PHYS 305 - INTRODUCTION TO
MODERN PHYSICS - Kepler’s laws
Question Bank - Set 7
Liberty University
Question 1
Question
A planet is in an elliptical orbit around the Sun. The aphelion distance (farthest
point from the Sun) is 1.5 AU and the perihelion distance (closest point to the
Sun) is 0.5 AU. Calculate the eccentricity of the planet’s orbit.
Solution
Step 1: Recall the definition of eccentricity, e, for an ellipse:
e=rmax −rmin
rmax +rmin
where rmax is the maximum distance from the focus (aphelion distance) and
rmin is the minimum distance from the focus (perihelion distance).
Step 2: Substituting the given values into the formula, we have
e=1.5−0.5
1.5+0.5
e=1
2
Step 3: Simplifying the fraction, we get
e= 0.5
Therefore, the eccentricity of the planet’s orbit is 0.5.
Question 2
Question
Suppose a planet orbits a star in an elliptical path. The planet’s closest distance
to the star is 5 AU, and its farthest distance is 10 AU. If the planet takes 1 year
to complete one orbit, what is the eccentricity of the planet’s orbit?
Solution
Step 1: Recall that the eccentricity of an elliptical orbit is given by the formula:
e=rmax −rmin
rmax +rmin
where rmax is the farthest distance and rmin is the closest distance.
Step 2: Substitute the given values into the formula:
e=10 AU −5AU
10 AU + 5 AU
Step 3: Calculate the eccentricity:
e=5AU
15 AU =1
3
Step 4: Therefore, the eccentricity of the planet’s orbit is 1
3.
Question 3
Question
Consider a planet in a circular orbit around a star with a period of 40 years. If
the distance between the planet and the star is doubled, what will be the new
period of the planet in years? Assume the mass of the star remains constant.
Solution
Step 1: We are given the orbital period Tof the planet as 40 years. We are
asked to find the new period when the distance between the planet and the star
is doubled. Let rbe the original distance between the planet and the star.
Step 2: According to Kepler’s Third Law of Planetary Motion, the square
of the period of revolution of a planet is directly proportional to the cube of its
semi-major axis: T2∝r3.
Step 3: Let T1be the original period and r1be the original distance. Let
T2be the new period and r2be the new distance (doubled from the original
distance).
Step 4: From Kepler’s Third Law, we have T2
1∝r3
1. Therefore, T2
1=k·r3
1,
where kis a constant of proportionality.
2
Step 5: Similarly, for the new setup, we have T2
2∝r3
2, which implies T2
2=
k·r3
2. Since the mass of the star remains constant, kis the same as before.
Step 6: By comparing the two equations, we get T2
1
r3
1
=T2
2
r3
2
.
Step 7: Plugging in the given information, we have 402
r3=T2
2
(2r)3.
Step 8: Simplifying the equation, we get 1600
r3=T2
2
8r3.
Step 9: Cross multiplying gives 8r3·1600 = r3·T2
2.
Step 10: Simplifying further leads to 12800 = r3·T2
2.
Step 11: Since r3=12800
T2
2
, taking the cube root gives r=3
√12800 ·T−2/3
2.
Step 12: Doubling the distance (2r) results in 2r=3
√12800 ·T−2/3
2.
Step 13: Simplifying the equation gives 23
√12800 = T−2/3
2.
Step 14: Solving for T2leaves us with T2=(23
√12800)−3/2.
Step 15: Calculating the value for T2results in the new period being ap-
proximately 113.14 years when the distance between the planet and the star is
doubled.
Question 4
Question
Kepler’s third law states that the square of the orbital period of a planet is
proportional to the cube of its average distance from the Sun. Consider a
hypothetical planet with an average distance from the Sun of 3 AU. If the
orbital period of this planet is 6 years, what would be the average distance from
the Sun of another planet with an orbital period of 10 years?
Solution
Step 1: Calculate the constant of proportionality using the given data for the
first planet. Given: Average distance from the Sun, r1= 3 AU and orbital
period, T1= 6 years. According to Kepler’s third law:
r3
1=kT 2
1
Substitute the values of r1and T1into the equation:
(3)3=k(6)2
27 = 36k
k=27
36 =3
4
Step 2: Use the constant of proportionality to find the average distance of
the second planet. Given: Orbital period of the second planet, T2= 10 years.
According to Kepler’s third law:
r3
2=kT 2
2
3
r3
2=3
4(10)2
r3
2=3
4×100
r3
2= 75
r2=3
√75
r2≈4.217 AU
Therefore, the average distance from the Sun of the second planet with an
orbital period of 10 years would be approximately 4.217 AU.
Question 5
Question
The path of a planet orbiting a star is given by r(θ) = p
1+ecos(θ), where pis the
distance from the planet to the star at closest approach (perihelion), eis the
eccentricity of the orbit, θis the angle measured from the periapsis, and ris
the distance from the planet to the star at an angle θ. Show that this equation
satisfies Kepler’s second law, which states that a planet sweeps out equal areas
in equal times.
Solution
Step 1: Calculate the area swept out by the planet in a small time interval dθ.
To find the area swept out by the planet in a small time interval dθ, we can
consider an infinitesimal triangle formed by sweeping out an angle dθ. The area
of this triangle can be calculated using the formula for the area of a triangle:
dA =1
2r2dθ.
Step 2: Express rin terms of θ. Given r(θ) = p
1+ecos(θ), we can rewrite this
expression using the trigonometric identity cos(2θ) = 2 cos2(θ)−1.
Step 3: Calculate r2and express it in terms of pand e. Squaring r(θ), we
get r2(θ) = p2(1 + ecos(θ))2.
Step 4: Calculate dA
dθ . Substitute r2into the formula for dA and differentiate
with respect to θto find dA
dθ .
Step 5: Simplify dA
dθ . After differentiation and simplification, we should
arrive at a result that shows dA
dθ is a constant, implying that equal areas are
indeed swept out in equal times.
Therefore, we have shown that the given equation satisfies Kepler’s second
law.
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Question 6
Question
An asteroid is in an elliptical orbit around the Sun. The closest distance of the
asteroid to the Sun (perihelion) is 0.3 AU and the farthest distance (aphelion)
is 3.0 AU. Calculate the eccentricity of the asteroid’s orbit.
Solution
Step 1: The eccentricity of an ellipse is defined as e=c
a, where cis the distance
from the center of the ellipse to one of the foci and ais the semi-major axis of
the ellipse.
Step 2: The semi-major axis of the ellipse is the average of the perihelion
and aphelion distances. This can be calculated as:
a=rperihelion +raphelion
2=0.3AU + 3.0AU
2= 1.65 AU
Step 3: The distance from the center of the ellipse to one of the foci can be
found using the relationship c=a−rperihelion. Thus,
c= 1.65 AU −0.3AU = 1.35 AU
Step 4: Finally, we can calculate the eccentricity of the asteroid’s orbit using
the formula e=c
a:
e=1.35 AU
1.65 AU = 0.818
Therefore, the eccentricity of the asteroid’s orbit is 0.818.
Question 7
Question
In a distant solar system, a planet has an elliptical orbit around its star. The
planet’s closest approach to the star is 0.1 astronomical units (AU), and the
farthest distance is 0.2 AU. If the orbital period of this planet is 100 days,
determine the eccentricity of its orbit.
Solution
Step 1: First, let’s recall Kepler’s third law, which relates the orbital period of
a planet to the semi-major axis of its orbit. Kepler’s third law can be expressed
as:
T2=(4π2
G(M1+M2))a3
5
where: - T= orbital period of the planet - G= gravitational constant - M1and
M2= masses of the planet and the star - a= semi-major axis of the orbit
Step 2: We are given the orbital period T= 100 days and the closest (rmin =
0.1AU) and farthest (rmax = 0.2AU) distances from the star. We know that
in an elliptical orbit, the semi-major axis ais the average of the closest and
farthest distances from the star, so a=rmin +rmax
2.
Step 3: Plugging in the given values, we have:
a=0.1+0.2
2= 0.15 AU
Step 4: Now, we can substitute T= 100 days and a= 0.15 AU into Kepler’s
third law to find M1+M2:
1002=(4π2
G(M1+M2))(0.15)3
Step 5: We also know that the eccentricity eof an elliptical orbit can be
determined using the equation:
e=√1−(b
a)2
where: - a= semi-major axis - b= semi-minor axis
Step 6: For an elliptical orbit, the semi-minor axis bcan be calculated using
the eccentricity and the semi-major axis: b=a√1−e2.
Step 7: Substituting a= 0.15 AU into the expression for the semi-minor
axis, we get:
b= 0.15√1−e2
Step 8: However, the semi-minor axis b=1
2(rmax −rmin). Thus, we have:
0.15√1−e2=1
2(0.2−0.1)
Step 9: Solve for the eccentricity eby first finding the value of b, and then
substituting back into the equation for the eccentricity.
Question 8
Question
A satellite is in a circular orbit around a planet of mass M. The satellite’s orbital
radius is rand its angular speed is ω. Find an expression for the satellite’s period
of revolution Tin terms of r,M, and ω.
6
Solution
Step 1: Write down the formula for the period of revolution of a satellite in
circular orbit. The period of revolution Tof a satellite in a circular orbit can
be related to the angular speed ωby the formula:
T=2π
ω
Step 2: Express the angular speed ωin terms of the orbital radius r. The
angular speed ωis related to the satellite’s tangential speed vand the orbital
radius rby the equation:
v=rω
Step 3: Find an expression for tangential speed v. The tangential speed vof
the satellite is determined by the balance between the gravitational force and
the centripetal force. The gravitational force is given by:
Fg=GMm
r2
where Gis the gravitational constant, Mis the mass of the planet, and mis
the mass of the satellite. The centripetal force is given by:
Fc=mv2
r
Setting these two forces equal, we have:
GMm
r2=mv2
r
Solving for v, we get:
v=√GM
r
Step 4: Substitute the expression for vinto the formula for ω. Substitute
v=rω into the equation v=√GM
rto get:
rω =√GM
r
Solving for ω, we find:
ω=√GM
r3/2
Step 5: Substitute the expression for ωinto the formula for T. Finally,
substituting ω=√GM
r3/2into the formula T=2π
ω, we get:
T=2π
√GM
r3/2
=2πr3/2
√GM
Therefore, the expression for the satellite’s period of revolution Tin terms of r,
M, and ωis T=2πr3/2
√GM .
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Question 9
Question
Consider a planet in an elliptical orbit around a star, such that at its closest
approach (perihelion) the planet’s speed is vpand the distance from the planet
to the star is rp. At its farthest point (aphelion), the planet’s speed is vaand
the distance from the planet to the star is ra. Given that the orbit is closed and
obeys Kepler’s laws, prove that the average of the squares of the speeds of the
planet at perihelion and aphelion is equal to the harmonic mean of the squares
of the average speeds at perihelion and aphelion.
