PHYS 305 - INTRODUCTION TO
MODERN PHYSICS - Kepler’s laws
Question Bank - Set 10
Liberty University
Question 1
Question
A planet orbits a star in an elliptical orbit with the star located at one of the
foci. The distance between the planet and the star at the closest approach is
0.5 AU and at the farthest point is 2 AU. If the period of the planet’s orbit is
2 years, determine the semi-major axis of the orbit.
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion, which relates the period
of a planet’s orbit to its semi-major axis:
T2=(4π2
GM )a3
where: T= period of the orbit, G= gravitational constant, M= mass of the
star, a= semi-major axis of the orbit.
Step 2: First, we need to convert the given distances from AU to meters. 1
AU is equivalent to 1.496 ×1011 meters. So, the closest approach distance is
0.5×1.496 ×1011 = 7.48 ×1010 meters, and the farthest distance is 2×1.496 ×
1011 = 2.992 ×1011 meters.
Step 3: Next, we can find the semi-major axis of the orbit using the formula:
a=1
2(rmin +rmax)
where rmin is the closest approach distance and rmax is the farthest distance.
Step 4: Plugging in the values, we get:
a=1
2(7.48 ×1010 + 2.992 ×1011)
a=1
2(3.74 ×1011)
a= 1.87 ×1011 meters
Step 5: Now we can substitute the period and semi-major axis into Kepler’s
Third Law to solve for the mass of the star:
(2 years)2=(4π2
G)(1.87 ×1011)3
Step 6: Solve for the gravitational constant G:
4 = (4π2
G)(1.87 ×1011)3
G=4π2
4×(1
1.87 ×1011 )3
G=π2
1.873×4.195 ×10−32
Step 7: Calculate the value of Gto find the semi-major axis of the orbit
around the star.
Question 2
Question
A planet orbits a star in an elliptical path with eccentricity e= 0.3. If the
distance between the planet and the star at the closest point of approach (per-
ihelion) is 100 million km, determine the distance between the planet and the
star at the farthest point of separation (aphelion).
Solution
Let rmin = 100 million km be the distance between the planet and the star at
perihelion. We can find the distance at aphelion, rmax, using Kepler’s second
law in combination with the fact that total mechanical energy is conserved for
an object moving under the influence of gravity.
Step 1: Find the semi-major axis of the elliptical orbit. The semi-major axis
aof an elliptical orbit is related to the distance at perihelion and the eccentricity
by the formula:
a=rmin
1−e.
Substitute rmin = 100 million km and e= 0.3into the formula:
a=100
1−0.3=100
0.7= 142.857 million km.
2
Step 2: Use Kepler’s second law to find the distance at aphelion. Kepler’s
second law states that the line joining the planet and the star sweeps out equal
areas in equal times. This implies that the time taken for the planet to travel
from perihelion to aphelion is the same as the time taken to travel from aphelion
back to perihelion. Since the planet moves faster when it is closer to the star,
the distance at aphelion, rmax, can be determined in terms of the semi-major
axis aand the distance at perihelion rmin:
rmax = 2a−rmin.
Substitute a= 142.857 million km and rmin = 100 million km into the formula:
rmax = 2 ×142.857 −100 = 285.714 −100 = 185.714 million km.
Therefore, the distance between the planet and the star at aphelion is 185.714
million km.
Question 3
Question
A planet orbiting a star follows a path that can be described by the equation
r= 2 cos(2θ), where ris the distance from the planet to the star and θis the
angle measured counterclockwise from the positive x-axis. Determine the shape
of the orbit and discuss which of Kepler’s laws apply in this case.
Solution
Step 1: Start by plotting the equation r= 2 cos(2θ)to visualize the shape of
the orbit.
Step 2: To plot the polar equation, it’s useful to first convert it to Cartesian
coordinates. Recall that x=rcos(θ)and y=rsin(θ). Substituting r=
2 cos(2θ):
x= 2 cos(2θ) cos(θ)and y= 2 cos(2θ) sin(θ)
Step 3: Simplify the expressions for xand y:
x= 2 cos(2θ) cos(θ) = cos(3θ) + cos(θ)
y= 2 cos(2θ) sin(θ) = sin(3θ)−sin(θ)
Step 4: Now, plot the Cartesian equation obtained in Step 3 to identify the
shape of the orbit. The shape of the orbit corresponds to a curve traced by the
planet.
Step 5: The shape of the orbit described by the equation r= 2 cos(2θ)is
an ellipse. It represents Kepler’s First Law, which states that planets move in
elliptical orbits with the star at one of the foci.
Step 6: The orbit’s parameters, such as the eccentricity and orientation, can
be further analyzed to determine if Kepler’s Second Law and Third Law are
applicable in this specific case.
3
Question 4
Question
An asteroid has an elliptical orbit around the Sun with semi-major axis a=
3.0×1011 m and eccentricity e= 0.4. Calculate the period of revolution of the
asteroid around the Sun.
Solution
Step 1: Recall Kepler’s Third Law which states that the square of the period of
revolution of a planet (or asteroid) is proportional to the cube of the semi-major
axis of its orbit. Mathematically, this can be written as:
T2∝a3
Step 2: To find the period of revolution T, we need to determine the pro-
portionality constant. Since we are dealing with astronomical distances, we will
use the gravitational constant Gand the mass of the Sun M⊙to find the exact
relationship. The formula for the period then becomes:
T= 2π√a3
GM⊙
Step 3: Substitute the given values into the formula:
T= 2π√(3.0×1011 m)3
6.67 ×10−11 m3kg−1s−2×1.99 ×1030 kg
Step 4: Calculate the period of revolution:
T= 2π√27 ×1033 m3
13.3×1019 m3kg−1s−2
T= 2π√2.03 ×1014 s2
T= 2π×1.43 ×107s
T≈9.00 ×107s
Therefore, the period of revolution of the asteroid around the Sun is approx-
imately 9.00 ×107seconds.
Question 5
Question
In accordance with Kepler’s laws of planetary motion, consider a hypothetical
planet orbiting a star with a semi-major axis of 2 AU. The planet completes
one full orbit around the star in 500 days. Determine the period of the planet’s
orbit if its semi-major axis were doubled to 4 AU.
4
Solution
To solve this problem, we can use Kepler’s third law which states that the square
of the period of an orbit is directly proportional to the cube of the semi-major
axis of the orbit. Mathematically, this can be expressed as:
T2
1
a3
1
=T2
2
a3
2
where T1and a1are the initial period and semi-major axis, and T2and a2
are the final period and semi-major axis.
Step 1: Substitute the given values into the equation:
Given: T1= 500 days, a1= 2 AU, a2= 4 AU.
Plugging into the formula:
5002
23=T2
2
43
Simplify:
5002
23=T2
2
64
Step 2: Solve for T2:
5002
8=T2
2
62500 = T2
2
T2=√62500
T2= 250 days
Therefore, if the semi-major axis of the planet’s orbit were doubled to 4 AU,
the period of its orbit would be 250 days.
Question 6
Question
A planet follows an elliptical orbit around the Sun. According to Kepler’s laws,
the square of the period of the planet’s orbit is proportional to the cube of the
semi-major axis of its elliptical orbit. Suppose a planet has a period of 10 years
and its semi-major axis is 3 AU (astronomical units). What would be the period
of another planet with a semi-major axis of 4 AU in the same system?
5
Solution
Step 1: Determine the proportionality relationship given by Kepler’s Third Law:
Let T1be the period of the first planet (with semi-major axis a1) and T2
be the period of the second planet (with semi-major axis a2). According to
Kepler’s Third Law,
T2
1
T2
2
=a3
1
a3
2
Step 2: Given that the period of the first planet is 10 years and its semi-major
axis is 3 AU, we have:
102
T2
2
=33
43
Step 3: Solve for T2:
100
T2
2
=27
64
Step 4: Cross multiply and solve for T2:
27T2
2= 6400
T2
2=6400
27
T2=√6400
27 ≈16.17 years
Therefore, the period of the second planet with a semi-major axis of 4 AU
would be approximately 16.17 years in the same system.
Question 7
Question
An asteroid orbits the Sun in an elliptical path. The closest distance of the
asteroid to the Sun is 0.8 AU, while the farthest distance is 2.4 AU. If the
period of the asteroid’s orbit is 3.5 years, determine the semi-major axis of the
orbit. Recall that one astronomical unit (AU) is the average distance between
the Earth and the Sun, approximately 1.5×1011 meters.
6
Solution
Step 1: Find the semi-major axis using Kepler’s third law, which relates the
period of the orbit to the semi-major axis of the orbit. The formula is given by:
T2=4π2
G(M1+M2)a3
where: T= period of the orbit, G= gravitational constant, M1and M2=
masses of the two objects (in this case, the asteroid and the Sun), a= semi-
major axis of the orbit.
Step 2: Convert the distances of the asteroid’s closest and farthest points
from AU to meters. Recall that 1AU is approximately 1.5×1011 meters.
rmin = 0.8AU ×1.5×1011 m/AU
rmax = 2.4AU ×1.5×1011 m/AU
Step 3: The semi-major axis of an elliptical orbit is the average of the closest
and farthest distances from the primary focus. Calculate the semi-major axis:
a=rmin +rmax
2
Step 4: Finally, substitute the known values into Kepler’s third law equation
and solve for the semi-major axis a. With G= 6.67 ×10−11 m3/kg s2and
MSun = 1.989 ×1030 kg.
3.52=4π2
G(MSun)a3
a=(4π2×6.67 ×10−11 m3/kg s2
1.989 ×1030 kg )−1/3
×3.5
Question 8
Question
An exoplanet is discovered orbiting a star in a distant galaxy. Observations show
that the exoplanet has an orbital period of 200 days and an average orbital radius
of 0.6 astronomical units (AU). If the star has a mass of 2×1030 kg, determine
the exoplanet’s orbital speed.
Solution
Step 1: Find the gravitational force between the star and the exoplanet using
Newton’s law of universal gravitation: The gravitational force between the star
and the exoplanet is given by
F=G·Mstar ·mplanet
r2
7
where Gis the gravitational constant, Mstar is the mass of the star, mplanet is
the mass of the planet (assuming negligible compared to the star), and ris the
average orbital radius of the exoplanet. For the exoplanet orbiting the star:
F=G·(2 ×1030 kg)·mplanet
(0.6AU ×1.5×1011 m/AU)2
F=6.67 ×10−11 N·m2/kg2·2×1030 kg ·mplanet
(0.9×1011 m)2
Step 2: Find the centripetal force acting on the exoplanet: The centripetal
force required to keep an object in a circular orbit is given by
F=mplanet ·v2
r
where mplanet is the mass of the exoplanet, vis its orbital speed, and ris the
average orbital radius of the exoplanet.
Step 3: Equate the gravitational force with the centripetal force: Set the
gravitational force equal to the centripetal force:
G·Mstar ·mplanet
r2=mplanet ·v2
r
Step 4: Solve for the orbital speed of the exoplanet: Rewrite the equation
in terms of the orbital speed v:
v=√G·Mstar
r
Step 5: Substitute the given values and calculate the orbital speed: Plugging
in the known values:
v=√6.67 ×10−11 N·m2/kg2·2×1030 kg
0.6AU ×1.5×1011 m/AU
v=√1.334 ×1020
9×1010 =√1.4822 ×109m/s ≈1.218 ×104m/s
Therefore, the orbital speed of the exoplanet is approximately 1.218 ×104
m/s.
Question 9
Question
In a distant planetary system, a planet orbits its star in an elliptical path. The
semi-major axis of the planet’s orbit is 2.5 AU and its eccentricity is 0.4. If
the planet’s closest approach to the star (perihelion) is 1.5 AU, determine the
farthest distance of the planet from the star (aphelion).
