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# Circular Planetary Orbits

by Ron Kurtus

Although the gravitational orbits of the planets around the Sun are slightly elliptical, they are typically considered circular for ease of calculations. This is also true for the orbits of moons around the planets, as well as the various satellites in orbit around the Earth.

If you view the orbits of two objects in space with respect to the center of mass (CM) or barycenter between them, they would each seem to be orbiting that CM. However, typically you view the orbit of the smaller object with respect to the larger object. This reduces two equations into one velocity equation. Knowing the masses and separation, you can calculate the tangential velocity of one object with respect to the other. When one object is much larger than the other, the orbital equation is simplified.

Examples of calculations are the velocity of the Moon with respect to the Earth, the Earth in orbit around the Sun and the planet Jupiter around the Sun.

(

See Derivation of Circular Orbits Around Center of Mass for more information.)

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. Useful tool: Units Conversion

## Moon orbits the Earth

The Moon and Earth are in orbit around the center of mass (CM) between them. That CM is within the surface of the Earth, such that the Earth appears to wobble about the center when viewed from outer space.

Instead of viewing the orbits from outer space, you can view the orbit of the Moon from the Earth. You simply calculate the orbital velocity of the Moon with respect to the Earth.

(

See Orbital Motion Relative to Other Object for more information.)

### Location of CM

The location of the CM with respect to the Earth is determined from the equation:

R_{M}= mR/(M + m)

where

**R**is the separation between the center of the Earth and the CM in km_{M}**m**is the mass of the Moon = 7.348*10^{22}kg = 0.073*10^{24}kg**M**is the mass of the Earth = 5.974*10^{24}kg**R**is the fixed separation between the Earth and the Moon = 3.844*10^{5}km

(

See Gravitation and Center of Mass for more information.)

The calculated value is:

R7.348*10_{M}=^{22}*3.844*10^{5}/(5.974*10^{24}+ 0.073*10^{24}) km

R28.246*10_{M}=^{27}/6.047*10^{24}km

R4.671*10_{M}=^{3}= 4671 km

Since the radius of the Earth is about 6371 km, the CM is almost 3/4 of the radius. This results in the Earth appearing to wobble about the CM, as seen from the Moon.

Earth orbits CM between it and the Moon

### View from Earth

The Moon's tangential velocity with respect to the Earth is:

v_{T}= √[G(M + m)/R]

where

**v**is the linear tangential velocity of the Moon with respect to Earth in km/s_{T}**G**is the Universal Gravitational Constant = 6.674*10^{−20 }km^{3}/kg-s^{2}

### Velocity for circular orbit

The calculation of the velocity of the Moon with respect to the Earth is:

v[(6.674*10_{T}= √^{−20})(6.047*10^{24})/(3.844*10^{5})^{}] km/s

v(1.050) km/s_{T}= √

v1.025 km/s_{T}=

Compare this result from the commonly seen approximation:

v_{a}= √(GM/R)

v_{a}= √[(6.674*10^{−20})(5.974*10^{24})/3.844*10^{5})]km/s

v1.018 km/s_{a}=

## Earth orbits the Sun

The Earth orbits the Sun in an elliptical orbit that is close to being circular. Since the mass of the Sun is so much greater than the mass of the Earth, the CM between them is almost at the geometric center of the Sun. This simplifies the orbital velocity equation.

Earth's tangential velocity while orbiting the Sun

### Sum of masses

To find the location of the CM, first consider the sum of the masses of the Sun and Earth:

M + m ≈ M

where

**M**is the mass of the Sun = 1.989*10^{30}kg**m**is the mass of the Earth = 5.974*10^{24}kg = 0.000005974*10^{30}kg**≈**means "approximately equal to"

M + m =1.989*10^{30}kg + 0.000005974*10^{30}kg

M + m ≈1.989*10^{30}kg

The sum of the masses is approximately the mass of the Sun.