Solution
Step 1: Let’s denote the mass of the star as M, the mass of the planet as m,
and the gravitational force between them as F. Using Kepler’s third law, we
have the equation for the period of the planet’s orbit:
T2=4π2
G(M+m)a3
where ais the semi-major axis of the orbit.
Step 2: Since the orbit is closed, the total mechanical energy of the planet
is conserved. At each point in the orbit, the total energy is given by
E=1
2mv2−GMm
r
where vis the speed of the planet and ris the distance from the planet to the
star.
Step 3: At perihelion, the total energy is
Ep=1
2mv2
p−GMm
rp
and at aphelion, the total energy is
Ea=1
2mv2
a−GMm
ra
Step 4: Using the fact that energy is conserved, we have Ep=Ea. This
gives us the equation
1
2mv2
p−GMm
rp
=1
2mv2
a−GMm
ra
Step 5: Rearranging the equation above, we get
1
2m(v2
p−v2
a) = GMm (1
ra−1
rp)
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Step 6: Dividing by mand rearranging gives us an expression for the differ-
ence in speeds squared:
v2
p−v2
a= 2GM (1
ra−1
rp)
Step 7: To find the average of the squares of the speeds, we calculate
v2
p+v2
a
2=
v2
p+v2
p−2GM (1
ra−1
rp)
2=v2
p−GM (1
ra−1
rp)
Step 8: Similarly, the average of the squares of the average speeds is given
by
√1
2((v2
p−2GM/rp)+(v2
a−2GM/ra)) = √1
2(2v2
p+ 2v2
a−2GM(1
ra
+1
rp
))
=√v2
p+v2
a−2GM(1
ra
+1
rp
)
Step 9: Since raand rpare the distances from the planet to the star at
aphelion and perihelion respectively, we notice that their sum is equal to twice
the semi-major axis a, which is a constant. Thus,
ra+rp= 2a
Step 10: Substituting this back into our expressions, we find that
v2
p−GM (2
rp−1
rp)=v2
p−GM (1
ra−1
rp)
v2
a−GM
Question 10
Question
A planet orbits a star in an elliptical orbit. The planet’s distance from the star
varies between 0.3 AU and 0.7 AU. If the planet’s closest distance to the star is
0.3 AU, determine its farthest distance. Assume the orbit is perfectly elliptical.
Solution
Step 1: Recall Kepler’s laws of planetary motion. In particular, Kepler’s first
law states that the orbit of a planet is an ellipse with the star at one of the two
foci.
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Step 2: The distance between the center of the ellipse (the star) and the
closest point on the ellipse (the planet at its closest distance) is called the semi-
major axis (a). In this case, the closest distance is 0.3 AU.
Step 3: The distance between the center of the ellipse and the farthest point
on the ellipse (the planet at its farthest distance) is also equal to the sum of the
semi-major axis and the distance between the two foci. This distance is also
known as the major axis (2a).
Step 4: Given that the closest distance (0.3 AU) is half of the major axis
(2a), we can solve for the value of a. Therefore, we have:
2a= 2(0.3AU) = 0.6AU
Step 5: Since the major axis (2a) is equal to the sum of the semi-major axis
(a) and the distance between the foci, and we know that the distance between
the foci (2ae) is 2(a-c), we can write:
0.6AU =a+ 2(a−c)
Step 6: From Kepler’s second law, we know that 2c is equal to the difference
between the closest and farthest distances from the star. We are given that the
closest distance is 0.3 AU and we need to find the farthest distance.
2c= 0.7AU −0.3AU = 0.4AU
Step 7: Substitute the value of 2cinto the equation from Step 5 and solve
for the farthest distance a+c. This yields:
0.6AU =a+ 2(0.4AU)
0.6AU =a+ 0.8AU
a= 0.6AU −0.8AU =−0.2AU
Step 8: Since a planet’s distance from a star cannot be negative, this solution
is extraneous. Therefore, we have made an error in our calculations. Let’s revisit
our previous steps to identify and correct our mistake.
Question 11
Question
Consider a comet in a highly eccentric orbit around the Sun. The aphelion
distance (furthest distance from the Sun) of this comet is known to be 4.5 AU,
while the perihelion distance (closest distance to the Sun) is 0.5 AU. Determine
the eccentricity of this comet’s orbit.
10
Solution
Step 1: Recall that the eccentricity of an orbit is defined as:
e=dap −dper
dap +dper
where dap is the aphelion distance and dper is the perihelion distance.
Step 2: Substitute the given values into the eccentricity formula:
e=4.5−0.5
4.5+0.5=4
5= 0.8
Step 3: Therefore, the eccentricity of the comet’s orbit is 0.8.
Question 12
Question
A comet takes 76 years to complete one orbit around the sun. If the distance
between the comet and the sun at its closest approach (perihelion) is 0.89 AU
(astronomical units), determine the semi-major axis of the comet’s elliptical
orbit in AU. Given that the semi-major axis of Earth’s orbit is approximately
1 AU.
Solution
Step 1: Let’s denote the semi-major axis of the comet’s elliptical orbit as a, and
the distance between the comet and the sun at its furthest point (aphelion) as
2a. We know that the period of the comet’s orbit is related to the semi-major
axis by Kepler’s third law:
T2
comet
a3
comet
=T2
Earth
a3
Earth
Given Tcomet = 76 years, aEarth = 1 AU, and TEarth = 1 year:
762
a3=1
1
Step 2: Solve for the semi-major axis a:
a3= 762
a=3
√762
Step 3: Calculate the value of a:
a≈3
√5776 ≈18.06 AU
11
Step 4: Finally, since the perihelion distance is 0.89 AU, the semi-major axis
of the comet’s elliptical orbit is:
a=0.89 + 2a
2
Substitute the value of a:
18.06 = 0.89 + 2(18.06)
2
Step 5: Solve for a:
18.06 = 0.89 + 36.12
2
18.06 = 37.01
2
a≈18.5AU
Therefore, the semi-major axis of the comet’s elliptical orbit is approximately
18.5 AU.
Question 13
Question
A comet has a highly eccentric orbit around the sun. At its perihelion, the
comet’s distance from the sun is 0.2 AU, and at its aphelion, the distance is 6.0
AU. Calculate the eccentricity of the comet’s orbit.
Solution
Step 1: We know that the eccentricity of an orbit is given by the formula:
e=rmax −rmin
rmax +rmin
Step 2: Given that the perihelion distance rmin = 0.2AU and the aphelion
distance rmax = 6.0AU, we can substitute these values into the formula:
e=6.0−0.2
6.0+0.2
Step 3: Calculating the eccentricity:
e=5.8
6.2= 0.9355
Step 4: Therefore, the eccentricity of the comet’s orbit is 0.9355 .
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Question 14
Question
An asteroid is known to have an orbital period around the Sun of 6.5 years and
an average distance from the Sun of 3.2 AU. Determine the asteroid’s semi-major
axis and eccentricity of its orbit.
Solution
Step 1: Calculate the semi-major axis (a) of the asteroid’s orbit using Kepler’s
Third Law:
T2=4π2a3
G(M1+M2)
where Tis the orbital period, Gis the gravitational constant, M1is the mass
of the Sun, and M2is the mass of the asteroid.
Step 2: Rearrange the equation to solve for a:
a=(G(M1+M2)T2
4π2)1/3
Step 3: Substitute the given values into the equation to find a:
a=(6.67 ×10−11 ·(1.99 ×1030 +m)·(6.5years)2
4π2)1/3
Step 4: Convert the orbital period from years to seconds:
T= 6.5years = 6.5×365 ×24 ×3600 seconds
Step 5: Calculate ain astronomical units (AU).
Step 6: Calculate the eccentricity (e) of the orbit using the equation:
e=√1−b2
a2
where bis the semi-minor axis of the ellipse.
Step 7: Calculate the semi-minor axis using the formula for an ellipse:
b=a√1−e2
Step 8: Substitute the known values of aand solve for b.
Step 9: Substitute the calculated values of aand binto the eccentricity
equation to find e.
13
Question 15
Question
State Kepler’s third law of planetary motion and derive the equation which
relates the orbital period of a planet (T) to the distance of the planet from the
sun (r).
Solution
Kepler’s Third Law: Kepler’s third law of planetary motion states that the
square of the orbital period of a planet is directly proportional to the cube of
the semi-major axis of its orbit.
Let Tbe the orbital period of a planet and rbe the distance of the planet
from the sun. According to Kepler’s third law, we have:
T2∝r3
To derive the equation relating Tand r, we introduce a constant of propor-
tionality, k. Thus, we can write:
T2=kr3
We can determine the value of kby considering a known planet, such as
Earth. The values of Tand rfor Earth are TEarth = 1 year and rEarth = 1
astronomical unit (AU, the average distance between the Earth and the Sun).
Substituting these values into the equation, we get:
(1)2=k(1)3=⇒k= 1
Therefore, the equation relating the orbital period of a planet (T) and the
distance of the planet from the sun (r) is:
T2=r3
This equation shows the quantitative relationship between the orbital period
and the distance of a planet from the sun based on Kepler’s third law.
Question 16
Question
A planet moves in an elliptical orbit around the Sun. The planet’s closest
distance to the Sun (perihelion) is 0.3 AU and its farthest distance (aphelion)
is 0.7 AU. Calculate the eccentricity of the planet’s orbit.
14
Solution
To find the eccentricity of the planet’s orbit, we can use the formula relating the
perihelion distance (rmin), the aphelion distance (rmax), and the eccentricity (e)
of an elliptical orbit:
e=rmax −rmin
rmax +rmin
Step 1: Substitute the given distances into the formula.
e=0.7−0.3
0.7+0.3
Step 2: Calculate the eccentricity.
e=0.4
1.0= 0.4
Therefore, the eccentricity of the planet’s orbit is 0.4.
Question 17
Question
At a certain planet, a satellite takes 10 hours to orbit the planet at a distance
of 2.0 Earth radii. What is the mass of the planet? (Assume the planet is
spherical)
Solution
Step 1: Calculate the orbital speed of the satellite. The orbital speed of a
satellite is given by:
v=2πr
T
where: v= orbital speed, r= radius of orbit, T= time period of orbit.
Substitute r= 2.0×REarth and T= 10 hours into the formula:
v=2π×2.0×REarth
10 ×3600 s/hour
Step 2: Calculate the gravitational force acting on the satellite. The gravi-
tational force between the planet and the satellite is given by:
F=GMm
r2
where: F= gravitational force, G= gravitational constant, M= mass of the
planet, m= mass of the satellite, r= distance between the planet and the
satellite.
15
Using Newton’s second law (F=m·a) and the formula for centripetal
acceleration (a=v2
r), we have:
GMm
r2=m·v2
r
Step 3: Calculate the mass of the planet. Substitute the given values of r
and vinto the equation above and solve for M:
M=v2·r
G
Question 18
Question
Consider a planet in an elliptical orbit around a star, with the star at one of
the foci of the ellipse. The planet covers equal areas in equal times according
to Kepler’s second law of planetary motion.
