1 AU.
Step 6: Therefore, the eccentricity eis:
e=0.4AU
1AU = 0.4
Step 7: Thus, the eccentricity of the orbit of Planet X is 0.4.
12
Question 16
Question
An exoplanet orbits a star with a semi-major axis of 0.7 AU. If the period of
the exoplanet’s orbit is 260 days, determine the approximate mass of the star
in solar masses (assuming a circular orbit).
Solution
To solve this problem, we can use Kepler’s Third Law, which states:
T2
a3=4π2
G(M1+M2)
Where: - T= period of the orbit (in seconds), - a= semi-major axis of the
orbit (in meters), - G= gravitational constant (6.67 ×10−11 m3/kg ·s2), - M1
= mass of the star (in kg), - M2= mass of the exoplanet (negligible compared
to the star).
Converting the period to seconds and the semi-major axis to meters, we
have:
T= 260 days×24 hours/day×60 minutes/hour×60 seconds/minute = 22,464,000 seconds
a= 0.7AU ×1.496 ×1011 meters/AU = 1.0442 ×1011 meters
Now we can substitute the values into Kepler’s Third Law:
(22,464,000 s)2
(1.0442 ×1011 m)3=4π2
6.67 ×10−11 m3/kg ·s2(M1)
5.034 ×1014 s2
1.237 ×1033 m3=4π2
6.67 ×10−11 (M1)
4.07 ×10−19 = 7.444 ×1010 ×M1
Solving for M1, we get:
M1≈4.07 ×10−19
7.444 ×1010 ≈5.46 ×10−30 kg
Finally, we can convert the mass of the star from kilograms to solar masses
by dividing by the mass of the Sun:
M1≈5.46 ×10−30 kg
1.989 ×1030 kg/solar mass ≈2.75 ×10−60 solar masses
So, the approximate mass of the star in solar masses is 2.75 ×10−60.
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Question 17
Question
Kepler’s third law of planetary motion states that the square of the orbital
period of a planet is proportional to the cube of its semi-major axis. Suppose a
planet has an orbital period of 4 years and a semi-major axis of 2 astronomical
units (AU). If another planet has an orbital period of 16 years, what is the
semi-major axis of its orbit in AU?
Solution
Step 1: Let’s denote the orbital period of the second planet as T2and the semi-
major axis of its orbit as a2. According to Kepler’s third law, we have the
following relation:
(T1
T2)2
=(a1
a2)3
Step 2: Substituting the given values for the first planet (T1= 4 years,
a1= 2 AU) into the equation, we get:
(4
16)2
=(2
a2)3
Step 3: Simplifying the left side of the equation gives us:
(1
4)2
=(2
a2)3
1
16 =(2
a2)3
Step 4: Taking the cube root of both sides to solve for a2, we have:
3
√1
16 =2
a2
Step 5: Simplifying the cube root gives:
1
2=2
a2
Step 6: Cross multiplying yields:
a2= 4 AU
Therefore, the semi-major axis of the second planet’s orbit is 4 AU.
14
Question 18
Question
A planet orbits a star in an elliptical orbit. The planet’s closest approach to the
star is 0.3 AU and its furthest distance from the star is 0.7 AU. If the period of
the planet’s orbit is 1 year, what is the eccentricity of the planet’s orbit?
Solution
Step 1: Recall the formula for the eccentricity (e) of an elliptical orbit:
e=ra−rp
ra+rp
where rais the aphelion distance (furthest distance from the star) and rpis the
perihelion distance (closest approach to the star).
Step 2: Substitute ra= 0.7AU and rp= 0.3AU into the formula:
e=0.7−0.3
0.7+0.3
Step 3: Simplify the expression:
e=0.4
1= 0.4
Step 4: Therefore, the eccentricity of the planet’s orbit is 0.4.
