PHSC 121 - INTRODUCTION TO
ASTRONOMY - Kepler’s Laws of
Planetary Motion
Question Bank - Set 9
Liberty University
Question 1
Question
Given that a planet has an orbital period of 12 years and a semi-major axis of
3 AU, find the mass of the central star if the gravitational constant is 6.67 ×
10−11 m3kg−1s−2.
Solution
Step 1: Recall Kepler’s third law of planetary motion, which relates the orbital
period (T) and the semi-major axis (a) of a planet to the mass (M) of the
central star:
T2=(4π2a3
GM )
Step 2: Given that the orbital period Tis 12 years and the semi-major axis
ais 3 AU, convert these values to SI units:
T= 12 ×365.25 days ×24 hours/day ×3600 s/hour
T= 3.78 ×108s
a= 3 ×1.496 ×1011 m
a= 4.49 ×1011 m
Step 3: Substitute the given values into Kepler’s third law equation:
(3.78 ×108s)2=4π2×(4.49 ×1011 m)3
6.67 ×10−11 m3kg−1s−2×M
Step 4: Solve for the mass Mof the central star:
M=4π2×(4.49 ×1011 m)3
6.67 ×10−11 m3kg−1s−2×(3.78 ×108s)2
Step 5: Calculate the mass of the central star using the above formula:
M≈1.98 ×1030 kg
Therefore, the mass of the central star is approximately 1.98 ×1030 kg.
Question 2
Question
An asteroid is orbiting the Sun in an elliptical orbit. The closest point of the
asteroid’s orbit to the Sun is 0.2 AU, and the farthest point is 0.8 AU. Calculate
the eccentricity of this orbit.
Solution
Step 1: The eccentricity of an elliptical orbit is given by the formula:
e=rmax −rmin
rmax +rmin
where rmax is the distance from the focus (the Sun) to the farthest point on the
orbit, and rmin is the distance from the focus to the closest point on the orbit.
Step 2: Given that rmin = 0.2AU and rmax = 0.8AU, we can substitute
these values into the formula for eccentricity:
e=0.8−0.2
0.8+0.2
Step 3: Simplifying the expression gives:
e=0.6
1= 0.6
Step 4: Therefore, the eccentricity of the asteroid’s orbit is 0.6.
Question 3
Question
In a planetary system, a planet has an orbital period of 5 years and an average
distance from the star of 2.5 astronomical units. Determine the mass of the star
(M) in terms of the mass of the planet (m) using Kepler’s third law.
2
Solution
To solve this problem, we can use Kepler’s third law of planetary motion, which
relates the orbital period of a planet (T), the average distance from the star
(r), the mass of the planet (m), and the mass of the star (M) in the following
equation:
T2=4π2
G(M+m)r3
where: - Tis the orbital period of the planet, - ris the average distance
from the star, - mis the mass of the planet, - Mis the mass of the star, - Gis
the gravitational constant.
Step 1: Write down the given information: - T= 5 years - r= 2.5astro-
nomical units (AU) - We will solve for Min terms of m.
Step 2: Plug in the known values into Kepler’s third law equation:
(5)2=4π2
G(M+m)(2.5)3
Step 3: Simplify the equation by squaring 5on the left side and cubing 2.5
on the right side:
25 = 4π2
G(M+m)(15.625)
Step 4: Further simplification gives:
25 = 62.5π2
G(M+m)
Step 5: Rearrange the equation to solve for M:
M=62.5π2
25G−m
Therefore, the mass of the star in terms of the mass of the planet is M=62.5π2
25G−m.
Question 4
Question
Two planets are in elliptical orbits around a star. Planet A has a semi-major
axis of 2 AU and an eccentricity of 0.4, while Planet B has a semi-major axis of
3 AU and an eccentricity of 0.6. Calculate the period of orbit for each planet in
Earth years and determine which planet has the longer orbital period.
3
Solution
Step 1: Calculate the period of orbit for Planet A. Given that the semi-major
axis of Planet A is aA= 2 AU and eccentricity is eA= 0.4, we can use Kepler’s
third law of planetary motion to calculate the orbital period PAin Earth years:
PA=√4π2a3
A
G(Mstar +MA)
where Gis the gravitational constant, Mstar is the mass of the star, MAis the
mass of Planet A (which we assume to be negligible compared to the star’s
mass), and aAis the semi-major axis.
Step 2: Substitute the given values into the equation. Plugging in aA= 2
AU and eA= 0.4into the formula, we get:
PA=√4π2(2)3
G(Mstar)
Step 3: Calculate the period of orbit for Planet B. Similarly, for Planet B
with aB= 3 AU and eB= 0.6, the orbital period PBcan be calculated using
the same formula:
PB=√4π2a3
B
G(Mstar +MB)
where MBis the mass of Planet B (assumed to be negligible).
Step 4: Substitute the given values into the equation. Plugging in aB= 3
AU and eB= 0.6into the formula, we get:
PB=√4π2(3)3
G(Mstar)
Step 5: Compare the periods of orbit for Planet A and Planet B. Now,
compare the calculated values of PAand PBto determine which planet has the
longer orbital period.
Question 5
Question
Using Kepler’s third law of planetary motion, calculate the orbital period of a
satellite that is orbiting a planet at a distance of 10,000 km from the planet’s
center. Assume the mass of the planet is 5×1024 kg.
4
Solution
Step 6: Identify the known values
The known values are: - Radius of the satellite from the center of the planet,
r= 10,000 km = 10,000 ×103m. - Mass of the planet, M= 5 ×1024 kg.
Step 7: Determine the orbital period using Kepler’s third law
Kepler’s third law states:
T2=4π2
G(M+m)r3
where: - T is the orbital period, - G is the gravitational constant, 6.67 ×
10−11N m2/kg2, - M is the mass of the planet, - m is the mass of the satel-
lite (not given in this case), - r is the radius of the satellite from the center of
the planet.
Since the mass of the satellite is usually much smaller than the mass of the
planet, we can neglect it in this calculation.
Plugging in the known values:
T2=4π2
6.67 ×10−11 ×5×1024 ×(10,000 ×103)3
Step 8: Calculate the orbital period
T2=4π2×10,0003
6.67 ×5
T2=4π2×1010
33.35
T2≈3.79 ×109
Taking the square root of both sides:
T=√3.79 ×109
T≈61,564 s
Therefore, the orbital period of the satellite is approximately 61,564 seconds.
Question 6
Question
The planet Mars has an orbital period of approximately 687 Earth days. Using
Kepler’s Third Law of Planetary Motion, determine the average distance of
Mars from the Sun in astronomical units (AU). (1 AU is the average distance
between the Earth and the Sun, approximately 1.496 ×1011 meters.)
5
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 its average distance from the
Sun. Mathematically, this can be expressed as:
T2
orbit =k×r3
average
Where: Torbit = orbital period of the planet raverage = average distance of
the planet from the Sun k= constant of proportionality
Given that Mars has an orbital period of 687 Earth days and the average
distance between the Earth and the Sun is 1 AU, we can set up the proportion
as follows:
(687)2=k×r3
average
We need to solve for raverage in AU. Since the Earth’s orbital period is 1 year
(365.25 days), we can set up the following proportion:
(687)2=k×(1)3
6872=k
Substituting k= 6872into the original equation:
(687)2= 6872×r3
average
r3
average = 1
raverage = 1 AU
Therefore, the average distance of Mars from the Sun is 1 astronomical unit.
Question 7
Question
The semi-major axis of a planet’s orbit around the Sun is 3.2 AU. If the eccen-
tricity of the orbit is 0.4, calculate the distance of the planet from the Sun when
it is closest and furthest from the Sun.
Solution
Step 1: We can use Kepler’s laws to find the distance of the planet from the Sun
when it is closest and furthest. Kepler’s laws state that the semi-major axis, a,
of an orbit and the eccentricity, e, determine the distance of the planet from the
focus (Sun) at closest approach (rmin) and furthest distance (rmax) using the
formulas:
rmin =a(1 −e)
6
rmax =a(1 + e)
Step 2: Given that the semi-major axis a= 3.2AU and the eccentricity
e= 0.4, we can now substitute these values into the formulas to find rmin and
rmax.
Step 3: Calculate rmin:
rmin = 3.2(1 −0.4) = 3.2(0.6) = 1.92 AU
Step 4: Calculate rmax:
rmax = 3.2(1 + 0.4) = 3.2(1.4) = 4.48 AU
Step 5: Therefore, the distance of the planet from the Sun when it is closest
is 1.92 AU, and when it is furthest is 4.48 AU.
Question 8
Question
The period of a satellite orbiting a planet is 5 hours. If the average distance
between the satellite and the planet is 20,000 km, determine the mass of the
planet.
(Hint: You may assume the satellite’s orbit is circular.)
Solution
Step 1: Recall Kepler’s Third Law, which states:
T2=(4π2
G(m1+m2))r3
where: - Tis the period of the orbit, - Gis the gravitational constant, - m1is
the mass of the planet, - m2is the mass of the satellite, and - ris the average
distance between the center of the planet and the center of the satellite’s orbit.
Step 2: In this scenario, m1is the unknown mass of the planet, and the mass
of the satellite (m2) can be assumed to be negligible compared to the planet’s
mass. Therefore, the equation can be simplified to:
T2=4π2
Gm1
r3
Step 3: Substituting the given values, we get:
(5hours)2=4π2
G·m1
(20,000km)3
Step 4: Convert the period to seconds:
5hours = 5 ×60 ×60 seconds = 18,000 seconds
7
Step 5: Simplify the equation:
18,0002=4π2
G·m1
(20,000)3
Step 6: Solve for the mass of the planet, m1:
m1=4π2·(20,000)3
G·18,0002
Step 7: Calculate m1:
m1=4π2·8×1015
6.67 ×10−11 ·3.24 ×108
Step 8: Simplify the calculation to find the mass of the planet, m1.
Question 9
Question
Consider a hypothetical solar system where a planet orbits a star in a perfect
circle. The radius of the planet’s orbit is 5 AU (astronomical units). The
planet takes 10 years to complete one full orbit around the star. Calculate the
gravitational force between the planet and the star in this system.
Solution
Step 1: Calculate the speed of the planet in its orbit. Given that the planet
orbits in a perfect circle, we can use the formula for the speed of an object
moving in a circle:
v=2πr
T
where, v= speed of the planet, r= radius of the orbit (5 AU = 7.5×1011
meters), T= time period of the orbit (10 years = 3.15 ×108seconds).
Substitute the values into the formula:
v=2π×7.5×1011
3.15 ×108
Calculate the speed of the planet.
Step 2: Calculate the mass of the star. To calculate the gravitational force,
we need the mass of the star. Let’s assume the mass of the planet is negligible
compared to the mass of the star. Using Kepler’s Third Law:
T2
1
r3
1
=T2
2
r3
2
8
Given T1= 10 years, r1= 5 AU, and T2=r2= 1, solving for the mass of the
star.