Given that the distance of the planet from the star at its closest approach
(perihelion) is r1and the distance at its farthest point (aphelion) is r2, prove
that the time taken for the planet to travel from aphelion to perihelion is the
same as the time taken to travel from perihelion to aphelion.
Solution
To prove that the time taken for the planet to travel from aphelion to perihelion
is the same as the time taken to travel from perihelion to aphelion, we need to
show that the areas swept out by the planet are equal for these two portions of
its orbit. We will make use of Kepler’s second law which states that equal areas
are swept out in equal times.
Step 1: Calculate the area swept out from aphelion to perihelion. Let’s
consider a small time interval from tto t+dt when the planet is at a distance r
from the star. The area swept out during this time interval can be approximated
as a small triangle with base rand height r dθ (where dθ is the angle swept out
by the radius vector). The area dA of this triangle is given by:
dA =1
2r2dθ
Integrating from θ1to θ2(the angles at aphelion and perihelion respectively),
we get:
Aaphelion to perihelion =∫θ2
θ1
1
2r2dθ
Step 2: Calculate the area swept out from perihelion to aphelion. Now,
consider the time interval from t+dt to twhen the planet moves from perihelion
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to aphelion. The area swept out during this time can be represented as a triangle
with the same base rand height r dθ. Hence, in this interval, the area swept
out is also dA =1
2r2dθ.
Integrating from θ2to θ1, we get:
Aperihelion to aphelion =∫θ1
θ2
1
2r2dθ
Step 3: Show that the areas are equal. Since the area swept out is the same
in each case, we have:
Aaphelion to perihelion =Aperihelion to aphelion
This shows that the time taken for the planet to travel from aphelion to
perihelion is the same as the time taken to travel from perihelion to aphelion,
consistent with Kepler’s second law.
Question 19
Question
An asteroid is in an elliptical orbit around the Sun with a semi-major axis of 2.5
AU. The asteroid’s closest approach to the Sun (perihelion) is 0.5 AU. Calculate
the eccentricity of the asteroid’s orbit.
Solution
Step 1: Recall that the eccentricity of an ellipse is defined as
e=distance from center to focus
semi-major axis
Step 2: In this case, the distance from the center to one focus is the differ-
ence between the semi-major axis and the perihelion distance. Therefore, the
eccentricity can be calculated as:
e=2.5−0.5
2.5
Step 3: Calculating this expression gives:
e=2
2.5=4
5= 0.8
Step 4: Therefore, the eccentricity of the asteroid’s orbit is 0.8.
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Question 20
Question
A comet is discovered that follows a highly elliptical orbit around the Sun.
By observing the comet over a period of time, astronomers determine that the
closest approach of the comet to the Sun (perihelion) is 0.2 AU and the farthest
distance from the Sun (aphelion) is 3.0 AU. If the comet’s period of revolution
is 12 years, calculate the eccentricity of the comet’s orbit.
Solution
Step 1: Use Kepler’s third law to find the semi-major axis of the comet’s orbit.
According to Kepler’s third law, the square of the period of revolution of a
planet or comet around the Sun is proportional to the cube of the semi-major
axis of its orbit.
T2=ka3
where Tis the period of revolution in years, ais the semi-major axis in astro-
nomical units (AU), and kis a constant. Since the period of revolution is 12
years:
122=ka3
144 = ka3
Step 2: Calculate the semi-major axis. Since the closest distance (perihelion)
is 0.2 AU and the farthest distance (aphelion) is 3.0 AU, the semi-major axis a
is the average of these two distances.
a=0.2+3.0
2= 1.6AU
Step 3: Calculate the eccentricity of the orbit. The eccentricity eof an
ellipse is related to the semi-major axis a, perihelion distance rp, and aphelion
distance raby the formula:
e=ra−rp
ra+rp
=3.0−0.2
3.0+0.2=2.8
3.2= 0.875
Therefore, the eccentricity of the comet’s orbit is 0.875.
Question 21
Question
A planet has an elliptical orbit around the Sun, with the Sun located at one of
the foci. The planet’s closest approach to the Sun (perihelion) is 0.3 AU and its
farthest distance from the Sun (aphelion) is 0.7 AU. Calculate the eccentricity
of the planet’s orbit.
18
Solution
Step 1: The eccentricity eof an ellipse is defined as the ratio of the distance
between the foci of the ellipse to the length of the major axis.
Step 2: In this case, the major axis of the ellipse is the distance between
the perihelion and aphelion, which is the sum of the distances to the Sun:
a= 0.3AU + 0.7AU = 1.0AU.
Step 3: The distance between the foci of the ellipse can be calculated using
the formula 2ae = 1.0AU,where ais the semi-major axis of the ellipse.
Step 4: Substituting the given perihelion (a(1 −e)) and aphelion (a(1 + e))
distances into the formula, we have 2ae = (0.3AU)(1 −e) + (0.7AU)(1 + e).
Step 5: Expanding and simplifying the equation, we get 2ae = 0.3AU −
0.3e+ 0.7AU + 0.7e.
Step 6: Rearranging terms, we have 2ae = 1.0AU + 0.4e.
Step 7: Solving for e, we get e=1.0AU
2a−0.4.
Step 8: Substituting a= 1.0AU into the equation, we find e=1.0AU
2(1.0AU)−0.4=
1.0AU
1.6= 0.625.
Step 9: Therefore, the eccentricity of the planet’s orbit is 0.625 .
Question 22
Question
A planet follows an elliptical orbit around the Sun. The distance between the
planet and the Sun at its closest approach is 0.5 AU (astronomical units), and
at its farthest distance, it is 2.0 AU. Calculate the eccentricity of the planet’s
orbit.
Solution
To calculate the eccentricity of the planet’s orbit, we can use the formula relating
the closest and farthest distances of a planet from the focus of its elliptical orbit
to the semi-major axis and eccentricity:
e=rmax −rmin
rmax +rmin
Step 1: Convert the closest and farthest distances from AU to the same
unit. Given:
rmin = 0.5AU
rmax = 2.0AU
Step 2: Substitute the values into the eccentricity formula. Plugging in the
values:
e=2.0−0.5
2.0+0.5
19
Step 3: Solve for eccentricity.
e=1.5
2.5= 0.6
Therefore, the eccentricity of the planet’s orbit is 0.6.
Question 23
Question
Consider a planet orbiting a star in an elliptical orbit. The planet’s farthest
distance from the star is 3.0 AU, and its closest distance to the star is 1.0 AU.
Determine the eccentricity of the planet’s orbit.
Solution
To find the eccentricity (e) of the planet’s orbit, we can use the following formula:
e=rmax −rmin
rmax +rmin
Step 1: Given that the planet’s farthest distance from the star is rmax = 3.0
AU and its closest distance to the star is rmin = 1.0AU, we can substitute these
values into the formula:
e=3.0−1.0
3.0+1.0
Step 2: Calculate the numerator and denominator separately:
e=2.0
4.0
Step 3: Simplify the fraction to find the eccentricity:
e= 0.5
Step 4: Therefore, the eccentricity of the planet’s orbit is 0.5.
Question 24
Question
An artificial satellite is in an elliptical orbit around the Earth. The aphelion
(farthest point from Earth) of the orbit is 40,000 km and the perihelion (closest
point to Earth) is 10,000 km. Calculate the eccentricity of the orbit. Given:
• Aphelion distance (maximum distance from Earth) = 40,000 km
• Perihelion distance (minimum distance from Earth) = 10,000 km
20
Solution
Step 1: Use the definition of eccentricity (e) in terms of aphelion and perihelion
distances:
e=Aphelion distance −Perihelion distance
Aphelion distance +Perihelion distance
Step 2: Substitute the given values into the formula:
e=40,000 −10,000
40,000 + 10,000
Step 3: Simplify the expression:
e=30,000
50,000 = 0.6
Therefore, the eccentricity of the orbit is 0.6.
21
Question 2
Question
Suppose a planet orbits a star in an elliptical path. The planet’s closest distance
to the star is 5 AU, and its farthest distance is 10 AU. If the planet takes 1 year
to complete one orbit, what is the eccentricity of the planet’s orbit?
Solution
Step 1: Recall that the eccentricity of an elliptical orbit is given by the formula:
e=rmax −rmin
rmax +rmin
where rmax is the farthest distance and rmin is the closest distance.
Step 2: Substitute the given values into the formula:
e=10 AU −5AU
10 AU + 5 AU
Step 3: Calculate the eccentricity:
e=5AU
15 AU =1
3
Step 4: Therefore, the eccentricity of the planet’s orbit is 1
3.
Question 3
Question
Consider a planet in a circular orbit around a star with a period of 40 years. If
the distance between the planet and the star is doubled, what will be the new
period of the planet in years? Assume the mass of the star remains constant.
Solution
Step 1: We are given the orbital period Tof the planet as 40 years. We are
asked to find the new period when the distance between the planet and the star
is doubled. Let rbe the original distance between the planet and the star.
Step 2: According to Kepler’s Third Law of Planetary Motion, the square
of the period of revolution of a planet is directly proportional to the cube of its
semi-major axis: T2∝r3.
Step 3: Let T1be the original period and r1be the original distance. Let
T2be the new period and r2be the new distance (doubled from the original
distance).
Step 4: From Kepler’s Third Law, we have T2
1∝r3
1. Therefore, T2
1=k·r3
1,
where kis a constant of proportionality.
2
Step 5: Similarly, for the new setup, we have T2
2∝r3
2, which implies T2
2=
k·r3
2. Since the mass of the star remains constant, kis the same as before.
Step 6: By comparing the two equations, we get T2
1
r3
1
=T2
2
r3
2
.
Step 7: Plugging in the given information, we have 402
r3=T2
2
(2r)3.
Step 8: Simplifying the equation, we get 1600
r3=T2
2
8r3.
Step 9: Cross multiplying gives 8r3·1600 = r3·T2
2.
Step 10: Simplifying further leads to 12800 = r3·T2
2.
Step 11: Since r3=12800
T2
2
, taking the cube root gives r=3
√12800 ·T−2/3
2.
Step 12: Doubling the distance (2r) results in 2r=3
√12800 ·T−2/3
2.
Step 13: Simplifying the equation gives 23
√12800 = T−2/3
2.
Step 14: Solving for T2leaves us with T2=(23
√12800)−3/2.
Step 15: Calculating the value for T2results in the new period being ap-
proximately 113.14 years when the distance between the planet and the star is
doubled.
Question 4
Question
Kepler’s third law states that the square of the orbital period of a planet is
proportional to the cube of its average distance from the Sun. Consider a
hypothetical planet with an average distance from the Sun of 3 AU. If the
orbital period of this planet is 6 years, what would be the average distance from
the Sun of another planet with an orbital period of 10 years?
