Circular Gravitational Orbit Examples
About 5 min read
In order for an object to go into a circular gravitational orbit around another object, it must have a specific tangential velocity with respect to the other object. That velocity is a function of the mass of each object and their separation.
In reality, each object is orbiting the center of mass (CM) or barycenter between them. Equations for the location of the CM, the individual velocities around the CM and the total velocity with respect to each other were derived in Circular Gravitational Orbits.
You can calculate the orbital velocity of the Moon around the Earth, the Earth around the Sun and the planet Jupiter around the Sun by finding the position of the CM and using the appropriate velocity equation.
Questions you may have include:
- What is the velocity of a Moon orbiting the Earth?
- What is the velocity of the Earth orbiting the Sun?
- What is the velocity of Jupiter orbiting the Sun?
This lesson will answer those questions. There is a mini-quiz near the end of the lesson.
Moon orbiting the Earth
Consider the orbit of the Moon around the Earth (or the orbits of the Moon and Earth around their CM). Find the Moon’s velocity with respect to the Earth.
Equation
The equation for the velocity of the Moon and Earth, revolving around their CM or barycenter is:
v = √[(G)(M + m)/R]
where
- G = 6.674*10−20 km3/kg-s2
- M = 5.974*1024 kg (mass of Earth)
- m = 7.348*1022 kg = 0.073*1024 kg (mass of Moon)
- R = 3.844*105 km (separation between centers)
Note: Because we are stating velocity in km/s, we converted the units of G from 6.674*10−11 N-m2/kg2 to 6.674*10−20 km3/kg-s2. Also, we consider R in km instead of meters.
Sum of masses
The sum of their masses is:
M + m = 5.974*1024 kg + 0.073*1024 kg
M + m = 6.047*1024 kg
Distance to CM
The separation between the Earth’s center and the CM is:
rM = mR/(M + m)
rM = (7.348*1022 kg)(3.844*105 km)/(6.047*1024 kg)
rM = 4.671*103 km = 4681 km
Since the radius of the Earth is 6370 km, the CM is within the Earth. However, it is at a sufficient separation to affect the orbit.
Velocity for circular orbit
The calculation of the velocity of the Moon with respect to the Earth is:
v = √[(G)(M + m)/R] km/s
v = √[(6.674*10−20)(6.047*1024)/(3.844*105)] km/s
v = √(1.050) km/s
v = 1.025 km/s
Compare this result from the commonly seen approximation:
va = √(GM/R)
va = √[(6.674*10−20)(5.974*1024)/3.844*105)] km/s
va = 1.018 km/s
Earth orbiting the Sun
Consider the orbit of the Earth around the Sun. Find the Earth’s velocity with respect to the Sun.
Equation
The equation for the velocity of the Earth and the Sun, revolving around their CM is:
v = √[(G)(M + m)/R]
where
- G = 6.674*10−20 km3/kg-s2
- M = 1.989*1030 kg (mass of Sun)
- m = 5.974*1024 kg (mass of Earth)
- R = 1.496*108 km (separation between centers)
Center of mass
The sum of their masses is:
M + m = 1.989*1030 kg + 5.974*1024 kg
Set exponents to equal values:
M + m = 1.989*1030 kg + 0.000005974*1030 kg
Since the mass of the Earth, m, is so small compared to the mass of the Sun, M, it can be considered negligible.
Velocity for circular orbit
Thus, the simple equation is used:
v = √(GM/R) km/s
v = √[(6.674*10−20)(1.989*1030)/(1.496*108)] km/s
v = √(8.873*102) km/s
The orbital velocity of the Earth with respect to the Sun is:
v = 29.793 km/s
This corresponds to listed values of the orbital velocity of the Earth.
Jupiter orbiting the Sun
Since Jupiter is the largest planet in our Solar System, the CM between it and the Sun is outside the Sun’s surface. Both Jupiter and the Sun revolve around their barycenter.
v = √[(G)(M + m)/R]
where
- G = 6.674*10−20 km3/kg-s2
- M = 1.989*1030 kg (mass of Sun)
- m = 1.899*1027 kg (mass of Jupiter)
- R = 7.785*108 km (separation between centers)
Sum of masses
M + m = 1.989*1030 kg + 1.899*1027 kg
Set exponents to be the same:
M + m = 1.989*1030 + 0.001899*1030
M + m = 1.991*1030 kg
Separation from CM
The separation between the Sun’s center and the CM is:
rM = mR/(M + m)
rM = (1.899*1027 kg)(7.785*108 km)/(1.991*1030 kg)
rM = 7.453*105 km
Since the radius of the Sun is 695,500 km = 6.955*105 km, the CM is outside the surface of the Sun.
Velocity for circular orbit
The calculation of the velocity of Jupiter with respect to the Sun is:
v = √[(G)(M + m)/R]
v = √[(6.674*10−20*1.991*1030/7.785*108)] km/s
v = √(1.707*102) km/s
v = 13.065 km/s
This corresponds to the measured mean orbital velocity of Jupiter.
Summary
Equations for the location of the CM, the individual velocities around the CM and the total velocity with respect to each other were derived in Circular Gravitational Orbits.
You can calculate the orbital velocity of the Moon around the Earth, the Earth around the Sun and the planet Jupiter around the Sun by finding the position of the CM and using the appropriate velocity equation.
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Resources
The following resources provide information on this subject:
Websites
Orbital Mechanics – Rocket & Space Technology by Robert A. Braeunig
How Satellites Work – HowStuffWorks.com
Orbit – Wikipedia
Circular orbit – Wikipedia
Acceleration due to Gravity Calculations – from Western Washington University
Gravity and Gravitation Resources
Mini-quiz to check your understanding
Self-check your understanding
How would the orbital velocity of the Moon be affected if it were further from the Earth?
Why is the simple equation for orbital velocity used in the Earth-Sun case?
Why is the CM between Jupiter and the Sun outside the Sun's surface?
Pick an answer to see instant feedback. Re-read the lesson if you miss one.
If you got all three correct, you are on your way to becoming a Champion in Physics. If you had problems, you had better look over the material again.
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