Step 3: Calculate the gravitational force between the planet and the star.
Now, we can use Newton’s law of universal gravitation to calculate the force:
F=G×M1×M2
r2
where, F= gravitational force, G= gravitational constant (6.67×10−11 m3kg−1s−2),
M1= mass of the planet, M2= mass of the star, r= distance between the planet
and the star.
Substitute the calculated values to find the gravitational force.
Question 10
Question
Consider a planet in a circular orbit around the Sun with a period of 2 years. If
the distance between the planet and the Sun is doubled, what will be the new
period of revolution of the planet?
Solution
To solve this problem, we can use Kepler’s third law of planetary motion, which
states that the square of the period of revolution of a planet is proportional to
the cube of its average distance from the Sun.
Step 1: Calculate the initial period of revolution using the given data. Let
the initial distance between the planet and the Sun be r, and the initial period
be T. According to Kepler’s third law, we have:
T2∝r3
Given that T= 2 years, we can solve for r:
22∝r3
4∝r3
r=3
√4 = 2 AU
Step 2: Calculate the new period of revolution when the distance is doubled.
Let the new distance between the planet and the Sun be 2r. Using Kepler’s third
law again:
(T′)2∝(2r)3
(T′)2∝8r3
Since r= 2 AU, we can now solve for the new period T′:
(T′)2∝8×(23)
9
(T′)2∝8×8
(T′)2∝64
T′=√64 = 8 years
Therefore, when the distance between the planet and the Sun is doubled,
the new period of revolution of the planet will be 8 years.
Question 11
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 semi-major axis. If the
orbital period of a planet is 8 years and the semi-major axis of its orbit is 3
AU (astronomical units), determine the orbital period of another planet with a
semi-major axis of 5 AU.
Solution
Step 1: Let T1be the orbital period of the first planet and a1be its semi-major
axis. Similarly, let T2be the orbital period of the second planet and a2be its
semi-major axis. According to Kepler’s Third Law,
T2
1
a3
1
=T2
2
a3
2
Step 2: Given that T1= 8 years and a1= 3 AU, we can substitute these
values into the equation:
82
33=T2
2
53
Step 3: Simplifying the equation, we have:
64
27 =T2
2
125
Step 4: Cross multiplying gives us:
64 ×125 = 27 ×T2
2
Step 5: Solving for T2, we have:
T2
2=64 ×125
27 =8000
27
Step 6: Taking the square root of both sides gives us:
T2=√8000
27 ≈15.85 years
Therefore, the orbital period of the second planet with a semi-major axis of
5 AU is approximately 15.85 years.
10
Question 12
Question
Consider a hypothetical planetary system in which a planet has an orbit with
an eccentricity of 0.6. If the semi-major axis of the planet’s orbit is 3 AU,
determine the distance of closest approach and the distance of farthest retreat
of the planet from its parent star.
Solution
To solve this problem, we will first recall the formula for the distance of closest
approach and the distance of farthest retreat of a planet from its parent star
in terms of the semi-major axis and eccentricity of the orbit. Then we will
substitute the given values to find the required distances.
Step 1: The formula for the distance of closest approach (rmin) and the
distance of farthest retreat (rmax) from the parent star in terms of the semi-
major axis (a) and eccentricity (e) of the orbit is:
rmin =a(1 −e)
rmax =a(1 + e)
Step 2: Given that the semi-major axis a= 3 AU and the eccentricity
e= 0.6, we can calculate the distances of closest approach and farthest retreat:
rmin = 3(1 −0.6) = 3(0.4) = 1.2AU
rmax = 3(1 + 0.6) = 3(1.6) = 4.8AU
Step 3: Therefore, the distance of closest approach of the planet from its
parent star is 1.2 AU, and the distance of farthest retreat is 4.8 AU.
Question 13
Question
Suppose a planet has an elliptical orbit around the Sun with a semi-major axis
of a= 2 AU and an eccentricity of e= 0.5. If the planet is at its aphelion,
which is the farthest point from the Sun, calculate its distance from the Sun.
Given that the distance from the Sun to the Earth is 1 AU.
Solution
Step 1: Recall the formula for the distance between the Sun and the planet in
an elliptical orbit:
r=a(1 −e)
11
Step 2: Substitute a= 2 AU and e= 0.5into the formula:
r= 2(1 −0.5)
Step 3: Calculate the value of r:
r= 2 ×0.5 = 1 AU
Step 4: Since the Earth is at 1 AU from the Sun, the planet at its aphelion
is also 1 AU from the Sun.
Therefore, when the planet is at its aphelion, its distance from the Sun is 1
AU.
Question 14
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 semi-major axis. Suppose a
hypothetical planet has an orbital period of 5 years and a semi-major axis of 2
AU (astronomical units). What would be the orbital period of another planet
with a semi-major axis of 4 AU?
Solution
Step 1: Write down Kepler’s third law in mathematical form: The relationship
between the orbital period (T) and the semi-major axis (a) of a planet is given
by:
T2∝a3
Step 2: Determine the ratio of the semi-major axes for the two planets: The
ratio of the semi-major axes of the two planets is:
(a2
a1)3
=(4
2)3
= 23= 8
Step 3: Use the ratio to find the orbital period of the second planet: Since
the ratio of the semi-major axes is 8, the orbital period of the second planet can
be found using this ratio. Let T2be the orbital period of the second planet:
T2
2∝(4)3
T2
2∝64
T2=√64
T2= 8 years
Therefore, the orbital period of the second planet with a semi-major axis of
4 AU would be 8 years.
12
Question 15
Question
Kepler’s second law of planetary motion states that a planet moves fastest when
it is closest to the sun and slowest when it is farthest from the sun. Consider
a planet with an elliptical orbit around the sun. If the planet takes 100 days
to travel from its closest point to the sun (perihelion) to its farthest point
(aphelion), and 200 days to travel from aphelion back to perihelion, determine
the ratio of the planet’s speed at aphelion to its speed at perihelion.
Solution
Step 1: Recall that Kepler’s second law states that a planet sweeps out equal
areas in equal times. This implies that the planet covers the same amount of
area in the same time interval regardless of its distance from the sun.
Step 2: Let the distance from the sun to the planet at perihelion be denoted
as rpand the distance at aphelion be denoted as ra. Then according to the law
of areas, we have:
r2
p·∆θp=r2
a·∆θa
where ∆θpis the angle swept out by the planet at perihelion, and ∆θais the
angle swept out by the planet at aphelion.
Step 3: Since the planet takes 100 days to travel from perihelion to aphelion
and 200 days to travel back, we can say that the area swept out by the planet
from perihelion to aphelion is equal to the area swept out during the return
journey:
r2
p·∆θp=r2
a·∆θa
Step 4: Since the distances from the sun are given by rpand ra, and the
time taken for each journey is given, we can relate the speeds of the planet at
these points as:
Speed at perihelion =2πrp
100
Speed at aphelion =2πra
200
Step 5: Using the equation from Step 3, we can substitute for rain terms
of rpto determine the ratio of speeds at aphelion to perihelion. Let’s calculate
this ratio:
Speed at aphelion
Speed at perihelion =rp
ra
=rp
√r2
p·100
200
=rp
rp·√1
2
=√2
Therefore, the ratio of the planet’s speed at aphelion to its speed at perihelion
is √2.
13
Question 16
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period (T) of a planet is proportional to the cube of its semi-major axis (a)
for any planet orbiting the Sun. Suppose there are two planets, Planet X and
Planet Y, orbiting the Sun. Planet X has an orbital period of 4 years and a
semi-major axis of 2 AU. Planet Y has an orbital period of 9 years. What is
the semi-major axis of Planet Y in astronomical units (AU)?
Solution
Let’s denote the orbital periods of Planet X and Planet Y as TXand TY, and
their respective semi-major axes as aXand aY. According to Kepler’s third law:
T2
X
T2
Y
=a3
X
a3
Y
Step 1: Plug in the given values for Planet X:
(4)2
92=(2)3
a3
Y
Step 2: Simplify the equation:
16
81 =8
a3
Y
Step 3: Cross multiply to solve for aY:
16 ·a3
Y= 81 ·8
Step 4: Calculate the value of aY:
a3
Y=81 ·8
16 = 40.5
Step 5: Take the cube root of both sides to find aY:
aY=3
√40.5≈3.53 AU
Therefore, the semi-major axis of Planet Y is approximately 3.53 AU.
Question 17
Question
Consider a planet in a nearly circular orbit around a Sun-like star with a semi-
major axis of 2 AU. If the orbital period of the planet is 5 years, calculate the
orbital speed of the planet in km/s.
14
Solution
Step 1: We can use Kepler’s third law of planetary motion to find the orbital
speed of the planet. Kepler’s third law states that the square of the orbital
period of a planet is directly proportional to the cube of its semi-major axis.
Mathematically, this can be expressed as:
T2=k·a3
where Tis the orbital period, ais the semi-major axis, and kis a constant.
Step 2: Given that T= 5 years and a= 2 AU, we can rewrite the equation
as:
52=k·23
Solving for k:
25 = 8k=⇒k=25
8= 3.125
Step 3: Now, we can use Kepler’s second law to find the orbital speed of the
planet. Kepler’s second law states that a planet sweeps out equal areas in equal
times. Mathematically, this can be expressed as:
dA
dt =constant
Step 4: The area swept by the planet in a small amount of time (dt) is equal
to the area of a small sector of the orbit. The area of a sector of a circle is given
by:
dA =1
2r2dθ
Step 5: The distance traveled by the planet in a small amount of time is
given by the arc length of the sector:
ds =rdθ
Step 6: The orbital speed (v) of the planet is the rate at which the planet
moves along its orbit, which can be expressed as:
v=ds
dt
Step 7: Combining the equations for dA,ds, and v, we get:
v=rdθ
dt
Step 8: Since the planet travels the full circumference of its orbit in one
period (T), we have:
2πa =v·T
15
Step 9: Substituting the values of aand T, we can solve for the orbital speed
(v):
2π·2 = v·5 =⇒v=4π
5
Step 10: Finally, we can calculate the numerical value of the orbital speed:
v=4π
5≈2.513 km/s
Therefore, the orbital speed of the planet is approximately 2.513 km/s.
Question 18
Question
Consider a hypothetical planetary system where a planet orbits a star with
a semi-major axis of 3.5 AU. If the orbital period of the planet is 8.2 years,
determine the mass of the star (in solar masses) using Kepler’s third law.
Solution
To determine the mass of the star in this hypothetical planetary system, we
will use Kepler’s third law, which relates the orbital period of a planet to the
semi-major axis of its orbit and the mass of the central star.
Step 1: Recall Kepler’s third law, which can be written as:
T2
a3=4π2
G(M1+M2)
where: T= orbital period of the planet, a= semi-major axis of the planet’s
orbit, G= gravitational constant (6.674 × 10−11m3kg−1s−2), M1= mass of
the star, M2= mass of the planet.