Solution
Step 1: Calculate the constant of proportionality using the given data for the
first planet. Given: Average distance from the Sun, r1= 3 AU and orbital
period, T1= 6 years. According to Kepler’s third law:
r3
1=kT 2
1
Substitute the values of r1and T1into the equation:
(3)3=k(6)2
27 = 36k
k=27
36 =3
4
Step 2: Use the constant of proportionality to find the average distance of
the second planet. Given: Orbital period of the second planet, T2= 10 years.
According to Kepler’s third law:
r3
2=kT 2
2
3
r3
2=3
4(10)2
r3
2=3
4×100
r3
2= 75
r2=3
√75
r2≈4.217 AU
Therefore, the average distance from the Sun of the second planet with an
orbital period of 10 years would be approximately 4.217 AU.
Question 5
Question
The path of a planet orbiting a star is given by r(θ) = p
1+ecos(θ), where pis the
distance from the planet to the star at closest approach (perihelion), eis the
eccentricity of the orbit, θis the angle measured from the periapsis, and ris
the distance from the planet to the star at an angle θ. Show that this equation
satisfies Kepler’s second law, which states that a planet sweeps out equal areas
in equal times.
Solution
Step 1: Calculate the area swept out by the planet in a small time interval dθ.
To find the area swept out by the planet in a small time interval dθ, we can
consider an infinitesimal triangle formed by sweeping out an angle dθ. The area
of this triangle can be calculated using the formula for the area of a triangle:
dA =1
2r2dθ.
Step 2: Express rin terms of θ. Given r(θ) = p
1+ecos(θ), we can rewrite this
expression using the trigonometric identity cos(2θ) = 2 cos2(θ)−1.
Step 3: Calculate r2and express it in terms of pand e. Squaring r(θ), we
get r2(θ) = p2(1 + ecos(θ))2.
Step 4: Calculate dA
dθ . Substitute r2into the formula for dA and differentiate
with respect to θto find dA
dθ .
Step 5: Simplify dA
dθ . After differentiation and simplification, we should
arrive at a result that shows dA
dθ is a constant, implying that equal areas are
indeed swept out in equal times.
Therefore, we have shown that the given equation satisfies Kepler’s second
law.
4
Question 6
Question
An asteroid is in an elliptical orbit around the Sun. The closest distance of the
asteroid to the Sun (perihelion) is 0.3 AU and the farthest distance (aphelion)
is 3.0 AU. Calculate the eccentricity of the asteroid’s orbit.
Solution
Step 1: The eccentricity of an ellipse is defined as e=c
a, where cis the distance
from the center of the ellipse to one of the foci and ais the semi-major axis of
the ellipse.
Step 2: The semi-major axis of the ellipse is the average of the perihelion
and aphelion distances. This can be calculated as:
a=rperihelion +raphelion
2=0.3AU + 3.0AU
2= 1.65 AU
Step 3: The distance from the center of the ellipse to one of the foci can be
found using the relationship c=a−rperihelion. Thus,
c= 1.65 AU −0.3AU = 1.35 AU
Step 4: Finally, we can calculate the eccentricity of the asteroid’s orbit using
the formula e=c
a:
e=1.35 AU
1.65 AU = 0.818
Therefore, the eccentricity of the asteroid’s orbit is 0.818.
Question 7
Question
In a distant solar system, a planet has an elliptical orbit around its star. The
planet’s closest approach to the star is 0.1 astronomical units (AU), and the
farthest distance is 0.2 AU. If the orbital period of this planet is 100 days,
determine the eccentricity of its orbit.
Solution
Step 1: First, let’s recall Kepler’s third law, which relates the orbital period of
a planet to the semi-major axis of its orbit. Kepler’s third law can be expressed
as:
T2=(4π2
G(M1+M2))a3
5
where: - T= orbital period of the planet - G= gravitational constant - M1and
M2= masses of the planet and the star - a= semi-major axis of the orbit
Step 2: We are given the orbital period T= 100 days and the closest (rmin =
0.1AU) and farthest (rmax = 0.2AU) distances from the star. We know that
in an elliptical orbit, the semi-major axis ais the average of the closest and
farthest distances from the star, so a=rmin +rmax
2.
Step 3: Plugging in the given values, we have:
a=0.1+0.2
2= 0.15 AU
Step 4: Now, we can substitute T= 100 days and a= 0.15 AU into Kepler’s
third law to find M1+M2:
1002=(4π2
G(M1+M2))(0.15)3
Step 5: We also know that the eccentricity eof an elliptical orbit can be
determined using the equation:
e=√1−(b
a)2
where: - a= semi-major axis - b= semi-minor axis
Step 6: For an elliptical orbit, the semi-minor axis bcan be calculated using
the eccentricity and the semi-major axis: b=a√1−e2.
Step 7: Substituting a= 0.15 AU into the expression for the semi-minor
axis, we get:
b= 0.15√1−e2
Step 8: However, the semi-minor axis b=1
2(rmax −rmin). Thus, we have:
0.15√1−e2=1
2(0.2−0.1)
Step 9: Solve for the eccentricity eby first finding the value of b, and then
substituting back into the equation for the eccentricity.
Question 8
Question
A satellite is in a circular orbit around a planet of mass M. The satellite’s orbital
radius is rand its angular speed is ω. Find an expression for the satellite’s period
of revolution Tin terms of r,M, and ω.
6
Solution
Step 1: Write down the formula for the period of revolution of a satellite in
circular orbit. The period of revolution Tof a satellite in a circular orbit can
be related to the angular speed ωby the formula:
T=2π
ω
Step 2: Express the angular speed ωin terms of the orbital radius r. The
angular speed ωis related to the satellite’s tangential speed vand the orbital
radius rby the equation:
v=rω
Step 3: Find an expression for tangential speed v. The tangential speed vof
the satellite is determined by the balance between the gravitational force and
the centripetal force. The gravitational force is given by:
Fg=GMm
r2
where Gis the gravitational constant, Mis the mass of the planet, and mis
the mass of the satellite. The centripetal force is given by:
Fc=mv2
r
Setting these two forces equal, we have:
GMm
r2=mv2
r
Solving for v, we get:
v=√GM
r
Step 4: Substitute the expression for vinto the formula for ω. Substitute
v=rω into the equation v=√GM
rto get:
rω =√GM
r
Solving for ω, we find:
ω=√GM
r3/2
Step 5: Substitute the expression for ωinto the formula for T. Finally,
substituting ω=√GM
r3/2into the formula T=2π
ω, we get:
T=2π
√GM
r3/2
=2πr3/2
√GM
Therefore, the expression for the satellite’s period of revolution Tin terms of r,
M, and ωis T=2πr3/2
√GM .
7
Question 9
Question
Consider a planet in an elliptical orbit around a star, such that at its closest
approach (perihelion) the planet’s speed is vpand the distance from the planet
to the star is rp. At its farthest point (aphelion), the planet’s speed is vaand
the distance from the planet to the star is ra. Given that the orbit is closed and
obeys Kepler’s laws, prove that the average of the squares of the speeds of the
planet at perihelion and aphelion is equal to the harmonic mean of the squares
of the average speeds at perihelion and aphelion.
Solution
Step 1: Let’s denote the mass of the star as M, the mass of the planet as m,
and the gravitational force between them as F. Using Kepler’s third law, we
have the equation for the period of the planet’s orbit:
T2=4π2
G(M+m)a3
where ais the semi-major axis of the orbit.
Step 2: Since the orbit is closed, the total mechanical energy of the planet
is conserved. At each point in the orbit, the total energy is given by
E=1
2mv2−GMm
r
where vis the speed of the planet and ris the distance from the planet to the
star.
Step 3: At perihelion, the total energy is
Ep=1
2mv2
p−GMm
rp
and at aphelion, the total energy is
Ea=1
2mv2
a−GMm
ra
Step 4: Using the fact that energy is conserved, we have Ep=Ea. This
gives us the equation
1
2mv2
p−GMm
rp
=1
2mv2
a−GMm
ra
Step 5: Rearranging the equation above, we get
1
2m(v2
p−v2
a) = GMm (1
ra−1
rp)
8
Step 6: Dividing by mand rearranging gives us an expression for the differ-
ence in speeds squared:
v2
p−v2
a= 2GM (1
ra−1
rp)
Step 7: To find the average of the squares of the speeds, we calculate
v2
p+v2
a
2=
v2
p+v2
p−2GM (1
ra−1
rp)
2=v2
p−GM (1
ra−1
rp)
Step 8: Similarly, the average of the squares of the average speeds is given
by
√1
2((v2
p−2GM/rp)+(v2
a−2GM/ra)) = √1
2(2v2
p+ 2v2
a−2GM(1
ra
+1
rp
))
=√v2
p+v2
a−2GM(1
ra
+1
rp
)
Step 9: Since raand rpare the distances from the planet to the star at
aphelion and perihelion respectively, we notice that their sum is equal to twice
the semi-major axis a, which is a constant. Thus,
ra+rp= 2a
Step 10: Substituting this back into our expressions, we find that
v2
p−GM (2
rp−1
rp)=v2
p−GM (1
ra−1
rp)
v2
a−GM
Question 10
Question
A planet orbits a star in an elliptical orbit. The planet’s distance from the star
varies between 0.3 AU and 0.7 AU. If the planet’s closest distance to the star is
0.3 AU, determine its farthest distance. Assume the orbit is perfectly elliptical.
Solution
Step 1: Recall Kepler’s laws of planetary motion. In particular, Kepler’s first
law states that the orbit of a planet is an ellipse with the star at one of the two
foci.
9
Step 2: The distance between the center of the ellipse (the star) and the
closest point on the ellipse (the planet at its closest distance) is called the semi-
major axis (a). In this case, the closest distance is 0.3 AU.
Step 3: The distance between the center of the ellipse and the farthest point
on the ellipse (the planet at its farthest distance) is also equal to the sum of the
semi-major axis and the distance between the two foci. This distance is also
known as the major axis (2a).
Step 4: Given that the closest distance (0.3 AU) is half of the major axis
(2a), we can solve for the value of a. Therefore, we have:
2a= 2(0.3AU) = 0.6AU
Step 5: Since the major axis (2a) is equal to the sum of the semi-major axis
(a) and the distance between the foci, and we know that the distance between
the foci (2ae) is 2(a-c), we can write:
0.6AU =a+ 2(a−c)
Step 6: From Kepler’s second law, we know that 2c is equal to the difference
between the closest and farthest distances from the star. We are given that the
closest distance is 0.3 AU and we need to find the farthest distance.
2c= 0.7AU −0.3AU = 0.4AU
Step 7: Substitute the value of 2cinto the equation from Step 5 and solve
for the farthest distance a+c. This yields:
0.6AU =a+ 2(0.4AU)
0.6AU =a+ 0.8AU
a= 0.6AU −0.8AU =−0.2AU
Step 8: Since a planet’s distance from a star cannot be negative, this solution
is extraneous. Therefore, we have made an error in our calculations. Let’s revisit
our previous steps to identify and correct our mistake.
