SweetStudy (HomeworkMarket.com)
  • chat0

PHSC 121 - INTRODUCTION TO ASTRONOMY Kepler's Laws of Planetary Motion Question Bank Set 10

Is there anything else you׳d like to ask?
Our top-rated tutors can help you.

Click here to post a question
Related Documents
1 / 67100%
PHSC 121 - INTRODUCTION TO
ASTRONOMY - Kepler’s Laws of
Planetary Motion
Question Bank - Set 10
Liberty University
Question 1
Question
Kepler’s third law of planetary motion states that the ratio of the square of
the period of revolution (T) of a planet to the cube of its average distance
from the Sun (R) is constant for all planets. If Earth takes 365.25 days to
complete one revolution around the Sun and its average distance from the Sun
is 1 astronomical unit (AU), calculate the value of the constant in this case.
Solution
Step 1: Write Kepler’s third law in mathematical form: Kepler’s third law can
be written as:
T2
Earth
R3
Earth
=T2
planet
R3
planet
where TEarth = 365.25 days, REarth = 1 AU, and the subscript ”planet” refers
to any other planet in the solar system.
Step 2: Find the value of the constant: Substitute the values for Earth
(TEarth = 365.25 days and REarth = 1 AU) into the equation to find the constant:
365.252
13=T2
planet
R3
planet
133225.0625 = T2
planetR3
planet
Step 3: Identify the value of the constant: Since Kepler’s third law states
that the ratio is constant for all planets, the value found in Step 2 (133225.0625)
is the constant in this case.
Question 2
Question
Two planets, A and B, are orbiting a star. The semi-major axis of planet A’s
orbit is twice as long as the semi-major axis of planet B’s orbit. If planet A
takes 15 years to complete one orbit, how long does planet B take to complete
one orbit?
Solution
Step 1: Recall Kepler’s third law which states that the square of the period of
an orbiting body is proportional to the cube of the semi-major axis of its orbit.
Step 2: Let TAbe the period of planet A and TBbe the period of planet
B. Let aAbe the semi-major axis of planet A and aBbe the semi-major axis of
planet B. We are given that aA= 2aBand TA= 15 years.
Step 3: By Kepler’s third law, we have the following relation:
(TA)2=k(aA)3
(TB)2=k(aB)3
Step 4: Since aA= 2aB, we can rewrite the first equation as:
(TA)2=k(2aB)3
(TA)2= 8ka3
B
Step 5: Since planet A takes 15 years to complete one orbit, we can substitute
TA= 15 into the equation:
152= 8ka3
B
225 = 8ka3
B
Step 6: Now, let’s solve for TBusing the second equation:
(TB)2=k(aB)3
(TB)2=1
8×225
(TB)2=225
8
TB=√225
8
TB=15√5
2
Step 7: Therefore, planet B takes 15√5
2years to complete one orbit.
2
Question 3
Question
According to Kepler’s laws of planetary motion, the square of the orbital period
of a planet is directly proportional to the cube of the semi-major axis of its
orbit. Given that Earth has an orbital period of approximately 365.25 days and
a semi-major axis of 1 astronomical unit (AU), determine the orbital period of
a new planet with a semi-major axis of 2 AU.
Solution
Step 1: Let Tbe the orbital period of the new planet and abe the semi-major
axis of its orbit. According to Kepler’s third law, we have:
T2∝a3
Step 2: For Earth, T2
Earth =a3
Earth. Thus, we can write:
T2
Earth = (365.25 days)2= (1 AU)3
Step 3: Now we can find the orbital period of the new planet by setting up
a proportion:
T2
Earth
a3
Earth
=T2
a3
Substitute TEarth = 365.25 days and aEarth = 1 AU:
(365.25 days)2
(1 AU)3=T2
(2 AU)3
Step 4: Solving for T2, we have:
T2=((365.25 days)2
(1 AU)3)×(2 AU)3
T2= (365.25)2×23days2
Step 5: Calculate Tto find the orbital period of the new planet:
T=√(365.25)2×23days
T≈√133225.5625 days
T≈365.25 ×√8days
T≈365.25 ×2.83 days
3
T≈1034.55 days
Therefore, the orbital period of the new planet with a semi-major axis of 2
AU is approximately 1034.55 days.
Question 4
Question
A planet orbits a star in a nearly circular orbit. The semi-major axis of the
planet’s orbit is 2.5 AU (astronomical units). The planet takes 1.8 years to
complete one orbit around the star. Determine the mass of the star in solar
masses (where 1 solar mass is the mass of our Sun).
Solution
Step 1: Recall Kepler’s third law of planetary motion, which states:
(T1
T2)2
=(a1
a2)3
where T1and T2are the periods of two planets orbiting the same star, and a1
and a2are their semi-major axes.
Step 2: Let’s denote the mass of the star as M, and the semi-major axis and
period of the Earth’s orbit as a(1 AU) and T(1 year) respectively.
Step 3: Applying Kepler’s third law to the Earth’s orbit, we have:
(T
1.8years )2
=(a
2.5AU )3
Step 4: Simplifying this equation gives:
(1
1.8)2
=(1
2.5)3
Step 5: Solving for the ratio of the Earth’s mass to the star’s mass:
(1
1.8)2
=(1
2.5)3
⇒1
3.24 =1
15.625 ⇒M
1=15.625
3.24
Step 6: Therefore, the mass of the star in solar masses is:
M=15.625
3.24 ≈4.81 solar masses
Step 7: The mass of the star is approximately 4.81 times the mass of our
Sun.
4
Question 5
Question
In the context of Kepler’s Laws of Planetary Motion, consider a planet in an
elliptical orbit around the Sun. Given that the semi-major axis of the planet’s
orbit is 2 AU and its eccentricity is 0.4, determine the distance of the planet
from the Sun when it is at perihelion and aphelion.
Solution
Step 1: Calculate the distance of the planet from the Sun at perihelion. Given
the semi-major axis a= 2 AU and eccentricity e= 0.4, the distance from the
Sun at perihelion can be calculated using the formula:
rmin =a(1 −e)
Step 2: Substitute the values of aand einto the formula.
rmin = 2(1 −0.4) = 2 ×0.6 = 1.2AU
Therefore, the distance of the planet from the Sun at perihelion is 1.2 AU.
Step 3: Calculate the distance of the planet from the Sun at aphelion. The
distance from the Sun at aphelion can be calculated using the formula:
rmax =a(1 + e)
Step 4: Substitute the values of aand einto the formula.
rmax = 2(1 + 0.4) = 2 ×1.4 = 2.8AU
Therefore, the distance of the planet from the Sun at aphelion is 2.8 AU.
Question 6
Question
A planet is observed to have an orbital period of 4.2 years and a semimajor axis
of 2.8 astronomical units (AU). Calculate the mass of the star around which the
planet orbits, given that the gravitational constant is 6.67 ×10−11 N m2/kg2.
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion:
T2=(4π2
G(M+m))a3
5
where: - Tis the orbital period of the planet, - Gis the gravitational constant,
-Mis the mass of the star, - mis the mass of the planet, - ais the semimajor
axis of the planet’s orbit.
Step 2: We are given: T= 4.2years, a= 2.8AU, G= 6.67 ×10−11 N
m2/kg.
Step 3: Convert the semimajor axis afrom AU to meters:
1AU = 1.496 ×1011 m
a= 2.8×1.496 ×1011
a= 4.1888 ×1011 m
Step 4: Plug in the known values into Kepler’s Third Law and solve for the
mass of the star M:
(4.2years)2=(4π2
6.67 ×10−11(M+m))(4.1888 ×1011)3
Step 5: Simplify and solve for M.
This calculation requires careful manipulation of the formula and under-
standing of astronomical units.
Question 7
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is proportional to the cube of its average distance from the
sun. Consider a hypothetical planet with an orbital period of 5 years and an
average distance from the sun of 10 astronomical units (AU). What would be
the orbital period of another planet with an average distance from the sun of
15 AU?
Solution
To solve this problem, we can use Kepler’s third law of planetary motion, which
states:
T2
1
T2
2
=r3
1
r3
2
where: - T1and T2are the orbital periods of the two planets, - r1and r2are
the average distances of the two planets from the sun.
Given that the first planet has an orbital period (T1) of 5 years and an
average distance from the sun (r1) of 10 AU, we can substitute these values into
the equation and solve for the orbital period of the second planet (T2) when its
average distance from the sun (r2) is 15 AU.
6
Step 1: Write down Kepler’s third law formula:
T2
1
T2
2
=r3
1
r3
2
Step 2: Substitute the given values:
52
T2
2
=103
153
Step 3: Simplify the equation:
25
T2
2
=1000
3375
Step 4: Cross multiply to solve for T2
2:
25 ·3375 = 1000 ·T2
2
84375 = 1000 ·T2
2
T2
2=84375
1000
T2
2= 84.375
Step 5: Take the square root of both sides to find T2:
T2=√84.375
T2≈9.19 years
Therefore, the orbital period of the second planet would be approximately
9.19 years.
Question 8
Question
At one point in its orbit, a planet is 0.25 AU from the Sun and has a speed of
28 km/s. If the eccentricity of the planet’s orbit is 0.4, calculate the distance of
the planet from the Sun when it is at the farthest point from it.
Solution
Step 1: Recall Kepler’s Second Law, which states that the line joining a planet
and the Sun sweeps out equal areas in equal times. This means that a planet
moves faster when it is closer to the Sun and slower when it is farther away.
Step 2: Let’s denote the closest point to the Sun as periapsis and the farthest
point as apoapsis. The distance from the Sun to the periapsis is the planet’s
7
perihelion distance (rp) and the distance from the Sun to the apoapsis is the
planet’s aphelion distance (ra).
Step 3: According to Kepler’s Second Law, the speed of the planet when it
is at periapsis is related to its speed at apoapsis by the following equation:
vp·rp=va·ra
where vpand vaare the speeds of the planet at periapsis and apoapsis, respec-
tively.
Step 4: We are given that the planet’s speed at periapsis, vp= 28 km/s. We
are also given the eccentricity of the orbit, e= 0.4. The relationship between
the perihelion distance (rp), aphelion distance (ra), and eccentricity (e) is:
e=ra−rp
ra+rp
Solving for ragives us:
ra=1 + e
1−e·rp
Step 5: Substituting vp= 28 km/s into the Kepler’s Second Law equation
gives:
28 ·0.25 = va·ra
Step 6: We can solve for va, the speed of the planet at apoapsis:
28 ·0.25 = va·(1+0.4
1−0.4·0.25)
Step 7: Simplifying the equation from Step 6 will give us the speed of the
planet at apoapsis. Once we have that value, we can use it to find the distance
of the planet from the Sun when it is at the farthest point from it.
Question 9
Question
Kepler’s third law of planetary motion states that the square of the period of
revolution of a planet around the sun is proportional to the cube of the semi-
major axis of its elliptical orbit. Suppose a planet has a period of revolution
around the sun of 10 years and a semi-major axis of 3 astronomical units (AU).