8
Solution
Step 1: Recall that for an elliptical orbit, the distance from the focus (star) to
any point on the orbit is constant and equal to the semi-major axis, denoted by
a.
Step 2: The distance from the center to the farthest point on the ellipse
(aphelion) is given by a(1 + e), where eis the eccentricity of the ellipse.
Step 3: Given that the semi-major axis a= 2.5AU and eccentricity e= 0.4,
we can calculate the aphelion distance using the formula a(1 + e).
Step 4: Substitute the values of aand einto the formula to find the aphelion
distance:
Aphelion distance = 2.5×(1 + 0.4) AU
Step 5: Perform the calculations:
Aphelion distance = 2.5×1.4AU = 3.5AU
Step 6: Therefore, the farthest distance of the planet from the star (aphelion)
is 3.5 AU.
Question 10
Question
A planet is in a circular orbit around a star with a period of 1 year. If the
planet’s mass is doubled, what will be the new period of the planet’s orbit,
assuming the distance between the planet and the star remains constant?
(Given: Kepler’s third law states that the square of the period of an orbit
is proportional to the cube of the average distance between the planet and the
star.)
Solution
Step 1: Let Tbe the original period of the planet’s orbit, and Mbe the original
mass of the planet. Let T′be the new period of the planet’s orbit after the mass
is doubled (i.e., 2M).
Step 2: According to Kepler’s third law,
T2∝r3
where ris the average distance between the planet and the star.
Step 3: Since the distance rbetween the planet and the star remains con-
stant, the cube of the distance will not change.
Step 4: We can set up a ratio using the original and new masses of the
planet:
T′2
T2=(2M)3
M3
9
Step 5: Simplifying the ratio, we get:
T′2
T2= 8
Step 6: Taking the square root of both sides gives:
T′
T=√8 = 2√2
Step 7: Therefore, the new period T′of the planet’s orbit is 2√2years.
Question 11
Question
A planet has an elliptical orbit around the Sun, with the eccentricity of the orbit
being 0.4. The distance between the Sun and the planet at its closest approach
is 0.6 AU. Find the distance between the Sun and the planet at its farthest
distance.
Solution
Step 1: Let’s denote the distance between the Sun and the planet at its farthest
distance as rmax, and the distance at its closest approach as rmin.
Step 2: According to Kepler’s second law, the area swept out by the line
joining the Sun and the planet is the same during equal intervals of time. This
means that the planet moves fastest when it is closest to the Sun, and slowest
when it is farthest away.
Step 3: We can use the fact that the area of an ellipse is given by πab,
where ais the semi-major axis and bis the semi-minor axis. Since the Sun is
at one of the foci of the ellipse, we have πab =1
2rminvmin =1
2rmaxvmax, where
vmin and vmax are the velocities of the planet at its closest and farthest points,
respectively.
Step 4: We know that the orbital energy of the planet is given by E=
−GMm
2a, where Mis the mass of the Sun and mis the mass of the planet. At
the closest point, the total energy is entirely kinetic, and at the farthest point,
it is entirely potential.
Step 5: By conservation of energy, we have E=1
2mv2−GMm
r, where vis
the speed of the planet and ris its distance from the Sun. Using this equation
at the closest and farthest points, we can set up two equations to solve for rmax.
Step 6: Solving the equations, we find rmax = 1.4AU. Therefore, the distance
between the Sun and the planet at its farthest distance is 1.4 AU.
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Question 12
Question
Suppose a comet is moving in an elliptical orbit around the Sun. The comet’s
closest distance to the Sun (perihelion) is 0.3 AU and its farthest distance from
the Sun (aphelion) is 6.0 AU. Calculate the eccentricity of the comet’s orbit.
Solution
Step 1: Recall the definition of eccentricity, e, for an ellipse:
e=rmax −rmin
rmax +rmin
where rmax is the distance from the center to the aphelion, and rmin is the
distance from the center to the perihelion.
Step 2: Substitute rmax = 6.0AU and rmin = 0.3AU into the formula:
e=6.0−0.3
6.0+0.3
Step 3: Calculate the numerator and the denominator separately:
e=5.7
6.3
Step 4: Simplify the expression to find the eccentricity:
e=57
63 =19
21
Thus, the eccentricity of the comet’s orbit is 19
21 .
Question 13
Question
An asteroid is orbiting the Sun in an elliptical orbit with an eccentricity of
0.6. If the asteroid is closest to the Sun at a distance of 0.5 AU, determine its
farthest distance from the Sun. Assume the Sun’s mass is 2×1030 kg and the
gravitational constant is 6.67 ×10−11 N m2/kg2.
Solution
Step 1: We can determine the semi-major axis of the asteroid’s orbit using the
given closest distance (rmin) and the eccentricity (e) of the orbit. The semi-
major axis (a) is related to the closest distance by the formula:
a=rmin
1−e
11
Step 2: Substituting rmin = 0.5AU and e= 0.6into the equation:
a=0.5
1−0.6=0.5
0.4= 1.25 AU
Step 3: The farthest distance (rmax) of the asteroid from the Sun occurs at
the aphelion, which is at a distance of a(1 + e)from the Sun.
rmax =a(1 + e) = 1.25(1 + 0.6) = 1.25(1.6) = 2 AU
Therefore, the farthest distance of the asteroid from the Sun is 2 AU.
Question 14
Question
A planet orbits a star in an elliptical orbit. The semi-major axis of the orbit is
2 AU and the eccentricity of the orbit is 0.6. If the planet is closest to the star
at perihelion and farthest from the star at aphelion, find the distances of the
planet from the star at perihelion and aphelion.
Solution
Step 1: Calculate the distance of the planet from the star at perihelion. At
perihelion, the planet is closest to the star. The distance from the star at
perihelion is given by:
rperihelion =a(1 −e)
where ais the semi-major axis and eis the eccentricity. Substituting a= 2 AU
and e= 0.6:
rperihelion = 2(1 −0.6) = 2(0.4) = 0.8AU
Step 2: Calculate the distance of the planet from the star at aphelion. At
aphelion, the planet is farthest from the star. The distance from the star at
aphelion is given by:
raphelion =a(1 + e)
Substituting a= 2 AU and e= 0.6:
raphelion = 2(1 + 0.6) = 2(1.6) = 3.2AU
Therefore, the planet is 0.8 AU from the star at perihelion and 3.2 AU from
the star at aphelion.
12
Question 15
Question
Consider a planet that follows a perfectly elliptical orbit around a star. The
planet takes 60 days to travel from its closest point to the star (perihelion) to
its farthest point from the star (aphelion). If the distance between the planet
and the star at perihelion is 0.4 AU, determine the distance between the planet
and the star at aphelion in AU.
Solution
Step 1: Recall Kepler’s second law, which states that a line segment joining a
planet and the sun sweeps out equal areas during equal intervals of time. This
law can be written as: dA
dt =constant
where dA is the area swept out by the line segment in time dt.
Step 2: In this case, since the planet follows an elliptical orbit, we know
that the area swept out at perihelion is equal to the area swept out at aphelion.
Let rpbe the distance between the planet and the star at perihelion, and let ra
be the distance at aphelion. Then, the area swept out at perihelion is 1
2r2
p∆θ,
and the area swept out at aphelion is 1
2r2
a∆θ, where ∆θis the change in angle
subtended by the planet at the star.
Step 3: Since the time taken to travel from perihelion to aphelion is 60 days,
we can relate the areas at perihelion and aphelion using the fact that they sweep
out equal areas in equal time intervals. Thus, we have:
1
2r2
p∆θ=1
2r2
a∆θ
r2
p=r2
a
ra=rp= 0.4AU
Therefore, the distance between the planet and the star at aphelion is also
0.4 AU.
Question 16
Question
In an exoplanetary system, a distant planet orbits a star in an elliptical path.
The planet’s closest distance to the star is 0.3 AU and its farthest distance is
0.9 AU. If the period of the planet’s orbit is 600 days, determine the eccentricity
of the planet’s orbit.
13
Solution
Step 1: Determine the semi-major axis of the planet’s orbit. Given that the
closest distance (rmin) is 0.3 AU and the farthest distance (rmax) is 0.9 AU, the
semi-major axis (a) of the elliptical orbit can be calculated as the average of
these two distances.
a=rmin +rmax
2=0.3+0.9
2= 0.6AU
Step 2: Determine the eccentricity of the planet’s orbit using Kepler’s third
law. Kepler’s third law relates the period of an orbiting body to the semi-major
axis of its orbit. It can be expressed as:
T2=4π2a3
G(M1+M2)
where Tis the period of the orbit, ais the semi-major axis, Gis the gravitational
constant, and M1and M2are the masses of the two bodies.
Given that T= 600 days, a= 0.6AU, and G≈6.67430 ×10−11 m3/kgs2,
we can rewrite the equation in terms of AU and days:
(600 days)2=4π2(0.6AU)3
G(M1+M2)
Step 3: Calculate the sum of the masses M1+M2. We can rewrite the
equation from step 2 in terms of the sum of the masses:
M1+M2=4π2(0.6AU)3
G(600 days)2
Step 4: Calculate the eccentricity of the orbit. The eccentricity (e) of an
elliptical orbit can be determined using the formula:
e=√1−b2
a2
where ais the semi-major axis and bis the semi-minor axis of the ellipse.
Since the orbit is elliptical, we need to find the value of b(semi-minor axis)
using the formula for an ellipse:
b=a√1−e2
Given that a= 0.6AU, we substitute the avalue calculated earlier to find
b.
Step 5: Calculate the eccentricity using the semi-major axis and semi-minor
axis. Substitute the values of aand binto the eccentricity formula to find the
eccentricity e.
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Question 17
Question
In a distant solar system, a planet has an orbital period of 600 Earth days and
an average orbital radius of 2 AU. Calculate the mass of the star around which
the planet orbits, given that the mass of the planet is 3.2×1025 kg. Assume
the orbit is nearly circular.
Solution
Let’s denote the mass of the star as Mand the orbital radius of the planet as r.
According to Kepler’s third law, the square of the orbital period of a planet is
proportional to the cube of its semi-major axis (in this case, the orbital radius):
T2=4π2r3
G(M+m)
where: - Tis the orbital period, - ris the orbital radius, - Gis the grav-
itational constant, - Mis the mass of the star, and - mis the mass of the
planet.
Given: - The orbital period T= 600 Earth days = 600 ×24 ×3600 seconds,
- The orbital radius r= 2 AU = 2 ×1.496 ×1011 m, - The mass of the planet
m= 3.2×1025 kg.
We can rearrange the formula to solve for the star’s mass M:
M=4π2r3
GT 2−m
Step 1: Convert the orbital period to seconds:
T= 600 ×24 ×3600 = 51840000 seconds
Step 2: Substitute the known values into the formula:
M=4π2(2 ×1.496 ×1011)3
6.67430 ×10−11 ×518400002−3.2×1025
Step 3: Calculate the mass of the star Musing a calculator:
M≈4π2(2 ×1.496 ×1011)3
6.67430 ×10−11 ×518400002−3.2×1025
M≈2.059 ×1030 kg
Therefore, the mass of the star around which the planet orbits is approxi-
mately 2.059 ×1030 kg.
15
Question 18
Question
Consider a star with a mass of 2×1030 kg and a planet with a mass of 6×1024
kg in a circular orbit around the star. The radius of the planet’s orbit is 2×1011
m. Determine the period of the planet’s orbit.
Solution
Step 1: Calculate the gravitational force between the star and the planet using
Newton’s law of universal gravitation:
F=G·M1·M2
R2
where Fis the gravitational force, Gis the gravitational constant (6.67 ×10−11
N m2/kg2), M1and M2are the masses of the star and planet respectively, and
Ris the radius of the planet’s orbit.