Question 11
Question
Consider a comet in a highly eccentric orbit around the Sun. The aphelion
distance (furthest distance from the Sun) of this comet is known to be 4.5 AU,
while the perihelion distance (closest distance to the Sun) is 0.5 AU. Determine
the eccentricity of this comet’s orbit.
10
Solution
Step 1: Recall that the eccentricity of an orbit is defined as:
e=dap −dper
dap +dper
where dap is the aphelion distance and dper is the perihelion distance.
Step 2: Substitute the given values into the eccentricity formula:
e=4.5−0.5
4.5+0.5=4
5= 0.8
Step 3: Therefore, the eccentricity of the comet’s orbit is 0.8.
Question 12
Question
A comet takes 76 years to complete one orbit around the sun. If the distance
between the comet and the sun at its closest approach (perihelion) is 0.89 AU
(astronomical units), determine the semi-major axis of the comet’s elliptical
orbit in AU. Given that the semi-major axis of Earth’s orbit is approximately
1 AU.
Solution
Step 1: Let’s denote the semi-major axis of the comet’s elliptical orbit as a, and
the distance between the comet and the sun at its furthest point (aphelion) as
2a. We know that the period of the comet’s orbit is related to the semi-major
axis by Kepler’s third law:
T2
comet
a3
comet
=T2
Earth
a3
Earth
Given Tcomet = 76 years, aEarth = 1 AU, and TEarth = 1 year:
762
a3=1
1
Step 2: Solve for the semi-major axis a:
a3= 762
a=3
√762
Step 3: Calculate the value of a:
a≈3
√5776 ≈18.06 AU
11
Step 4: Finally, since the perihelion distance is 0.89 AU, the semi-major axis
of the comet’s elliptical orbit is:
a=0.89 + 2a
2
Substitute the value of a:
18.06 = 0.89 + 2(18.06)
2
Step 5: Solve for a:
18.06 = 0.89 + 36.12
2
18.06 = 37.01
2
a≈18.5AU
Therefore, the semi-major axis of the comet’s elliptical orbit is approximately
18.5 AU.
Question 13
Question
A comet has a highly eccentric orbit around the sun. At its perihelion, the
comet’s distance from the sun is 0.2 AU, and at its aphelion, the distance is 6.0
AU. Calculate the eccentricity of the comet’s orbit.
Solution
Step 1: We know that the eccentricity of an orbit is given by the formula:
e=rmax −rmin
rmax +rmin
Step 2: Given that the perihelion distance rmin = 0.2AU and the aphelion
distance rmax = 6.0AU, we can substitute these values into the formula:
e=6.0−0.2
6.0+0.2
Step 3: Calculating the eccentricity:
e=5.8
6.2= 0.9355
Step 4: Therefore, the eccentricity of the comet’s orbit is 0.9355 .
12
Question 14
Question
An asteroid is known to have an orbital period around the Sun of 6.5 years and
an average distance from the Sun of 3.2 AU. Determine the asteroid’s semi-major
axis and eccentricity of its orbit.
Solution
Step 1: Calculate the semi-major axis (a) of the asteroid’s orbit using Kepler’s
Third Law:
T2=4π2a3
G(M1+M2)
where Tis the orbital period, Gis the gravitational constant, M1is the mass
of the Sun, and M2is the mass of the asteroid.
Step 2: Rearrange the equation to solve for a:
a=(G(M1+M2)T2
4π2)1/3
Step 3: Substitute the given values into the equation to find a:
a=(6.67 ×10−11 ·(1.99 ×1030 +m)·(6.5years)2
4π2)1/3
Step 4: Convert the orbital period from years to seconds:
T= 6.5years = 6.5×365 ×24 ×3600 seconds
Step 5: Calculate ain astronomical units (AU).
Step 6: Calculate the eccentricity (e) of the orbit using the equation:
e=√1−b2
a2
where bis the semi-minor axis of the ellipse.
Step 7: Calculate the semi-minor axis using the formula for an ellipse:
b=a√1−e2
Step 8: Substitute the known values of aand solve for b.
Step 9: Substitute the calculated values of aand binto the eccentricity
equation to find e.
13
Question 15
Question
State Kepler’s third law of planetary motion and derive the equation which
relates the orbital period of a planet (T) to the distance of the planet from the
sun (r).
Solution
Kepler’s Third Law: Kepler’s third law of planetary motion states that the
square of the orbital period of a planet is directly proportional to the cube of
the semi-major axis of its orbit.
Let Tbe the orbital period of a planet and rbe the distance of the planet
from the sun. According to Kepler’s third law, we have:
T2∝r3
To derive the equation relating Tand r, we introduce a constant of propor-
tionality, k. Thus, we can write:
T2=kr3
We can determine the value of kby considering a known planet, such as
Earth. The values of Tand rfor Earth are TEarth = 1 year and rEarth = 1
astronomical unit (AU, the average distance between the Earth and the Sun).
Substituting these values into the equation, we get:
(1)2=k(1)3=⇒k= 1
Therefore, the equation relating the orbital period of a planet (T) and the
distance of the planet from the sun (r) is:
T2=r3
This equation shows the quantitative relationship between the orbital period
and the distance of a planet from the sun based on Kepler’s third law.
Question 16
Question
A planet moves in an elliptical orbit around the Sun. The planet’s closest
distance to the Sun (perihelion) is 0.3 AU and its farthest distance (aphelion)
is 0.7 AU. Calculate the eccentricity of the planet’s orbit.
14
Solution
To find the eccentricity of the planet’s orbit, we can use the formula relating the
perihelion distance (rmin), the aphelion distance (rmax), and the eccentricity (e)
of an elliptical orbit:
e=rmax −rmin
rmax +rmin
Step 1: Substitute the given distances into the formula.
e=0.7−0.3
0.7+0.3
Step 2: Calculate the eccentricity.
e=0.4
1.0= 0.4
Therefore, the eccentricity of the planet’s orbit is 0.4.
Question 17
Question
At a certain planet, a satellite takes 10 hours to orbit the planet at a distance
of 2.0 Earth radii. What is the mass of the planet? (Assume the planet is
spherical)
Solution
Step 1: Calculate the orbital speed of the satellite. The orbital speed of a
satellite is given by:
v=2πr
T
where: v= orbital speed, r= radius of orbit, T= time period of orbit.
Substitute r= 2.0×REarth and T= 10 hours into the formula:
v=2π×2.0×REarth
10 ×3600 s/hour
Step 2: Calculate the gravitational force acting on the satellite. The gravi-
tational force between the planet and the satellite is given by:
F=GMm
r2
where: F= gravitational force, G= gravitational constant, M= mass of the
planet, m= mass of the satellite, r= distance between the planet and the
satellite.
15
Using Newton’s second law (F=m·a) and the formula for centripetal
acceleration (a=v2
r), we have:
GMm
r2=m·v2
r
Step 3: Calculate the mass of the planet. Substitute the given values of r
and vinto the equation above and solve for M:
M=v2·r
G
Question 18
Question
Consider a planet in an elliptical orbit around a star, with the star at one of
the foci of the ellipse. The planet covers equal areas in equal times according
to Kepler’s second law of planetary motion.
Given that the distance of the planet from the star at its closest approach
(perihelion) is r1and the distance at its farthest point (aphelion) is r2, prove
that the time taken for the planet to travel from aphelion to perihelion is the
same as the time taken to travel from perihelion to aphelion.
Solution
To prove that the time taken for the planet to travel from aphelion to perihelion
is the same as the time taken to travel from perihelion to aphelion, we need to
show that the areas swept out by the planet are equal for these two portions of
its orbit. We will make use of Kepler’s second law which states that equal areas
are swept out in equal times.
Step 1: Calculate the area swept out from aphelion to perihelion. Let’s
consider a small time interval from tto t+dt when the planet is at a distance r
from the star. The area swept out during this time interval can be approximated
as a small triangle with base rand height r dθ (where dθ is the angle swept out
by the radius vector). The area dA of this triangle is given by:
dA =1
2r2dθ
Integrating from θ1to θ2(the angles at aphelion and perihelion respectively),
we get:
Aaphelion to perihelion =∫θ2
θ1
1
2r2dθ
Step 2: Calculate the area swept out from perihelion to aphelion. Now,
consider the time interval from t+dt to twhen the planet moves from perihelion
16
to aphelion. The area swept out during this time can be represented as a triangle
with the same base rand height r dθ. Hence, in this interval, the area swept
out is also dA =1
2r2dθ.
Integrating from θ2to θ1, we get:
Aperihelion to aphelion =∫θ1
θ2
1
2r2dθ
Step 3: Show that the areas are equal. Since the area swept out is the same
in each case, we have:
Aaphelion to perihelion =Aperihelion to aphelion
This shows that the time taken for the planet to travel from aphelion to
perihelion is the same as the time taken to travel from perihelion to aphelion,
consistent with Kepler’s second law.
Question 19
Question
An asteroid is in an elliptical orbit around the Sun with a semi-major axis of 2.5
AU. The asteroid’s closest approach to the Sun (perihelion) is 0.5 AU. Calculate
the eccentricity of the asteroid’s orbit.
Solution
Step 1: Recall that the eccentricity of an ellipse is defined as
e=distance from center to focus
semi-major axis
Step 2: In this case, the distance from the center to one focus is the differ-
ence between the semi-major axis and the perihelion distance. Therefore, the
eccentricity can be calculated as:
e=2.5−0.5
2.5
Step 3: Calculating this expression gives:
e=2
2.5=4
5= 0.8
Step 4: Therefore, the eccentricity of the asteroid’s orbit is 0.8.
17
Question 20
Question
A comet is discovered that follows a highly elliptical orbit around the Sun.
By observing the comet over a period of time, astronomers determine that the
closest approach of the comet to the Sun (perihelion) is 0.2 AU and the farthest
distance from the Sun (aphelion) is 3.0 AU. If the comet’s period of revolution
is 12 years, calculate the eccentricity of the comet’s orbit.
Solution
Step 1: Use Kepler’s third law to find the semi-major axis of the comet’s orbit.
According to Kepler’s third law, the square of the period of revolution of a
planet or comet around the Sun is proportional to the cube of the semi-major
axis of its orbit.
T2=ka3
where Tis the period of revolution in years, ais the semi-major axis in astro-
nomical units (AU), and kis a constant. Since the period of revolution is 12
years:
122=ka3
144 = ka3
Step 2: Calculate the semi-major axis. Since the closest distance (perihelion)
is 0.2 AU and the farthest distance (aphelion) is 3.0 AU, the semi-major axis a
is the average of these two distances.
a=0.2+3.0
2= 1.6AU
Step 3: Calculate the eccentricity of the orbit. The eccentricity eof an
ellipse is related to the semi-major axis a, perihelion distance rp, and aphelion
distance raby the formula:
e=ra−rp
ra+rp
=3.0−0.2
3.0+0.2=2.8
3.2= 0.875
Therefore, the eccentricity of the comet’s orbit is 0.875.