Calculate the corresponding values for a different planet’s period of revolution
and semi-major axis if the square of the period is proportional to the cube of
the semi-major axis.
8
Solution
Step 1: Write down the ratio of periods and semi-major axes for the two planets:
Let T1be the period of revolution of the first planet, a1be the semi-major axis
of the first planet, T2be the period of revolution of the second planet, and a2
be the semi-major axis of the second planet. According to Kepler’s third law,
we have: T2
1
a3
1
=T2
2
a3
2
Step 2: Substitute the given values for the first planet into the equation:
Given that T1= 10 years and a1= 3 AU, we have:
102
33=T2
2
a3
2
Step 3: Simplify the equation by solving for the ratio T2
2
a3
2
:
100
27 =T2
2
a3
2
Step 4: Cross multiply to find the values for the second planet:
100 ·a3
2= 27 ·T2
2
Step 5: Given that this equation must hold true, we can set T2equal to x
and express a2in terms of x:
100 ·a3
2= 27 ·x2
Step 6: Since a2=x/10, substitute a2back into the equation:
100 ·(x/10)3= 27 ·x2
Step 7: Solve for xto find the period of revolution of the second planet:
100x3
1000 = 27x2
x= 27 years
Step 8: Finally, find the semi-major axis of the second planet using a2=
x/10:
a2=27
10 = 2.7AU
Therefore, the period of revolution of the second planet is 27 years and its
semi-major axis is 2.7 AU.
9
Question 10
Question
Kepler’s Third Law of Planetary Motion states that the square of the orbital
period (T) of a planet is proportional to the cube of the semi-major axis (a) of
its orbit. Suppose a planet has an orbital period of 4 years and a semi-major
axis of 3 AU (astronomical units). Calculate the orbital period of another planet
whose semi-major axis is 5 AU.
Solution
Step 1: Calculate the ratio of the cubes of the semi-major axes. Let the or-
bital period of the second planet be T2years and its semi-major axis be 5 AU.
According to Kepler’s Third Law:
T2
1
T2
2
=a3
1
a3
2
Substitute T1= 4 years, a1= 3 AU, a2= 5 AU:
42
T2
2
=33
53
Step 2: Calculate the orbital period of the second planet. Solving for T2:
16 = 27
125 ×T2
2
T2
2=125 ×16
27
T2
2=2000
27
T2=√2000
27 ≈7.09 years
Therefore, the orbital period of the second planet with a semi-major axis of
5 AU is approximately 7.09 years.
Question 11
Question
According to Kepler’s Third Law of Planetary Motion, the square of the orbital
period (T, in years) of a planet is proportional to the cube of its average distance
from the Sun (r, in astronomical units). For Earth, the average distance from
the Sun is about 1 AU and the orbital period is about 1 year.
If a newly discovered planet has an average distance from the Sun of 3 AU,
what is the predicted orbital period of this planet?
10
Solution
Step 1: Let’s express Kepler’s Third Law mathematically:
T2∝r3
Step 2: We can write the general form of Kepler’s Third Law as:
T2=k·r3
where kis the constant of proportionality.
Step 3: To find the value of the constant k, we can use the information given
for Earth:
(1)2=k·(1)3
Solving for k, we find k= 1.
Step 4: Now that we have the constant k, we can use it to predict the orbital
period of the newly discovered planet with r= 3 AU:
T2= 1 ·(3)3
T2= 27
Step 5: Taking the square root of both sides to find T, we get:
T=√27
T= 3√3
Therefore, the predicted orbital period of the newly discovered planet would
be approximately 3√3years.
Question 12
Question
Kepler’s third law relates the period of a planet’s orbit to its average distance
from the Sun. For two planets in the same solar system, planet X has an average
distance from the Sun that is 3 times greater than planet Y. If planet X takes
6 years to orbit the Sun, how long does it take planet Y to complete one orbit?
Solution
Step 1: Let TXbe the period of planet X and TYbe the period of planet Y.
Let rXand rYbe the average distances of planets X and Y from the Sun,
respectively. Given that rX= 3rYand TX= 6 years, we need to determine TY.
Step 2: According to Kepler’s third law, the square of the period of each
planet is proportional to the cube of its average distance from the Sun. Math-
ematically, this relationship can be expressed as:
(TX
TY)2
=(rX
rY)3
11
Step 3: Substituting TX= 6 years and rX= 3rYinto the equation from
Step 2, we get:
(6
TY)2
=(3rY
rY)3
Step 4: Simplifying the expression further, we have:
(6
TY)2
= 27
Step 5: Solving for TY, we get:
(6
TY)2
= 27
36
T2
Y
= 27
T2
Y=36
27 =4
3
TY=√4
3=2
√3=2√3
3
Step 6: Therefore, it takes planet Y 2√3
3years to complete one orbit around
the Sun.
Question 13
Question
The asteroid Pallas orbits the Sun at an average distance of 414 million kilo-
meters. Calculate the period of Pallas’ orbit in Earth years. (Hint: You may
assume that Pallas’ orbit is approximately circular.)
Solution
Step 1: We can use Kepler’s third law to relate the period of an orbit with the
semi-major axis of the orbit:
T2=4π2
G(M1+M2)a3
where: T= period of the orbit, G= gravitational constant (6.674×10−11 m3kg−1s−2),
M1= mass of the asteroid Pallas, M2= mass of the Sun, and a= semi-major
axis of the orbit.
Step 2: Since Pallas is an asteroid, we can assume its mass is negligible
compared to the Sun’s mass. Therefore, M1≈0. The mass of the Sun is
approximately 1.989 ×1030 kg.
12
Step 3: We can plug in the known values into Kepler’s third law:
T2=4π2
G(M1+M2)a3=4π2
6.674 ×10−11 ×1.989 ×1030 (414 ×109)3
Step 4: Calculating the period, we have:
T2=4π2
6.674 ×1.989(4143) = 4π2
13.281(7.096794 ×1019)
Step 5: Solving for T:
T=√4π2
13.281(7.096794 ×1019)
Step 6: After evaluating the expression, we find:
T≈√2.5993 ×1020 ≈5.0979 ×1010 s
Step 7: Converting the period to Earth years:
Tyears =5.0979 ×1010 s
60 ×60 ×24 ×365.25 ≈1621 years
Therefore, the period of Pallas’ orbit in Earth years is approximately 1621
years.
Question 14
Question
The period of a satellite orbiting a planet is 8 hours. If the planet’s radius is 2
times the radius of the satellite’s orbit, what is the orbital radius of the satellite
in terms of the planet’s radius?
Solution
Step 1: Let’s denote the radius of the satellite’s orbit as rand the radius of the
planet as R. We are given that the period of the satellite is 8 hours. According
to Kepler’s third law of planetary motion, the square of the period of an orbiting
body is proportional to the cube of the semi-major axis of its orbit. Therefore,
we have:
T2
1
r3
1
=T2
2
r3
2
where T1= 8 hours, r1=r,T2is the period of the planet (which is irrelevant
in this case), and r2=R.
Step 2: Substituting the given values into the formula, we get:
13
82
r3=T2
2
R3
Solving for r, we find:
r3= 64R3
r=3
√64 ·R
r= 4R
Therefore, the orbital radius of the satellite is 4 times the radius of the
planet.
Question 15
Question
A hypothetical planet orbits a star in an elliptical orbit, with the star located
at one of the foci of the ellipse. The semi-major axis of the planet’s orbit is 2
AU, and the eccentricity of the ellipse is 0.4. Calculate the distance between
the star and the planet when the planet is at its closest point to the star.
Solution
Step 1: Recall the formula to calculate the distance between a planet and the
star at any point in the orbit:
r=a(1 −e2)
1 + ecos(θ)
where: - ais the semi-major axis of the orbit, - eis the eccentricity of the orbit,
-θis the true anomaly (angle between the perihelion and the planet’s current
position), - ris the distance between the planet and the star at that point.
Step 2: Since we are looking for the distance when the planet is at its closest
point to the star, we need to find the distance when the planet is at perihelion,
where θ= 0◦.
Step 3: Substitute the given values into the formula:
r=2(1 −0.42)
1+0.4 cos(0◦)
Step 4: Simplify the equation:
r=2(1 −0.16)
1+0.4=2(0.84)
1.4=1.68
1.4= 1.2AU
Step 5: Therefore, the distance between the star and the planet when the
planet is at its closest point to the star is 1.2 AU.
14
Question 16
Question
Consider a hypothetical planetary system where a planet has an elliptical orbit
around a star. The semi-major axis of the planet’s orbit is 3 AU and the
eccentricity of the orbit is 0.5. Determine the distance between the closest
approach of the planet to the star (perihelion) and the farthest point from the
star (aphelion).
Solution
Step 1: Recall the formula for the distance between the closest approach and
the farthest point on an elliptical orbit, given the semi-major axis (a) and the
eccentricity (e):
Perihelion distance =a(1 −e)
Aphelion distance =a(1 + e)
Step 2: Substitute the given values into the formulas:
Perihelion distance = 3 AU(1 −0.5)
Perihelion distance = 3 AU ×0.5 = 1.5AU
Step 3: Calculate the aphelion distance:
Aphelion distance = 3 AU(1 + 0.5)
Aphelion distance = 3 AU ×1.5 = 4.5AU
Therefore, the distance between the closest approach of the planet to the
star (perihelion) is 1.5 AU, and the distance between the farthest point from
the star (aphelion) is 4.5 AU.
Question 17
Question
Consider a planet orbiting around a star in an elliptical path. The semi-major
axis of the planet’s orbit is 2.5 AU. If the eccentricity of the orbit is 0.3, deter-
mine the semi-minor axis of the orbit in AU.
Solution
Step 1: Recall the relationship between the semi-major axis (a), semi-minor
axis(b), and the eccentricity(e) of an ellipse:
a2=b2(1 −e2)
15
Step 2: Given that the semi-major axis (a) is 2.5 AU and the eccentricity(e)
is 0.3, we can substitute these values into the equation:
2.52=b2(1 −0.32)
Step 3: Solve for bby rearranging the equation:
b=√a2
1−e2
Step 4: Substitute a= 2.5AU and e= 0.3into the equation:
b=√2.52
1−0.32
Step 5: Calculate the value of b:
b=√6.25
1−0.09 =√6.25
0.91 ≈√6.8681 ≈2.622 AU
Therefore, the semi-minor axis of the orbit is approximately 2.622 AU.
Question 18
Question
In the study of planetary motion, Kepler’s Third Law states that the square
of a planet’s orbital period is proportional to the cube of its semi-major axis.
Suppose a new planet is discovered with a semi-major axis of 2.5 AU (astro-
nomical units). If the orbital period of this planet is 7 years, what would be the
semi-major axis of another planet with an orbital period of 20 years?