F=(6.67 ×10−11 N m2/kg2)·(2 ×1030 kg)·(6 ×1024 kg)
(2 ×1011 m)2
F=8.004 ×1044
4×1022
F= 2.001 ×1022 N
Step 2: Use the centripetal force formula to find the velocity of the planet:
Fcentripetal =M·V2
R
where Mis the mass of the planet and Vis the velocity of the planet.
Setting the gravitational force equal to the centripetal force, we have:
M1·M2
R2=M·V2
R
Solving for V:
V=√M1·M2·R
M
V=√2×1030 ×6×1024 ×2×1011
6×1024
V=√2.4×1065
6×1024
V=√4×1040
16
V= 2 ×1020 m/s
Step 3: Calculate the period of the planet’s orbit using the formula for
circular motion:
V=2πR
T
where Tis the period of the orbit.
Solving for T:
T=2πR
V
T=2π×2×1011
2×1020
T=4π×1011
2×1020
T= 2 ×10−9s
Therefore, the period of the planet’s orbit is 2×10−9seconds.
Question 19
Question
A planet orbits a star in a nearly circular orbit with a radius of 2.5 AU. The
orbital period of the planet is 4 years. Determine the mass of the star in terms
of the mass of the Sun (M⊙).
Solution
Step 1: Recall Kepler’s Third Law, which states that the square of the period
of an orbit is proportional to the cube of the semi-major axis of the orbit.
Mathematically, this is expressed as:
T2
1
a3
1
=T2
2
a3
2
where Tis the orbital period and ais the semi-major axis.
Step 2: Let’s denote the mass of the star as Msand the mass of the planet
as Mp. According to Newton’s version of Kepler’s third law, the sum of the
masses of the star and the planet is involved:
Ms+Mp=4π2a3
GT 2
where Gis the gravitational constant.
Step 3: Given that the star is much more massive than the planet, we can
assume Ms≈Mp.
17
Step 4: Using the information given in the question, we can substitute a=
2.5AU and T= 4 years into the equation to solve for Msin terms of M⊙:
Ms≈4π2(2.5AU)3
G(4 years)2=4π2(2.5×1.496 ×1011 m)3
G(4 ×365.25 ×24 ×3600 s)2M⊙
Step 5: Calculating the mass of the star in terms of the mass of the Sun will
give us the final answer.
Question 20
Question
An asteroid orbits a distant star with a period of 10 years and a semi-major axis
of 3 AU. If the eccentricity of the orbit is 0.4, calculate the asteroid’s closest
distance to the star (perihelion) in AU.
(Note: AU stands for astronomical unit and is the average distance between
the Earth and the Sun, approximately 149.6 million kilometers.)
Solution
Step 1: Calculate the eccentricity distance eusing the semi-major axis aand
the perihelion distance r1.
eccentricity e= 1 −r1
a
Given that a= 3 AU and e= 0.4:
0.4 = 1 −r1
3
r1= 3(1 −0.4) = 3 ×0.6 = 1.8AU
Therefore, the perihelion distance of the asteroid from the star is 1.8 AU.
Question 21
Question
A planet orbits a star in a highly eccentric ellipse. The semi-major axis of the
ellipse is 2 AU, while the eccentricity is 0.8. If the period of the planet’s orbit
is 5 years, what is the closest distance of the planet from the star?
18
Solution
Step 1: Find the closest distance from the star to the planet using the formula
for distance from the focus of an ellipse:
r=a(1 −e)
where: - ris the distance from the star to the closest point on the planet’s orbit,
-ais the semi-major axis of the ellipse, and - eis the eccentricity of the ellipse.
Step 2: Substitute the given values into the formula:
r= 2 AU ×(1 −0.8) = 2 AU ×0.2 = 0.4AU
Therefore, the closest distance of the planet from the star is 0.4 AU.
Question 22
Question
Consider a planet with mass Mmoving in an elliptical orbit around a star with
mass 2M. The semi-major axis of the planet’s orbit is aand the semi-minor
axis is b. Given that the period of revolution of the planet is T, find the total
energy of the planet-star system.
Solution
Let’s denote the gravitational constant as G, the distance between the planet
and the star as r, and the eccentricity of the planet’s orbit as e.
Step 1: Find the relationship between a,b, and e. The semi-major axis a
and the semi-minor axis bof an elliptical orbit are related to the eccentricity e
by the formula:
b=a√1−e2
Step 2: Find the relationship between a,b, and r. The distance between
the planet and the star ris related to the semi-major axis aand the eccentricity
eby the formula:
r=a(1 −e)
Step 3: Find the orbital speed of the planet in terms of rand T. The
orbital speed of the planet is given by:
v=2πr
T
Step 4: Find the potential energy of the planet-star system. The potential
energy of the planet-star system is given by:
U=−GM(2M)
r
19
Step 5: Find the kinetic energy of the planet. The kinetic energy of the
planet is given by:
K=1
2Mv2
Step 6: Find the total energy of the planet-star system using the conserva-
tion of energy. The total energy Eof the system is the sum of the kinetic and
potential energies:
E=K+U
Therefore, the total energy of the planet-star system can be expressed in
terms of the given variables G,M,a,T, and e.
Question 23
Question
Consider a planet that orbits a star in an elliptical orbit. The planet is at its
closest point to the star (perihelion) at a distance of 0.3 AU and at its farthest
point from the star (aphelion) at a distance of 0.7 AU. If the period of the
planet’s orbit is 1.5 years, determine:
1. The semi-major axis of the planet’s orbit.
2. The eccentricity of the planet’s orbit.
Solution
Let’s denote the distance from the star to the perihelion as rp= 0.3 AU and
the distance from the star to the aphelion as ra= 0.7 AU. The period of the
planet’s orbit is given as T= 1.5 years.
Step 1: Calculate the semi-major axis The semi-major axis aof an
elliptical orbit is given by the formula:
a=rp+ra
2
Substitute rp= 0.3AU and ra= 0.7AU into the formula:
a=0.3+0.7
2
a=1
2
a= 0.5AU
Therefore, the semi-major axis of the planet’s orbit is 0.5 AU.
20
Step 2: Calculate the eccentricity The eccentricity eof an elliptical
orbit is related to the distance from the star to the perihelion and aphelion by
the formula:
e=ra−rp
ra+rp
Substitute rp= 0.3AU and ra= 0.7AU into the formula:
e=0.7−0.3
0.7+0.3
e=0.4
1
e= 0.4
Therefore, the eccentricity of the planet’s orbit is 0.4.
Question 24
Question
A planet orbits a star in an elliptical path. At its closest approach to the star
(perihelion), the planet has a speed of 30 km/s, and at its farthest distance from
the star (aphelion), the speed is 15 km/s.
1. Calculate the eccentricity of the planet’s orbit.
2. Determine the speeds of the planet when it is at a distance from the star
equal to half the distance at perihelion.
Solution
Let’s denote the speed of the planet at perihelion as vp= 30 km/s, the speed at
aphelion as va= 15 km/s, and the distance at perihelion as rp.
Step 1: Calculate the eccentricity of the planet’s orbit.
The speed of the planet can be related to its distance from the star using
Kepler’s second law of planetary motion, which states that the line connecting
a planet to the Sun sweeps out equal areas in equal times. This implies that
the angular momentum of the planet (L) is constant.
The specific angular momentum of the planet is given by L=r×mv, which
is constant.
At perihelion:
Lp=rp×m×vp
At aphelion:
La=ra×m×va
Since the specific angular momentum Lis constant, Lp=La.
21
Step 2: Determine the speeds of the planet when it is at a distance from
the star equal to half the distance at perihelion.
Let r=rp
2be the distance from the star. The speed of the planet at this
distance can be calculated using the conservation of angular momentum.
Applying the law of conservation of angular momentum at distance r:
r×m×v=rp×m×vp
Solving for vgives the speed of the planet at a distance r=rp
2.
Question 25
Question
A planet follows an elliptical orbit with the Sun located at one of the foci. The
planet travels from its closest point to the Sun (perihelion) to its farthest point
(aphelion). If the ratio of the distance of the planet from the Sun at perihelion
to the distance at aphelion is 1:2, calculate the eccentricity of the planet’s orbit.
Solution
Step 1: Recall that the eccentricity eof an elliptical orbit can be found using
the formula:
e=ra−rp
ra+rp
where rais the distance of the planet from the Sun at aphelion and rpis the
distance at perihelion.
Step 2: Given that the ratio of the distance from the Sun at perihelion to
aphelion is 1:2, let’s denote the distance at perihelion as dunits, which means
the distance at aphelion is 2dunits.
Step 3: Substitute ra= 2dand rp=dinto the eccentricity formula:
e=2d−d
2d+d=d
3d=1
3
Step 4: Therefore, the eccentricity of the planet’s orbit is 1
3.
22
a=1
2(3.74 ×1011)
a= 1.87 ×1011 meters
Step 5: Now we can substitute the period and semi-major axis into Kepler’s
Third Law to solve for the mass of the star:
(2 years)2=(4π2
G)(1.87 ×1011)3
Step 6: Solve for the gravitational constant G:
4 = (4π2
G)(1.87 ×1011)3
G=4π2
4×(1
1.87 ×1011 )3
G=π2
1.873×4.195 ×10−32
Step 7: Calculate the value of Gto find the semi-major axis of the orbit
around the star.
Question 2
Question
A planet orbits a star in an elliptical path with eccentricity e= 0.3. If the
distance between the planet and the star at the closest point of approach (per-
ihelion) is 100 million km, determine the distance between the planet and the
star at the farthest point of separation (aphelion).
Solution
Let rmin = 100 million km be the distance between the planet and the star at
perihelion. We can find the distance at aphelion, rmax, using Kepler’s second
law in combination with the fact that total mechanical energy is conserved for
an object moving under the influence of gravity.
Step 1: Find the semi-major axis of the elliptical orbit. The semi-major axis
aof an elliptical orbit is related to the distance at perihelion and the eccentricity
by the formula:
a=rmin
1−e.
Substitute rmin = 100 million km and e= 0.3into the formula:
a=100
1−0.3=100
0.7= 142.857 million km.
2
Step 2: Use Kepler’s second law to find the distance at aphelion. Kepler’s
second law states that the line joining the planet and the star sweeps out equal
areas in equal times. This implies that the time taken for the planet to travel
from perihelion to aphelion is the same as the time taken to travel from aphelion
back to perihelion. Since the planet moves faster when it is closer to the star,
the distance at aphelion, rmax, can be determined in terms of the semi-major
axis aand the distance at perihelion rmin:
rmax = 2a−rmin.
Substitute a= 142.857 million km and rmin = 100 million km into the formula:
rmax = 2 ×142.857 −100 = 285.714 −100 = 185.714 million km.
Therefore, the distance between the planet and the star at aphelion is 185.714
million km.
Question 3
Question
A planet orbiting a star follows a path that can be described by the equation
r= 2 cos(2θ), where ris the distance from the planet to the star and θis the
angle measured counterclockwise from the positive x-axis. Determine the shape
of the orbit and discuss which of Kepler’s laws apply in this case.
Solution
Step 1: Start by plotting the equation r= 2 cos(2θ)to visualize the shape of
the orbit.
Step 2: To plot the polar equation, it’s useful to first convert it to Cartesian
coordinates. Recall that x=rcos(θ)and y=rsin(θ). Substituting r=
2 cos(2θ):
x= 2 cos(2θ) cos(θ)and y= 2 cos(2θ) sin(θ)
Step 3: Simplify the expressions for xand y:
x= 2 cos(2θ) cos(θ) = cos(3θ) + cos(θ)
y= 2 cos(2θ) sin(θ) = sin(3θ)−sin(θ)
Step 4: Now, plot the Cartesian equation obtained in Step 3 to identify the
shape of the orbit. The shape of the orbit corresponds to a curve traced by the
planet.