Question 21
Question
A planet has an elliptical orbit around the Sun, with the Sun located at one of
the foci. The planet’s closest approach to the Sun (perihelion) is 0.3 AU and its
farthest distance from the Sun (aphelion) is 0.7 AU. Calculate the eccentricity
of the planet’s orbit.
18
Solution
Step 1: The eccentricity eof an ellipse is defined as the ratio of the distance
between the foci of the ellipse to the length of the major axis.
Step 2: In this case, the major axis of the ellipse is the distance between
the perihelion and aphelion, which is the sum of the distances to the Sun:
a= 0.3AU + 0.7AU = 1.0AU.
Step 3: The distance between the foci of the ellipse can be calculated using
the formula 2ae = 1.0AU,where ais the semi-major axis of the ellipse.
Step 4: Substituting the given perihelion (a(1 −e)) and aphelion (a(1 + e))
distances into the formula, we have 2ae = (0.3AU)(1 −e) + (0.7AU)(1 + e).
Step 5: Expanding and simplifying the equation, we get 2ae = 0.3AU −
0.3e+ 0.7AU + 0.7e.
Step 6: Rearranging terms, we have 2ae = 1.0AU + 0.4e.
Step 7: Solving for e, we get e=1.0AU
2a−0.4.
Step 8: Substituting a= 1.0AU into the equation, we find e=1.0AU
2(1.0AU)−0.4=
1.0AU
1.6= 0.625.
Step 9: Therefore, the eccentricity of the planet’s orbit is 0.625 .
Question 22
Question
A planet follows an elliptical orbit around the Sun. The distance between the
planet and the Sun at its closest approach is 0.5 AU (astronomical units), and
at its farthest distance, it is 2.0 AU. Calculate the eccentricity of the planet’s
orbit.
Solution
To calculate the eccentricity of the planet’s orbit, we can use the formula relating
the closest and farthest distances of a planet from the focus of its elliptical orbit
to the semi-major axis and eccentricity:
e=rmax −rmin
rmax +rmin
Step 1: Convert the closest and farthest distances from AU to the same
unit. Given:
rmin = 0.5AU
rmax = 2.0AU
Step 2: Substitute the values into the eccentricity formula. Plugging in the
values:
e=2.0−0.5
2.0+0.5
19
Step 3: Solve for eccentricity.
e=1.5
2.5= 0.6
Therefore, the eccentricity of the planet’s orbit is 0.6.
Question 23
Question
Consider a planet orbiting a star in an elliptical orbit. The planet’s farthest
distance from the star is 3.0 AU, and its closest distance to the star is 1.0 AU.
Determine the eccentricity of the planet’s orbit.
Solution
To find the eccentricity (e) of the planet’s orbit, we can use the following formula:
e=rmax −rmin
rmax +rmin
Step 1: Given that the planet’s farthest distance from the star is rmax = 3.0
AU and its closest distance to the star is rmin = 1.0AU, we can substitute these
values into the formula:
e=3.0−1.0
3.0+1.0
Step 2: Calculate the numerator and denominator separately:
e=2.0
4.0
Step 3: Simplify the fraction to find the eccentricity:
e= 0.5
Step 4: Therefore, the eccentricity of the planet’s orbit is 0.5.
Question 24
Question
An artificial satellite is in an elliptical orbit around the Earth. The aphelion
(farthest point from Earth) of the orbit is 40,000 km and the perihelion (closest
point to Earth) is 10,000 km. Calculate the eccentricity of the orbit. Given:
• Aphelion distance (maximum distance from Earth) = 40,000 km
• Perihelion distance (minimum distance from Earth) = 10,000 km
20
Solution
Step 1: Use the definition of eccentricity (e) in terms of aphelion and perihelion
distances:
e=Aphelion distance −Perihelion distance
Aphelion distance +Perihelion distance
Step 2: Substitute the given values into the formula:
e=40,000 −10,000
40,000 + 10,000
Step 3: Simplify the expression:
e=30,000
50,000 = 0.6
Therefore, the eccentricity of the orbit is 0.6.
21
Question 2
Question
Suppose a planet orbits a star in an elliptical path. The planet’s closest distance
to the star is 5 AU, and its farthest distance is 10 AU. If the planet takes 1 year
to complete one orbit, what is the eccentricity of the planet’s orbit?
Solution
Step 1: Recall that the eccentricity of an elliptical orbit is given by the formula:
e=rmax −rmin
rmax +rmin
where rmax is the farthest distance and rmin is the closest distance.
Step 2: Substitute the given values into the formula:
e=10 AU −5AU
10 AU + 5 AU
Step 3: Calculate the eccentricity:
e=5AU
15 AU =1
3
Step 4: Therefore, the eccentricity of the planet’s orbit is 1
3.
Question 3
Question
Consider a planet in a circular orbit around a star with a period of 40 years. If
the distance between the planet and the star is doubled, what will be the new
period of the planet in years? Assume the mass of the star remains constant.
Solution
Step 1: We are given the orbital period Tof the planet as 40 years. We are
asked to find the new period when the distance between the planet and the star
is doubled. Let rbe the original distance between the planet and the star.
Step 2: According to Kepler’s Third Law of Planetary Motion, the square
of the period of revolution of a planet is directly proportional to the cube of its
semi-major axis: T2∝r3.
Step 3: Let T1be the original period and r1be the original distance. Let
T2be the new period and r2be the new distance (doubled from the original
distance).
Step 4: From Kepler’s Third Law, we have T2
1∝r3
1. Therefore, T2
1=k·r3
1,
where kis a constant of proportionality.
2
Step 5: Similarly, for the new setup, we have T2
2∝r3
2, which implies T2
2=
k·r3
2. Since the mass of the star remains constant, kis the same as before.
Step 6: By comparing the two equations, we get T2
1
r3
1
=T2
2
r3
2
.
Step 7: Plugging in the given information, we have 402
r3=T2
2
(2r)3.
Step 8: Simplifying the equation, we get 1600
r3=T2
2
8r3.
Step 9: Cross multiplying gives 8r3·1600 = r3·T2
2.
Step 10: Simplifying further leads to 12800 = r3·T2
2.
Step 11: Since r3=12800
T2
2
, taking the cube root gives r=3
√12800 ·T−2/3
2.
Step 12: Doubling the distance (2r) results in 2r=3
√12800 ·T−2/3
2.
Step 13: Simplifying the equation gives 23
√12800 = T−2/3
2.
Step 14: Solving for T2leaves us with T2=(23
√12800)−3/2.
Step 15: Calculating the value for T2results in the new period being ap-
proximately 113.14 years when the distance between the planet and the star is
doubled.
Question 4
Question
Kepler’s third law states that the square of the orbital period of a planet is
proportional to the cube of its average distance from the Sun. Consider a
hypothetical planet with an average distance from the Sun of 3 AU. If the
orbital period of this planet is 6 years, what would be the average distance from
the Sun of another planet with an orbital period of 10 years?
Solution
Step 1: Calculate the constant of proportionality using the given data for the
first planet. Given: Average distance from the Sun, r1= 3 AU and orbital
period, T1= 6 years. According to Kepler’s third law:
r3
1=kT 2
1
Substitute the values of r1and T1into the equation:
(3)3=k(6)2
27 = 36k
k=27
36 =3
4
Step 2: Use the constant of proportionality to find the average distance of
the second planet. Given: Orbital period of the second planet, T2= 10 years.
According to Kepler’s third law:
r3
2=kT 2
2
3
r3
2=3
4(10)2
r3
2=3
4×100
r3
2= 75
r2=3
√75
r2≈4.217 AU
Therefore, the average distance from the Sun of the second planet with an
orbital period of 10 years would be approximately 4.217 AU.
Question 5
Question
The path of a planet orbiting a star is given by r(θ) = p
1+ecos(θ), where pis the
distance from the planet to the star at closest approach (perihelion), eis the
eccentricity of the orbit, θis the angle measured from the periapsis, and ris
the distance from the planet to the star at an angle θ. Show that this equation
satisfies Kepler’s second law, which states that a planet sweeps out equal areas
in equal times.
Solution
Step 1: Calculate the area swept out by the planet in a small time interval dθ.
To find the area swept out by the planet in a small time interval dθ, we can
consider an infinitesimal triangle formed by sweeping out an angle dθ. The area
of this triangle can be calculated using the formula for the area of a triangle:
dA =1
2r2dθ.
Step 2: Express rin terms of θ. Given r(θ) = p
1+ecos(θ), we can rewrite this
expression using the trigonometric identity cos(2θ) = 2 cos2(θ)−1.
Step 3: Calculate r2and express it in terms of pand e. Squaring r(θ), we
get r2(θ) = p2(1 + ecos(θ))2.
Step 4: Calculate dA
dθ . Substitute r2into the formula for dA and differentiate
with respect to θto find dA
dθ .
Step 5: Simplify dA
dθ . After differentiation and simplification, we should
arrive at a result that shows dA
dθ is a constant, implying that equal areas are
indeed swept out in equal times.
Therefore, we have shown that the given equation satisfies Kepler’s second
law.
4
Question 6
Question
An asteroid is in an elliptical orbit around the Sun. The closest distance of the
asteroid to the Sun (perihelion) is 0.3 AU and the farthest distance (aphelion)
is 3.0 AU. Calculate the eccentricity of the asteroid’s orbit.
Solution
Step 1: The eccentricity of an ellipse is defined as e=c
a, where cis the distance
from the center of the ellipse to one of the foci and ais the semi-major axis of
the ellipse.
Step 2: The semi-major axis of the ellipse is the average of the perihelion
and aphelion distances. This can be calculated as:
a=rperihelion +raphelion
2=0.3AU + 3.0AU
2= 1.65 AU
Step 3: The distance from the center of the ellipse to one of the foci can be
found using the relationship c=a−rperihelion. Thus,
c= 1.65 AU −0.3AU = 1.35 AU
Step 4: Finally, we can calculate the eccentricity of the asteroid’s orbit using
the formula e=c
a:
e=1.35 AU
1.65 AU = 0.818
Therefore, the eccentricity of the asteroid’s orbit is 0.818.
Question 7
Question
In a distant solar system, a planet has an elliptical orbit around its star. The
planet’s closest approach to the star is 0.1 astronomical units (AU), and the
farthest distance is 0.2 AU. If the orbital period of this planet is 100 days,
determine the eccentricity of its orbit.
Solution
Step 1: First, let’s recall Kepler’s third law, which relates the orbital period of
a planet to the semi-major axis of its orbit. Kepler’s third law can be expressed
as:
T2=(4π2
G(M1+M2))a3
5
where: - T= orbital period of the planet - G= gravitational constant - M1and
M2= masses of the planet and the star - a= semi-major axis of the orbit
Step 2: We are given the orbital period T= 100 days and the closest (rmin =
0.1AU) and farthest (rmax = 0.2AU) distances from the star. We know that
in an elliptical orbit, the semi-major axis ais the average of the closest and
farthest distances from the star, so a=rmin +rmax
2.