Solution
Step 1: Express Kepler’s Third Law mathematically: Let T1and a1represent
the orbital period and semi-major axis of the new planet, and let T2and a2
represent the orbital period and semi-major axis of the second planet. According
to Kepler’s Third Law, we have:
T2
1
T2
2
=a3
1
a3
2
Step 2: Plug in the given values: We are given that a1= 2.5AU, T1= 7
years, and T2= 20 years. We can now plug these values into the equation:
72
202=2.53
a3
2
16
Step 3: Solve for a2:49
400 =15.625
a3
2
a3
2=15.625 ×400
49 = 128.06
a2=3
√128.06 ≈5.03
Therefore, the semi-major axis of another planet with an orbital period of
20 years would be approximately 5.03 AU.
Question 19
Question
Consider a planet that moves in an elliptical orbit around the Sun. The planet
is at aphelion (its farthest point from the Sun) and travels at a speed of 15
km/s. If the planet’s speed at perihelion (its closest point to the Sun) is 25
km/s, calculate the average speed of the planet over a complete orbit.
Solution
Step 1: Recall Kepler’s second law, which states that a planet sweeps out equal
areas in equal times. This means that the planet moves faster when it is closer
to the Sun (at perihelion) and slower when it is farther away (at aphelion).
Step 2: Let vave be the average speed of the planet over a complete orbit. We
can calculate this average speed by finding the average of the speeds at aphelion
and perihelion.
Step 3: The average speed vave is given by:
vave =vaph +vper
2,
where vaph is the speed at aphelion and vper is the speed at perihelion.
Step 4: Substituting the given values for vaph = 15 km/s and vper = 25 km/s
into the formula, we have:
vave =15 + 25
2=40
2= 20 km/s.
Step 5: Therefore, the average speed of the planet over a complete orbit is
20 km/s.
Question 20
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period (T) of a planet is proportional to the cube of its semi-major axis (a).
17
Suppose a planet has an orbital period of 8 years and a semi-major axis of 3
AU (astronomical units). What would be the orbital period of a different planet
with a semi-major axis of 6 AU?
Solution
Step 1: Let’s denote the orbital period of the second planet as T2and its semi-
major axis as a2. According to Kepler’s third law, we have:
(T2
T)2
=(a2
a)3
Step 2: Given that T= 8 years and a= 3 AU, we can substitute these
values into the equation:
(T2
8)2
=(6
3)3
Step 3: Simplifying the equation gives:
(T2
8)2
= 23
Step 4: Further simplify to find the value of T2:
(T2
8)2
= 8 ⇒T2
8=√8
Step 5: Solving for T2gives:
T2= 8 ×√8 = 8 ×2√2 = 16√2years
Therefore, the orbital period of the second planet with a semi-major axis of
6 AU would be 16√2years.
Question 21
Question
Suppose an exoplanet is orbiting around a star in a nearly circular orbit, with
a semi-major axis of 0.5 AU. If the exoplanet takes 200 days to complete one
full orbit, calculate the mass of the central star in units of solar mass.
Solution
Step 1: Calculate the orbital period of the exoplanet using Kepler’s Third Law.
Kepler’s Third Law: P2=4π2
G(M1+M2)a3
18
where Pis the orbital period, Gis the gravitational constant, M1is the mass
of the star, M2is the mass of the exoplanet, and ais the semi-major axis of the
orbit.
Given: P= 200 days, a= 0.5AU, G= 6.67 ×10−11 m3kg−1s−2
Substitute the known values into the equation:
2002=4π2
G(M1)(0.5)3
Step 2: Solve for the mass of the central star M1.
M1=4π2(0.5)3
G×1
2002
M1=4π2(0.125)
G×40000
Step 3: Simplify the expression and convert the mass to solar mass units.
M1=0.5π2
107×6.67
M1=0.5π2
6.67 ×107
M1≈0.5×9.87
6.67 ×107
M1≈4.935
6.67 ×107
M1≈7.40 ×10−8solar mass
Therefore, the mass of the central star in solar mass units is approximately
7.40 ×10−8solar masses.
Question 22
Question
In the context of Kepler’s Laws of Planetary Motion, a planet orbits a star in
an elliptical path. The semi-major axis of the orbit is 2.5 AU. Given that the
eccentricity of the orbit is 0.4, determine the semi-minor axis of the orbit in
astronomical units.
19
Solution
Step 1: Recall the relationship between the semi-major axis (a), semi-minor axis
(b), and eccentricity (e) in an elliptical orbit:
b=a√1−e2
Step 2: Substitute the given values into the formula:
b= 2.5√1−0.42
Step 3: Calculate the value inside the square root first:
1−0.42= 1 −0.16 = 0.84
Step 4: Substitute the value back into the formula:
b= 2.5√0.84
Step 5: Calculate the square root of 0.84:
√0.84 ≈0.9165
Step 6: Multiply the calculated square root by 2.5 to find the semi-minor
axis in AU:
b≈2.5×0.9165 ≈2.29125 AU
Therefore, the semi-minor axis of the orbit is approximately 2.29125 AU.
Question 23
Question
Consider a planet with mass Morbiting a star with mass m. If the planet orbits
the star in a circular orbit with radius rand period T, derive an expression for
the orbital speed of the planet using Kepler’s Third Law.
Solution
Step 1: We begin by stating Kepler’s Third Law, which relates the period of an
orbiting body to the semi-major axis of its orbit. This law can be expressed as:
T2=(4π2
G(m+M))r3
where Gis the gravitational constant, mis the mass of the star, Mis the mass
of the planet, ris the radius of the orbit, and Tis the period of the orbit.
20
Step 2: Next, we rewrite the expression for T2in terms of the period T:
T=√(4π2
G(m+M))r3
Step 3: The orbital speed vof the planet can be calculated as the distance
traveled (2πr) divided by the time taken for one complete orbit (T):
v=2πr
T
Step 4: Substitute the expression for Tinto the equation for v:
v=2πr
√(4π2
G(m+M))r3
Step 5: Simplify the expression for v:
v=2π
√(4π2
G(m+M))r
Step 6: Further simplify the expression to obtain the final result:
v=√G(m+M)
r
Therefore, the orbital speed of the planet in a circular orbit can be calculated
using the expression v=√G(m+M)
r.
Question 24
Question
Kepler’s Third Law of Planetary Motion states that the square of the period of
revolution of a planet around the sun is proportional to the cube of its average
distance from the sun. Suppose a newly-discovered planet has a period of revo-
lution of 8 years and an average distance from the sun of 10 astronomical units
(AU). If another planet has a period of revolution of 2 years, what is its average
distance from the sun in AU?
Solution
Step 1: Let’s denote the period of revolution of the second planet as T(in years)
and its average distance from the sun as d(in AU). According to Kepler’s Third
Law, we have:
T2∝d3
21
Step 2: For the first planet, with a period of 8 years and an average distance
of 10 AU, we have the following relationship:
(8)2∝(10)3
64 ∝1000
Step 3: Now, for the second planet with a period of 2 years, we can set up
the following proportion:
T2
1=T2
2
d3
1=d3
2
Step 4: Since 64 ∝1000, we can set up a proportion using this relation:
22∝d3
2
4∝d3
2
Step 5: Taking the cube root of both sides, we find:
d2=3
√4
d2= 1.5874 AU
Therefore, the average distance from the sun for the second planet is ap-
proximately 1.5874 AU.
Question 25
Question
Consider a planet that follows an elliptical orbit around a star with the follow-
ing properties: semi-major axis length a= 2.5AU and eccentricity e= 0.6.
Determine the planet’s closest approach and furthest distance from the star in
AU.
Solution
Step 1: Calculate the distances at closest approach and furthest distance from
the star using the semi-major axis length and eccentricity formula.
Closest distance =a(1 −e)
Furthest distance =a(1 + e)
Step 2: Substitute the values a= 2.5AU and e= 0.6into the formulas.
Closest distance = 2.5(1 −0.6)
Furthest distance = 2.5(1 + 0.6)
22
Step 3: Calculate the distances.
Closest distance = 2.5(0.4) = 1.0AU
Furthest distance = 2.5(1.6) = 4.0AU
Therefore, the planet’s closest approach to the star is at a distance of 1.0
AU, while its furthest distance is at 4.0 AU.
23
Question 2
Question
Two planets, A and B, are orbiting a star. The semi-major axis of planet A’s
orbit is twice as long as the semi-major axis of planet B’s orbit. If planet A
takes 15 years to complete one orbit, how long does planet B take to complete
one orbit?
Solution
Step 1: Recall Kepler’s third law which states that the square of the period of
an orbiting body is proportional to the cube of the semi-major axis of its orbit.
Step 2: Let TAbe the period of planet A and TBbe the period of planet
B. Let aAbe the semi-major axis of planet A and aBbe the semi-major axis of
planet B. We are given that aA= 2aBand TA= 15 years.
Step 3: By Kepler’s third law, we have the following relation:
(TA)2=k(aA)3
(TB)2=k(aB)3
Step 4: Since aA= 2aB, we can rewrite the first equation as:
(TA)2=k(2aB)3
(TA)2= 8ka3
B
Step 5: Since planet A takes 15 years to complete one orbit, we can substitute
TA= 15 into the equation:
152= 8ka3
B
225 = 8ka3
B
Step 6: Now, let’s solve for TBusing the second equation:
(TB)2=k(aB)3
(TB)2=1
8×225
(TB)2=225
8
TB=√225
8
TB=15√5
2
Step 7: Therefore, planet B takes 15√5
2years to complete one orbit.
2
Question 3
Question
According to Kepler’s laws of planetary motion, the square of the orbital period
of a planet is directly proportional to the cube of the semi-major axis of its
orbit. Given that Earth has an orbital period of approximately 365.25 days and
a semi-major axis of 1 astronomical unit (AU), determine the orbital period of
a new planet with a semi-major axis of 2 AU.
Solution
Step 1: Let Tbe the orbital period of the new planet and abe the semi-major
axis of its orbit. According to Kepler’s third law, we have:
T2∝a3
Step 2: For Earth, T2
Earth =a3
Earth. Thus, we can write:
T2
Earth = (365.25 days)2= (1 AU)3
Step 3: Now we can find the orbital period of the new planet by setting up
a proportion:
T2
Earth
a3
Earth
=T2
a3
Substitute TEarth = 365.25 days and aEarth = 1 AU:
(365.25 days)2
(1 AU)3=T2
(2 AU)3
Step 4: Solving for T2, we have:
T2=((365.25 days)2
(1 AU)3)×(2 AU)3
T2= (365.25)2×23days2
Step 5: Calculate Tto find the orbital period of the new planet:
T=√(365.25)2×23days
T≈√133225.5625 days
T≈365.25 ×√8days
T≈365.25 ×2.83 days
3
T≈1034.55 days
Therefore, the orbital period of the new planet with a semi-major axis of 2
AU is approximately 1034.55 days.