Step 5: The shape of the orbit described by the equation r= 2 cos(2θ)is
an ellipse. It represents Kepler’s First Law, which states that planets move in
elliptical orbits with the star at one of the foci.
Step 6: The orbit’s parameters, such as the eccentricity and orientation, can
be further analyzed to determine if Kepler’s Second Law and Third Law are
applicable in this specific case.
3
Question 4
Question
An asteroid has an elliptical orbit around the Sun with semi-major axis a=
3.0×1011 m and eccentricity e= 0.4. Calculate the period of revolution of the
asteroid around the Sun.
Solution
Step 1: Recall Kepler’s Third Law which states that the square of the period of
revolution of a planet (or asteroid) is proportional to the cube of the semi-major
axis of its orbit. Mathematically, this can be written as:
T2∝a3
Step 2: To find the period of revolution T, we need to determine the pro-
portionality constant. Since we are dealing with astronomical distances, we will
use the gravitational constant Gand the mass of the Sun M⊙to find the exact
relationship. The formula for the period then becomes:
T= 2π√a3
GM⊙
Step 3: Substitute the given values into the formula:
T= 2π√(3.0×1011 m)3
6.67 ×10−11 m3kg−1s−2×1.99 ×1030 kg
Step 4: Calculate the period of revolution:
T= 2π√27 ×1033 m3
13.3×1019 m3kg−1s−2
T= 2π√2.03 ×1014 s2
T= 2π×1.43 ×107s
T≈9.00 ×107s
Therefore, the period of revolution of the asteroid around the Sun is approx-
imately 9.00 ×107seconds.
Question 5
Question
In accordance with Kepler’s laws of planetary motion, consider a hypothetical
planet orbiting a star with a semi-major axis of 2 AU. The planet completes
one full orbit around the star in 500 days. Determine the period of the planet’s
orbit if its semi-major axis were doubled to 4 AU.
4
Solution
To solve this problem, we can use Kepler’s third law which states that the square
of the period of an orbit is directly proportional to the cube of the semi-major
axis of the orbit. Mathematically, this can be expressed as:
T2
1
a3
1
=T2
2
a3
2
where T1and a1are the initial period and semi-major axis, and T2and a2
are the final period and semi-major axis.
Step 1: Substitute the given values into the equation:
Given: T1= 500 days, a1= 2 AU, a2= 4 AU.
Plugging into the formula:
5002
23=T2
2
43
Simplify:
5002
23=T2
2
64
Step 2: Solve for T2:
5002
8=T2
2
62500 = T2
2
T2=√62500
T2= 250 days
Therefore, if the semi-major axis of the planet’s orbit were doubled to 4 AU,
the period of its orbit would be 250 days.
Question 6
Question
A planet follows an elliptical orbit around the Sun. According to Kepler’s laws,
the square of the period of the planet’s orbit is proportional to the cube of the
semi-major axis of its elliptical orbit. Suppose a planet has a period of 10 years
and its semi-major axis is 3 AU (astronomical units). What would be the period
of another planet with a semi-major axis of 4 AU in the same system?
5
Solution
Step 1: Determine the proportionality relationship given by Kepler’s Third Law:
Let T1be the period of the first planet (with semi-major axis a1) and T2
be the period of the second planet (with semi-major axis a2). According to
Kepler’s Third Law,
T2
1
T2
2
=a3
1
a3
2
Step 2: Given that the period of the first planet is 10 years and its semi-major
axis is 3 AU, we have:
102
T2
2
=33
43
Step 3: Solve for T2:
100
T2
2
=27
64
Step 4: Cross multiply and solve for T2:
27T2
2= 6400
T2
2=6400
27
T2=√6400
27 ≈16.17 years
Therefore, the period of the second planet with a semi-major axis of 4 AU
would be approximately 16.17 years in the same system.
Question 7
Question
An asteroid orbits the Sun in an elliptical path. The closest distance of the
asteroid to the Sun is 0.8 AU, while the farthest distance is 2.4 AU. If the
period of the asteroid’s orbit is 3.5 years, determine the semi-major axis of the
orbit. Recall that one astronomical unit (AU) is the average distance between
the Earth and the Sun, approximately 1.5×1011 meters.
6
Solution
Step 1: Find the semi-major axis using Kepler’s third law, which relates the
period of the orbit to the semi-major axis of the orbit. The formula is given by:
T2=4π2
G(M1+M2)a3
where: T= period of the orbit, G= gravitational constant, M1and M2=
masses of the two objects (in this case, the asteroid and the Sun), a= semi-
major axis of the orbit.
Step 2: Convert the distances of the asteroid’s closest and farthest points
from AU to meters. Recall that 1AU is approximately 1.5×1011 meters.
rmin = 0.8AU ×1.5×1011 m/AU
rmax = 2.4AU ×1.5×1011 m/AU
Step 3: The semi-major axis of an elliptical orbit is the average of the closest
and farthest distances from the primary focus. Calculate the semi-major axis:
a=rmin +rmax
2
Step 4: Finally, substitute the known values into Kepler’s third law equation
and solve for the semi-major axis a. With G= 6.67 ×10−11 m3/kg s2and
MSun = 1.989 ×1030 kg.
3.52=4π2
G(MSun)a3
a=(4π2×6.67 ×10−11 m3/kg s2
1.989 ×1030 kg )−1/3
×3.5
Question 8
Question
An exoplanet is discovered orbiting a star in a distant galaxy. Observations show
that the exoplanet has an orbital period of 200 days and an average orbital radius
of 0.6 astronomical units (AU). If the star has a mass of 2×1030 kg, determine
the exoplanet’s orbital speed.
Solution
Step 1: Find the gravitational force between the star and the exoplanet using
Newton’s law of universal gravitation: The gravitational force between the star
and the exoplanet is given by
F=G·Mstar ·mplanet
r2
7
where Gis the gravitational constant, Mstar is the mass of the star, mplanet is
the mass of the planet (assuming negligible compared to the star), and ris the
average orbital radius of the exoplanet. For the exoplanet orbiting the star:
F=G·(2 ×1030 kg)·mplanet
(0.6AU ×1.5×1011 m/AU)2
F=6.67 ×10−11 N·m2/kg2·2×1030 kg ·mplanet
(0.9×1011 m)2
Step 2: Find the centripetal force acting on the exoplanet: The centripetal
force required to keep an object in a circular orbit is given by
F=mplanet ·v2
r
where mplanet is the mass of the exoplanet, vis its orbital speed, and ris the
average orbital radius of the exoplanet.
Step 3: Equate the gravitational force with the centripetal force: Set the
gravitational force equal to the centripetal force:
G·Mstar ·mplanet
r2=mplanet ·v2
r
Step 4: Solve for the orbital speed of the exoplanet: Rewrite the equation
in terms of the orbital speed v:
v=√G·Mstar
r
Step 5: Substitute the given values and calculate the orbital speed: Plugging
in the known values:
v=√6.67 ×10−11 N·m2/kg2·2×1030 kg
0.6AU ×1.5×1011 m/AU
v=√1.334 ×1020
9×1010 =√1.4822 ×109m/s ≈1.218 ×104m/s
Therefore, the orbital speed of the exoplanet is approximately 1.218 ×104
m/s.
Question 9
Question
In a distant planetary system, a planet orbits its star in an elliptical path. The
semi-major axis of the planet’s orbit is 2.5 AU and its eccentricity is 0.4. If
the planet’s closest approach to the star (perihelion) is 1.5 AU, determine the
farthest distance of the planet from the star (aphelion).
8
Solution
Step 1: Recall that for an elliptical orbit, the distance from the focus (star) to
any point on the orbit is constant and equal to the semi-major axis, denoted by
a.
Step 2: The distance from the center to the farthest point on the ellipse
(aphelion) is given by a(1 + e), where eis the eccentricity of the ellipse.
Step 3: Given that the semi-major axis a= 2.5AU and eccentricity e= 0.4,
we can calculate the aphelion distance using the formula a(1 + e).
Step 4: Substitute the values of aand einto the formula to find the aphelion
distance:
Aphelion distance = 2.5×(1 + 0.4) AU
Step 5: Perform the calculations:
Aphelion distance = 2.5×1.4AU = 3.5AU
Step 6: Therefore, the farthest distance of the planet from the star (aphelion)
is 3.5 AU.
Question 10
Question
A planet is in a circular orbit around a star with a period of 1 year. If the
planet’s mass is doubled, what will be the new period of the planet’s orbit,
assuming the distance between the planet and the star remains constant?
(Given: Kepler’s third law states that the square of the period of an orbit
is proportional to the cube of the average distance between the planet and the
star.)
Solution
Step 1: Let Tbe the original period of the planet’s orbit, and Mbe the original
mass of the planet. Let T′be the new period of the planet’s orbit after the mass
is doubled (i.e., 2M).
Step 2: According to Kepler’s third law,
T2∝r3
where ris the average distance between the planet and the star.
Step 3: Since the distance rbetween the planet and the star remains con-
stant, the cube of the distance will not change.
Step 4: We can set up a ratio using the original and new masses of the
planet:
T′2
T2=(2M)3
M3
9
Step 5: Simplifying the ratio, we get:
T′2
T2= 8
Step 6: Taking the square root of both sides gives:
T′
T=√8 = 2√2
Step 7: Therefore, the new period T′of the planet’s orbit is 2√2years.
Question 11
Question
A planet has an elliptical orbit around the Sun, with the eccentricity of the orbit
being 0.4. The distance between the Sun and the planet at its closest approach
is 0.6 AU. Find the distance between the Sun and the planet at its farthest
distance.
Solution
Step 1: Let’s denote the distance between the Sun and the planet at its farthest
distance as rmax, and the distance at its closest approach as rmin.
Step 2: According to Kepler’s second law, the area swept out by the line
joining the Sun and the planet is the same during equal intervals of time. This
means that the planet moves fastest when it is closest to the Sun, and slowest
when it is farthest away.
Step 3: We can use the fact that the area of an ellipse is given by πab,
where ais the semi-major axis and bis the semi-minor axis. Since the Sun is
at one of the foci of the ellipse, we have πab =1
2rminvmin =1
2rmaxvmax, where
vmin and vmax are the velocities of the planet at its closest and farthest points,
respectively.
Step 4: We know that the orbital energy of the planet is given by E=
−GMm
2a, where Mis the mass of the Sun and mis the mass of the planet. At
the closest point, the total energy is entirely kinetic, and at the farthest point,
it is entirely potential.
Step 5: By conservation of energy, we have E=1
2mv2−GMm
r, where vis
the speed of the planet and ris its distance from the Sun. Using this equation
at the closest and farthest points, we can set up two equations to solve for rmax.
Step 6: Solving the equations, we find rmax = 1.4AU. Therefore, the distance
between the Sun and the planet at its farthest distance is 1.4 AU.
10
Question 12
Question
Suppose a comet is moving in an elliptical orbit around the Sun. The comet’s
closest distance to the Sun (perihelion) is 0.3 AU and its farthest distance from
the Sun (aphelion) is 6.0 AU. Calculate the eccentricity of the comet’s orbit.
Solution
Step 1: Recall the definition of eccentricity, e, for an ellipse:
e=rmax −rmin
rmax +rmin
where rmax is the distance from the center to the aphelion, and rmin is the
distance from the center to the perihelion.
Step 2: Substitute rmax = 6.0AU and rmin = 0.3AU into the formula:
e=6.0−0.3
6.0+0.3
Step 3: Calculate the numerator and the denominator separately:
e=5.7
6.3
Step 4: Simplify the expression to find the eccentricity:
e=57
63 =19
21
Thus, the eccentricity of the comet’s orbit is 19
21 .
Question 13
Question
An asteroid is orbiting the Sun in an elliptical orbit with an eccentricity of
0.6. If the asteroid is closest to the Sun at a distance of 0.5 AU, determine its
farthest distance from the Sun. Assume the Sun’s mass is 2×1030 kg and the
gravitational constant is 6.67 ×10−11 N m2/kg2.