Question 19
Question
According to Kepler’s Third Law of Planetary Motion, the square of the period
of revolution of a planet (T) is proportional to the cube of the semimajor axis
of its orbit (a). Suppose a new planet is discovered with a semimajor axis of 2.5
AU (astronomical units). If the period of revolution of this planet is 7.5 years,
find the period of revolution of a different planet with a semimajor axis of 5 AU.
Solution
Step 1: Let T1and a1represent the period and semimajor axis of the newly dis-
covered planet, and let T2and a2represent the unknown period and semimajor
axis of the different planet, respectively. According to Kepler’s Third Law, we
have the equation:
T2
1
a3
1
=T2
2
a3
2
15
Step 2: Substituting the values T1= 7.5years and a1= 2.5AU into the
equation, we get:
7.52
2.53=T2
2
53
Step 3: Solving for T2
2, we have:
7.52
2.53·53=T2
2
T2
2= 225
Step 4: Taking the square root of both sides, we find the period of revolution
of the different planet:
T2=√225
T2= 15
Therefore, the period of revolution of the different planet with a semimajor
axis of 5 AU is 15 years.
Question 20
Question
An asteroid orbits around the Sun in an elliptical path such that its closest
distance to the Sun (perihelion) is 0.8 AU and its farthest distance from the
Sun (aphelion) is 2.4 AU. Calculate the eccentricity of the asteroid’s orbit.
Solution
Step 1: Recall the definition of eccentricity for an elliptical orbit. The eccen-
tricity of an ellipse is a measure of how ”stretched out” the ellipse is, with a
value between 0 and 1 given by the formula:
e=rmax −rmin
rmax +rmin
where eis the eccentricity, rmax is the distance from the center of the ellipse
to the farthest point along the major axis (aphelion), and rmin is the distance
from the center to the closest point along the major axis (perihelion).
Step 2: Substitute the given values into the eccentricity formula. In this
case, rmin = 0.8AU and rmax = 2.4AU:
e=2.4−0.8
2.4+0.8
e=1.6
3.2
e= 0.5
Step 3: Therefore, the eccentricity of the asteroid’s orbit is 0.5.
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Question 21
Question
On which law of planetary motion does the following statement align: ”The
square of the orbital period of a planet is directly proportional to the cube of
the semi-major axis of its orbit”?
Solution
Step 1: Kepler’s third law of planetary motion 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.
Therefore, the given statement aligns with Kepler’s third law of planetary
motion.
Question 22
Question
In a distant solar system, a planet has an orbital period of 230 days. If the
average distance from the planet to the star is 0.8 AU, determine the mass of
the star in terms of solar masses. (1 AU is the average distance between the
Earth and the Sun, which is approximately 1.496 ×1011 meters.)
Solution
Step 1: First, let’s find the orbital speed of the planet using Kepler’s third law,
which states: T2
r3=4π2
GM . Here, Tis the orbital period, ris the average distance,
Gis the gravitational constant, and Mis the mass of the star.
Given: Orbital period, T= 230 days Average distance, r= 0.8AU =
0.8×1.496 ×1011 meters
Step 2: Let’s convert the orbital period to seconds: T= 230 days = 230 ×
24 ×60 ×60 seconds = 19,872,000 seconds.
Step 3: Now, let’s plug in the values into the equation and solve for the
orbital speed: (19,872,000)2
(0.8×1.496×1011 )3=4π2
GM .
Step 4: Calculate the orbital speed: (19,872,000)2= 394,906,590,000,000
(0.8×1.496 ×1011)3= 2.860672
The equation becomes: 394,906,590,000,000
2,860672 =4π2
GM .
Step 5: Solve for Mto find the mass of the star: M=4π2×2.860672
394,906,590,000,000 ≈
0.003003 solar masses.
Therefore, the mass of the star in terms of solar masses is approximately
0.003003.
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Question 23
Question
According to Kepler’s third law of planetary motion, the ratio of the square of
a planet’s orbital period (T) to the cube of its average distance from the sun
(r) is constant.