Question 4
Question
A planet orbits a star in a nearly circular orbit. The semi-major axis of the
planet’s orbit is 2.5 AU (astronomical units). The planet takes 1.8 years to
complete one orbit around the star. Determine the mass of the star in solar
masses (where 1 solar mass is the mass of our Sun).
Solution
Step 1: Recall Kepler’s third law of planetary motion, which states:
(T1
T2)2
=(a1
a2)3
where T1and T2are the periods of two planets orbiting the same star, and a1
and a2are their semi-major axes.
Step 2: Let’s denote the mass of the star as M, and the semi-major axis and
period of the Earth’s orbit as a(1 AU) and T(1 year) respectively.
Step 3: Applying Kepler’s third law to the Earth’s orbit, we have:
(T
1.8years )2
=(a
2.5AU )3
Step 4: Simplifying this equation gives:
(1
1.8)2
=(1
2.5)3
Step 5: Solving for the ratio of the Earth’s mass to the star’s mass:
(1
1.8)2
=(1
2.5)3
⇒1
3.24 =1
15.625 ⇒M
1=15.625
3.24
Step 6: Therefore, the mass of the star in solar masses is:
M=15.625
3.24 ≈4.81 solar masses
Step 7: The mass of the star is approximately 4.81 times the mass of our
Sun.
4
Question 5
Question
In the context of Kepler’s Laws of Planetary Motion, consider a planet in an
elliptical orbit around the Sun. Given that the semi-major axis of the planet’s
orbit is 2 AU and its eccentricity is 0.4, determine the distance of the planet
from the Sun when it is at perihelion and aphelion.
Solution
Step 1: Calculate the distance of the planet from the Sun at perihelion. Given
the semi-major axis a= 2 AU and eccentricity e= 0.4, the distance from the
Sun at perihelion can be calculated using the formula:
rmin =a(1 −e)
Step 2: Substitute the values of aand einto the formula.
rmin = 2(1 −0.4) = 2 ×0.6 = 1.2AU
Therefore, the distance of the planet from the Sun at perihelion is 1.2 AU.
Step 3: Calculate the distance of the planet from the Sun at aphelion. The
distance from the Sun at aphelion can be calculated using the formula:
rmax =a(1 + e)
Step 4: Substitute the values of aand einto the formula.
rmax = 2(1 + 0.4) = 2 ×1.4 = 2.8AU
Therefore, the distance of the planet from the Sun at aphelion is 2.8 AU.
Question 6
Question
A planet is observed to have an orbital period of 4.2 years and a semimajor axis
of 2.8 astronomical units (AU). Calculate the mass of the star around which the
planet orbits, given that the gravitational constant is 6.67 ×10−11 N m2/kg2.
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion:
T2=(4π2
G(M+m))a3
5
where: - Tis the orbital period of the planet, - Gis the gravitational constant,
-Mis the mass of the star, - mis the mass of the planet, - ais the semimajor
axis of the planet’s orbit.
Step 2: We are given: T= 4.2years, a= 2.8AU, G= 6.67 ×10−11 N
m2/kg.
Step 3: Convert the semimajor axis afrom AU to meters:
1AU = 1.496 ×1011 m
a= 2.8×1.496 ×1011
a= 4.1888 ×1011 m
Step 4: Plug in the known values into Kepler’s Third Law and solve for the
mass of the star M:
(4.2years)2=(4π2
6.67 ×10−11(M+m))(4.1888 ×1011)3
Step 5: Simplify and solve for M.
This calculation requires careful manipulation of the formula and under-
standing of astronomical units.
Question 7
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is proportional to the cube of its average distance from the
sun. Consider a hypothetical planet with an orbital period of 5 years and an
average distance from the sun of 10 astronomical units (AU). What would be
the orbital period of another planet with an average distance from the sun of
15 AU?
Solution
To solve this problem, we can use Kepler’s third law of planetary motion, which
states:
T2
1
T2
2
=r3
1
r3
2
where: - T1and T2are the orbital periods of the two planets, - r1and r2are
the average distances of the two planets from the sun.
Given that the first planet has an orbital period (T1) of 5 years and an
average distance from the sun (r1) of 10 AU, we can substitute these values into
the equation and solve for the orbital period of the second planet (T2) when its
average distance from the sun (r2) is 15 AU.
6
Step 1: Write down Kepler’s third law formula:
T2
1
T2
2
=r3
1
r3
2
Step 2: Substitute the given values:
52
T2
2
=103
153
Step 3: Simplify the equation:
25
T2
2
=1000
3375
Step 4: Cross multiply to solve for T2
2:
25 ·3375 = 1000 ·T2
2
84375 = 1000 ·T2
2
T2
2=84375
1000
T2
2= 84.375
Step 5: Take the square root of both sides to find T2:
T2=√84.375
T2≈9.19 years
Therefore, the orbital period of the second planet would be approximately
9.19 years.
Question 8
Question
At one point in its orbit, a planet is 0.25 AU from the Sun and has a speed of
28 km/s. If the eccentricity of the planet’s orbit is 0.4, calculate the distance of
the planet from the Sun when it is at the farthest point from it.
Solution
Step 1: Recall Kepler’s Second Law, which states that the line joining a planet
and the Sun sweeps out equal areas in equal times. This means that a planet
moves faster when it is closer to the Sun and slower when it is farther away.
Step 2: Let’s denote the closest point to the Sun as periapsis and the farthest
point as apoapsis. The distance from the Sun to the periapsis is the planet’s
7
perihelion distance (rp) and the distance from the Sun to the apoapsis is the
planet’s aphelion distance (ra).
Step 3: According to Kepler’s Second Law, the speed of the planet when it
is at periapsis is related to its speed at apoapsis by the following equation:
vp·rp=va·ra
where vpand vaare the speeds of the planet at periapsis and apoapsis, respec-
tively.
Step 4: We are given that the planet’s speed at periapsis, vp= 28 km/s. We
are also given the eccentricity of the orbit, e= 0.4. The relationship between
the perihelion distance (rp), aphelion distance (ra), and eccentricity (e) is:
e=ra−rp
ra+rp
Solving for ragives us:
ra=1 + e
1−e·rp
Step 5: Substituting vp= 28 km/s into the Kepler’s Second Law equation
gives:
28 ·0.25 = va·ra
Step 6: We can solve for va, the speed of the planet at apoapsis:
28 ·0.25 = va·(1+0.4
1−0.4·0.25)
Step 7: Simplifying the equation from Step 6 will give us the speed of the
planet at apoapsis. Once we have that value, we can use it to find the distance
of the planet from the Sun when it is at the farthest point from it.
Question 9
Question
Kepler’s third law of planetary motion states that the square of the period of
revolution of a planet around the sun is proportional to the cube of the semi-
major axis of its elliptical orbit. Suppose a planet has a period of revolution
around the sun of 10 years and a semi-major axis of 3 astronomical units (AU).
Calculate the corresponding values for a different planet’s period of revolution
and semi-major axis if the square of the period is proportional to the cube of
the semi-major axis.
8
Solution
Step 1: Write down the ratio of periods and semi-major axes for the two planets:
Let T1be the period of revolution of the first planet, a1be the semi-major axis
of the first planet, T2be the period of revolution of the second planet, and a2
be the semi-major axis of the second planet. According to Kepler’s third law,
we have: T2
1
a3
1
=T2
2
a3
2
Step 2: Substitute the given values for the first planet into the equation:
Given that T1= 10 years and a1= 3 AU, we have:
102
33=T2
2
a3
2
Step 3: Simplify the equation by solving for the ratio T2
2
a3
2
:
100
27 =T2
2
a3
2
Step 4: Cross multiply to find the values for the second planet:
100 ·a3
2= 27 ·T2
2
Step 5: Given that this equation must hold true, we can set T2equal to x
and express a2in terms of x:
100 ·a3
2= 27 ·x2
Step 6: Since a2=x/10, substitute a2back into the equation:
100 ·(x/10)3= 27 ·x2
Step 7: Solve for xto find the period of revolution of the second planet:
100x3
1000 = 27x2
x= 27 years
Step 8: Finally, find the semi-major axis of the second planet using a2=
x/10:
a2=27
10 = 2.7AU
Therefore, the period of revolution of the second planet is 27 years and its
semi-major axis is 2.7 AU.
9
Question 10
Question
Kepler’s Third Law of Planetary Motion states that the square of the orbital
period (T) of a planet is proportional to the cube of the semi-major axis (a) of
its orbit. Suppose a planet has an orbital period of 4 years and a semi-major
axis of 3 AU (astronomical units). Calculate the orbital period of another planet
whose semi-major axis is 5 AU.
Solution
Step 1: Calculate the ratio of the cubes of the semi-major axes. Let the or-
bital period of the second planet be T2years and its semi-major axis be 5 AU.
According to Kepler’s Third Law:
T2
1
T2
2
=a3
1
a3
2
Substitute T1= 4 years, a1= 3 AU, a2= 5 AU:
42
T2
2
=33
53
Step 2: Calculate the orbital period of the second planet. Solving for T2:
16 = 27
125 ×T2
2
T2
2=125 ×16
27
T2
2=2000
27
T2=√2000
27 ≈7.09 years
Therefore, the orbital period of the second planet with a semi-major axis of
5 AU is approximately 7.09 years.
Question 11
Question
According to Kepler’s Third Law of Planetary Motion, the square of the orbital
period (T, in years) of a planet is proportional to the cube of its average distance
from the Sun (r, in astronomical units). For Earth, the average distance from
the Sun is about 1 AU and the orbital period is about 1 year.
If a newly discovered planet has an average distance from the Sun of 3 AU,
what is the predicted orbital period of this planet?
10
Solution
Step 1: Let’s express Kepler’s Third Law mathematically:
T2∝r3
Step 2: We can write the general form of Kepler’s Third Law as:
T2=k·r3
where kis the constant of proportionality.
Step 3: To find the value of the constant k, we can use the information given
for Earth:
(1)2=k·(1)3
Solving for k, we find k= 1.
Step 4: Now that we have the constant k, we can use it to predict the orbital
period of the newly discovered planet with r= 3 AU:
T2= 1 ·(3)3
T2= 27
Step 5: Taking the square root of both sides to find T, we get:
T=√27
T= 3√3
Therefore, the predicted orbital period of the newly discovered planet would
be approximately 3√3years.
Question 12
Question
Kepler’s third law relates the period of a planet’s orbit to its average distance
from the Sun. For two planets in the same solar system, planet X has an average
distance from the Sun that is 3 times greater than planet Y. If planet X takes
6 years to orbit the Sun, how long does it take planet Y to complete one orbit?
Solution
Step 1: Let TXbe the period of planet X and TYbe the period of planet Y.
Let rXand rYbe the average distances of planets X and Y from the Sun,
respectively. Given that rX= 3rYand TX= 6 years, we need to determine TY.