Solution
Step 1: We can determine the semi-major axis of the asteroid’s orbit using the
given closest distance (rmin) and the eccentricity (e) of the orbit. The semi-
major axis (a) is related to the closest distance by the formula:
a=rmin
1−e
11
Step 2: Substituting rmin = 0.5AU and e= 0.6into the equation:
a=0.5
1−0.6=0.5
0.4= 1.25 AU
Step 3: The farthest distance (rmax) of the asteroid from the Sun occurs at
the aphelion, which is at a distance of a(1 + e)from the Sun.
rmax =a(1 + e) = 1.25(1 + 0.6) = 1.25(1.6) = 2 AU
Therefore, the farthest distance of the asteroid from the Sun is 2 AU.
Question 14
Question
A planet orbits a star in an elliptical orbit. The semi-major axis of the orbit is
2 AU and the eccentricity of the orbit is 0.6. If the planet is closest to the star
at perihelion and farthest from the star at aphelion, find the distances of the
planet from the star at perihelion and aphelion.
Solution
Step 1: Calculate the distance of the planet from the star at perihelion. At
perihelion, the planet is closest to the star. The distance from the star at
perihelion is given by:
rperihelion =a(1 −e)
where ais the semi-major axis and eis the eccentricity. Substituting a= 2 AU
and e= 0.6:
rperihelion = 2(1 −0.6) = 2(0.4) = 0.8AU
Step 2: Calculate the distance of the planet from the star at aphelion. At
aphelion, the planet is farthest from the star. The distance from the star at
aphelion is given by:
raphelion =a(1 + e)
Substituting a= 2 AU and e= 0.6:
raphelion = 2(1 + 0.6) = 2(1.6) = 3.2AU
Therefore, the planet is 0.8 AU from the star at perihelion and 3.2 AU from
the star at aphelion.
12
Question 15
Question
Consider a planet that follows a perfectly elliptical orbit around a star. The
planet takes 60 days to travel from its closest point to the star (perihelion) to
its farthest point from the star (aphelion). If the distance between the planet
and the star at perihelion is 0.4 AU, determine the distance between the planet
and the star at aphelion in AU.
Solution
Step 1: Recall Kepler’s second law, which states that a line segment joining a
planet and the sun sweeps out equal areas during equal intervals of time. This
law can be written as: dA
dt =constant
where dA is the area swept out by the line segment in time dt.
Step 2: In this case, since the planet follows an elliptical orbit, we know
that the area swept out at perihelion is equal to the area swept out at aphelion.
Let rpbe the distance between the planet and the star at perihelion, and let ra
be the distance at aphelion. Then, the area swept out at perihelion is 1
2r2
p∆θ,
and the area swept out at aphelion is 1
2r2
a∆θ, where ∆θis the change in angle
subtended by the planet at the star.
Step 3: Since the time taken to travel from perihelion to aphelion is 60 days,
we can relate the areas at perihelion and aphelion using the fact that they sweep
out equal areas in equal time intervals. Thus, we have:
1
2r2
p∆θ=1
2r2
a∆θ
r2
p=r2
a
ra=rp= 0.4AU
Therefore, the distance between the planet and the star at aphelion is also
0.4 AU.
Question 16
Question
In an exoplanetary system, a distant planet orbits a star in an elliptical path.
The planet’s closest distance to the star is 0.3 AU and its farthest distance is
0.9 AU. If the period of the planet’s orbit is 600 days, determine the eccentricity
of the planet’s orbit.
13
Solution
Step 1: Determine the semi-major axis of the planet’s orbit. Given that the
closest distance (rmin) is 0.3 AU and the farthest distance (rmax) is 0.9 AU, the
semi-major axis (a) of the elliptical orbit can be calculated as the average of
these two distances.
a=rmin +rmax
2=0.3+0.9
2= 0.6AU
Step 2: Determine the eccentricity of the planet’s orbit using Kepler’s third
law. Kepler’s third law relates the period of an orbiting body to the semi-major
axis of its orbit. It can be expressed as:
T2=4π2a3
G(M1+M2)
where Tis the period of the orbit, ais the semi-major axis, Gis the gravitational
constant, and M1and M2are the masses of the two bodies.
Given that T= 600 days, a= 0.6AU, and G≈6.67430 ×10−11 m3/kgs2,
we can rewrite the equation in terms of AU and days:
(600 days)2=4π2(0.6AU)3
G(M1+M2)
Step 3: Calculate the sum of the masses M1+M2. We can rewrite the
equation from step 2 in terms of the sum of the masses:
M1+M2=4π2(0.6AU)3
G(600 days)2
Step 4: Calculate the eccentricity of the orbit. The eccentricity (e) of an
elliptical orbit can be determined using the formula:
e=√1−b2
a2
where ais the semi-major axis and bis the semi-minor axis of the ellipse.
Since the orbit is elliptical, we need to find the value of b(semi-minor axis)
using the formula for an ellipse:
b=a√1−e2
Given that a= 0.6AU, we substitute the avalue calculated earlier to find
b.
Step 5: Calculate the eccentricity using the semi-major axis and semi-minor
axis. Substitute the values of aand binto the eccentricity formula to find the
eccentricity e.
14
Question 17
Question
In a distant solar system, a planet has an orbital period of 600 Earth days and
an average orbital radius of 2 AU. Calculate the mass of the star around which
the planet orbits, given that the mass of the planet is 3.2×1025 kg. Assume
the orbit is nearly circular.
Solution
Let’s denote the mass of the star as Mand the orbital radius of the planet as r.
According to Kepler’s third law, the square of the orbital period of a planet is
proportional to the cube of its semi-major axis (in this case, the orbital radius):
T2=4π2r3
G(M+m)
where: - Tis the orbital period, - ris the orbital radius, - Gis the grav-
itational constant, - Mis the mass of the star, and - mis the mass of the
planet.
Given: - The orbital period T= 600 Earth days = 600 ×24 ×3600 seconds,
- The orbital radius r= 2 AU = 2 ×1.496 ×1011 m, - The mass of the planet
m= 3.2×1025 kg.
We can rearrange the formula to solve for the star’s mass M:
M=4π2r3
GT 2−m
Step 1: Convert the orbital period to seconds:
T= 600 ×24 ×3600 = 51840000 seconds
Step 2: Substitute the known values into the formula:
M=4π2(2 ×1.496 ×1011)3
6.67430 ×10−11 ×518400002−3.2×1025
Step 3: Calculate the mass of the star Musing a calculator:
M≈4π2(2 ×1.496 ×1011)3
6.67430 ×10−11 ×518400002−3.2×1025
M≈2.059 ×1030 kg
Therefore, the mass of the star around which the planet orbits is approxi-
mately 2.059 ×1030 kg.
15
Question 18
Question
Consider a star with a mass of 2×1030 kg and a planet with a mass of 6×1024
kg in a circular orbit around the star. The radius of the planet’s orbit is 2×1011
m. Determine the period of the planet’s orbit.
Solution
Step 1: Calculate the gravitational force between the star and the planet using
Newton’s law of universal gravitation:
F=G·M1·M2
R2
where Fis the gravitational force, Gis the gravitational constant (6.67 ×10−11
N m2/kg2), M1and M2are the masses of the star and planet respectively, and
Ris the radius of the planet’s orbit.
F=(6.67 ×10−11 N m2/kg2)·(2 ×1030 kg)·(6 ×1024 kg)
(2 ×1011 m)2
F=8.004 ×1044
4×1022
F= 2.001 ×1022 N
Step 2: Use the centripetal force formula to find the velocity of the planet:
Fcentripetal =M·V2
R
where Mis the mass of the planet and Vis the velocity of the planet.
Setting the gravitational force equal to the centripetal force, we have:
M1·M2
R2=M·V2
R
Solving for V:
V=√M1·M2·R
M
V=√2×1030 ×6×1024 ×2×1011
6×1024
V=√2.4×1065
6×1024
V=√4×1040
16
V= 2 ×1020 m/s
Step 3: Calculate the period of the planet’s orbit using the formula for
circular motion:
V=2πR
T
where Tis the period of the orbit.
Solving for T:
T=2πR
V
T=2π×2×1011
2×1020
T=4π×1011
2×1020
T= 2 ×10−9s
Therefore, the period of the planet’s orbit is 2×10−9seconds.
Question 19
Question
A planet orbits a star in a nearly circular orbit with a radius of 2.5 AU. The
orbital period of the planet is 4 years. Determine the mass of the star in terms
of the mass of the Sun (M⊙).
Solution
Step 1: Recall Kepler’s Third Law, which states that the square of the period
of an orbit is proportional to the cube of the semi-major axis of the orbit.
Mathematically, this is expressed as:
T2
1
a3
1
=T2
2
a3
2
where Tis the orbital period and ais the semi-major axis.
Step 2: Let’s denote the mass of the star as Msand the mass of the planet
as Mp. According to Newton’s version of Kepler’s third law, the sum of the
masses of the star and the planet is involved:
Ms+Mp=4π2a3
GT 2
where Gis the gravitational constant.
Step 3: Given that the star is much more massive than the planet, we can
assume Ms≈Mp.
17
Step 4: Using the information given in the question, we can substitute a=
2.5AU and T= 4 years into the equation to solve for Msin terms of M⊙:
Ms≈4π2(2.5AU)3
G(4 years)2=4π2(2.5×1.496 ×1011 m)3
G(4 ×365.25 ×24 ×3600 s)2M⊙
Step 5: Calculating the mass of the star in terms of the mass of the Sun will
give us the final answer.
Question 20
Question
An asteroid orbits a distant star with a period of 10 years and a semi-major axis
of 3 AU. If the eccentricity of the orbit is 0.4, calculate the asteroid’s closest
distance to the star (perihelion) in AU.
(Note: AU stands for astronomical unit and is the average distance between
the Earth and the Sun, approximately 149.6 million kilometers.)
Solution
Step 1: Calculate the eccentricity distance eusing the semi-major axis aand
the perihelion distance r1.
eccentricity e= 1 −r1
a
Given that a= 3 AU and e= 0.4:
0.4 = 1 −r1
3
r1= 3(1 −0.4) = 3 ×0.6 = 1.8AU
Therefore, the perihelion distance of the asteroid from the star is 1.8 AU.
Question 21
Question
A planet orbits a star in a highly eccentric ellipse. The semi-major axis of the
ellipse is 2 AU, while the eccentricity is 0.8. If the period of the planet’s orbit
is 5 years, what is the closest distance of the planet from the star?
18
Solution
Step 1: Find the closest distance from the star to the planet using the formula
for distance from the focus of an ellipse:
r=a(1 −e)
where: - ris the distance from the star to the closest point on the planet’s orbit,
-ais the semi-major axis of the ellipse, and - eis the eccentricity of the ellipse.
Step 2: Substitute the given values into the formula:
r= 2 AU ×(1 −0.8) = 2 AU ×0.2 = 0.4AU
Therefore, the closest distance of the planet from the star is 0.4 AU.
Question 22
Question
Consider a planet with mass Mmoving in an elliptical orbit around a star with
mass 2M. The semi-major axis of the planet’s orbit is aand the semi-minor
axis is b. Given that the period of revolution of the planet is T, find the total
energy of the planet-star system.
Solution
Let’s denote the gravitational constant as G, the distance between the planet
and the star as r, and the eccentricity of the planet’s orbit as e.