### Location of CM

The equation for the separation between the center of the Earth and the CM is:

R_{m}= MR/(M + m)

where

**R**is the separation between the center of the Earth and the CM_{m}**R**is the separation between the centers of the Earth and Sun =

1.496*10^{8}km.

Since **M + m ≈ M**, the equation reduces to:

R_{m}= MR/M

R_{m}= R

### Velocity for orbit

The equation for the tangential velocity of the Earth orbiting the Sun reduces to:

v_{T}= √(GM/R)

v(6.674*10_{T}= √[^{−20})*(1.989*10^{30})/(1.496*10^{8})]km/s

v(8.873*10_{T}= √^{2}) km/s

The tangential orbital velocity of the Earth with respect to the Sun is:

v29.793 km/s_{T}=

This corresponds to listed values of the orbital velocity of the Earth. This velocity can also be used to calculate the length of a year or time it takes for one revolution of the Sun.

(

See Length of Year for Planets in Gravitational Orbit for more information.)

## Jupiter orbits the Sun

Jupiter is the largest planet in our Solar System. Although the mass of the Sun is much greater than that of Jupiter, the CM or barycenter between Jupiter and the Sun lies outside the Sun's surface. This means that they would both appear to orbit the CM when viewed from outside the Solar System.

However, the view of from the Earth would be that the Sun is a fixed point and that Jupiter's orbit is with respect to the Sun.

### Location of CM

The separation between the Sun's center and the CM is:

R_{M}= mR/(M + m)

where

**R**is the separation between the Sun's center and the barycenter_{M}**R**is the separation between the centers of the Sun and Jupiter =

7.785*10^{8}km**M**is the mass of the Sun = 1.989*10^{30}kg**m**is the mass of Jupiter = 1.899*10^{27}kg

Determine **M + m**:

M + m =1.989*10^{30}kg + 1.899*10^{27}kg

Set exponents to be the same:

M + m =1.989*10^{30}+ 0.001899*10^{30}kg

M + m =1.991*10^{30}kg

Calculate **R _{M}**:

R(1.899*10_{M}=^{27}kg)(7.785*10^{8}km)/(1.991*10^{30}kg)

R7.453*10_{M}=^{5}km

Since the radius of the Sun is 695,500 km = 6.955*10^{5} km, the CM is outside the surface of the Sun.

### Velocity for circular orbit

The calculation of the tangential velocity of Jupiter with respect to the Sun is:

v_{T}= √[G(M + m)/R]

v(6.674*10_{T}= √[^{−20}*1.991*10^{30}/7.785*10^{8})]km/s

v(1.707*10_{T}= √^{2}) km/s

v13.065 km/s_{T}=

This corresponds to the measured mean orbital velocity of Jupiter.

### Velocity of Jupiter around the CM

The velocity of Jupiter around the CM between Jupiter and the Sun is calculated from the equation:

vkm/s_{Tm}= √[GM^{2}/R(M + m)]

v(6.674*10_{Tm}= √[^{−20}*{1.989*10^{30}}^{2}/7.785*10^{8}*1.991*10^{30})]km/s

v26.403*10_{Tm}= √(^{40}/15.500*10^{38})

v170.342_{Tm}= √()

v13.052 km/s_{Tm}=

This value is slightly different that what is observed.

## Summary

The gravitational orbits of the planets around the Sun can be considered circular for ease of calculations. This is also true for the orbits of moons around the planets, as well as the various satellites in orbit around the Earth.

Although, each object is orbiting the center of mass (CM) or barycenter between them, by knowing their masses and separation, you can calculate the tangential velocity of one object with respect to the other. This includes the velocity of the Moon with respect to the Earth, the Earth in orbit around the Sun and the planet Jupiter around the Sun.

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## Resources and references

### 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

### Books

(Notice: The *School for Champions* may earn commissions from book purchases)

**Top-rated books on Gravitation**

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## Circular Planetary Orbits