Step 2: According to Kepler’s third law, the square of the period of each
planet is proportional to the cube of its average distance from the Sun. Math-
ematically, this relationship can be expressed as:
(TX
TY)2
=(rX
rY)3
11
Step 3: Substituting TX= 6 years and rX= 3rYinto the equation from
Step 2, we get:
(6
TY)2
=(3rY
rY)3
Step 4: Simplifying the expression further, we have:
(6
TY)2
= 27
Step 5: Solving for TY, we get:
(6
TY)2
= 27
36
T2
Y
= 27
T2
Y=36
27 =4
3
TY=√4
3=2
√3=2√3
3
Step 6: Therefore, it takes planet Y 2√3
3years to complete one orbit around
the Sun.
Question 13
Question
The asteroid Pallas orbits the Sun at an average distance of 414 million kilo-
meters. Calculate the period of Pallas’ orbit in Earth years. (Hint: You may
assume that Pallas’ orbit is approximately circular.)
Solution
Step 1: We can use Kepler’s third law to relate the period of an orbit with the
semi-major axis of the orbit:
T2=4π2
G(M1+M2)a3
where: T= period of the orbit, G= gravitational constant (6.674×10−11 m3kg−1s−2),
M1= mass of the asteroid Pallas, M2= mass of the Sun, and a= semi-major
axis of the orbit.
Step 2: Since Pallas is an asteroid, we can assume its mass is negligible
compared to the Sun’s mass. Therefore, M1≈0. The mass of the Sun is
approximately 1.989 ×1030 kg.
12
Step 3: We can plug in the known values into Kepler’s third law:
T2=4π2
G(M1+M2)a3=4π2
6.674 ×10−11 ×1.989 ×1030 (414 ×109)3
Step 4: Calculating the period, we have:
T2=4π2
6.674 ×1.989(4143) = 4π2
13.281(7.096794 ×1019)
Step 5: Solving for T:
T=√4π2
13.281(7.096794 ×1019)
Step 6: After evaluating the expression, we find:
T≈√2.5993 ×1020 ≈5.0979 ×1010 s
Step 7: Converting the period to Earth years:
Tyears =5.0979 ×1010 s
60 ×60 ×24 ×365.25 ≈1621 years
Therefore, the period of Pallas’ orbit in Earth years is approximately 1621
years.
Question 14
Question
The period of a satellite orbiting a planet is 8 hours. If the planet’s radius is 2
times the radius of the satellite’s orbit, what is the orbital radius of the satellite
in terms of the planet’s radius?
Solution
Step 1: Let’s denote the radius of the satellite’s orbit as rand the radius of the
planet as R. We are given that the period of the satellite is 8 hours. According
to Kepler’s third law of planetary motion, the square of the period of an orbiting
body is proportional to the cube of the semi-major axis of its orbit. Therefore,
we have:
T2
1
r3
1
=T2
2
r3
2
where T1= 8 hours, r1=r,T2is the period of the planet (which is irrelevant
in this case), and r2=R.
Step 2: Substituting the given values into the formula, we get:
13
82
r3=T2
2
R3
Solving for r, we find:
r3= 64R3
r=3
√64 ·R
r= 4R
Therefore, the orbital radius of the satellite is 4 times the radius of the
planet.
Question 15
Question
A hypothetical planet orbits a star in an elliptical orbit, with the star located
at one of the foci of the ellipse. The semi-major axis of the planet’s orbit is 2
AU, and the eccentricity of the ellipse is 0.4. Calculate the distance between
the star and the planet when the planet is at its closest point to the star.
Solution
Step 1: Recall the formula to calculate the distance between a planet and the
star at any point in the orbit:
r=a(1 −e2)
1 + ecos(θ)
where: - ais the semi-major axis of the orbit, - eis the eccentricity of the orbit,
-θis the true anomaly (angle between the perihelion and the planet’s current
position), - ris the distance between the planet and the star at that point.
Step 2: Since we are looking for the distance when the planet is at its closest
point to the star, we need to find the distance when the planet is at perihelion,
where θ= 0◦.
Step 3: Substitute the given values into the formula:
r=2(1 −0.42)
1+0.4 cos(0◦)
Step 4: Simplify the equation:
r=2(1 −0.16)
1+0.4=2(0.84)
1.4=1.68
1.4= 1.2AU
Step 5: Therefore, the distance between the star and the planet when the
planet is at its closest point to the star is 1.2 AU.
14
Question 16
Question
Consider a hypothetical planetary system where a planet has an elliptical orbit
around a star. The semi-major axis of the planet’s orbit is 3 AU and the
eccentricity of the orbit is 0.5. Determine the distance between the closest
approach of the planet to the star (perihelion) and the farthest point from the
star (aphelion).
Solution
Step 1: Recall the formula for the distance between the closest approach and
the farthest point on an elliptical orbit, given the semi-major axis (a) and the
eccentricity (e):
Perihelion distance =a(1 −e)
Aphelion distance =a(1 + e)
Step 2: Substitute the given values into the formulas:
Perihelion distance = 3 AU(1 −0.5)
Perihelion distance = 3 AU ×0.5 = 1.5AU
Step 3: Calculate the aphelion distance:
Aphelion distance = 3 AU(1 + 0.5)
Aphelion distance = 3 AU ×1.5 = 4.5AU
Therefore, the distance between the closest approach of the planet to the
star (perihelion) is 1.5 AU, and the distance between the farthest point from
the star (aphelion) is 4.5 AU.
Question 17
Question
Consider a planet orbiting around a star in an elliptical path. The semi-major
axis of the planet’s orbit is 2.5 AU. If the eccentricity of the orbit is 0.3, deter-
mine the semi-minor axis of the orbit in AU.
Solution
Step 1: Recall the relationship between the semi-major axis (a), semi-minor
axis(b), and the eccentricity(e) of an ellipse:
a2=b2(1 −e2)
15
Step 2: Given that the semi-major axis (a) is 2.5 AU and the eccentricity(e)
is 0.3, we can substitute these values into the equation:
2.52=b2(1 −0.32)
Step 3: Solve for bby rearranging the equation:
b=√a2
1−e2
Step 4: Substitute a= 2.5AU and e= 0.3into the equation:
b=√2.52
1−0.32
Step 5: Calculate the value of b:
b=√6.25
1−0.09 =√6.25
0.91 ≈√6.8681 ≈2.622 AU
Therefore, the semi-minor axis of the orbit is approximately 2.622 AU.
Question 18
Question
In the study of planetary motion, Kepler’s Third Law states that the square
of a planet’s orbital period is proportional to the cube of its semi-major axis.
Suppose a new planet is discovered with a semi-major axis of 2.5 AU (astro-
nomical units). If the orbital period of this planet is 7 years, what would be the
semi-major axis of another planet with an orbital period of 20 years?
Solution
Step 1: Express Kepler’s Third Law mathematically: Let T1and a1represent
the orbital period and semi-major axis of the new planet, and let T2and a2
represent the orbital period and semi-major axis of the second planet. According
to Kepler’s Third Law, we have:
T2
1
T2
2
=a3
1
a3
2
Step 2: Plug in the given values: We are given that a1= 2.5AU, T1= 7
years, and T2= 20 years. We can now plug these values into the equation:
72
202=2.53
a3
2
16
Step 3: Solve for a2:49
400 =15.625
a3
2
a3
2=15.625 ×400
49 = 128.06
a2=3
√128.06 ≈5.03
Therefore, the semi-major axis of another planet with an orbital period of
20 years would be approximately 5.03 AU.
Question 19
Question
Consider a planet that moves in an elliptical orbit around the Sun. The planet
is at aphelion (its farthest point from the Sun) and travels at a speed of 15
km/s. If the planet’s speed at perihelion (its closest point to the Sun) is 25
km/s, calculate the average speed of the planet over a complete orbit.
Solution
Step 1: Recall Kepler’s second law, which states that a planet sweeps out equal
areas in equal times. This means that the planet moves faster when it is closer
to the Sun (at perihelion) and slower when it is farther away (at aphelion).
Step 2: Let vave be the average speed of the planet over a complete orbit. We
can calculate this average speed by finding the average of the speeds at aphelion
and perihelion.
Step 3: The average speed vave is given by:
vave =vaph +vper
2,
where vaph is the speed at aphelion and vper is the speed at perihelion.
Step 4: Substituting the given values for vaph = 15 km/s and vper = 25 km/s
into the formula, we have:
vave =15 + 25
2=40
2= 20 km/s.
Step 5: Therefore, the average speed of the planet over a complete orbit is
20 km/s.
Question 20
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period (T) of a planet is proportional to the cube of its semi-major axis (a).
17
Suppose a planet has an orbital period of 8 years and a semi-major axis of 3
AU (astronomical units). What would be the orbital period of a different planet
with a semi-major axis of 6 AU?
Solution
Step 1: Let’s denote the orbital period of the second planet as T2and its semi-
major axis as a2. According to Kepler’s third law, we have:
(T2
T)2
=(a2
a)3
Step 2: Given that T= 8 years and a= 3 AU, we can substitute these
values into the equation:
(T2
8)2
=(6
3)3
Step 3: Simplifying the equation gives:
(T2
8)2
= 23
Step 4: Further simplify to find the value of T2:
(T2
8)2
= 8 ⇒T2
8=√8
Step 5: Solving for T2gives:
T2= 8 ×√8 = 8 ×2√2 = 16√2years
Therefore, the orbital period of the second planet with a semi-major axis of
6 AU would be 16√2years.
Question 21
Question
Suppose an exoplanet is orbiting around a star in a nearly circular orbit, with
a semi-major axis of 0.5 AU. If the exoplanet takes 200 days to complete one
full orbit, calculate the mass of the central star in units of solar mass.
Solution
Step 1: Calculate the orbital period of the exoplanet using Kepler’s Third Law.
Kepler’s Third Law: P2=4π2
G(M1+M2)a3
18
where Pis the orbital period, Gis the gravitational constant, M1is the mass
of the star, M2is the mass of the exoplanet, and ais the semi-major axis of the
orbit.
Given: P= 200 days, a= 0.5AU, G= 6.67 ×10−11 m3kg−1s−2
Substitute the known values into the equation:
2002=4π2
G(M1)(0.5)3
Step 2: Solve for the mass of the central star M1.
M1=4π2(0.5)3
G×1
2002
M1=4π2(0.125)
G×40000
Step 3: Simplify the expression and convert the mass to solar mass units.
M1=0.5π2
107×6.67
M1=0.5π2
6.67 ×107
M1≈0.5×9.87
6.67 ×107
M1≈4.935
6.67 ×107
M1≈7.40 ×10−8solar mass
Therefore, the mass of the central star in solar mass units is approximately
7.40 ×10−8solar masses.