Step 1: Find the relationship between a,b, and e. The semi-major axis a
and the semi-minor axis bof an elliptical orbit are related to the eccentricity e
by the formula:
b=a√1−e2
Step 2: Find the relationship between a,b, and r. The distance between
the planet and the star ris related to the semi-major axis aand the eccentricity
eby the formula:
r=a(1 −e)
Step 3: Find the orbital speed of the planet in terms of rand T. The
orbital speed of the planet is given by:
v=2πr
T
Step 4: Find the potential energy of the planet-star system. The potential
energy of the planet-star system is given by:
U=−GM(2M)
r
19
Step 5: Find the kinetic energy of the planet. The kinetic energy of the
planet is given by:
K=1
2Mv2
Step 6: Find the total energy of the planet-star system using the conserva-
tion of energy. The total energy Eof the system is the sum of the kinetic and
potential energies:
E=K+U
Therefore, the total energy of the planet-star system can be expressed in
terms of the given variables G,M,a,T, and e.
Question 23
Question
Consider a planet that orbits a star in an elliptical orbit. The planet is at its
closest point to the star (perihelion) at a distance of 0.3 AU and at its farthest
point from the star (aphelion) at a distance of 0.7 AU. If the period of the
planet’s orbit is 1.5 years, determine:
1. The semi-major axis of the planet’s orbit.
2. The eccentricity of the planet’s orbit.
Solution
Let’s denote the distance from the star to the perihelion as rp= 0.3 AU and
the distance from the star to the aphelion as ra= 0.7 AU. The period of the
planet’s orbit is given as T= 1.5 years.
Step 1: Calculate the semi-major axis The semi-major axis aof an
elliptical orbit is given by the formula:
a=rp+ra
2
Substitute rp= 0.3AU and ra= 0.7AU into the formula:
a=0.3+0.7
2
a=1
2
a= 0.5AU
Therefore, the semi-major axis of the planet’s orbit is 0.5 AU.
20
Step 2: Calculate the eccentricity The eccentricity eof an elliptical
orbit is related to the distance from the star to the perihelion and aphelion by
the formula:
e=ra−rp
ra+rp
Substitute rp= 0.3AU and ra= 0.7AU into the formula:
e=0.7−0.3
0.7+0.3
e=0.4
1
e= 0.4
Therefore, the eccentricity of the planet’s orbit is 0.4.
Question 24
Question
A planet orbits a star in an elliptical path. At its closest approach to the star
(perihelion), the planet has a speed of 30 km/s, and at its farthest distance from
the star (aphelion), the speed is 15 km/s.
1. Calculate the eccentricity of the planet’s orbit.
2. Determine the speeds of the planet when it is at a distance from the star
equal to half the distance at perihelion.
Solution
Let’s denote the speed of the planet at perihelion as vp= 30 km/s, the speed at
aphelion as va= 15 km/s, and the distance at perihelion as rp.
Step 1: Calculate the eccentricity of the planet’s orbit.
The speed of the planet can be related to its distance from the star using
Kepler’s second law of planetary motion, which states that the line connecting
a planet to the Sun sweeps out equal areas in equal times. This implies that
the angular momentum of the planet (L) is constant.
The specific angular momentum of the planet is given by L=r×mv, which
is constant.
At perihelion:
Lp=rp×m×vp
At aphelion:
La=ra×m×va
Since the specific angular momentum Lis constant, Lp=La.
21
Step 2: Determine the speeds of the planet when it is at a distance from
the star equal to half the distance at perihelion.
Let r=rp
2be the distance from the star. The speed of the planet at this
distance can be calculated using the conservation of angular momentum.
Applying the law of conservation of angular momentum at distance r:
r×m×v=rp×m×vp
Solving for vgives the speed of the planet at a distance r=rp
2.
Question 25
Question
A planet follows an elliptical orbit with the Sun located at one of the foci. The
planet travels from its closest point to the Sun (perihelion) to its farthest point
(aphelion). If the ratio of the distance of the planet from the Sun at perihelion
to the distance at aphelion is 1:2, calculate the eccentricity of the planet’s orbit.
Solution
Step 1: Recall that the eccentricity eof an elliptical orbit can be found using
the formula:
e=ra−rp
ra+rp
where rais the distance of the planet from the Sun at aphelion and rpis the
distance at perihelion.
Step 2: Given that the ratio of the distance from the Sun at perihelion to
aphelion is 1:2, let’s denote the distance at perihelion as dunits, which means
the distance at aphelion is 2dunits.
Step 3: Substitute ra= 2dand rp=dinto the eccentricity formula:
e=2d−d
2d+d=d
3d=1
3
Step 4: Therefore, the eccentricity of the planet’s orbit is 1
3.
22
a=1
2(3.74 ×1011)
a= 1.87 ×1011 meters
Step 5: Now we can substitute the period and semi-major axis into Kepler’s
Third Law to solve for the mass of the star:
(2 years)2=(4π2
G)(1.87 ×1011)3
Step 6: Solve for the gravitational constant G:
4 = (4π2
G)(1.87 ×1011)3
G=4π2
4×(1
1.87 ×1011 )3
G=π2
1.873×4.195 ×10−32
Step 7: Calculate the value of Gto find the semi-major axis of the orbit
around the star.
Question 2
Question
A planet orbits a star in an elliptical path with eccentricity e= 0.3. If the
distance between the planet and the star at the closest point of approach (per-
ihelion) is 100 million km, determine the distance between the planet and the
star at the farthest point of separation (aphelion).
Solution
Let rmin = 100 million km be the distance between the planet and the star at
perihelion. We can find the distance at aphelion, rmax, using Kepler’s second
law in combination with the fact that total mechanical energy is conserved for
an object moving under the influence of gravity.
Step 1: Find the semi-major axis of the elliptical orbit. The semi-major axis
aof an elliptical orbit is related to the distance at perihelion and the eccentricity
by the formula:
a=rmin
1−e.
Substitute rmin = 100 million km and e= 0.3into the formula:
a=100
1−0.3=100
0.7= 142.857 million km.
2
Step 2: Use Kepler’s second law to find the distance at aphelion. Kepler’s
second law states that the line joining the planet and the star sweeps out equal
areas in equal times. This implies that the time taken for the planet to travel
from perihelion to aphelion is the same as the time taken to travel from aphelion
back to perihelion. Since the planet moves faster when it is closer to the star,
the distance at aphelion, rmax, can be determined in terms of the semi-major
axis aand the distance at perihelion rmin:
rmax = 2a−rmin.
Substitute a= 142.857 million km and rmin = 100 million km into the formula:
rmax = 2 ×142.857 −100 = 285.714 −100 = 185.714 million km.
Therefore, the distance between the planet and the star at aphelion is 185.714
million km.
Question 3
Question
A planet orbiting a star follows a path that can be described by the equation
r= 2 cos(2θ), where ris the distance from the planet to the star and θis the
angle measured counterclockwise from the positive x-axis. Determine the shape
of the orbit and discuss which of Kepler’s laws apply in this case.
Solution
Step 1: Start by plotting the equation r= 2 cos(2θ)to visualize the shape of
the orbit.
Step 2: To plot the polar equation, it’s useful to first convert it to Cartesian
coordinates. Recall that x=rcos(θ)and y=rsin(θ). Substituting r=
2 cos(2θ):
x= 2 cos(2θ) cos(θ)and y= 2 cos(2θ) sin(θ)
Step 3: Simplify the expressions for xand y:
x= 2 cos(2θ) cos(θ) = cos(3θ) + cos(θ)
y= 2 cos(2θ) sin(θ) = sin(3θ)−sin(θ)
Step 4: Now, plot the Cartesian equation obtained in Step 3 to identify the
shape of the orbit. The shape of the orbit corresponds to a curve traced by the
planet.
Step 5: The shape of the orbit described by the equation r= 2 cos(2θ)is
an ellipse. It represents Kepler’s First Law, which states that planets move in
elliptical orbits with the star at one of the foci.
Step 6: The orbit’s parameters, such as the eccentricity and orientation, can
be further analyzed to determine if Kepler’s Second Law and Third Law are
applicable in this specific case.
3
Question 4
Question
An asteroid has an elliptical orbit around the Sun with semi-major axis a=
3.0×1011 m and eccentricity e= 0.4. Calculate the period of revolution of the
asteroid around the Sun.
Solution
Step 1: Recall Kepler’s Third Law which states that the square of the period of
revolution of a planet (or asteroid) is proportional to the cube of the semi-major
axis of its orbit. Mathematically, this can be written as:
T2∝a3
Step 2: To find the period of revolution T, we need to determine the pro-
portionality constant. Since we are dealing with astronomical distances, we will
use the gravitational constant Gand the mass of the Sun M⊙to find the exact
relationship. The formula for the period then becomes:
T= 2π√a3
GM⊙
Step 3: Substitute the given values into the formula:
T= 2π√(3.0×1011 m)3
6.67 ×10−11 m3kg−1s−2×1.99 ×1030 kg
Step 4: Calculate the period of revolution:
T= 2π√27 ×1033 m3
13.3×1019 m3kg−1s−2
T= 2π√2.03 ×1014 s2
T= 2π×1.43 ×107s
T≈9.00 ×107s
Therefore, the period of revolution of the asteroid around the Sun is approx-
imately 9.00 ×107seconds.
Question 5
Question
In accordance with Kepler’s laws of planetary motion, consider a hypothetical
planet orbiting a star with a semi-major axis of 2 AU. The planet completes
one full orbit around the star in 500 days. Determine the period of the planet’s
orbit if its semi-major axis were doubled to 4 AU.
4
Solution
To solve this problem, we can use Kepler’s third law which states that the square
of the period of an orbit is directly proportional to the cube of the semi-major
axis of the orbit. Mathematically, this can be expressed as:
T2
1
a3
1
=T2
2
a3
2
where T1and a1are the initial period and semi-major axis, and T2and a2
are the final period and semi-major axis.
Step 1: Substitute the given values into the equation:
Given: T1= 500 days, a1= 2 AU, a2= 4 AU.
Plugging into the formula:
5002
23=T2
2
43
Simplify:
5002
23=T2
2
64
Step 2: Solve for T2:
5002
8=T2
2
62500 = T2
2
T2=√62500
T2= 250 days
Therefore, if the semi-major axis of the planet’s orbit were doubled to 4 AU,
the period of its orbit would be 250 days.
Question 6
Question
A planet follows an elliptical orbit around the Sun. According to Kepler’s laws,
the square of the period of the planet’s orbit is proportional to the cube of the
semi-major axis of its elliptical orbit. Suppose a planet has a period of 10 years
and its semi-major axis is 3 AU (astronomical units). What would be the period
of another planet with a semi-major axis of 4 AU in the same system?
5
Solution
Step 1: Determine the proportionality relationship given by Kepler’s Third Law:
Let T1be the period of the first planet (with semi-major axis a1) and T2
be the period of the second planet (with semi-major axis a2). According to
Kepler’s Third Law,
T2
1
T2
2
=a3
1
a3
2
Step 2: Given that the period of the first planet is 10 years and its semi-major
axis is 3 AU, we have:
102
T2
2
=33
43
Step 3: Solve for T2:
100
T2
2
=27
64
Step 4: Cross multiply and solve for T2:
27T2
2= 6400
T2
2=6400
27
T2=√6400
27 ≈16.17 years
Therefore, the period of the second planet with a semi-major axis of 4 AU
would be approximately 16.17 years in the same system.
Question 7
Question
An asteroid orbits the Sun in an elliptical path. The closest distance of the
asteroid to the Sun is 0.8 AU, while the farthest distance is 2.4 AU. If the
period of the asteroid’s orbit is 3.5 years, determine the semi-major axis of the
orbit. Recall that one astronomical unit (AU) is the average distance between
the Earth and the Sun, approximately 1.5×1011 meters.
6
Solution
Step 1: Find the semi-major axis using Kepler’s third law, which relates the
period of the orbit to the semi-major axis of the orbit. The formula is given by:
T2=4π2
G(M1+M2)a3
where: T= period of the orbit, G= gravitational constant, M1and M2=
masses of the two objects (in this case, the asteroid and the Sun), a= semi-
major axis of the orbit.