Step 2: Substitute the given values into Kepler’s third law:
(8.2years)2
(3.5AU)3=4π2
G(M1+M2)
Step 3: Convert the units into SI units:
(8.2×365.25 ×24 ×3600 s)2
(3.5×1.496 ×1011 m)3=4π2
6.674 ×10−11(M1+M2)
Step 4: Simplify the equation and solve for the mass of the star M1:
M1=(4π2
6.674 ×10−11 )((8.2×365.25 ×24 ×3600)2
(3.5×1.496 ×1011)3)−M2
16
Step 5: Assume the mass of the planet is negligible compared to the mass
of the star, so M1≈M2and thus we have:
M1≈(4π2
6.674 ×10−11 )((8.2×365.25 ×24 ×3600)2
(3.5×1.496 ×1011)3)
Therefore, the mass of the star in solar masses is approximately equal to the
calculated value.
Question 19
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 semi-major axis. A spacecraft
is in a circular orbit around a planet with a semi-major axis of 2.5 AU. If the
spacecraft completes one orbit in 1.5 years, what would be the semi-major axis
of another planet if that planet’s spacecraft completes one orbit in 5 years?
Solution
Let’s denote the orbital period of the first planet as T1= 1.5years and the semi-
major axis of the first planet as a1= 2.5AU. We are looking for the semi-major
axis a2of the second planet.
According to Kepler’s Third Law, the ratio of the squares of the orbital
periods to the cubes of the semi-major axes is constant. Mathematically, we
have:
T2
1
a3
1
=T2
2
a3
2
Given T1= 1.5years and T2= 5 years, we can substitute these values into
the equation above:
(1.5)2
(2.5)3=(5)2
a3
2
2.25
15.625 =25
a3
2
a3
2=25 ×15.625
2.25
a3
2= 173.61
a2=3
√173.61
17
a2≈5.5AU
Therefore, the semi-major axis of another planet would be approximately
5.5 AU if the spacecraft completes one orbit in 5 years.
Question 20
Question
Examine the following table which lists the mean distances (in astronomical
units) from the sun for four different planets and their respective orbital periods
(in Earth years). Use Kepler’s third law to determine the value of the constant
k.
Planet Mean distance (AU) Orbital period (years)
Mercury 0.39 0.24
Earth 1.00 1.00
Mars 1.52 1.88
Jupiter 5.20 11.86
Solution
Step 1: Recall Kepler’s third law, 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, this can be expressed as
T2=k×a3
where Tis the orbital period of the planet, ais the mean distance of the planet
from the sun, and kis a constant.
Step 2: We can use the data provided for Earth to find the value of the
constant k. For Earth:
12=k×13
1 = k
So, k= 1.
Step 3: Now, we can find the constant kfor the other planets using the
information provided in the table. Let’s calculate the values of kfor Mercury,
Mars, and Jupiter.
For Mercury:
0.242= 1 ×0.393
0.0576 = 0.0639
Thus, the value of kfor Mercury is approximately 0.0576.
For Mars:
1.882= 1 ×1.523
18
3.5344 = 3.011008
Thus, the value of kfor Mars is approximately 3.5344.
For Jupiter:
11.862= 1 ×5.203
140.6596 = 140.608
Thus, the value of kfor Jupiter is approximately 140.6596.
Question 21
Question
According to Kepler’s laws of planetary motion, an astronomical body moves
in an elliptical orbit with the sun at one of the foci. Consider a planet with
semi-major axis a= 2.5AU and eccentricity e= 0.4. Determine the semi-minor
axis bof the planet’s orbit.
Solution
Step 1: Recall the relationship between the semi-major axis a, semi-minor axis
b, and eccentricity eof an elliptical orbit:
b=a·√1−e2
Step 2: Substitute the given values of a= 2.5AU and e= 0.4into the
formula:
b= 2.5·√1−0.42
Step 3: Calculate 1−0.42:
1−0.42= 1 −0.16 = 0.84
Step 4: Substitute 0.84 back into the formula:
b= 2.5·√0.84
Step 5: Calculate √0.84:
√0.84 = 0.917
Step 6: Substitute 0.917 back into the formula to find the semi-minor axis
b:
b= 2.5·0.917 = 2.2925
Therefore, the semi-minor axis of the planet’s orbit is b= 2.2925 AU.
19
Question 22
Question
In the context of Kepler’s Laws of Planetary Motion, state the three laws and
briefly explain each one.
Solution
Step 1: Kepler’s First Law (Law of Ellipses): Kepler’s First Law states
that the orbit of a planet is an ellipse with the Sun at one of the two foci. This
means that planets move in an elliptical path around the Sun rather than a
perfect circle.
Step 2: Kepler’s Second Law (Law of Equal Areas): Kepler’s Second
Law states that a line segment joining a planet and the Sun sweeps out equal
areas during equal intervals of time. This means that planets move faster when
they are closer to the Sun along their elliptical orbit.
Step 3: Kepler’s Third Law (Harmonic Law): Kepler’s Third Law
states that the square of the period of any planet is proportional to the cube of
the semi-major axis of its orbit. Mathematically, this can be expressed as:
T2∝a3
where Tis the orbital period of the planet and ais the semi-major axis of its
orbit.
Question 23
Question
Given a planet with a semimajor axis of 2.5 AU, determine the time it takes for
this planet to complete one full orbit around the Sun.
Solution
Step 1: Recall Kepler’s third law of planetary motion, which relates the orbital
period (T) of a planet to its semimajor axis (a). The formula is:
T2=(4π2
G(M1+M2))a3
where Tis the orbital period, ais the semimajor axis, Gis the gravitational
constant, and M1and M2are the masses of the two bodies interacting (in this
case, the Sun and the planet).
Step 2: The Sun’s mass is much larger than the planet’s mass, so we can
neglect the planet’s mass in this case. The mass of the Sun is 1.989 ×1030 kg.
20
Step 3: Plug in the values into the formula:
T2=(4π2
6.67430 ×10−11 ×1.989 ×1030 )(2.5)3
Step 4: Calculate the orbital period (T):
T2=(4π2
1.32678 ×1020 )(15.625)
T2≈0.0029
T≈√0.0029
T≈0.054 years
Step 5: Hence, the time it takes for the planet to complete one full orbit
around the Sun is approximately 0.054 years (or about 19.7 days).
Question 24
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 Earth’s average distance from the sun is 1 astronomical
unit (AU) and its period of revolution is 1 year, calculate the period of revolution
for a planet with an average distance from the sun of 3 AU.
Solution
Step 1: Let Tbe the period of revolution for the planet with an average distance
of 3 AU from the sun. Let rbe the distance of the planet from the sun. By
Kepler’s third law, we have:
T2∝r3
Step 2: For Earth, with r= 1 AU and T= 1 year, we have:
12= 13
1 = 1
Step 3: Now, for the planet with r= 3 AU, we have:
T2∝33
T2∝27
Step 4: To find T, we take the square root of both sides:
T=√27
T= 3√3
Thus, the period of revolution for a planet with an average distance of 3 AU
from the sun is 3√3years.
21
Question 25
Question
According to Kepler’s laws of planetary motion, the square of the period of
revolution of a planet is proportional to the cube of its average distance from
the Sun. Suppose a planet has a period of revolution of 10 years and an average
distance from the Sun of 4 astronomical units (AU). If another planet has an
average distance of 9 AU from the Sun, what would be the period of revolution
of the second planet?
Solution
Let’s denote the period of revolution of the second planet as T(in years) and
its average distance from the Sun as r(in AU). According to Kepler’s third law
of planetary motion:
T2
1
r3
1
=T2
2
r3
2
Substitute in the values for the first planet (T1= 10 years, r1= 4 AU) and
the second planet (r2= 9 AU):
102
43=T2
93
100
64 =T2
729
100
64 ×729 = T2
T2= 1134.375
T=√1134.375 ≈33.67
Therefore, the period of revolution of the second planet would be approxi-
mately 33.67 years.
22
Step 4: Solve for the mass Mof the central star:
M=4π2×(4.49 ×1011 m)3
6.67 ×10−11 m3kg−1s−2×(3.78 ×108s)2
Step 5: Calculate the mass of the central star using the above formula:
M≈1.98 ×1030 kg
Therefore, the mass of the central star is approximately 1.98 ×1030 kg.
Question 2
Question
An asteroid is orbiting the Sun in an elliptical orbit. The closest point of the
asteroid’s orbit to the Sun is 0.2 AU, and the farthest point is 0.8 AU. Calculate
the eccentricity of this orbit.
Solution
Step 1: The eccentricity of an elliptical orbit is given by the formula:
e=rmax −rmin
rmax +rmin
where rmax is the distance from the focus (the Sun) to the farthest point on the
orbit, and rmin is the distance from the focus to the closest point on the orbit.
Step 2: Given that rmin = 0.2AU and rmax = 0.8AU, we can substitute
these values into the formula for eccentricity:
e=0.8−0.2
0.8+0.2
Step 3: Simplifying the expression gives:
e=0.6
1= 0.6
Step 4: Therefore, the eccentricity of the asteroid’s orbit is 0.6.
Question 3
Question
In a planetary system, a planet has an orbital period of 5 years and an average
distance from the star of 2.5 astronomical units. Determine the mass of the star
(M) in terms of the mass of the planet (m) using Kepler’s third law.
2
Solution
To solve this problem, we can use Kepler’s third law of planetary motion, which
relates the orbital period of a planet (T), the average distance from the star
(r), the mass of the planet (m), and the mass of the star (M) in the following
equation:
T2=4π2
G(M+m)r3
where: - Tis the orbital period of the planet, - ris the average distance
from the star, - mis the mass of the planet, - Mis the mass of the star, - Gis
the gravitational constant.
Step 1: Write down the given information: - T= 5 years - r= 2.5astro-
nomical units (AU) - We will solve for Min terms of m.
Step 2: Plug in the known values into Kepler’s third law equation:
(5)2=4π2
G(M+m)(2.5)3
Step 3: Simplify the equation by squaring 5on the left side and cubing 2.5
on the right side:
25 = 4π2
G(M+m)(15.625)
Step 4: Further simplification gives:
25 = 62.5π2
G(M+m)
Step 5: Rearrange the equation to solve for M:
M=62.5π2
25G−m
Therefore, the mass of the star in terms of the mass of the planet is M=62.5π2
25G−m.
Question 4
Question
Two planets are in elliptical orbits around a star. Planet A has a semi-major
axis of 2 AU and an eccentricity of 0.4, while Planet B has a semi-major axis of
3 AU and an eccentricity of 0.6. Calculate the period of orbit for each planet in
Earth years and determine which planet has the longer orbital period.