Step 3: Plugging in the given values, we have:
a=0.1+0.2
2= 0.15 AU
Step 4: Now, we can substitute T= 100 days and a= 0.15 AU into Kepler’s
third law to find M1+M2:
1002=(4π2
G(M1+M2))(0.15)3
Step 5: We also know that the eccentricity eof an elliptical orbit can be
determined using the equation:
e=√1−(b
a)2
where: - a= semi-major axis - b= semi-minor axis
Step 6: For an elliptical orbit, the semi-minor axis bcan be calculated using
the eccentricity and the semi-major axis: b=a√1−e2.
Step 7: Substituting a= 0.15 AU into the expression for the semi-minor
axis, we get:
b= 0.15√1−e2
Step 8: However, the semi-minor axis b=1
2(rmax −rmin). Thus, we have:
0.15√1−e2=1
2(0.2−0.1)
Step 9: Solve for the eccentricity eby first finding the value of b, and then
substituting back into the equation for the eccentricity.
Question 8
Question
A satellite is in a circular orbit around a planet of mass M. The satellite’s orbital
radius is rand its angular speed is ω. Find an expression for the satellite’s period
of revolution Tin terms of r,M, and ω.
6
Solution
Step 1: Write down the formula for the period of revolution of a satellite in
circular orbit. The period of revolution Tof a satellite in a circular orbit can
be related to the angular speed ωby the formula:
T=2π
ω
Step 2: Express the angular speed ωin terms of the orbital radius r. The
angular speed ωis related to the satellite’s tangential speed vand the orbital
radius rby the equation:
v=rω
Step 3: Find an expression for tangential speed v. The tangential speed vof
the satellite is determined by the balance between the gravitational force and
the centripetal force. The gravitational force is given by:
Fg=GMm
r2
where Gis the gravitational constant, Mis the mass of the planet, and mis
the mass of the satellite. The centripetal force is given by:
Fc=mv2
r
Setting these two forces equal, we have:
GMm
r2=mv2
r
Solving for v, we get:
v=√GM
r
Step 4: Substitute the expression for vinto the formula for ω. Substitute
v=rω into the equation v=√GM
rto get:
rω =√GM
r
Solving for ω, we find:
ω=√GM
r3/2
Step 5: Substitute the expression for ωinto the formula for T. Finally,
substituting ω=√GM
r3/2into the formula T=2π
ω, we get:
T=2π
√GM
r3/2
=2πr3/2
√GM
Therefore, the expression for the satellite’s period of revolution Tin terms of r,
M, and ωis T=2πr3/2
√GM .
7
Question 9
Question
Consider a planet in an elliptical orbit around a star, such that at its closest
approach (perihelion) the planet’s speed is vpand the distance from the planet
to the star is rp. At its farthest point (aphelion), the planet’s speed is vaand
the distance from the planet to the star is ra. Given that the orbit is closed and
obeys Kepler’s laws, prove that the average of the squares of the speeds of the
planet at perihelion and aphelion is equal to the harmonic mean of the squares
of the average speeds at perihelion and aphelion.
Solution
Step 1: Let’s denote the mass of the star as M, the mass of the planet as m,
and the gravitational force between them as F. Using Kepler’s third law, we
have the equation for the period of the planet’s orbit:
T2=4π2
G(M+m)a3
where ais the semi-major axis of the orbit.
Step 2: Since the orbit is closed, the total mechanical energy of the planet
is conserved. At each point in the orbit, the total energy is given by
E=1
2mv2−GMm
r
where vis the speed of the planet and ris the distance from the planet to the
star.
Step 3: At perihelion, the total energy is
Ep=1
2mv2
p−GMm
rp
and at aphelion, the total energy is
Ea=1
2mv2
a−GMm
ra
Step 4: Using the fact that energy is conserved, we have Ep=Ea. This
gives us the equation
1
2mv2
p−GMm
rp
=1
2mv2
a−GMm
ra
Step 5: Rearranging the equation above, we get
1
2m(v2
p−v2
a) = GMm (1
ra−1
rp)
8
Step 6: Dividing by mand rearranging gives us an expression for the differ-
ence in speeds squared:
v2
p−v2
a= 2GM (1
ra−1
rp)
Step 7: To find the average of the squares of the speeds, we calculate
v2
p+v2
a
2=
v2
p+v2
p−2GM (1
ra−1
rp)
2=v2
p−GM (1
ra−1
rp)
Step 8: Similarly, the average of the squares of the average speeds is given
by
√1
2((v2
p−2GM/rp)+(v2
a−2GM/ra)) = √1
2(2v2
p+ 2v2
a−2GM(1
ra
+1
rp
))
=√v2
p+v2
a−2GM(1
ra
+1
rp
)
Step 9: Since raand rpare the distances from the planet to the star at
aphelion and perihelion respectively, we notice that their sum is equal to twice
the semi-major axis a, which is a constant. Thus,
ra+rp= 2a
Step 10: Substituting this back into our expressions, we find that
v2
p−GM (2
rp−1
rp)=v2
p−GM (1
ra−1
rp)
v2
a−GM
Question 10
Question
A planet orbits a star in an elliptical orbit. The planet’s distance from the star
varies between 0.3 AU and 0.7 AU. If the planet’s closest distance to the star is
0.3 AU, determine its farthest distance. Assume the orbit is perfectly elliptical.
Solution
Step 1: Recall Kepler’s laws of planetary motion. In particular, Kepler’s first
law states that the orbit of a planet is an ellipse with the star at one of the two
foci.
9
Step 2: The distance between the center of the ellipse (the star) and the
closest point on the ellipse (the planet at its closest distance) is called the semi-
major axis (a). In this case, the closest distance is 0.3 AU.
Step 3: The distance between the center of the ellipse and the farthest point
on the ellipse (the planet at its farthest distance) is also equal to the sum of the
semi-major axis and the distance between the two foci. This distance is also
known as the major axis (2a).
Step 4: Given that the closest distance (0.3 AU) is half of the major axis
(2a), we can solve for the value of a. Therefore, we have:
2a= 2(0.3AU) = 0.6AU
Step 5: Since the major axis (2a) is equal to the sum of the semi-major axis
(a) and the distance between the foci, and we know that the distance between
the foci (2ae) is 2(a-c), we can write:
0.6AU =a+ 2(a−c)
Step 6: From Kepler’s second law, we know that 2c is equal to the difference
between the closest and farthest distances from the star. We are given that the
closest distance is 0.3 AU and we need to find the farthest distance.
2c= 0.7AU −0.3AU = 0.4AU
Step 7: Substitute the value of 2cinto the equation from Step 5 and solve
for the farthest distance a+c. This yields:
0.6AU =a+ 2(0.4AU)
0.6AU =a+ 0.8AU
a= 0.6AU −0.8AU =−0.2AU
Step 8: Since a planet’s distance from a star cannot be negative, this solution
is extraneous. Therefore, we have made an error in our calculations. Let’s revisit
our previous steps to identify and correct our mistake.
Question 11
Question
Consider a comet in a highly eccentric orbit around the Sun. The aphelion
distance (furthest distance from the Sun) of this comet is known to be 4.5 AU,
while the perihelion distance (closest distance to the Sun) is 0.5 AU. Determine
the eccentricity of this comet’s orbit.
10
Solution
Step 1: Recall that the eccentricity of an orbit is defined as:
e=dap −dper
dap +dper
where dap is the aphelion distance and dper is the perihelion distance.
Step 2: Substitute the given values into the eccentricity formula:
e=4.5−0.5
4.5+0.5=4
5= 0.8
Step 3: Therefore, the eccentricity of the comet’s orbit is 0.8.
Question 12
Question
A comet takes 76 years to complete one orbit around the sun. If the distance
between the comet and the sun at its closest approach (perihelion) is 0.89 AU
(astronomical units), determine the semi-major axis of the comet’s elliptical
orbit in AU. Given that the semi-major axis of Earth’s orbit is approximately
1 AU.
Solution
Step 1: Let’s denote the semi-major axis of the comet’s elliptical orbit as a, and
the distance between the comet and the sun at its furthest point (aphelion) as
2a. We know that the period of the comet’s orbit is related to the semi-major
axis by Kepler’s third law:
T2
comet
a3
comet
=T2
Earth
a3
Earth
Given Tcomet = 76 years, aEarth = 1 AU, and TEarth = 1 year:
762
a3=1
1
Step 2: Solve for the semi-major axis a:
a3= 762
a=3
√762
Step 3: Calculate the value of a:
a≈3
√5776 ≈18.06 AU
11
Step 4: Finally, since the perihelion distance is 0.89 AU, the semi-major axis
of the comet’s elliptical orbit is:
a=0.89 + 2a
2
Substitute the value of a:
18.06 = 0.89 + 2(18.06)
2
Step 5: Solve for a:
18.06 = 0.89 + 36.12
2
18.06 = 37.01
2
a≈18.5AU
Therefore, the semi-major axis of the comet’s elliptical orbit is approximately
18.5 AU.
Question 13
Question
A comet has a highly eccentric orbit around the sun. At its perihelion, the
comet’s distance from the sun is 0.2 AU, and at its aphelion, the distance is 6.0
AU. Calculate the eccentricity of the comet’s orbit.
Solution
Step 1: We know that the eccentricity of an orbit is given by the formula:
e=rmax −rmin
rmax +rmin
Step 2: Given that the perihelion distance rmin = 0.2AU and the aphelion
distance rmax = 6.0AU, we can substitute these values into the formula:
e=6.0−0.2
6.0+0.2
Step 3: Calculating the eccentricity:
e=5.8
6.2= 0.9355
Step 4: Therefore, the eccentricity of the comet’s orbit is 0.9355 .
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Question 14
Question
An asteroid is known to have an orbital period around the Sun of 6.5 years and
an average distance from the Sun of 3.2 AU. Determine the asteroid’s semi-major
axis and eccentricity of its orbit.
Solution
Step 1: Calculate the semi-major axis (a) of the asteroid’s orbit using Kepler’s
Third Law:
T2=4π2a3
G(M1+M2)
where Tis the orbital period, Gis the gravitational constant, M1is the mass
of the Sun, and M2is the mass of the asteroid.
Step 2: Rearrange the equation to solve for a:
a=(G(M1+M2)T2
4π2)1/3
Step 3: Substitute the given values into the equation to find a:
a=(6.67 ×10−11 ·(1.99 ×1030 +m)·(6.5years)2
4π2)1/3
Step 4: Convert the orbital period from years to seconds:
T= 6.5years = 6.5×365 ×24 ×3600 seconds
Step 5: Calculate ain astronomical units (AU).
Step 6: Calculate the eccentricity (e) of the orbit using the equation:
e=√1−b2
a2
where bis the semi-minor axis of the ellipse.
Step 7: Calculate the semi-minor axis using the formula for an ellipse:
b=a√1−e2
Step 8: Substitute the known values of aand solve for b.