Question 22
Question
In the context of Kepler’s Laws of Planetary Motion, a planet orbits a star in
an elliptical path. The semi-major axis of the orbit is 2.5 AU. Given that the
eccentricity of the orbit is 0.4, determine the semi-minor axis of the orbit in
astronomical units.
19
Solution
Step 1: Recall the relationship between the semi-major axis (a), semi-minor axis
(b), and eccentricity (e) in an elliptical orbit:
b=a√1−e2
Step 2: Substitute the given values into the formula:
b= 2.5√1−0.42
Step 3: Calculate the value inside the square root first:
1−0.42= 1 −0.16 = 0.84
Step 4: Substitute the value back into the formula:
b= 2.5√0.84
Step 5: Calculate the square root of 0.84:
√0.84 ≈0.9165
Step 6: Multiply the calculated square root by 2.5 to find the semi-minor
axis in AU:
b≈2.5×0.9165 ≈2.29125 AU
Therefore, the semi-minor axis of the orbit is approximately 2.29125 AU.
Question 23
Question
Consider a planet with mass Morbiting a star with mass m. If the planet orbits
the star in a circular orbit with radius rand period T, derive an expression for
the orbital speed of the planet using Kepler’s Third Law.
Solution
Step 1: We begin by stating Kepler’s Third Law, which relates the period of an
orbiting body to the semi-major axis of its orbit. This law can be expressed as:
T2=(4π2
G(m+M))r3
where Gis the gravitational constant, mis the mass of the star, Mis the mass
of the planet, ris the radius of the orbit, and Tis the period of the orbit.
20
Step 2: Next, we rewrite the expression for T2in terms of the period T:
T=√(4π2
G(m+M))r3
Step 3: The orbital speed vof the planet can be calculated as the distance
traveled (2πr) divided by the time taken for one complete orbit (T):
v=2πr
T
Step 4: Substitute the expression for Tinto the equation for v:
v=2πr
√(4π2
G(m+M))r3
Step 5: Simplify the expression for v:
v=2π
√(4π2
G(m+M))r
Step 6: Further simplify the expression to obtain the final result:
v=√G(m+M)
r
Therefore, the orbital speed of the planet in a circular orbit can be calculated
using the expression v=√G(m+M)
r.
Question 24
Question
Kepler’s Third Law of Planetary Motion states that the square of the period of
revolution of a planet around the sun is proportional to the cube of its average
distance from the sun. Suppose a newly-discovered planet has a period of revo-
lution of 8 years and an average distance from the sun of 10 astronomical units
(AU). If another planet has a period of revolution of 2 years, what is its average
distance from the sun in AU?
Solution
Step 1: Let’s denote the period of revolution of the second planet as T(in years)
and its average distance from the sun as d(in AU). According to Kepler’s Third
Law, we have:
T2∝d3
21
Step 2: For the first planet, with a period of 8 years and an average distance
of 10 AU, we have the following relationship:
(8)2∝(10)3
64 ∝1000
Step 3: Now, for the second planet with a period of 2 years, we can set up
the following proportion:
T2
1=T2
2
d3
1=d3
2
Step 4: Since 64 ∝1000, we can set up a proportion using this relation:
22∝d3
2
4∝d3
2
Step 5: Taking the cube root of both sides, we find:
d2=3
√4
d2= 1.5874 AU
Therefore, the average distance from the sun for the second planet is ap-
proximately 1.5874 AU.
Question 25
Question
Consider a planet that follows an elliptical orbit around a star with the follow-
ing properties: semi-major axis length a= 2.5AU and eccentricity e= 0.6.
Determine the planet’s closest approach and furthest distance from the star in
AU.
Solution
Step 1: Calculate the distances at closest approach and furthest distance from
the star using the semi-major axis length and eccentricity formula.
Closest distance =a(1 −e)
Furthest distance =a(1 + e)
Step 2: Substitute the values a= 2.5AU and e= 0.6into the formulas.
Closest distance = 2.5(1 −0.6)
Furthest distance = 2.5(1 + 0.6)
22
Step 3: Calculate the distances.
Closest distance = 2.5(0.4) = 1.0AU
Furthest distance = 2.5(1.6) = 4.0AU
Therefore, the planet’s closest approach to the star is at a distance of 1.0
AU, while its furthest distance is at 4.0 AU.
23
Question 2
Question
Two planets, A and B, are orbiting a star. The semi-major axis of planet A’s
orbit is twice as long as the semi-major axis of planet B’s orbit. If planet A
takes 15 years to complete one orbit, how long does planet B take to complete
one orbit?
Solution
Step 1: Recall Kepler’s third law which states that the square of the period of
an orbiting body is proportional to the cube of the semi-major axis of its orbit.
Step 2: Let TAbe the period of planet A and TBbe the period of planet
B. Let aAbe the semi-major axis of planet A and aBbe the semi-major axis of
planet B. We are given that aA= 2aBand TA= 15 years.
Step 3: By Kepler’s third law, we have the following relation:
(TA)2=k(aA)3
(TB)2=k(aB)3
Step 4: Since aA= 2aB, we can rewrite the first equation as:
(TA)2=k(2aB)3
(TA)2= 8ka3
B
Step 5: Since planet A takes 15 years to complete one orbit, we can substitute
TA= 15 into the equation:
152= 8ka3
B
225 = 8ka3
B
Step 6: Now, let’s solve for TBusing the second equation:
(TB)2=k(aB)3
(TB)2=1
8×225
(TB)2=225
8
TB=√225
8
TB=15√5
2
Step 7: Therefore, planet B takes 15√5
2years to complete one orbit.
2
Question 3
Question
According to Kepler’s laws of planetary motion, the square of the orbital period
of a planet is directly proportional to the cube of the semi-major axis of its
orbit. Given that Earth has an orbital period of approximately 365.25 days and
a semi-major axis of 1 astronomical unit (AU), determine the orbital period of
a new planet with a semi-major axis of 2 AU.
Solution
Step 1: Let Tbe the orbital period of the new planet and abe the semi-major
axis of its orbit. According to Kepler’s third law, we have:
T2∝a3
Step 2: For Earth, T2
Earth =a3
Earth. Thus, we can write:
T2
Earth = (365.25 days)2= (1 AU)3
Step 3: Now we can find the orbital period of the new planet by setting up
a proportion:
T2
Earth
a3
Earth
=T2
a3
Substitute TEarth = 365.25 days and aEarth = 1 AU:
(365.25 days)2
(1 AU)3=T2
(2 AU)3
Step 4: Solving for T2, we have:
T2=((365.25 days)2
(1 AU)3)×(2 AU)3
T2= (365.25)2×23days2
Step 5: Calculate Tto find the orbital period of the new planet:
T=√(365.25)2×23days
T≈√133225.5625 days
T≈365.25 ×√8days
T≈365.25 ×2.83 days
3
T≈1034.55 days
Therefore, the orbital period of the new planet with a semi-major axis of 2
AU is approximately 1034.55 days.
Question 4
Question
A planet orbits a star in a nearly circular orbit. The semi-major axis of the
planet’s orbit is 2.5 AU (astronomical units). The planet takes 1.8 years to
complete one orbit around the star. Determine the mass of the star in solar
masses (where 1 solar mass is the mass of our Sun).
Solution
Step 1: Recall Kepler’s third law of planetary motion, which states:
(T1
T2)2
=(a1
a2)3
where T1and T2are the periods of two planets orbiting the same star, and a1
and a2are their semi-major axes.
Step 2: Let’s denote the mass of the star as M, and the semi-major axis and
period of the Earth’s orbit as a(1 AU) and T(1 year) respectively.
Step 3: Applying Kepler’s third law to the Earth’s orbit, we have:
(T
1.8years )2
=(a
2.5AU )3
Step 4: Simplifying this equation gives:
(1
1.8)2
=(1
2.5)3
Step 5: Solving for the ratio of the Earth’s mass to the star’s mass:
(1
1.8)2
=(1
2.5)3
⇒1
3.24 =1
15.625 ⇒M
1=15.625
3.24
Step 6: Therefore, the mass of the star in solar masses is:
M=15.625
3.24 ≈4.81 solar masses
Step 7: The mass of the star is approximately 4.81 times the mass of our
Sun.
4
Question 5
Question
In the context of Kepler’s Laws of Planetary Motion, consider a planet in an
elliptical orbit around the Sun. Given that the semi-major axis of the planet’s
orbit is 2 AU and its eccentricity is 0.4, determine the distance of the planet
from the Sun when it is at perihelion and aphelion.
Solution
Step 1: Calculate the distance of the planet from the Sun at perihelion. Given
the semi-major axis a= 2 AU and eccentricity e= 0.4, the distance from the
Sun at perihelion can be calculated using the formula:
rmin =a(1 −e)
Step 2: Substitute the values of aand einto the formula.
rmin = 2(1 −0.4) = 2 ×0.6 = 1.2AU
Therefore, the distance of the planet from the Sun at perihelion is 1.2 AU.
Step 3: Calculate the distance of the planet from the Sun at aphelion. The
distance from the Sun at aphelion can be calculated using the formula:
rmax =a(1 + e)
Step 4: Substitute the values of aand einto the formula.
rmax = 2(1 + 0.4) = 2 ×1.4 = 2.8AU
Therefore, the distance of the planet from the Sun at aphelion is 2.8 AU.
Question 6
Question
A planet is observed to have an orbital period of 4.2 years and a semimajor axis
of 2.8 astronomical units (AU). Calculate the mass of the star around which the
planet orbits, given that the gravitational constant is 6.67 ×10−11 N m2/kg2.
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion:
T2=(4π2
G(M+m))a3
5
where: - Tis the orbital period of the planet, - Gis the gravitational constant,
-Mis the mass of the star, - mis the mass of the planet, - ais the semimajor
axis of the planet’s orbit.
Step 2: We are given: T= 4.2years, a= 2.8AU, G= 6.67 ×10−11 N
m2/kg.
Step 3: Convert the semimajor axis afrom AU to meters:
1AU = 1.496 ×1011 m
a= 2.8×1.496 ×1011
a= 4.1888 ×1011 m
Step 4: Plug in the known values into Kepler’s Third Law and solve for the
mass of the star M:
(4.2years)2=(4π2
6.67 ×10−11(M+m))(4.1888 ×1011)3
Step 5: Simplify and solve for M.
This calculation requires careful manipulation of the formula and under-
standing of astronomical units.
Question 7
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is proportional to the cube of its average distance from the
sun. Consider a hypothetical planet with an orbital period of 5 years and an
average distance from the sun of 10 astronomical units (AU). What would be
the orbital period of another planet with an average distance from the sun of
15 AU?
Solution
To solve this problem, we can use Kepler’s third law of planetary motion, which
states:
T2
1
T2
2
=r3
1
r3
2
where: - T1and T2are the orbital periods of the two planets, - r1and r2are
the average distances of the two planets from the sun.