3
Solution
Step 1: Calculate the period of orbit for Planet A. Given that the semi-major
axis of Planet A is aA= 2 AU and eccentricity is eA= 0.4, we can use Kepler’s
third law of planetary motion to calculate the orbital period PAin Earth years:
PA=√4π2a3
A
G(Mstar +MA)
where Gis the gravitational constant, Mstar is the mass of the star, MAis the
mass of Planet A (which we assume to be negligible compared to the star’s
mass), and aAis the semi-major axis.
Step 2: Substitute the given values into the equation. Plugging in aA= 2
AU and eA= 0.4into the formula, we get:
PA=√4π2(2)3
G(Mstar)
Step 3: Calculate the period of orbit for Planet B. Similarly, for Planet B
with aB= 3 AU and eB= 0.6, the orbital period PBcan be calculated using
the same formula:
PB=√4π2a3
B
G(Mstar +MB)
where MBis the mass of Planet B (assumed to be negligible).
Step 4: Substitute the given values into the equation. Plugging in aB= 3
AU and eB= 0.6into the formula, we get:
PB=√4π2(3)3
G(Mstar)
Step 5: Compare the periods of orbit for Planet A and Planet B. Now,
compare the calculated values of PAand PBto determine which planet has the
longer orbital period.
Question 5
Question
Using Kepler’s third law of planetary motion, calculate the orbital period of a
satellite that is orbiting a planet at a distance of 10,000 km from the planet’s
center. Assume the mass of the planet is 5×1024 kg.
4
Solution
Step 6: Identify the known values
The known values are: - Radius of the satellite from the center of the planet,
r= 10,000 km = 10,000 ×103m. - Mass of the planet, M= 5 ×1024 kg.
Step 7: Determine the orbital period using Kepler’s third law
Kepler’s third law states:
T2=4π2
G(M+m)r3
where: - T is the orbital period, - G is the gravitational constant, 6.67 ×
10−11N m2/kg2, - M is the mass of the planet, - m is the mass of the satel-
lite (not given in this case), - r is the radius of the satellite from the center of
the planet.
Since the mass of the satellite is usually much smaller than the mass of the
planet, we can neglect it in this calculation.
Plugging in the known values:
T2=4π2
6.67 ×10−11 ×5×1024 ×(10,000 ×103)3
Step 8: Calculate the orbital period
T2=4π2×10,0003
6.67 ×5
T2=4π2×1010
33.35
T2≈3.79 ×109
Taking the square root of both sides:
T=√3.79 ×109
T≈61,564 s
Therefore, the orbital period of the satellite is approximately 61,564 seconds.
Question 6
Question
The planet Mars has an orbital period of approximately 687 Earth days. Using
Kepler’s Third Law of Planetary Motion, determine the average distance of
Mars from the Sun in astronomical units (AU). (1 AU is the average distance
between the Earth and the Sun, approximately 1.496 ×1011 meters.)
5
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 its average distance from the
Sun. Mathematically, this can be expressed as:
T2
orbit =k×r3
average
Where: Torbit = orbital period of the planet raverage = average distance of
the planet from the Sun k= constant of proportionality
Given that Mars has an orbital period of 687 Earth days and the average
distance between the Earth and the Sun is 1 AU, we can set up the proportion
as follows:
(687)2=k×r3
average
We need to solve for raverage in AU. Since the Earth’s orbital period is 1 year
(365.25 days), we can set up the following proportion:
(687)2=k×(1)3
6872=k
Substituting k= 6872into the original equation:
(687)2= 6872×r3
average
r3
average = 1
raverage = 1 AU
Therefore, the average distance of Mars from the Sun is 1 astronomical unit.
Question 7
Question
The semi-major axis of a planet’s orbit around the Sun is 3.2 AU. If the eccen-
tricity of the orbit is 0.4, calculate the distance of the planet from the Sun when
it is closest and furthest from the Sun.
Solution
Step 1: We can use Kepler’s laws to find the distance of the planet from the Sun
when it is closest and furthest. Kepler’s laws state that the semi-major axis, a,
of an orbit and the eccentricity, e, determine the distance of the planet from the
focus (Sun) at closest approach (rmin) and furthest distance (rmax) using the
formulas:
rmin =a(1 −e)
6
rmax =a(1 + e)
Step 2: Given that the semi-major axis a= 3.2AU and the eccentricity
e= 0.4, we can now substitute these values into the formulas to find rmin and
rmax.
Step 3: Calculate rmin:
rmin = 3.2(1 −0.4) = 3.2(0.6) = 1.92 AU
Step 4: Calculate rmax:
rmax = 3.2(1 + 0.4) = 3.2(1.4) = 4.48 AU
Step 5: Therefore, the distance of the planet from the Sun when it is closest
is 1.92 AU, and when it is furthest is 4.48 AU.
Question 8
Question
The period of a satellite orbiting a planet is 5 hours. If the average distance
between the satellite and the planet is 20,000 km, determine the mass of the
planet.
(Hint: You may assume the satellite’s orbit is circular.)
Solution
Step 1: Recall Kepler’s Third Law, which states:
T2=(4π2
G(m1+m2))r3
where: - Tis the period of the orbit, - Gis the gravitational constant, - m1is
the mass of the planet, - m2is the mass of the satellite, and - ris the average
distance between the center of the planet and the center of the satellite’s orbit.
Step 2: In this scenario, m1is the unknown mass of the planet, and the mass
of the satellite (m2) can be assumed to be negligible compared to the planet’s
mass. Therefore, the equation can be simplified to:
T2=4π2
Gm1
r3
Step 3: Substituting the given values, we get:
(5hours)2=4π2
G·m1
(20,000km)3
Step 4: Convert the period to seconds:
5hours = 5 ×60 ×60 seconds = 18,000 seconds
7
Step 5: Simplify the equation:
18,0002=4π2
G·m1
(20,000)3
Step 6: Solve for the mass of the planet, m1:
m1=4π2·(20,000)3
G·18,0002
Step 7: Calculate m1:
m1=4π2·8×1015
6.67 ×10−11 ·3.24 ×108
Step 8: Simplify the calculation to find the mass of the planet, m1.
Question 9
Question
Consider a hypothetical solar system where a planet orbits a star in a perfect
circle. The radius of the planet’s orbit is 5 AU (astronomical units). The
planet takes 10 years to complete one full orbit around the star. Calculate the
gravitational force between the planet and the star in this system.
Solution
Step 1: Calculate the speed of the planet in its orbit. Given that the planet
orbits in a perfect circle, we can use the formula for the speed of an object
moving in a circle:
v=2πr
T
where, v= speed of the planet, r= radius of the orbit (5 AU = 7.5×1011
meters), T= time period of the orbit (10 years = 3.15 ×108seconds).
Substitute the values into the formula:
v=2π×7.5×1011
3.15 ×108
Calculate the speed of the planet.
Step 2: Calculate the mass of the star. To calculate the gravitational force,
we need the mass of the star. Let’s assume the mass of the planet is negligible
compared to the mass of the star. Using Kepler’s Third Law:
T2
1
r3
1
=T2
2
r3
2
8
Given T1= 10 years, r1= 5 AU, and T2=r2= 1, solving for the mass of the
star.
Step 3: Calculate the gravitational force between the planet and the star.
Now, we can use Newton’s law of universal gravitation to calculate the force:
F=G×M1×M2
r2
where, F= gravitational force, G= gravitational constant (6.67×10−11 m3kg−1s−2),
M1= mass of the planet, M2= mass of the star, r= distance between the planet
and the star.
Substitute the calculated values to find the gravitational force.
Question 10
Question
Consider a planet in a circular orbit around the Sun with a period of 2 years. If
the distance between the planet and the Sun is doubled, what will be the new
period of revolution of the planet?
Solution
To solve this problem, we can use Kepler’s third law of planetary motion, which
states that the square of the period of revolution of a planet is proportional to
the cube of its average distance from the Sun.
Step 1: Calculate the initial period of revolution using the given data. Let
the initial distance between the planet and the Sun be r, and the initial period
be T. According to Kepler’s third law, we have:
T2∝r3
Given that T= 2 years, we can solve for r:
22∝r3
4∝r3
r=3
√4 = 2 AU
Step 2: Calculate the new period of revolution when the distance is doubled.
Let the new distance between the planet and the Sun be 2r. Using Kepler’s third
law again:
(T′)2∝(2r)3
(T′)2∝8r3
Since r= 2 AU, we can now solve for the new period T′:
(T′)2∝8×(23)
9
(T′)2∝8×8
(T′)2∝64
T′=√64 = 8 years
Therefore, when the distance between the planet and the Sun is doubled,
the new period of revolution of the planet will be 8 years.
Question 11
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 semi-major axis. If the
orbital period of a planet is 8 years and the semi-major axis of its orbit is 3
AU (astronomical units), determine the orbital period of another planet with a
semi-major axis of 5 AU.
Solution
Step 1: Let T1be the orbital period of the first planet and a1be its semi-major
axis. Similarly, let T2be the orbital period of the second planet and a2be its
semi-major axis. According to Kepler’s Third Law,
T2
1
a3
1
=T2
2
a3
2
Step 2: Given that T1= 8 years and a1= 3 AU, we can substitute these
values into the equation:
82
33=T2
2
53
Step 3: Simplifying the equation, we have:
64
27 =T2
2
125
Step 4: Cross multiplying gives us:
64 ×125 = 27 ×T2
2
Step 5: Solving for T2, we have:
T2
2=64 ×125
27 =8000
27
Step 6: Taking the square root of both sides gives us:
T2=√8000
27 ≈15.85 years
Therefore, the orbital period of the second planet with a semi-major axis of
5 AU is approximately 15.85 years.
10
Question 12
Question
Consider a hypothetical planetary system in which a planet has an orbit with
an eccentricity of 0.6. If the semi-major axis of the planet’s orbit is 3 AU,
determine the distance of closest approach and the distance of farthest retreat
of the planet from its parent star.
Solution
To solve this problem, we will first recall the formula for the distance of closest
approach and the distance of farthest retreat of a planet from its parent star
in terms of the semi-major axis and eccentricity of the orbit. Then we will
substitute the given values to find the required distances.
Step 1: The formula for the distance of closest approach (rmin) and the
distance of farthest retreat (rmax) from the parent star in terms of the semi-
major axis (a) and eccentricity (e) of the orbit is:
rmin =a(1 −e)
rmax =a(1 + e)
Step 2: Given that the semi-major axis a= 3 AU and the eccentricity
e= 0.6, we can calculate the distances of closest approach and farthest retreat:
rmin = 3(1 −0.6) = 3(0.4) = 1.2AU
rmax = 3(1 + 0.6) = 3(1.6) = 4.8AU
Step 3: Therefore, the distance of closest approach of the planet from its
parent star is 1.2 AU, and the distance of farthest retreat is 4.8 AU.
Question 13
Question
Suppose a planet has an elliptical orbit around the Sun with a semi-major axis
of a= 2 AU and an eccentricity of e= 0.5. If the planet is at its aphelion,
which is the farthest point from the Sun, calculate its distance from the Sun.