Step 2: Convert the distances of the asteroid’s closest and farthest points
from AU to meters. Recall that 1AU is approximately 1.5×1011 meters.
rmin = 0.8AU ×1.5×1011 m/AU
rmax = 2.4AU ×1.5×1011 m/AU
Step 3: The semi-major axis of an elliptical orbit is the average of the closest
and farthest distances from the primary focus. Calculate the semi-major axis:
a=rmin +rmax
2
Step 4: Finally, substitute the known values into Kepler’s third law equation
and solve for the semi-major axis a. With G= 6.67 ×10−11 m3/kg s2and
MSun = 1.989 ×1030 kg.
3.52=4π2
G(MSun)a3
a=(4π2×6.67 ×10−11 m3/kg s2
1.989 ×1030 kg )−1/3
×3.5
Question 8
Question
An exoplanet is discovered orbiting a star in a distant galaxy. Observations show
that the exoplanet has an orbital period of 200 days and an average orbital radius
of 0.6 astronomical units (AU). If the star has a mass of 2×1030 kg, determine
the exoplanet’s orbital speed.
Solution
Step 1: Find the gravitational force between the star and the exoplanet using
Newton’s law of universal gravitation: The gravitational force between the star
and the exoplanet is given by
F=G·Mstar ·mplanet
r2
7
where Gis the gravitational constant, Mstar is the mass of the star, mplanet is
the mass of the planet (assuming negligible compared to the star), and ris the
average orbital radius of the exoplanet. For the exoplanet orbiting the star:
F=G·(2 ×1030 kg)·mplanet
(0.6AU ×1.5×1011 m/AU)2
F=6.67 ×10−11 N·m2/kg2·2×1030 kg ·mplanet
(0.9×1011 m)2
Step 2: Find the centripetal force acting on the exoplanet: The centripetal
force required to keep an object in a circular orbit is given by
F=mplanet ·v2
r
where mplanet is the mass of the exoplanet, vis its orbital speed, and ris the
average orbital radius of the exoplanet.
Step 3: Equate the gravitational force with the centripetal force: Set the
gravitational force equal to the centripetal force:
G·Mstar ·mplanet
r2=mplanet ·v2
r
Step 4: Solve for the orbital speed of the exoplanet: Rewrite the equation
in terms of the orbital speed v:
v=√G·Mstar
r
Step 5: Substitute the given values and calculate the orbital speed: Plugging
in the known values:
v=√6.67 ×10−11 N·m2/kg2·2×1030 kg
0.6AU ×1.5×1011 m/AU
v=√1.334 ×1020
9×1010 =√1.4822 ×109m/s ≈1.218 ×104m/s
Therefore, the orbital speed of the exoplanet is approximately 1.218 ×104
m/s.
Question 9
Question
In a distant planetary system, a planet orbits its star in an elliptical path. The
semi-major axis of the planet’s orbit is 2.5 AU and its eccentricity is 0.4. If
the planet’s closest approach to the star (perihelion) is 1.5 AU, determine the
farthest distance of the planet from the star (aphelion).
8
Solution
Step 1: Recall that for an elliptical orbit, the distance from the focus (star) to
any point on the orbit is constant and equal to the semi-major axis, denoted by
a.
Step 2: The distance from the center to the farthest point on the ellipse
(aphelion) is given by a(1 + e), where eis the eccentricity of the ellipse.
Step 3: Given that the semi-major axis a= 2.5AU and eccentricity e= 0.4,
we can calculate the aphelion distance using the formula a(1 + e).
Step 4: Substitute the values of aand einto the formula to find the aphelion
distance:
Aphelion distance = 2.5×(1 + 0.4) AU
Step 5: Perform the calculations:
Aphelion distance = 2.5×1.4AU = 3.5AU
Step 6: Therefore, the farthest distance of the planet from the star (aphelion)
is 3.5 AU.
Question 10
Question
A planet is in a circular orbit around a star with a period of 1 year. If the
planet’s mass is doubled, what will be the new period of the planet’s orbit,
assuming the distance between the planet and the star remains constant?
(Given: Kepler’s third law states that the square of the period of an orbit
is proportional to the cube of the average distance between the planet and the
star.)
Solution
Step 1: Let Tbe the original period of the planet’s orbit, and Mbe the original
mass of the planet. Let T′be the new period of the planet’s orbit after the mass
is doubled (i.e., 2M).
Step 2: According to Kepler’s third law,
T2∝r3
where ris the average distance between the planet and the star.
Step 3: Since the distance rbetween the planet and the star remains con-
stant, the cube of the distance will not change.
Step 4: We can set up a ratio using the original and new masses of the
planet:
T′2
T2=(2M)3
M3
9
Step 5: Simplifying the ratio, we get:
T′2
T2= 8
Step 6: Taking the square root of both sides gives:
T′
T=√8 = 2√2
Step 7: Therefore, the new period T′of the planet’s orbit is 2√2years.
Question 11
Question
A planet has an elliptical orbit around the Sun, with the eccentricity of the orbit
being 0.4. The distance between the Sun and the planet at its closest approach
is 0.6 AU. Find the distance between the Sun and the planet at its farthest
distance.
Solution
Step 1: Let’s denote the distance between the Sun and the planet at its farthest
distance as rmax, and the distance at its closest approach as rmin.
Step 2: According to Kepler’s second law, the area swept out by the line
joining the Sun and the planet is the same during equal intervals of time. This
means that the planet moves fastest when it is closest to the Sun, and slowest
when it is farthest away.
Step 3: We can use the fact that the area of an ellipse is given by πab,
where ais the semi-major axis and bis the semi-minor axis. Since the Sun is
at one of the foci of the ellipse, we have πab =1
2rminvmin =1
2rmaxvmax, where
vmin and vmax are the velocities of the planet at its closest and farthest points,
respectively.
Step 4: We know that the orbital energy of the planet is given by E=
−GMm
2a, where Mis the mass of the Sun and mis the mass of the planet. At
the closest point, the total energy is entirely kinetic, and at the farthest point,
it is entirely potential.
Step 5: By conservation of energy, we have E=1
2mv2−GMm
r, where vis
the speed of the planet and ris its distance from the Sun. Using this equation
at the closest and farthest points, we can set up two equations to solve for rmax.
Step 6: Solving the equations, we find rmax = 1.4AU. Therefore, the distance
between the Sun and the planet at its farthest distance is 1.4 AU.
10
Question 12
Question
Suppose a comet is moving in an elliptical orbit around the Sun. The comet’s
closest distance to the Sun (perihelion) is 0.3 AU and its farthest distance from
the Sun (aphelion) is 6.0 AU. Calculate the eccentricity of the comet’s orbit.
Solution
Step 1: Recall the definition of eccentricity, e, for an ellipse:
e=rmax −rmin
rmax +rmin
where rmax is the distance from the center to the aphelion, and rmin is the
distance from the center to the perihelion.
Step 2: Substitute rmax = 6.0AU and rmin = 0.3AU into the formula:
e=6.0−0.3
6.0+0.3
Step 3: Calculate the numerator and the denominator separately:
e=5.7
6.3
Step 4: Simplify the expression to find the eccentricity:
e=57
63 =19
21
Thus, the eccentricity of the comet’s orbit is 19
21 .
Question 13
Question
An asteroid is orbiting the Sun in an elliptical orbit with an eccentricity of
0.6. If the asteroid is closest to the Sun at a distance of 0.5 AU, determine its
farthest distance from the Sun. Assume the Sun’s mass is 2×1030 kg and the
gravitational constant is 6.67 ×10−11 N m2/kg2.
Solution
Step 1: We can determine the semi-major axis of the asteroid’s orbit using the
given closest distance (rmin) and the eccentricity (e) of the orbit. The semi-
major axis (a) is related to the closest distance by the formula:
a=rmin
1−e
11
Step 2: Substituting rmin = 0.5AU and e= 0.6into the equation:
a=0.5
1−0.6=0.5
0.4= 1.25 AU
Step 3: The farthest distance (rmax) of the asteroid from the Sun occurs at
the aphelion, which is at a distance of a(1 + e)from the Sun.
rmax =a(1 + e) = 1.25(1 + 0.6) = 1.25(1.6) = 2 AU
Therefore, the farthest distance of the asteroid from the Sun is 2 AU.
Question 14
Question
A planet orbits a star in an elliptical orbit. The semi-major axis of the orbit is
2 AU and the eccentricity of the orbit is 0.6. If the planet is closest to the star
at perihelion and farthest from the star at aphelion, find the distances of the
planet from the star at perihelion and aphelion.
Solution
Step 1: Calculate the distance of the planet from the star at perihelion. At
perihelion, the planet is closest to the star. The distance from the star at
perihelion is given by:
rperihelion =a(1 −e)
where ais the semi-major axis and eis the eccentricity. Substituting a= 2 AU
and e= 0.6:
rperihelion = 2(1 −0.6) = 2(0.4) = 0.8AU
Step 2: Calculate the distance of the planet from the star at aphelion. At
aphelion, the planet is farthest from the star. The distance from the star at
aphelion is given by:
raphelion =a(1 + e)
Substituting a= 2 AU and e= 0.6:
raphelion = 2(1 + 0.6) = 2(1.6) = 3.2AU
Therefore, the planet is 0.8 AU from the star at perihelion and 3.2 AU from
the star at aphelion.
12
Question 15
Question
Consider a planet that follows a perfectly elliptical orbit around a star. The
planet takes 60 days to travel from its closest point to the star (perihelion) to
its farthest point from the star (aphelion). If the distance between the planet
and the star at perihelion is 0.4 AU, determine the distance between the planet
and the star at aphelion in AU.
Solution
Step 1: Recall Kepler’s second law, which states that a line segment joining a
planet and the sun sweeps out equal areas during equal intervals of time. This
law can be written as: dA
dt =constant
where dA is the area swept out by the line segment in time dt.
Step 2: In this case, since the planet follows an elliptical orbit, we know
that the area swept out at perihelion is equal to the area swept out at aphelion.
Let rpbe the distance between the planet and the star at perihelion, and let ra
be the distance at aphelion. Then, the area swept out at perihelion is 1
2r2
p∆θ,
and the area swept out at aphelion is 1
2r2
a∆θ, where ∆θis the change in angle
subtended by the planet at the star.
Step 3: Since the time taken to travel from perihelion to aphelion is 60 days,
we can relate the areas at perihelion and aphelion using the fact that they sweep
out equal areas in equal time intervals. Thus, we have:
1
2r2
p∆θ=1
2r2
a∆θ
r2
p=r2
a
ra=rp= 0.4AU
Therefore, the distance between the planet and the star at aphelion is also
0.4 AU.
Question 16
Question
In an exoplanetary system, a distant planet orbits a star in an elliptical path.
The planet’s closest distance to the star is 0.3 AU and its farthest distance is
0.9 AU. If the period of the planet’s orbit is 600 days, determine the eccentricity
of the planet’s orbit.
13
Solution
Step 1: Determine the semi-major axis of the planet’s orbit. Given that the
closest distance (rmin) is 0.3 AU and the farthest distance (rmax) is 0.9 AU, the
semi-major axis (a) of the elliptical orbit can be calculated as the average of
these two distances.
a=rmin +rmax
2=0.3+0.9
2= 0.6AU
Step 2: Determine the eccentricity of the planet’s orbit using Kepler’s third
law. Kepler’s third law relates the period of an orbiting body to the semi-major
axis of its orbit. It can be expressed as:
T2=4π2a3
G(M1+M2)
where Tis the period of the orbit, ais the semi-major axis, Gis the gravitational
constant, and M1and M2are the masses of the two bodies.