Given that the first planet has an orbital period (T1) of 5 years and an
average distance from the sun (r1) of 10 AU, we can substitute these values into
the equation and solve for the orbital period of the second planet (T2) when its
average distance from the sun (r2) is 15 AU.
6
Step 1: Write down Kepler’s third law formula:
T2
1
T2
2
=r3
1
r3
2
Step 2: Substitute the given values:
52
T2
2
=103
153
Step 3: Simplify the equation:
25
T2
2
=1000
3375
Step 4: Cross multiply to solve for T2
2:
25 ·3375 = 1000 ·T2
2
84375 = 1000 ·T2
2
T2
2=84375
1000
T2
2= 84.375
Step 5: Take the square root of both sides to find T2:
T2=√84.375
T2≈9.19 years
Therefore, the orbital period of the second planet would be approximately
9.19 years.
Question 8
Question
At one point in its orbit, a planet is 0.25 AU from the Sun and has a speed of
28 km/s. If the eccentricity of the planet’s orbit is 0.4, calculate the distance of
the planet from the Sun when it is at the farthest point from it.
Solution
Step 1: Recall Kepler’s Second Law, which states that the line joining a planet
and the Sun sweeps out equal areas in equal times. This means that a planet
moves faster when it is closer to the Sun and slower when it is farther away.
Step 2: Let’s denote the closest point to the Sun as periapsis and the farthest
point as apoapsis. The distance from the Sun to the periapsis is the planet’s
7
perihelion distance (rp) and the distance from the Sun to the apoapsis is the
planet’s aphelion distance (ra).
Step 3: According to Kepler’s Second Law, the speed of the planet when it
is at periapsis is related to its speed at apoapsis by the following equation:
vp·rp=va·ra
where vpand vaare the speeds of the planet at periapsis and apoapsis, respec-
tively.
Step 4: We are given that the planet’s speed at periapsis, vp= 28 km/s. We
are also given the eccentricity of the orbit, e= 0.4. The relationship between
the perihelion distance (rp), aphelion distance (ra), and eccentricity (e) is:
e=ra−rp
ra+rp
Solving for ragives us:
ra=1 + e
1−e·rp
Step 5: Substituting vp= 28 km/s into the Kepler’s Second Law equation
gives:
28 ·0.25 = va·ra
Step 6: We can solve for va, the speed of the planet at apoapsis:
28 ·0.25 = va·(1+0.4
1−0.4·0.25)
Step 7: Simplifying the equation from Step 6 will give us the speed of the
planet at apoapsis. Once we have that value, we can use it to find the distance
of the planet from the Sun when it is at the farthest point from it.
Question 9
Question
Kepler’s third law of planetary motion states that the square of the period of
revolution of a planet around the sun is proportional to the cube of the semi-
major axis of its elliptical orbit. Suppose a planet has a period of revolution
around the sun of 10 years and a semi-major axis of 3 astronomical units (AU).
Calculate the corresponding values for a different planet’s period of revolution
and semi-major axis if the square of the period is proportional to the cube of
the semi-major axis.
8
Solution
Step 1: Write down the ratio of periods and semi-major axes for the two planets:
Let T1be the period of revolution of the first planet, a1be the semi-major axis
of the first planet, T2be the period of revolution of the second planet, and a2
be the semi-major axis of the second planet. According to Kepler’s third law,
we have: T2
1
a3
1
=T2
2
a3
2
Step 2: Substitute the given values for the first planet into the equation:
Given that T1= 10 years and a1= 3 AU, we have:
102
33=T2
2
a3
2
Step 3: Simplify the equation by solving for the ratio T2
2
a3
2
:
100
27 =T2
2
a3
2
Step 4: Cross multiply to find the values for the second planet:
100 ·a3
2= 27 ·T2
2
Step 5: Given that this equation must hold true, we can set T2equal to x
and express a2in terms of x:
100 ·a3
2= 27 ·x2
Step 6: Since a2=x/10, substitute a2back into the equation:
100 ·(x/10)3= 27 ·x2
Step 7: Solve for xto find the period of revolution of the second planet:
100x3
1000 = 27x2
x= 27 years
Step 8: Finally, find the semi-major axis of the second planet using a2=
x/10:
a2=27
10 = 2.7AU
Therefore, the period of revolution of the second planet is 27 years and its
semi-major axis is 2.7 AU.
9
Question 10
Question
Kepler’s Third Law of Planetary Motion states that the square of the orbital
period (T) of a planet is proportional to the cube of the semi-major axis (a) of
its orbit. Suppose a planet has an orbital period of 4 years and a semi-major
axis of 3 AU (astronomical units). Calculate the orbital period of another planet
whose semi-major axis is 5 AU.
Solution
Step 1: Calculate the ratio of the cubes of the semi-major axes. Let the or-
bital period of the second planet be T2years and its semi-major axis be 5 AU.
According to Kepler’s Third Law:
T2
1
T2
2
=a3
1
a3
2
Substitute T1= 4 years, a1= 3 AU, a2= 5 AU:
42
T2
2
=33
53
Step 2: Calculate the orbital period of the second planet. Solving for T2:
16 = 27
125 ×T2
2
T2
2=125 ×16
27
T2
2=2000
27
T2=√2000
27 ≈7.09 years
Therefore, the orbital period of the second planet with a semi-major axis of
5 AU is approximately 7.09 years.
Question 11
Question
According to Kepler’s Third Law of Planetary Motion, the square of the orbital
period (T, in years) of a planet is proportional to the cube of its average distance
from the Sun (r, in astronomical units). For Earth, the average distance from
the Sun is about 1 AU and the orbital period is about 1 year.
If a newly discovered planet has an average distance from the Sun of 3 AU,
what is the predicted orbital period of this planet?
10
Solution
Step 1: Let’s express Kepler’s Third Law mathematically:
T2∝r3
Step 2: We can write the general form of Kepler’s Third Law as:
T2=k·r3
where kis the constant of proportionality.
Step 3: To find the value of the constant k, we can use the information given
for Earth:
(1)2=k·(1)3
Solving for k, we find k= 1.
Step 4: Now that we have the constant k, we can use it to predict the orbital
period of the newly discovered planet with r= 3 AU:
T2= 1 ·(3)3
T2= 27
Step 5: Taking the square root of both sides to find T, we get:
T=√27
T= 3√3
Therefore, the predicted orbital period of the newly discovered planet would
be approximately 3√3years.
Question 12
Question
Kepler’s third law relates the period of a planet’s orbit to its average distance
from the Sun. For two planets in the same solar system, planet X has an average
distance from the Sun that is 3 times greater than planet Y. If planet X takes
6 years to orbit the Sun, how long does it take planet Y to complete one orbit?
Solution
Step 1: Let TXbe the period of planet X and TYbe the period of planet Y.
Let rXand rYbe the average distances of planets X and Y from the Sun,
respectively. Given that rX= 3rYand TX= 6 years, we need to determine TY.
Step 2: According to Kepler’s third law, the square of the period of each
planet is proportional to the cube of its average distance from the Sun. Math-
ematically, this relationship can be expressed as:
(TX
TY)2
=(rX
rY)3
11
Step 3: Substituting TX= 6 years and rX= 3rYinto the equation from
Step 2, we get:
(6
TY)2
=(3rY
rY)3
Step 4: Simplifying the expression further, we have:
(6
TY)2
= 27
Step 5: Solving for TY, we get:
(6
TY)2
= 27
36
T2
Y
= 27
T2
Y=36
27 =4
3
TY=√4
3=2
√3=2√3
3
Step 6: Therefore, it takes planet Y 2√3
3years to complete one orbit around
the Sun.
Question 13
Question
The asteroid Pallas orbits the Sun at an average distance of 414 million kilo-
meters. Calculate the period of Pallas’ orbit in Earth years. (Hint: You may
assume that Pallas’ orbit is approximately circular.)
Solution
Step 1: We can use Kepler’s third law to relate the period of an orbit with the
semi-major axis of the orbit:
T2=4π2
G(M1+M2)a3
where: T= period of the orbit, G= gravitational constant (6.674×10−11 m3kg−1s−2),
M1= mass of the asteroid Pallas, M2= mass of the Sun, and a= semi-major
axis of the orbit.
Step 2: Since Pallas is an asteroid, we can assume its mass is negligible
compared to the Sun’s mass. Therefore, M1≈0. The mass of the Sun is
approximately 1.989 ×1030 kg.
12
Step 3: We can plug in the known values into Kepler’s third law:
T2=4π2
G(M1+M2)a3=4π2
6.674 ×10−11 ×1.989 ×1030 (414 ×109)3
Step 4: Calculating the period, we have:
T2=4π2
6.674 ×1.989(4143) = 4π2
13.281(7.096794 ×1019)
Step 5: Solving for T:
T=√4π2
13.281(7.096794 ×1019)
Step 6: After evaluating the expression, we find:
T≈√2.5993 ×1020 ≈5.0979 ×1010 s
Step 7: Converting the period to Earth years:
Tyears =5.0979 ×1010 s
60 ×60 ×24 ×365.25 ≈1621 years
Therefore, the period of Pallas’ orbit in Earth years is approximately 1621
years.
Question 14
Question
The period of a satellite orbiting a planet is 8 hours. If the planet’s radius is 2
times the radius of the satellite’s orbit, what is the orbital radius of the satellite
in terms of the planet’s radius?
Solution
Step 1: Let’s denote the radius of the satellite’s orbit as rand the radius of the
planet as R. We are given that the period of the satellite is 8 hours. According
to Kepler’s third law of planetary motion, the square of the period of an orbiting
body is proportional to the cube of the semi-major axis of its orbit. Therefore,
we have:
T2
1
r3
1
=T2
2
r3
2
where T1= 8 hours, r1=r,T2is the period of the planet (which is irrelevant
in this case), and r2=R.
Step 2: Substituting the given values into the formula, we get:
13
82
r3=T2
2
R3
Solving for r, we find:
r3= 64R3
r=3
√64 ·R
r= 4R
Therefore, the orbital radius of the satellite is 4 times the radius of the
planet.
Question 15
Question
A hypothetical planet orbits a star in an elliptical orbit, with the star located
at one of the foci of the ellipse. The semi-major axis of the planet’s orbit is 2
AU, and the eccentricity of the ellipse is 0.4. Calculate the distance between
the star and the planet when the planet is at its closest point to the star.
Solution
Step 1: Recall the formula to calculate the distance between a planet and the
star at any point in the orbit:
r=a(1 −e2)
1 + ecos(θ)
where: - ais the semi-major axis of the orbit, - eis the eccentricity of the orbit,
-θis the true anomaly (angle between the perihelion and the planet’s current
position), - ris the distance between the planet and the star at that point.
Step 2: Since we are looking for the distance when the planet is at its closest
point to the star, we need to find the distance when the planet is at perihelion,
where θ= 0◦.