Given that the distance from the Sun to the Earth is 1 AU.
Solution
Step 1: Recall the formula for the distance between the Sun and the planet in
an elliptical orbit:
r=a(1 −e)
11
Step 2: Substitute a= 2 AU and e= 0.5into the formula:
r= 2(1 −0.5)
Step 3: Calculate the value of r:
r= 2 ×0.5 = 1 AU
Step 4: Since the Earth is at 1 AU from the Sun, the planet at its aphelion
is also 1 AU from the Sun.
Therefore, when the planet is at its aphelion, its distance from the Sun is 1
AU.
Question 14
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 semi-major axis. Suppose a
hypothetical planet has an orbital period of 5 years and a semi-major axis of 2
AU (astronomical units). What would be the orbital period of another planet
with a semi-major axis of 4 AU?
Solution
Step 1: Write down Kepler’s third law in mathematical form: The relationship
between the orbital period (T) and the semi-major axis (a) of a planet is given
by:
T2∝a3
Step 2: Determine the ratio of the semi-major axes for the two planets: The
ratio of the semi-major axes of the two planets is:
(a2
a1)3
=(4
2)3
= 23= 8
Step 3: Use the ratio to find the orbital period of the second planet: Since
the ratio of the semi-major axes is 8, the orbital period of the second planet can
be found using this ratio. Let T2be the orbital period of the second planet:
T2
2∝(4)3
T2
2∝64
T2=√64
T2= 8 years
Therefore, the orbital period of the second planet with a semi-major axis of
4 AU would be 8 years.
12
Question 15
Question
Kepler’s second law of planetary motion states that a planet moves fastest when
it is closest to the sun and slowest when it is farthest from the sun. Consider
a planet with an elliptical orbit around the sun. If the planet takes 100 days
to travel from its closest point to the sun (perihelion) to its farthest point
(aphelion), and 200 days to travel from aphelion back to perihelion, determine
the ratio of the planet’s speed at aphelion to its speed at perihelion.
Solution
Step 1: Recall that Kepler’s second law states that a planet sweeps out equal
areas in equal times. This implies that the planet covers the same amount of
area in the same time interval regardless of its distance from the sun.
Step 2: Let the distance from the sun to the planet at perihelion be denoted
as rpand the distance at aphelion be denoted as ra. Then according to the law
of areas, we have:
r2
p·∆θp=r2
a·∆θa
where ∆θpis the angle swept out by the planet at perihelion, and ∆θais the
angle swept out by the planet at aphelion.
Step 3: Since the planet takes 100 days to travel from perihelion to aphelion
and 200 days to travel back, we can say that the area swept out by the planet
from perihelion to aphelion is equal to the area swept out during the return
journey:
r2
p·∆θp=r2
a·∆θa
Step 4: Since the distances from the sun are given by rpand ra, and the
time taken for each journey is given, we can relate the speeds of the planet at
these points as:
Speed at perihelion =2πrp
100
Speed at aphelion =2πra
200
Step 5: Using the equation from Step 3, we can substitute for rain terms
of rpto determine the ratio of speeds at aphelion to perihelion. Let’s calculate
this ratio:
Speed at aphelion
Speed at perihelion =rp
ra
=rp
√r2
p·100
200
=rp
rp·√1
2
=√2
Therefore, the ratio of the planet’s speed at aphelion to its speed at perihelion
is √2.
13
Question 16
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period (T) of a planet is proportional to the cube of its semi-major axis (a)
for any planet orbiting the Sun. Suppose there are two planets, Planet X and
Planet Y, orbiting the Sun. Planet X has an orbital period of 4 years and a
semi-major axis of 2 AU. Planet Y has an orbital period of 9 years. What is
the semi-major axis of Planet Y in astronomical units (AU)?
Solution
Let’s denote the orbital periods of Planet X and Planet Y as TXand TY, and
their respective semi-major axes as aXand aY. According to Kepler’s third law:
T2
X
T2
Y
=a3
X
a3
Y
Step 1: Plug in the given values for Planet X:
(4)2
92=(2)3
a3
Y
Step 2: Simplify the equation:
16
81 =8
a3
Y
Step 3: Cross multiply to solve for aY:
16 ·a3
Y= 81 ·8
Step 4: Calculate the value of aY:
a3
Y=81 ·8
16 = 40.5
Step 5: Take the cube root of both sides to find aY:
aY=3
√40.5≈3.53 AU
Therefore, the semi-major axis of Planet Y is approximately 3.53 AU.
Question 17
Question
Consider a planet in a nearly circular orbit around a Sun-like star with a semi-
major axis of 2 AU. If the orbital period of the planet is 5 years, calculate the
orbital speed of the planet in km/s.
14
Solution
Step 1: We can use Kepler’s third law of planetary motion to find the orbital
speed of the planet. Kepler’s third law states that the square of the orbital
period of a planet is directly proportional to the cube of its semi-major axis.
Mathematically, this can be expressed as:
T2=k·a3
where Tis the orbital period, ais the semi-major axis, and kis a constant.
Step 2: Given that T= 5 years and a= 2 AU, we can rewrite the equation
as:
52=k·23
Solving for k:
25 = 8k=⇒k=25
8= 3.125
Step 3: Now, we can use Kepler’s second law to find the orbital speed of the
planet. Kepler’s second law states that a planet sweeps out equal areas in equal
times. Mathematically, this can be expressed as:
dA
dt =constant
Step 4: The area swept by the planet in a small amount of time (dt) is equal
to the area of a small sector of the orbit. The area of a sector of a circle is given
by:
dA =1
2r2dθ
Step 5: The distance traveled by the planet in a small amount of time is
given by the arc length of the sector:
ds =rdθ
Step 6: The orbital speed (v) of the planet is the rate at which the planet
moves along its orbit, which can be expressed as:
v=ds
dt
Step 7: Combining the equations for dA,ds, and v, we get:
v=rdθ
dt
Step 8: Since the planet travels the full circumference of its orbit in one
period (T), we have:
2πa =v·T
15
Step 9: Substituting the values of aand T, we can solve for the orbital speed
(v):
2π·2 = v·5 =⇒v=4π
5
Step 10: Finally, we can calculate the numerical value of the orbital speed:
v=4π
5≈2.513 km/s
Therefore, the orbital speed of the planet is approximately 2.513 km/s.
Question 18
Question
Consider a hypothetical planetary system where a planet orbits a star with
a semi-major axis of 3.5 AU. If the orbital period of the planet is 8.2 years,
determine the mass of the star (in solar masses) using Kepler’s third law.
Solution
To determine the mass of the star in this hypothetical planetary system, we
will use Kepler’s third law, which relates the orbital period of a planet to the
semi-major axis of its orbit and the mass of the central star.
Step 1: Recall Kepler’s third law, which can be written as:
T2
a3=4π2
G(M1+M2)
where: T= orbital period of the planet, a= semi-major axis of the planet’s
orbit, G= gravitational constant (6.674 × 10−11m3kg−1s−2), M1= mass of
the star, M2= mass of the planet.
Step 2: Substitute the given values into Kepler’s third law:
(8.2years)2
(3.5AU)3=4π2
G(M1+M2)
Step 3: Convert the units into SI units:
(8.2×365.25 ×24 ×3600 s)2
(3.5×1.496 ×1011 m)3=4π2
6.674 ×10−11(M1+M2)
Step 4: Simplify the equation and solve for the mass of the star M1:
M1=(4π2
6.674 ×10−11 )((8.2×365.25 ×24 ×3600)2
(3.5×1.496 ×1011)3)−M2
16
Step 5: Assume the mass of the planet is negligible compared to the mass
of the star, so M1≈M2and thus we have:
M1≈(4π2
6.674 ×10−11 )((8.2×365.25 ×24 ×3600)2
(3.5×1.496 ×1011)3)
Therefore, the mass of the star in solar masses is approximately equal to the
calculated value.
Question 19
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 semi-major axis. A spacecraft
is in a circular orbit around a planet with a semi-major axis of 2.5 AU. If the
spacecraft completes one orbit in 1.5 years, what would be the semi-major axis
of another planet if that planet’s spacecraft completes one orbit in 5 years?
Solution
Let’s denote the orbital period of the first planet as T1= 1.5years and the semi-
major axis of the first planet as a1= 2.5AU. We are looking for the semi-major
axis a2of the second planet.
According to Kepler’s Third Law, the ratio of the squares of the orbital
periods to the cubes of the semi-major axes is constant. Mathematically, we
have:
T2
1
a3
1
=T2
2
a3
2
Given T1= 1.5years and T2= 5 years, we can substitute these values into
the equation above:
(1.5)2
(2.5)3=(5)2
a3
2
2.25
15.625 =25
a3
2
a3
2=25 ×15.625
2.25
a3
2= 173.61
a2=3
√173.61
17
a2≈5.5AU
Therefore, the semi-major axis of another planet would be approximately
5.5 AU if the spacecraft completes one orbit in 5 years.
Question 20
Question
Examine the following table which lists the mean distances (in astronomical
units) from the sun for four different planets and their respective orbital periods
(in Earth years). Use Kepler’s third law to determine the value of the constant
k.
Planet Mean distance (AU) Orbital period (years)
Mercury 0.39 0.24
Earth 1.00 1.00
Mars 1.52 1.88
Jupiter 5.20 11.86
Solution
Step 1: Recall Kepler’s third law, 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, this can be expressed as
T2=k×a3
where Tis the orbital period of the planet, ais the mean distance of the planet
from the sun, and kis a constant.
Step 2: We can use the data provided for Earth to find the value of the
constant k. For Earth:
12=k×13
1 = k
So, k= 1.
Step 3: Now, we can find the constant kfor the other planets using the
information provided in the table. Let’s calculate the values of kfor Mercury,
Mars, and Jupiter.
For Mercury:
0.242= 1 ×0.393
0.0576 = 0.0639
Thus, the value of kfor Mercury is approximately 0.0576.
For Mars:
1.882= 1 ×1.523
18
3.5344 = 3.011008
Thus, the value of kfor Mars is approximately 3.5344.
For Jupiter:
11.862= 1 ×5.203
140.6596 = 140.608
Thus, the value of kfor Jupiter is approximately 140.6596.
Question 21
Question
According to Kepler’s laws of planetary motion, an astronomical body moves
in an elliptical orbit with the sun at one of the foci. Consider a planet with
semi-major axis a= 2.5AU and eccentricity e= 0.4. Determine the semi-minor
axis bof the planet’s orbit.
Solution
Step 1: Recall the relationship between the semi-major axis a, semi-minor axis
b, and eccentricity eof an elliptical orbit:
b=a·√1−e2
Step 2: Substitute the given values of a= 2.5AU and e= 0.4into the
formula:
b= 2.5·√1−0.42
Step 3: Calculate 1−0.42:
1−0.42= 1 −0.16 = 0.84
Step 4: Substitute 0.84 back into the formula:
b= 2.5·√0.84
Step 5: Calculate √0.84:
√0.84 = 0.917
Step 6: Substitute 0.917 back into the formula to find the semi-minor axis
b:
b= 2.5·0.917 = 2.2925
Therefore, the semi-minor axis of the planet’s orbit is b= 2.2925 AU.