Given that T= 600 days, a= 0.6AU, and G≈6.67430 ×10−11 m3/kgs2,
we can rewrite the equation in terms of AU and days:
(600 days)2=4π2(0.6AU)3
G(M1+M2)
Step 3: Calculate the sum of the masses M1+M2. We can rewrite the
equation from step 2 in terms of the sum of the masses:
M1+M2=4π2(0.6AU)3
G(600 days)2
Step 4: Calculate the eccentricity of the orbit. The eccentricity (e) of an
elliptical orbit can be determined using the formula:
e=√1−b2
a2
where ais the semi-major axis and bis the semi-minor axis of the ellipse.
Since the orbit is elliptical, we need to find the value of b(semi-minor axis)
using the formula for an ellipse:
b=a√1−e2
Given that a= 0.6AU, we substitute the avalue calculated earlier to find
b.
Step 5: Calculate the eccentricity using the semi-major axis and semi-minor
axis. Substitute the values of aand binto the eccentricity formula to find the
eccentricity e.
14
Question 17
Question
In a distant solar system, a planet has an orbital period of 600 Earth days and
an average orbital radius of 2 AU. Calculate the mass of the star around which
the planet orbits, given that the mass of the planet is 3.2×1025 kg. Assume
the orbit is nearly circular.
Solution
Let’s denote the mass of the star as Mand the orbital radius of the planet as r.
According to Kepler’s third law, the square of the orbital period of a planet is
proportional to the cube of its semi-major axis (in this case, the orbital radius):
T2=4π2r3
G(M+m)
where: - Tis the orbital period, - ris the orbital radius, - Gis the grav-
itational constant, - Mis the mass of the star, and - mis the mass of the
planet.
Given: - The orbital period T= 600 Earth days = 600 ×24 ×3600 seconds,
- The orbital radius r= 2 AU = 2 ×1.496 ×1011 m, - The mass of the planet
m= 3.2×1025 kg.
We can rearrange the formula to solve for the star’s mass M:
M=4π2r3
GT 2−m
Step 1: Convert the orbital period to seconds:
T= 600 ×24 ×3600 = 51840000 seconds
Step 2: Substitute the known values into the formula:
M=4π2(2 ×1.496 ×1011)3
6.67430 ×10−11 ×518400002−3.2×1025
Step 3: Calculate the mass of the star Musing a calculator:
M≈4π2(2 ×1.496 ×1011)3
6.67430 ×10−11 ×518400002−3.2×1025
M≈2.059 ×1030 kg
Therefore, the mass of the star around which the planet orbits is approxi-
mately 2.059 ×1030 kg.
15
Question 18
Question
Consider a star with a mass of 2×1030 kg and a planet with a mass of 6×1024
kg in a circular orbit around the star. The radius of the planet’s orbit is 2×1011
m. Determine the period of the planet’s orbit.
Solution
Step 1: Calculate the gravitational force between the star and the planet using
Newton’s law of universal gravitation:
F=G·M1·M2
R2
where Fis the gravitational force, Gis the gravitational constant (6.67 ×10−11
N m2/kg2), M1and M2are the masses of the star and planet respectively, and
Ris the radius of the planet’s orbit.
F=(6.67 ×10−11 N m2/kg2)·(2 ×1030 kg)·(6 ×1024 kg)
(2 ×1011 m)2
F=8.004 ×1044
4×1022
F= 2.001 ×1022 N
Step 2: Use the centripetal force formula to find the velocity of the planet:
Fcentripetal =M·V2
R
where Mis the mass of the planet and Vis the velocity of the planet.
Setting the gravitational force equal to the centripetal force, we have:
M1·M2
R2=M·V2
R
Solving for V:
V=√M1·M2·R
M
V=√2×1030 ×6×1024 ×2×1011
6×1024
V=√2.4×1065
6×1024
V=√4×1040
16
V= 2 ×1020 m/s
Step 3: Calculate the period of the planet’s orbit using the formula for
circular motion:
V=2πR
T
where Tis the period of the orbit.
Solving for T:
T=2πR
V
T=2π×2×1011
2×1020
T=4π×1011
2×1020
T= 2 ×10−9s
Therefore, the period of the planet’s orbit is 2×10−9seconds.
Question 19
Question
A planet orbits a star in a nearly circular orbit with a radius of 2.5 AU. The
orbital period of the planet is 4 years. Determine the mass of the star in terms
of the mass of the Sun (M⊙).
Solution
Step 1: Recall Kepler’s Third Law, which states that the square of the period
of an orbit is proportional to the cube of the semi-major axis of the orbit.
Mathematically, this is expressed as:
T2
1
a3
1
=T2
2
a3
2
where Tis the orbital period and ais the semi-major axis.
Step 2: Let’s denote the mass of the star as Msand the mass of the planet
as Mp. According to Newton’s version of Kepler’s third law, the sum of the
masses of the star and the planet is involved:
Ms+Mp=4π2a3
GT 2
where Gis the gravitational constant.
Step 3: Given that the star is much more massive than the planet, we can
assume Ms≈Mp.
17
Step 4: Using the information given in the question, we can substitute a=
2.5AU and T= 4 years into the equation to solve for Msin terms of M⊙:
Ms≈4π2(2.5AU)3
G(4 years)2=4π2(2.5×1.496 ×1011 m)3
G(4 ×365.25 ×24 ×3600 s)2M⊙
Step 5: Calculating the mass of the star in terms of the mass of the Sun will
give us the final answer.
Question 20
Question
An asteroid orbits a distant star with a period of 10 years and a semi-major axis
of 3 AU. If the eccentricity of the orbit is 0.4, calculate the asteroid’s closest
distance to the star (perihelion) in AU.
(Note: AU stands for astronomical unit and is the average distance between
the Earth and the Sun, approximately 149.6 million kilometers.)
Solution
Step 1: Calculate the eccentricity distance eusing the semi-major axis aand
the perihelion distance r1.
eccentricity e= 1 −r1
a
Given that a= 3 AU and e= 0.4:
0.4 = 1 −r1
3
r1= 3(1 −0.4) = 3 ×0.6 = 1.8AU
Therefore, the perihelion distance of the asteroid from the star is 1.8 AU.
Question 21
Question
A planet orbits a star in a highly eccentric ellipse. The semi-major axis of the
ellipse is 2 AU, while the eccentricity is 0.8. If the period of the planet’s orbit
is 5 years, what is the closest distance of the planet from the star?
18
Solution
Step 1: Find the closest distance from the star to the planet using the formula
for distance from the focus of an ellipse:
r=a(1 −e)
where: - ris the distance from the star to the closest point on the planet’s orbit,
-ais the semi-major axis of the ellipse, and - eis the eccentricity of the ellipse.
Step 2: Substitute the given values into the formula:
r= 2 AU ×(1 −0.8) = 2 AU ×0.2 = 0.4AU
Therefore, the closest distance of the planet from the star is 0.4 AU.
Question 22
Question
Consider a planet with mass Mmoving in an elliptical orbit around a star with
mass 2M. The semi-major axis of the planet’s orbit is aand the semi-minor
axis is b. Given that the period of revolution of the planet is T, find the total
energy of the planet-star system.
Solution
Let’s denote the gravitational constant as G, the distance between the planet
and the star as r, and the eccentricity of the planet’s orbit as e.
Step 1: Find the relationship between a,b, and e. The semi-major axis a
and the semi-minor axis bof an elliptical orbit are related to the eccentricity e
by the formula:
b=a√1−e2
Step 2: Find the relationship between a,b, and r. The distance between
the planet and the star ris related to the semi-major axis aand the eccentricity
eby the formula:
r=a(1 −e)
Step 3: Find the orbital speed of the planet in terms of rand T. The
orbital speed of the planet is given by:
v=2πr
T
Step 4: Find the potential energy of the planet-star system. The potential
energy of the planet-star system is given by:
U=−GM(2M)
r
19
Step 5: Find the kinetic energy of the planet. The kinetic energy of the
planet is given by:
K=1
2Mv2
Step 6: Find the total energy of the planet-star system using the conserva-
tion of energy. The total energy Eof the system is the sum of the kinetic and
potential energies:
E=K+U
Therefore, the total energy of the planet-star system can be expressed in
terms of the given variables G,M,a,T, and e.
Question 23
Question
Consider a planet that orbits a star in an elliptical orbit. The planet is at its
closest point to the star (perihelion) at a distance of 0.3 AU and at its farthest
point from the star (aphelion) at a distance of 0.7 AU. If the period of the
planet’s orbit is 1.5 years, determine:
1. The semi-major axis of the planet’s orbit.
2. The eccentricity of the planet’s orbit.
Solution
Let’s denote the distance from the star to the perihelion as rp= 0.3 AU and
the distance from the star to the aphelion as ra= 0.7 AU. The period of the
planet’s orbit is given as T= 1.5 years.
Step 1: Calculate the semi-major axis The semi-major axis aof an
elliptical orbit is given by the formula:
a=rp+ra
2
Substitute rp= 0.3AU and ra= 0.7AU into the formula:
a=0.3+0.7
2
a=1
2
a= 0.5AU
Therefore, the semi-major axis of the planet’s orbit is 0.5 AU.
20
Step 2: Calculate the eccentricity The eccentricity eof an elliptical
orbit is related to the distance from the star to the perihelion and aphelion by
the formula:
e=ra−rp
ra+rp
Substitute rp= 0.3AU and ra= 0.7AU into the formula:
e=0.7−0.3
0.7+0.3
e=0.4
1
e= 0.4
Therefore, the eccentricity of the planet’s orbit is 0.4.
Question 24
Question
A planet orbits a star in an elliptical path. At its closest approach to the star
(perihelion), the planet has a speed of 30 km/s, and at its farthest distance from
the star (aphelion), the speed is 15 km/s.
1. Calculate the eccentricity of the planet’s orbit.
2. Determine the speeds of the planet when it is at a distance from the star
equal to half the distance at perihelion.
Solution
Let’s denote the speed of the planet at perihelion as vp= 30 km/s, the speed at
aphelion as va= 15 km/s, and the distance at perihelion as rp.
Step 1: Calculate the eccentricity of the planet’s orbit.
The speed of the planet can be related to its distance from the star using
Kepler’s second law of planetary motion, which states that the line connecting
a planet to the Sun sweeps out equal areas in equal times. This implies that
the angular momentum of the planet (L) is constant.
The specific angular momentum of the planet is given by L=r×mv, which
is constant.
At perihelion:
Lp=rp×m×vp
At aphelion:
La=ra×m×va
Since the specific angular momentum Lis constant, Lp=La.
21
Step 2: Determine the speeds of the planet when it is at a distance from
the star equal to half the distance at perihelion.
Let r=rp
2be the distance from the star. The speed of the planet at this
distance can be calculated using the conservation of angular momentum.
Applying the law of conservation of angular momentum at distance r:
r×m×v=rp×m×vp
Solving for vgives the speed of the planet at a distance r=rp
2.
Question 25
Question
A planet follows an elliptical orbit with the Sun located at one of the foci. The
planet travels from its closest point to the Sun (perihelion) to its farthest point
(aphelion). If the ratio of the distance of the planet from the Sun at perihelion
to the distance at aphelion is 1:2, calculate the eccentricity of the planet’s orbit.
Solution
Step 1: Recall that the eccentricity eof an elliptical orbit can be found using
the formula:
e=ra−rp
ra+rp
where rais the distance of the planet from the Sun at aphelion and rpis the
distance at perihelion.
Step 2: Given that the ratio of the distance from the Sun at perihelion to
aphelion is 1:2, let’s denote the distance at perihelion as dunits, which means
the distance at aphelion is 2dunits.
Step 3: Substitute ra= 2dand rp=dinto the eccentricity formula:
e=2d−d
2d+d=d
3d=1
3
Step 4: Therefore, the eccentricity of the planet’s orbit is 1
3.
22