Step 3: Substitute the given values into the formula:
r=2(1 −0.42)
1+0.4 cos(0◦)
Step 4: Simplify the equation:
r=2(1 −0.16)
1+0.4=2(0.84)
1.4=1.68
1.4= 1.2AU
Step 5: Therefore, the distance between the star and the planet when the
planet is at its closest point to the star is 1.2 AU.
14
Question 16
Question
Consider a hypothetical planetary system where a planet has an elliptical orbit
around a star. The semi-major axis of the planet’s orbit is 3 AU and the
eccentricity of the orbit is 0.5. Determine the distance between the closest
approach of the planet to the star (perihelion) and the farthest point from the
star (aphelion).
Solution
Step 1: Recall the formula for the distance between the closest approach and
the farthest point on an elliptical orbit, given the semi-major axis (a) and the
eccentricity (e):
Perihelion distance =a(1 −e)
Aphelion distance =a(1 + e)
Step 2: Substitute the given values into the formulas:
Perihelion distance = 3 AU(1 −0.5)
Perihelion distance = 3 AU ×0.5 = 1.5AU
Step 3: Calculate the aphelion distance:
Aphelion distance = 3 AU(1 + 0.5)
Aphelion distance = 3 AU ×1.5 = 4.5AU
Therefore, the distance between the closest approach of the planet to the
star (perihelion) is 1.5 AU, and the distance between the farthest point from
the star (aphelion) is 4.5 AU.
Question 17
Question
Consider a planet orbiting around a star in an elliptical path. The semi-major
axis of the planet’s orbit is 2.5 AU. If the eccentricity of the orbit is 0.3, deter-
mine the semi-minor axis of the orbit in AU.
Solution
Step 1: Recall the relationship between the semi-major axis (a), semi-minor
axis(b), and the eccentricity(e) of an ellipse:
a2=b2(1 −e2)
15
Step 2: Given that the semi-major axis (a) is 2.5 AU and the eccentricity(e)
is 0.3, we can substitute these values into the equation:
2.52=b2(1 −0.32)
Step 3: Solve for bby rearranging the equation:
b=√a2
1−e2
Step 4: Substitute a= 2.5AU and e= 0.3into the equation:
b=√2.52
1−0.32
Step 5: Calculate the value of b:
b=√6.25
1−0.09 =√6.25
0.91 ≈√6.8681 ≈2.622 AU
Therefore, the semi-minor axis of the orbit is approximately 2.622 AU.
Question 18
Question
In the study of planetary motion, Kepler’s Third Law states that the square
of a planet’s orbital period is proportional to the cube of its semi-major axis.
Suppose a new planet is discovered with a semi-major axis of 2.5 AU (astro-
nomical units). If the orbital period of this planet is 7 years, what would be the
semi-major axis of another planet with an orbital period of 20 years?
Solution
Step 1: Express Kepler’s Third Law mathematically: Let T1and a1represent
the orbital period and semi-major axis of the new planet, and let T2and a2
represent the orbital period and semi-major axis of the second planet. According
to Kepler’s Third Law, we have:
T2
1
T2
2
=a3
1
a3
2
Step 2: Plug in the given values: We are given that a1= 2.5AU, T1= 7
years, and T2= 20 years. We can now plug these values into the equation:
72
202=2.53
a3
2
16
Step 3: Solve for a2:49
400 =15.625
a3
2
a3
2=15.625 ×400
49 = 128.06
a2=3
√128.06 ≈5.03
Therefore, the semi-major axis of another planet with an orbital period of
20 years would be approximately 5.03 AU.
Question 19
Question
Consider a planet that moves in an elliptical orbit around the Sun. The planet
is at aphelion (its farthest point from the Sun) and travels at a speed of 15
km/s. If the planet’s speed at perihelion (its closest point to the Sun) is 25
km/s, calculate the average speed of the planet over a complete orbit.
Solution
Step 1: Recall Kepler’s second law, which states that a planet sweeps out equal
areas in equal times. This means that the planet moves faster when it is closer
to the Sun (at perihelion) and slower when it is farther away (at aphelion).
Step 2: Let vave be the average speed of the planet over a complete orbit. We
can calculate this average speed by finding the average of the speeds at aphelion
and perihelion.
Step 3: The average speed vave is given by:
vave =vaph +vper
2,
where vaph is the speed at aphelion and vper is the speed at perihelion.
Step 4: Substituting the given values for vaph = 15 km/s and vper = 25 km/s
into the formula, we have:
vave =15 + 25
2=40
2= 20 km/s.
Step 5: Therefore, the average speed of the planet over a complete orbit is
20 km/s.
Question 20
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period (T) of a planet is proportional to the cube of its semi-major axis (a).
17
Suppose a planet has an orbital period of 8 years and a semi-major axis of 3
AU (astronomical units). What would be the orbital period of a different planet
with a semi-major axis of 6 AU?
Solution
Step 1: Let’s denote the orbital period of the second planet as T2and its semi-
major axis as a2. According to Kepler’s third law, we have:
(T2
T)2
=(a2
a)3
Step 2: Given that T= 8 years and a= 3 AU, we can substitute these
values into the equation:
(T2
8)2
=(6
3)3
Step 3: Simplifying the equation gives:
(T2
8)2
= 23
Step 4: Further simplify to find the value of T2:
(T2
8)2
= 8 ⇒T2
8=√8
Step 5: Solving for T2gives:
T2= 8 ×√8 = 8 ×2√2 = 16√2years
Therefore, the orbital period of the second planet with a semi-major axis of
6 AU would be 16√2years.
Question 21
Question
Suppose an exoplanet is orbiting around a star in a nearly circular orbit, with
a semi-major axis of 0.5 AU. If the exoplanet takes 200 days to complete one
full orbit, calculate the mass of the central star in units of solar mass.
Solution
Step 1: Calculate the orbital period of the exoplanet using Kepler’s Third Law.
Kepler’s Third Law: P2=4π2
G(M1+M2)a3
18
where Pis the orbital period, Gis the gravitational constant, M1is the mass
of the star, M2is the mass of the exoplanet, and ais the semi-major axis of the
orbit.
Given: P= 200 days, a= 0.5AU, G= 6.67 ×10−11 m3kg−1s−2
Substitute the known values into the equation:
2002=4π2
G(M1)(0.5)3
Step 2: Solve for the mass of the central star M1.
M1=4π2(0.5)3
G×1
2002
M1=4π2(0.125)
G×40000
Step 3: Simplify the expression and convert the mass to solar mass units.
M1=0.5π2
107×6.67
M1=0.5π2
6.67 ×107
M1≈0.5×9.87
6.67 ×107
M1≈4.935
6.67 ×107
M1≈7.40 ×10−8solar mass
Therefore, the mass of the central star in solar mass units is approximately
7.40 ×10−8solar masses.
Question 22
Question
In the context of Kepler’s Laws of Planetary Motion, a planet orbits a star in
an elliptical path. The semi-major axis of the orbit is 2.5 AU. Given that the
eccentricity of the orbit is 0.4, determine the semi-minor axis of the orbit in
astronomical units.
19
Solution
Step 1: Recall the relationship between the semi-major axis (a), semi-minor axis
(b), and eccentricity (e) in an elliptical orbit:
b=a√1−e2
Step 2: Substitute the given values into the formula:
b= 2.5√1−0.42
Step 3: Calculate the value inside the square root first:
1−0.42= 1 −0.16 = 0.84
Step 4: Substitute the value back into the formula:
b= 2.5√0.84
Step 5: Calculate the square root of 0.84:
√0.84 ≈0.9165
Step 6: Multiply the calculated square root by 2.5 to find the semi-minor
axis in AU:
b≈2.5×0.9165 ≈2.29125 AU
Therefore, the semi-minor axis of the orbit is approximately 2.29125 AU.
Question 23
Question
Consider a planet with mass Morbiting a star with mass m. If the planet orbits
the star in a circular orbit with radius rand period T, derive an expression for
the orbital speed of the planet using Kepler’s Third Law.
Solution
Step 1: We begin by stating Kepler’s Third Law, which relates the period of an
orbiting body to the semi-major axis of its orbit. This law can be expressed as:
T2=(4π2
G(m+M))r3
where Gis the gravitational constant, mis the mass of the star, Mis the mass
of the planet, ris the radius of the orbit, and Tis the period of the orbit.
20
Step 2: Next, we rewrite the expression for T2in terms of the period T:
T=√(4π2
G(m+M))r3
Step 3: The orbital speed vof the planet can be calculated as the distance
traveled (2πr) divided by the time taken for one complete orbit (T):
v=2πr
T
Step 4: Substitute the expression for Tinto the equation for v:
v=2πr
√(4π2
G(m+M))r3
Step 5: Simplify the expression for v:
v=2π
√(4π2
G(m+M))r
Step 6: Further simplify the expression to obtain the final result:
v=√G(m+M)
r
Therefore, the orbital speed of the planet in a circular orbit can be calculated
using the expression v=√G(m+M)
r.
Question 24
Question
Kepler’s Third Law of Planetary Motion states that the square of the period of
revolution of a planet around the sun is proportional to the cube of its average
distance from the sun. Suppose a newly-discovered planet has a period of revo-
lution of 8 years and an average distance from the sun of 10 astronomical units
(AU). If another planet has a period of revolution of 2 years, what is its average
distance from the sun in AU?
Solution
Step 1: Let’s denote the period of revolution of the second planet as T(in years)
and its average distance from the sun as d(in AU). According to Kepler’s Third
Law, we have:
T2∝d3
21
Step 2: For the first planet, with a period of 8 years and an average distance
of 10 AU, we have the following relationship:
(8)2∝(10)3
64 ∝1000
Step 3: Now, for the second planet with a period of 2 years, we can set up
the following proportion:
T2
1=T2
2
d3
1=d3
2
Step 4: Since 64 ∝1000, we can set up a proportion using this relation:
22∝d3
2
4∝d3
2
Step 5: Taking the cube root of both sides, we find:
d2=3
√4
d2= 1.5874 AU
Therefore, the average distance from the sun for the second planet is ap-
proximately 1.5874 AU.
Question 25
Question
Consider a planet that follows an elliptical orbit around a star with the follow-
ing properties: semi-major axis length a= 2.5AU and eccentricity e= 0.6.
Determine the planet’s closest approach and furthest distance from the star in
AU.
Solution
Step 1: Calculate the distances at closest approach and furthest distance from
the star using the semi-major axis length and eccentricity formula.
Closest distance =a(1 −e)
Furthest distance =a(1 + e)
Step 2: Substitute the values a= 2.5AU and e= 0.6into the formulas.
Closest distance = 2.5(1 −0.6)
Furthest distance = 2.5(1 + 0.6)
22
Step 3: Calculate the distances.
Closest distance = 2.5(0.4) = 1.0AU
Furthest distance = 2.5(1.6) = 4.0AU
Therefore, the planet’s closest approach to the star is at a distance of 1.0
AU, while its furthest distance is at 4.0 AU.
23
Students also viewed