19
Question 22
Question
In the context of Kepler’s Laws of Planetary Motion, state the three laws and
briefly explain each one.
Solution
Step 1: Kepler’s First Law (Law of Ellipses): Kepler’s First Law states
that the orbit of a planet is an ellipse with the Sun at one of the two foci. This
means that planets move in an elliptical path around the Sun rather than a
perfect circle.
Step 2: Kepler’s Second Law (Law of Equal Areas): Kepler’s Second
Law states that a line segment joining a planet and the Sun sweeps out equal
areas during equal intervals of time. This means that planets move faster when
they are closer to the Sun along their elliptical orbit.
Step 3: Kepler’s Third Law (Harmonic Law): Kepler’s Third Law
states that the square of the period of any planet is proportional to the cube of
the semi-major axis of its orbit. Mathematically, this can be expressed as:
T2∝a3
where Tis the orbital period of the planet and ais the semi-major axis of its
orbit.
Question 23
Question
Given a planet with a semimajor axis of 2.5 AU, determine the time it takes for
this planet to complete one full orbit around the Sun.
Solution
Step 1: Recall Kepler’s third law of planetary motion, which relates the orbital
period (T) of a planet to its semimajor axis (a). The formula is:
T2=(4π2
G(M1+M2))a3
where Tis the orbital period, ais the semimajor axis, Gis the gravitational
constant, and M1and M2are the masses of the two bodies interacting (in this
case, the Sun and the planet).
Step 2: The Sun’s mass is much larger than the planet’s mass, so we can
neglect the planet’s mass in this case. The mass of the Sun is 1.989 ×1030 kg.
20
Step 3: Plug in the values into the formula:
T2=(4π2
6.67430 ×10−11 ×1.989 ×1030 )(2.5)3
Step 4: Calculate the orbital period (T):
T2=(4π2
1.32678 ×1020 )(15.625)
T2≈0.0029
T≈√0.0029
T≈0.054 years
Step 5: Hence, the time it takes for the planet to complete one full orbit
around the Sun is approximately 0.054 years (or about 19.7 days).
Question 24
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 Earth’s average distance from the sun is 1 astronomical
unit (AU) and its period of revolution is 1 year, calculate the period of revolution
for a planet with an average distance from the sun of 3 AU.
Solution
Step 1: Let Tbe the period of revolution for the planet with an average distance
of 3 AU from the sun. Let rbe the distance of the planet from the sun. By
Kepler’s third law, we have:
T2∝r3
Step 2: For Earth, with r= 1 AU and T= 1 year, we have:
12= 13
1 = 1
Step 3: Now, for the planet with r= 3 AU, we have:
T2∝33
T2∝27
Step 4: To find T, we take the square root of both sides:
T=√27
T= 3√3
Thus, the period of revolution for a planet with an average distance of 3 AU
from the sun is 3√3years.
21
Question 25
Question
According to Kepler’s laws of planetary motion, the square of the period of
revolution of a planet is proportional to the cube of its average distance from
the Sun. Suppose a planet has a period of revolution of 10 years and an average
distance from the Sun of 4 astronomical units (AU). If another planet has an
average distance of 9 AU from the Sun, what would be the period of revolution
of the second planet?
Solution
Let’s denote the period of revolution of the second planet as T(in years) and
its average distance from the Sun as r(in AU). According to Kepler’s third law
of planetary motion:
T2
1
r3
1
=T2
2
r3
2
Substitute in the values for the first planet (T1= 10 years, r1= 4 AU) and
the second planet (r2= 9 AU):
102
43=T2
93
100
64 =T2
729
100
64 ×729 = T2
T2= 1134.375
T=√1134.375 ≈33.67
Therefore, the period of revolution of the second planet would be approxi-
mately 33.67 years.
22
Step 4: Solve for the mass Mof the central star:
M=4π2×(4.49 ×1011 m)3
6.67 ×10−11 m3kg−1s−2×(3.78 ×108s)2
Step 5: Calculate the mass of the central star using the above formula:
M≈1.98 ×1030 kg
Therefore, the mass of the central star is approximately 1.98 ×1030 kg.
Question 2
Question
An asteroid is orbiting the Sun in an elliptical orbit. The closest point of the
asteroid’s orbit to the Sun is 0.2 AU, and the farthest point is 0.8 AU. Calculate
the eccentricity of this orbit.
Solution
Step 1: The eccentricity of an elliptical orbit is given by the formula:
e=rmax −rmin
rmax +rmin
where rmax is the distance from the focus (the Sun) to the farthest point on the
orbit, and rmin is the distance from the focus to the closest point on the orbit.
Step 2: Given that rmin = 0.2AU and rmax = 0.8AU, we can substitute
these values into the formula for eccentricity:
e=0.8−0.2
0.8+0.2
Step 3: Simplifying the expression gives:
e=0.6
1= 0.6
Step 4: Therefore, the eccentricity of the asteroid’s orbit is 0.6.
Question 3
Question
In a planetary system, a planet has an orbital period of 5 years and an average
distance from the star of 2.5 astronomical units. Determine the mass of the star
(M) in terms of the mass of the planet (m) using Kepler’s third law.
2
Solution
To solve this problem, we can use Kepler’s third law of planetary motion, which
relates the orbital period of a planet (T), the average distance from the star
(r), the mass of the planet (m), and the mass of the star (M) in the following
equation:
T2=4π2
G(M+m)r3
where: - Tis the orbital period of the planet, - ris the average distance
from the star, - mis the mass of the planet, - Mis the mass of the star, - Gis
the gravitational constant.
Step 1: Write down the given information: - T= 5 years - r= 2.5astro-
nomical units (AU) - We will solve for Min terms of m.
Step 2: Plug in the known values into Kepler’s third law equation:
(5)2=4π2
G(M+m)(2.5)3
Step 3: Simplify the equation by squaring 5on the left side and cubing 2.5
on the right side:
25 = 4π2
G(M+m)(15.625)
Step 4: Further simplification gives:
25 = 62.5π2
G(M+m)
Step 5: Rearrange the equation to solve for M:
M=62.5π2
25G−m
Therefore, the mass of the star in terms of the mass of the planet is M=62.5π2
25G−m.
Question 4
Question
Two planets are in elliptical orbits around a star. Planet A has a semi-major
axis of 2 AU and an eccentricity of 0.4, while Planet B has a semi-major axis of
3 AU and an eccentricity of 0.6. Calculate the period of orbit for each planet in
Earth years and determine which planet has the longer orbital period.
3
Solution
Step 1: Calculate the period of orbit for Planet A. Given that the semi-major
axis of Planet A is aA= 2 AU and eccentricity is eA= 0.4, we can use Kepler’s
third law of planetary motion to calculate the orbital period PAin Earth years:
PA=√4π2a3
A
G(Mstar +MA)
where Gis the gravitational constant, Mstar is the mass of the star, MAis the
mass of Planet A (which we assume to be negligible compared to the star’s
mass), and aAis the semi-major axis.
Step 2: Substitute the given values into the equation. Plugging in aA= 2
AU and eA= 0.4into the formula, we get:
PA=√4π2(2)3
G(Mstar)
Step 3: Calculate the period of orbit for Planet B. Similarly, for Planet B
with aB= 3 AU and eB= 0.6, the orbital period PBcan be calculated using
the same formula:
PB=√4π2a3
B
G(Mstar +MB)
where MBis the mass of Planet B (assumed to be negligible).
Step 4: Substitute the given values into the equation. Plugging in aB= 3
AU and eB= 0.6into the formula, we get:
PB=√4π2(3)3
G(Mstar)
Step 5: Compare the periods of orbit for Planet A and Planet B. Now,
compare the calculated values of PAand PBto determine which planet has the
longer orbital period.
Question 5
Question
Using Kepler’s third law of planetary motion, calculate the orbital period of a
satellite that is orbiting a planet at a distance of 10,000 km from the planet’s
center. Assume the mass of the planet is 5×1024 kg.
4
Solution
Step 6: Identify the known values
The known values are: - Radius of the satellite from the center of the planet,
r= 10,000 km = 10,000 ×103m. - Mass of the planet, M= 5 ×1024 kg.
Step 7: Determine the orbital period using Kepler’s third law
Kepler’s third law states:
T2=4π2
G(M+m)r3
where: - T is the orbital period, - G is the gravitational constant, 6.67 ×
10−11N m2/kg2, - M is the mass of the planet, - m is the mass of the satel-
lite (not given in this case), - r is the radius of the satellite from the center of
the planet.
Since the mass of the satellite is usually much smaller than the mass of the
planet, we can neglect it in this calculation.
Plugging in the known values:
T2=4π2
6.67 ×10−11 ×5×1024 ×(10,000 ×103)3
Step 8: Calculate the orbital period
T2=4π2×10,0003
6.67 ×5
T2=4π2×1010
33.35
T2≈3.79 ×109
Taking the square root of both sides:
T=√3.79 ×109
T≈61,564 s
Therefore, the orbital period of the satellite is approximately 61,564 seconds.
Question 6
Question
The planet Mars has an orbital period of approximately 687 Earth days. Using
Kepler’s Third Law of Planetary Motion, determine the average distance of
Mars from the Sun in astronomical units (AU). (1 AU is the average distance
between the Earth and the Sun, approximately 1.496 ×1011 meters.)
5
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 its average distance from the
Sun. Mathematically, this can be expressed as:
T2
orbit =k×r3
average
Where: Torbit = orbital period of the planet raverage = average distance of
the planet from the Sun k= constant of proportionality
Given that Mars has an orbital period of 687 Earth days and the average
distance between the Earth and the Sun is 1 AU, we can set up the proportion
as follows:
(687)2=k×r3
average
We need to solve for raverage in AU. Since the Earth’s orbital period is 1 year
(365.25 days), we can set up the following proportion:
(687)2=k×(1)3
6872=k
Substituting k= 6872into the original equation:
(687)2= 6872×r3
average
r3
average = 1
raverage = 1 AU
Therefore, the average distance of Mars from the Sun is 1 astronomical unit.
Question 7
Question
The semi-major axis of a planet’s orbit around the Sun is 3.2 AU. If the eccen-
tricity of the orbit is 0.4, calculate the distance of the planet from the Sun when
it is closest and furthest from the Sun.
Solution
Step 1: We can use Kepler’s laws to find the distance of the planet from the Sun
when it is closest and furthest. Kepler’s laws state that the semi-major axis, a,
of an orbit and the eccentricity, e, determine the distance of the planet from the
focus (Sun) at closest approach (rmin) and furthest distance (rmax) using the
formulas:
rmin =a(1 −e)
6
rmax =a(1 + e)
Step 2: Given that the semi-major axis a= 3.2AU and the eccentricity
e= 0.4, we can now substitute these values into the formulas to find rmin and
rmax.