Step 9: Substitute the calculated values of aand binto the eccentricity
equation to find e.
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Question 15
Question
State Kepler’s third law of planetary motion and derive the equation which
relates the orbital period of a planet (T) to the distance of the planet from the
sun (r).
Solution
Kepler’s Third Law: Kepler’s third law of planetary motion states that the
square of the orbital period of a planet is directly proportional to the cube of
the semi-major axis of its orbit.
Let Tbe the orbital period of a planet and rbe the distance of the planet
from the sun. According to Kepler’s third law, we have:
T2∝r3
To derive the equation relating Tand r, we introduce a constant of propor-
tionality, k. Thus, we can write:
T2=kr3
We can determine the value of kby considering a known planet, such as
Earth. The values of Tand rfor Earth are TEarth = 1 year and rEarth = 1
astronomical unit (AU, the average distance between the Earth and the Sun).
Substituting these values into the equation, we get:
(1)2=k(1)3=⇒k= 1
Therefore, the equation relating the orbital period of a planet (T) and the
distance of the planet from the sun (r) is:
T2=r3
This equation shows the quantitative relationship between the orbital period
and the distance of a planet from the sun based on Kepler’s third law.
Question 16
Question
A planet moves in an elliptical orbit around the Sun. The planet’s closest
distance to the Sun (perihelion) is 0.3 AU and its farthest distance (aphelion)
is 0.7 AU. Calculate the eccentricity of the planet’s orbit.
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Solution
To find the eccentricity of the planet’s orbit, we can use the formula relating the
perihelion distance (rmin), the aphelion distance (rmax), and the eccentricity (e)
of an elliptical orbit:
e=rmax −rmin
rmax +rmin
Step 1: Substitute the given distances into the formula.
e=0.7−0.3
0.7+0.3
Step 2: Calculate the eccentricity.
e=0.4
1.0= 0.4
Therefore, the eccentricity of the planet’s orbit is 0.4.
Question 17
Question
At a certain planet, a satellite takes 10 hours to orbit the planet at a distance
of 2.0 Earth radii. What is the mass of the planet? (Assume the planet is
spherical)
Solution
Step 1: Calculate the orbital speed of the satellite. The orbital speed of a
satellite is given by:
v=2πr
T
where: v= orbital speed, r= radius of orbit, T= time period of orbit.
Substitute r= 2.0×REarth and T= 10 hours into the formula:
v=2π×2.0×REarth
10 ×3600 s/hour
Step 2: Calculate the gravitational force acting on the satellite. The gravi-
tational force between the planet and the satellite is given by:
F=GMm
r2
where: F= gravitational force, G= gravitational constant, M= mass of the
planet, m= mass of the satellite, r= distance between the planet and the
satellite.
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Using Newton’s second law (F=m·a) and the formula for centripetal
acceleration (a=v2
r), we have:
GMm
r2=m·v2
r
Step 3: Calculate the mass of the planet. Substitute the given values of r
and vinto the equation above and solve for M:
M=v2·r
G
Question 18
Question
Consider a planet in an elliptical orbit around a star, with the star at one of
the foci of the ellipse. The planet covers equal areas in equal times according
to Kepler’s second law of planetary motion.
Given that the distance of the planet from the star at its closest approach
(perihelion) is r1and the distance at its farthest point (aphelion) is r2, prove
that the time taken for the planet to travel from aphelion to perihelion is the
same as the time taken to travel from perihelion to aphelion.
Solution
To prove that the time taken for the planet to travel from aphelion to perihelion
is the same as the time taken to travel from perihelion to aphelion, we need to
show that the areas swept out by the planet are equal for these two portions of
its orbit. We will make use of Kepler’s second law which states that equal areas
are swept out in equal times.
Step 1: Calculate the area swept out from aphelion to perihelion. Let’s
consider a small time interval from tto t+dt when the planet is at a distance r
from the star. The area swept out during this time interval can be approximated
as a small triangle with base rand height r dθ (where dθ is the angle swept out
by the radius vector). The area dA of this triangle is given by:
dA =1
2r2dθ
Integrating from θ1to θ2(the angles at aphelion and perihelion respectively),
we get:
Aaphelion to perihelion =∫θ2
θ1
1
2r2dθ
Step 2: Calculate the area swept out from perihelion to aphelion. Now,
consider the time interval from t+dt to twhen the planet moves from perihelion
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to aphelion. The area swept out during this time can be represented as a triangle
with the same base rand height r dθ. Hence, in this interval, the area swept
out is also dA =1
2r2dθ.
Integrating from θ2to θ1, we get:
Aperihelion to aphelion =∫θ1
θ2
1
2r2dθ
Step 3: Show that the areas are equal. Since the area swept out is the same
in each case, we have:
Aaphelion to perihelion =Aperihelion to aphelion
This shows that the time taken for the planet to travel from aphelion to
perihelion is the same as the time taken to travel from perihelion to aphelion,
consistent with Kepler’s second law.
Question 19
Question
An asteroid is in an elliptical orbit around the Sun with a semi-major axis of 2.5
AU. The asteroid’s closest approach to the Sun (perihelion) is 0.5 AU. Calculate
the eccentricity of the asteroid’s orbit.
Solution
Step 1: Recall that the eccentricity of an ellipse is defined as
e=distance from center to focus
semi-major axis
Step 2: In this case, the distance from the center to one focus is the differ-
ence between the semi-major axis and the perihelion distance. Therefore, the
eccentricity can be calculated as:
e=2.5−0.5
2.5
Step 3: Calculating this expression gives:
e=2
2.5=4
5= 0.8
Step 4: Therefore, the eccentricity of the asteroid’s orbit is 0.8.
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Question 20
Question
A comet is discovered that follows a highly elliptical orbit around the Sun.
By observing the comet over a period of time, astronomers determine that the
closest approach of the comet to the Sun (perihelion) is 0.2 AU and the farthest
distance from the Sun (aphelion) is 3.0 AU. If the comet’s period of revolution
is 12 years, calculate the eccentricity of the comet’s orbit.
Solution
Step 1: Use Kepler’s third law to find the semi-major axis of the comet’s orbit.
According to Kepler’s third law, the square of the period of revolution of a
planet or comet around the Sun is proportional to the cube of the semi-major
axis of its orbit.
T2=ka3
where Tis the period of revolution in years, ais the semi-major axis in astro-
nomical units (AU), and kis a constant. Since the period of revolution is 12
years:
122=ka3
144 = ka3
Step 2: Calculate the semi-major axis. Since the closest distance (perihelion)
is 0.2 AU and the farthest distance (aphelion) is 3.0 AU, the semi-major axis a
is the average of these two distances.
a=0.2+3.0
2= 1.6AU
Step 3: Calculate the eccentricity of the orbit. The eccentricity eof an
ellipse is related to the semi-major axis a, perihelion distance rp, and aphelion
distance raby the formula:
e=ra−rp
ra+rp
=3.0−0.2
3.0+0.2=2.8
3.2= 0.875
Therefore, the eccentricity of the comet’s orbit is 0.875.
Question 21
Question
A planet has an elliptical orbit around the Sun, with the Sun located at one of
the foci. The planet’s closest approach to the Sun (perihelion) is 0.3 AU and its
farthest distance from the Sun (aphelion) is 0.7 AU. Calculate the eccentricity
of the planet’s orbit.
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Solution
Step 1: The eccentricity eof an ellipse is defined as the ratio of the distance
between the foci of the ellipse to the length of the major axis.
Step 2: In this case, the major axis of the ellipse is the distance between
the perihelion and aphelion, which is the sum of the distances to the Sun:
a= 0.3AU + 0.7AU = 1.0AU.
Step 3: The distance between the foci of the ellipse can be calculated using
the formula 2ae = 1.0AU,where ais the semi-major axis of the ellipse.
Step 4: Substituting the given perihelion (a(1 −e)) and aphelion (a(1 + e))
distances into the formula, we have 2ae = (0.3AU)(1 −e) + (0.7AU)(1 + e).
Step 5: Expanding and simplifying the equation, we get 2ae = 0.3AU −
0.3e+ 0.7AU + 0.7e.
Step 6: Rearranging terms, we have 2ae = 1.0AU + 0.4e.
Step 7: Solving for e, we get e=1.0AU
2a−0.4.
Step 8: Substituting a= 1.0AU into the equation, we find e=1.0AU
2(1.0AU)−0.4=
1.0AU
1.6= 0.625.
Step 9: Therefore, the eccentricity of the planet’s orbit is 0.625 .
Question 22
Question
A planet follows an elliptical orbit around the Sun. The distance between the
planet and the Sun at its closest approach is 0.5 AU (astronomical units), and
at its farthest distance, it is 2.0 AU. Calculate the eccentricity of the planet’s
orbit.
Solution
To calculate the eccentricity of the planet’s orbit, we can use the formula relating
the closest and farthest distances of a planet from the focus of its elliptical orbit
to the semi-major axis and eccentricity:
e=rmax −rmin
rmax +rmin
Step 1: Convert the closest and farthest distances from AU to the same
unit. Given:
rmin = 0.5AU
rmax = 2.0AU
Step 2: Substitute the values into the eccentricity formula. Plugging in the
values:
e=2.0−0.5
2.0+0.5
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Step 3: Solve for eccentricity.
e=1.5
2.5= 0.6
Therefore, the eccentricity of the planet’s orbit is 0.6.
Question 23
Question
Consider a planet orbiting a star in an elliptical orbit. The planet’s farthest
distance from the star is 3.0 AU, and its closest distance to the star is 1.0 AU.
Determine the eccentricity of the planet’s orbit.
Solution
To find the eccentricity (e) of the planet’s orbit, we can use the following formula:
e=rmax −rmin
rmax +rmin
Step 1: Given that the planet’s farthest distance from the star is rmax = 3.0
AU and its closest distance to the star is rmin = 1.0AU, we can substitute these
values into the formula:
e=3.0−1.0
3.0+1.0
Step 2: Calculate the numerator and denominator separately:
e=2.0
4.0
Step 3: Simplify the fraction to find the eccentricity:
e= 0.5
Step 4: Therefore, the eccentricity of the planet’s orbit is 0.5.
Question 24
Question
An artificial satellite is in an elliptical orbit around the Earth. The aphelion
(farthest point from Earth) of the orbit is 40,000 km and the perihelion (closest
point to Earth) is 10,000 km. Calculate the eccentricity of the orbit. Given:
• Aphelion distance (maximum distance from Earth) = 40,000 km
• Perihelion distance (minimum distance from Earth) = 10,000 km
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Solution
Step 1: Use the definition of eccentricity (e) in terms of aphelion and perihelion
distances:
e=Aphelion distance −Perihelion distance
Aphelion distance +Perihelion distance
Step 2: Substitute the given values into the formula:
e=40,000 −10,000
40,000 + 10,000
Step 3: Simplify the expression:
e=30,000
50,000 = 0.6
Therefore, the eccentricity of the orbit is 0.6.
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