If the average distance from the sun to Earth is approximately 1 astronom-
ical unit (AU) and the orbital period of Earth is 1 year, determine the average
distance from the sun to Mars given that Mars has an orbital period of approx-
imately 1.88 years.
Solution
Step 1: Let the average distance from the sun to Mars be represented by rMars.
According to Kepler’s third law, we have:
(TEarth
r3
Earth )=(TMars
r3
Mars )
Step 2: Substitute the values for Earth’s orbital period TEarth, Earth’s av-
erage distance from the sun rEarth, and Mars’s orbital period TMars into the
equation:
(1year
1AU3)=(1.88 years
r3
Mars )
Step 3: Simplify the equation by squaring the values on the left side:
(1
1)2
=(1.88
r3
Mars )
1 = 1.88
r3
Mars
Step 4: Solve for rMars:
r3
Mars = 1.88
rMars =3
√1.88 ≈1.24 AU
Therefore, the average distance from the sun to Mars is approximately 1.24
astronomical units.
Question 24
Question
According to Kepler’s Third Law of Planetary Motion, the square of the period
of a planet’s orbit is proportional to the cube of its average distance from the
Sun. Consider two planets, Planet A and Planet B, with orbital periods of 5
18
years and 10 years, respectively. If Planet A is located 1 AU from the Sun, how
far is Planet B from the Sun? (1 AU is the average distance between the Earth
and the Sun.)
Solution
To find the average distance of Planet B from the Sun, we can use Kepler’s
Third Law of Planetary Motion:
T2
A
r3
A
=T2
B
r3
B
where TAand rAare the period and average distance of Planet A, and TB
and rBare the period and average distance of Planet B, respectively.
Given that TA= 5 years, rA= 1 AU, and TB= 10 years, we can substitute
these values into the equation and solve for rB:
52
13=102
r3
B
25 = 100
r3
B
25r3
B= 100
r3
B=100
25 = 4
rB=3
√4 = 2
The average distance of Planet B from the Sun is 2 AU.
Question 25
Question
An extrasolar planet, Kepler-186f, has an orbital period of approximately 129.9
Earth days. If Kepler-186f is located at a distance of 0.36 AU from its host star,
determine the approximate mass of the star in terms of solar masses. (Note: 1
AU is the average distance from the Earth to the Sun, approximately 1.5×1011
meters, and the mass of the Sun is approximately 2×1030 kg.)
19
Solution
Step 1: Calculate the orbital radius of Kepler-186f in meters. Given that the
distance of Kepler-186f from its host star is 0.36 AU, we can calculate the
distance in meters:
(0.36 AU)×(1.5×1011 m/AU) = 5.4×1010 m
Step 2: Use Kepler’s Third Law to determine the star’s mass. Kepler’s Third
Law states that for any planet, the square of its orbital period is proportional
to the cube of its semi-major axis (orbital radius). This can be represented as:
T2=4π2
G(M⋆+Mp)a3
where: - Tis the orbital period of the planet, - Gis the gravitational constant,
-M⋆is the mass of the host star, - Mpis the mass of the planet, - ais the
semi-major axis of the planet’s orbit.
Since the mass of the planet is much smaller than the mass of the star, we
can consider M⋆≫Mp.
Substitute T= 129.9Earth days (converted to seconds) and a= 5.4×
1010 m into Kepler’s Third Law and solve for M⋆:
(129.9×24 ×3600)2=4π2
G
a3
M⋆
(129.9×24 ×3600)2×M⋆=4π2
G×a3
M⋆=4π2
G×a3∇ · (129.9×24 ×3600)2
Step 3: Calculate the mass of the star in terms of solar masses. Substitute
G= 6.674 ×10−11 m3kg−1s−2into the equation and convert the mass into
solar masses:
M⋆=4π2
6.674 ×10−11 ×(5.4×1010)3∇ · (129.9×24 ×3600)2
M⋆≈0.48 solar masses
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