Step 3: Calculate rmin:
rmin = 3.2(1 −0.4) = 3.2(0.6) = 1.92 AU
Step 4: Calculate rmax:
rmax = 3.2(1 + 0.4) = 3.2(1.4) = 4.48 AU
Step 5: Therefore, the distance of the planet from the Sun when it is closest
is 1.92 AU, and when it is furthest is 4.48 AU.
Question 8
Question
The period of a satellite orbiting a planet is 5 hours. If the average distance
between the satellite and the planet is 20,000 km, determine the mass of the
planet.
(Hint: You may assume the satellite’s orbit is circular.)
Solution
Step 1: Recall Kepler’s Third Law, which states:
T2=(4π2
G(m1+m2))r3
where: - Tis the period of the orbit, - Gis the gravitational constant, - m1is
the mass of the planet, - m2is the mass of the satellite, and - ris the average
distance between the center of the planet and the center of the satellite’s orbit.
Step 2: In this scenario, m1is the unknown mass of the planet, and the mass
of the satellite (m2) can be assumed to be negligible compared to the planet’s
mass. Therefore, the equation can be simplified to:
T2=4π2
Gm1
r3
Step 3: Substituting the given values, we get:
(5hours)2=4π2
G·m1
(20,000km)3
Step 4: Convert the period to seconds:
5hours = 5 ×60 ×60 seconds = 18,000 seconds
7
Step 5: Simplify the equation:
18,0002=4π2
G·m1
(20,000)3
Step 6: Solve for the mass of the planet, m1:
m1=4π2·(20,000)3
G·18,0002
Step 7: Calculate m1:
m1=4π2·8×1015
6.67 ×10−11 ·3.24 ×108
Step 8: Simplify the calculation to find the mass of the planet, m1.
Question 9
Question
Consider a hypothetical solar system where a planet orbits a star in a perfect
circle. The radius of the planet’s orbit is 5 AU (astronomical units). The
planet takes 10 years to complete one full orbit around the star. Calculate the
gravitational force between the planet and the star in this system.
Solution
Step 1: Calculate the speed of the planet in its orbit. Given that the planet
orbits in a perfect circle, we can use the formula for the speed of an object
moving in a circle:
v=2πr
T
where, v= speed of the planet, r= radius of the orbit (5 AU = 7.5×1011
meters), T= time period of the orbit (10 years = 3.15 ×108seconds).
Substitute the values into the formula:
v=2π×7.5×1011
3.15 ×108
Calculate the speed of the planet.
Step 2: Calculate the mass of the star. To calculate the gravitational force,
we need the mass of the star. Let’s assume the mass of the planet is negligible
compared to the mass of the star. Using Kepler’s Third Law:
T2
1
r3
1
=T2
2
r3
2
8
Given T1= 10 years, r1= 5 AU, and T2=r2= 1, solving for the mass of the
star.
Step 3: Calculate the gravitational force between the planet and the star.
Now, we can use Newton’s law of universal gravitation to calculate the force:
F=G×M1×M2
r2
where, F= gravitational force, G= gravitational constant (6.67×10−11 m3kg−1s−2),
M1= mass of the planet, M2= mass of the star, r= distance between the planet
and the star.
Substitute the calculated values to find the gravitational force.
Question 10
Question
Consider a planet in a circular orbit around the Sun with a period of 2 years. If
the distance between the planet and the Sun is doubled, what will be the new
period of revolution of the planet?
Solution
To solve this problem, we can use Kepler’s third law of planetary motion, which
states that the square of the period of revolution of a planet is proportional to
the cube of its average distance from the Sun.
Step 1: Calculate the initial period of revolution using the given data. Let
the initial distance between the planet and the Sun be r, and the initial period
be T. According to Kepler’s third law, we have:
T2∝r3
Given that T= 2 years, we can solve for r:
22∝r3
4∝r3
r=3
√4 = 2 AU
Step 2: Calculate the new period of revolution when the distance is doubled.
Let the new distance between the planet and the Sun be 2r. Using Kepler’s third
law again:
(T′)2∝(2r)3
(T′)2∝8r3
Since r= 2 AU, we can now solve for the new period T′:
(T′)2∝8×(23)
9
(T′)2∝8×8
(T′)2∝64
T′=√64 = 8 years
Therefore, when the distance between the planet and the Sun is doubled,
the new period of revolution of the planet will be 8 years.
Question 11
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 semi-major axis. If the
orbital period of a planet is 8 years and the semi-major axis of its orbit is 3
AU (astronomical units), determine the orbital period of another planet with a
semi-major axis of 5 AU.
Solution
Step 1: Let T1be the orbital period of the first planet and a1be its semi-major
axis. Similarly, let T2be the orbital period of the second planet and a2be its
semi-major axis. According to Kepler’s Third Law,
T2
1
a3
1
=T2
2
a3
2
Step 2: Given that T1= 8 years and a1= 3 AU, we can substitute these
values into the equation:
82
33=T2
2
53
Step 3: Simplifying the equation, we have:
64
27 =T2
2
125
Step 4: Cross multiplying gives us:
64 ×125 = 27 ×T2
2
Step 5: Solving for T2, we have:
T2
2=64 ×125
27 =8000
27
Step 6: Taking the square root of both sides gives us:
T2=√8000
27 ≈15.85 years
Therefore, the orbital period of the second planet with a semi-major axis of
5 AU is approximately 15.85 years.
10
Question 12
Question
Consider a hypothetical planetary system in which a planet has an orbit with
an eccentricity of 0.6. If the semi-major axis of the planet’s orbit is 3 AU,
determine the distance of closest approach and the distance of farthest retreat
of the planet from its parent star.
Solution
To solve this problem, we will first recall the formula for the distance of closest
approach and the distance of farthest retreat of a planet from its parent star
in terms of the semi-major axis and eccentricity of the orbit. Then we will
substitute the given values to find the required distances.
Step 1: The formula for the distance of closest approach (rmin) and the
distance of farthest retreat (rmax) from the parent star in terms of the semi-
major axis (a) and eccentricity (e) of the orbit is:
rmin =a(1 −e)
rmax =a(1 + e)
Step 2: Given that the semi-major axis a= 3 AU and the eccentricity
e= 0.6, we can calculate the distances of closest approach and farthest retreat:
rmin = 3(1 −0.6) = 3(0.4) = 1.2AU
rmax = 3(1 + 0.6) = 3(1.6) = 4.8AU
Step 3: Therefore, the distance of closest approach of the planet from its
parent star is 1.2 AU, and the distance of farthest retreat is 4.8 AU.
Question 13
Question
Suppose a planet has an elliptical orbit around the Sun with a semi-major axis
of a= 2 AU and an eccentricity of e= 0.5. If the planet is at its aphelion,
which is the farthest point from the Sun, calculate its distance from the Sun.
Given that the distance from the Sun to the Earth is 1 AU.
Solution
Step 1: Recall the formula for the distance between the Sun and the planet in
an elliptical orbit:
r=a(1 −e)
11
Step 2: Substitute a= 2 AU and e= 0.5into the formula:
r= 2(1 −0.5)
Step 3: Calculate the value of r:
r= 2 ×0.5 = 1 AU
Step 4: Since the Earth is at 1 AU from the Sun, the planet at its aphelion
is also 1 AU from the Sun.
Therefore, when the planet is at its aphelion, its distance from the Sun is 1
AU.
Question 14
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 semi-major axis. Suppose a
hypothetical planet has an orbital period of 5 years and a semi-major axis of 2
AU (astronomical units). What would be the orbital period of another planet
with a semi-major axis of 4 AU?
Solution
Step 1: Write down Kepler’s third law in mathematical form: The relationship
between the orbital period (T) and the semi-major axis (a) of a planet is given
by:
T2∝a3
Step 2: Determine the ratio of the semi-major axes for the two planets: The
ratio of the semi-major axes of the two planets is:
(a2
a1)3
=(4
2)3
= 23= 8
Step 3: Use the ratio to find the orbital period of the second planet: Since
the ratio of the semi-major axes is 8, the orbital period of the second planet can
be found using this ratio. Let T2be the orbital period of the second planet:
T2
2∝(4)3
T2
2∝64
T2=√64
T2= 8 years
Therefore, the orbital period of the second planet with a semi-major axis of
4 AU would be 8 years.
12
Question 15
Question
Kepler’s second law of planetary motion states that a planet moves fastest when
it is closest to the sun and slowest when it is farthest from the sun. Consider
a planet with an elliptical orbit around the sun. If the planet takes 100 days
to travel from its closest point to the sun (perihelion) to its farthest point
(aphelion), and 200 days to travel from aphelion back to perihelion, determine
the ratio of the planet’s speed at aphelion to its speed at perihelion.
Solution
Step 1: Recall that Kepler’s second law states that a planet sweeps out equal
areas in equal times. This implies that the planet covers the same amount of
area in the same time interval regardless of its distance from the sun.
Step 2: Let the distance from the sun to the planet at perihelion be denoted
as rpand the distance at aphelion be denoted as ra. Then according to the law
of areas, we have:
r2
p·∆θp=r2
a·∆θa
where ∆θpis the angle swept out by the planet at perihelion, and ∆θais the
angle swept out by the planet at aphelion.
Step 3: Since the planet takes 100 days to travel from perihelion to aphelion
and 200 days to travel back, we can say that the area swept out by the planet
from perihelion to aphelion is equal to the area swept out during the return
journey:
r2
p·∆θp=r2
a·∆θa
Step 4: Since the distances from the sun are given by rpand ra, and the
time taken for each journey is given, we can relate the speeds of the planet at
these points as:
Speed at perihelion =2πrp
100
Speed at aphelion =2πra
200
Step 5: Using the equation from Step 3, we can substitute for rain terms
of rpto determine the ratio of speeds at aphelion to perihelion. Let’s calculate
this ratio:
Speed at aphelion
Speed at perihelion =rp
ra
=rp
√r2
p·100
200
=rp
rp·√1
2
=√2
Therefore, the ratio of the planet’s speed at aphelion to its speed at perihelion
is √2.
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Question 16
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period (T) of a planet is proportional to the cube of its semi-major axis (a)
for any planet orbiting the Sun. Suppose there are two planets, Planet X and
Planet Y, orbiting the Sun. Planet X has an orbital period of 4 years and a
semi-major axis of 2 AU. Planet Y has an orbital period of 9 years. What is
the semi-major axis of Planet Y in astronomical units (AU)?
Solution
Let’s denote the orbital periods of Planet X and Planet Y as TXand TY, and
their respective semi-major axes as aXand aY. According to Kepler’s third law:
T2
X
T2
Y
=a3
X
a3
Y
Step 1: Plug in the given values for Planet X:
(4)2
92=(2)3
a3
Y
Step 2: Simplify the equation:
16
81 =8
a3
Y
Step 3: Cross multiply to solve for aY:
16 ·a3
Y= 81 ·8
Step 4: Calculate the value of aY:
a3
Y=81 ·8
16 = 40.5
Step 5: Take the cube root of both sides to find aY:
aY=3
√40.5≈3.53 AU
Therefore, the semi-major axis of Planet Y is approximately 3.53 AU.
Question 17
Question
Consider a planet in a nearly circular orbit around a Sun-like star with a semi-
major axis of 2 AU. If the orbital period of the planet is 5 years, calculate the
orbital speed of the planet in km/s.
14
Solution
Step 1: We can use Kepler’s third law of planetary motion to find the orbital
speed of the planet. Kepler’s third law states that the square of the orbital
period of a planet is directly proportional to the cube of its semi-major axis.
Mathematically, this can be expressed as:
T2=k·a3
where Tis the orbital period, ais the semi-major axis, and kis a constant.
Step 2: Given that T= 5 years and a= 2 AU, we can rewrite the equation
as:
52=k·23
Solving for k:
25 = 8k=⇒k=25
8= 3.125
Step 3: Now, we can use Kepler’s second law to find the orbital speed of the
planet. Kepler’s second law states that a planet sweeps out equal areas in equal
times. Mathematically, this can be expressed as:
dA
dt =constant
Step 4: The area swept by the planet in a small amount of time (dt) is equal
to the area of a small sector of the orbit. The area of a sector of a circle is given
by:
dA =1
2r2dθ
Step 5: The distance traveled by the planet in a small amount of time is
given by the arc length of the sector:
ds =rdθ
Step 6: The orbital speed (v) of the planet is the rate at which the planet
moves along its orbit, which can be expressed as:
v=ds
dt
Step 7: Combining the equations for dA,ds, and v, we get:
v=rdθ
dt
Step 8: Since the planet travels the full circumference of its orbit in one
period (T), we have:
2πa =v·T
15
Step 9: Substituting the values of aand T, we can solve for the orbital speed
(v):
2π·2 = v·5 =⇒v=4π
5
Step 10: Finally, we can calculate the numerical value of the orbital speed:
v=4π
5≈2.513 km/s
Therefore, the orbital speed of the planet is approximately 2.513 km/s.
Question 18
Question
Consider a hypothetical planetary system where a planet orbits a star with
a semi-major axis of 3.5 AU. If the orbital period of the planet is 8.2 years,
determine the mass of the star (in solar masses) using Kepler’s third law.
Solution
To determine the mass of the star in this hypothetical planetary system, we
will use Kepler’s third law, which relates the orbital period of a planet to the
semi-major axis of its orbit and the mass of the central star.
Step 1: Recall Kepler’s third law, which can be written as:
T2
a3=4π2
G(M1+M2)
where: T= orbital period of the planet, a= semi-major axis of the planet’s
orbit, G= gravitational constant (6.674 × 10−11m3kg−1s−2), M1= mass of
the star, M2= mass of the planet.
Step 2: Substitute the given values into Kepler’s third law:
(8.2years)2
(3.5AU)3=4π2
G(M1+M2)
Step 3: Convert the units into SI units:
(8.2×365.25 ×24 ×3600 s)2
(3.5×1.496 ×1011 m)3=4π2
6.674 ×10−11(M1+M2)
Step 4: Simplify the equation and solve for the mass of the star M1:
M1=(4π2
6.674 ×10−11 )((8.2×365.25 ×24 ×3600)2
(3.5×1.496 ×1011)3)−M2
16
Step 5: Assume the mass of the planet is negligible compared to the mass
of the star, so M1≈M2and thus we have:
M1≈(4π2
6.674 ×10−11 )((8.2×365.25 ×24 ×3600)2
(3.5×1.496 ×1011)3)
Therefore, the mass of the star in solar masses is approximately equal to the
calculated value.
Question 19
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 semi-major axis. A spacecraft
is in a circular orbit around a planet with a semi-major axis of 2.5 AU. If the
spacecraft completes one orbit in 1.5 years, what would be the semi-major axis
of another planet if that planet’s spacecraft completes one orbit in 5 years?
Solution
Let’s denote the orbital period of the first planet as T1= 1.5years and the semi-
major axis of the first planet as a1= 2.5AU. We are looking for the semi-major
axis a2of the second planet.
According to Kepler’s Third Law, the ratio of the squares of the orbital
periods to the cubes of the semi-major axes is constant. Mathematically, we
have:
T2
1
a3
1
=T2
2
a3
2
Given T1= 1.5years and T2= 5 years, we can substitute these values into
the equation above:
(1.5)2
(2.5)3=(5)2
a3
2
2.25
15.625 =25
a3
2
a3
2=25 ×15.625
2.25
a3
2= 173.61
a2=3
√173.61
17
a2≈5.5AU
Therefore, the semi-major axis of another planet would be approximately
5.5 AU if the spacecraft completes one orbit in 5 years.
Question 20
Question
Examine the following table which lists the mean distances (in astronomical
units) from the sun for four different planets and their respective orbital periods
(in Earth years). Use Kepler’s third law to determine the value of the constant
k.
Planet Mean distance (AU) Orbital period (years)
Mercury 0.39 0.24
Earth 1.00 1.00
Mars 1.52 1.88
Jupiter 5.20 11.86
Solution
Step 1: Recall Kepler’s third law, 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, this can be expressed as
T2=k×a3
where Tis the orbital period of the planet, ais the mean distance of the planet
from the sun, and kis a constant.
Step 2: We can use the data provided for Earth to find the value of the
constant k. For Earth:
12=k×13
1 = k
So, k= 1.
Step 3: Now, we can find the constant kfor the other planets using the
information provided in the table. Let’s calculate the values of kfor Mercury,
Mars, and Jupiter.
For Mercury:
0.242= 1 ×0.393
0.0576 = 0.0639
Thus, the value of kfor Mercury is approximately 0.0576.
For Mars:
1.882= 1 ×1.523
18
3.5344 = 3.011008
Thus, the value of kfor Mars is approximately 3.5344.
For Jupiter:
11.862= 1 ×5.203
140.6596 = 140.608
Thus, the value of kfor Jupiter is approximately 140.6596.
Question 21
Question
According to Kepler’s laws of planetary motion, an astronomical body moves
in an elliptical orbit with the sun at one of the foci. Consider a planet with
semi-major axis a= 2.5AU and eccentricity e= 0.4. Determine the semi-minor
axis bof the planet’s orbit.
Solution
Step 1: Recall the relationship between the semi-major axis a, semi-minor axis
b, and eccentricity eof an elliptical orbit:
b=a·√1−e2
Step 2: Substitute the given values of a= 2.5AU and e= 0.4into the
formula:
b= 2.5·√1−0.42
Step 3: Calculate 1−0.42:
1−0.42= 1 −0.16 = 0.84
Step 4: Substitute 0.84 back into the formula:
b= 2.5·√0.84
Step 5: Calculate √0.84:
√0.84 = 0.917
Step 6: Substitute 0.917 back into the formula to find the semi-minor axis
b:
b= 2.5·0.917 = 2.2925
Therefore, the semi-minor axis of the planet’s orbit is b= 2.2925 AU.
19
Question 22
Question
In the context of Kepler’s Laws of Planetary Motion, state the three laws and
briefly explain each one.
Solution
Step 1: Kepler’s First Law (Law of Ellipses): Kepler’s First Law states
that the orbit of a planet is an ellipse with the Sun at one of the two foci. This
means that planets move in an elliptical path around the Sun rather than a
perfect circle.
Step 2: Kepler’s Second Law (Law of Equal Areas): Kepler’s Second
Law states that a line segment joining a planet and the Sun sweeps out equal
areas during equal intervals of time. This means that planets move faster when
they are closer to the Sun along their elliptical orbit.
Step 3: Kepler’s Third Law (Harmonic Law): Kepler’s Third Law
states that the square of the period of any planet is proportional to the cube of
the semi-major axis of its orbit. Mathematically, this can be expressed as:
T2∝a3
where Tis the orbital period of the planet and ais the semi-major axis of its
orbit.
Question 23
Question
Given a planet with a semimajor axis of 2.5 AU, determine the time it takes for
this planet to complete one full orbit around the Sun.
Solution
Step 1: Recall Kepler’s third law of planetary motion, which relates the orbital
period (T) of a planet to its semimajor axis (a). The formula is:
T2=(4π2
G(M1+M2))a3
where Tis the orbital period, ais the semimajor axis, Gis the gravitational
constant, and M1and M2are the masses of the two bodies interacting (in this
case, the Sun and the planet).
Step 2: The Sun’s mass is much larger than the planet’s mass, so we can
neglect the planet’s mass in this case. The mass of the Sun is 1.989 ×1030 kg.
20
Step 3: Plug in the values into the formula:
T2=(4π2
6.67430 ×10−11 ×1.989 ×1030 )(2.5)3
Step 4: Calculate the orbital period (T):
T2=(4π2
1.32678 ×1020 )(15.625)
T2≈0.0029
T≈√0.0029
T≈0.054 years
Step 5: Hence, the time it takes for the planet to complete one full orbit
around the Sun is approximately 0.054 years (or about 19.7 days).
Question 24
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 Earth’s average distance from the sun is 1 astronomical
unit (AU) and its period of revolution is 1 year, calculate the period of revolution
for a planet with an average distance from the sun of 3 AU.
Solution
Step 1: Let Tbe the period of revolution for the planet with an average distance
of 3 AU from the sun. Let rbe the distance of the planet from the sun. By
Kepler’s third law, we have:
T2∝r3
Step 2: For Earth, with r= 1 AU and T= 1 year, we have:
12= 13
1 = 1
Step 3: Now, for the planet with r= 3 AU, we have:
T2∝33
T2∝27
Step 4: To find T, we take the square root of both sides:
T=√27
T= 3√3
Thus, the period of revolution for a planet with an average distance of 3 AU
from the sun is 3√3years.
21
Question 25
Question
According to Kepler’s laws of planetary motion, the square of the period of
revolution of a planet is proportional to the cube of its average distance from
the Sun. Suppose a planet has a period of revolution of 10 years and an average
distance from the Sun of 4 astronomical units (AU). If another planet has an
average distance of 9 AU from the Sun, what would be the period of revolution
of the second planet?
Solution
Let’s denote the period of revolution of the second planet as T(in years) and
its average distance from the Sun as r(in AU). According to Kepler’s third law
of planetary motion:
T2
1
r3
1
=T2
2
r3
2
Substitute in the values for the first planet (T1= 10 years, r1= 4 AU) and
the second planet (r2= 9 AU):
102
43=T2
93
100
64 =T2
729
100
64 ×729 = T2
T2= 1134.375
T=√1134.375 ≈33.67
Therefore, the period of revolution of the second planet would be approxi-
mately 33.67 years.
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