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# Derivation of Circular Orbits Around Center of Mass

by Ron Kurtus

** Circular orbits** of two objects around the

*(CM) between them require tangential velocities that equalize the gravitational attraction between the objects.*

**center of mass**Tangential velocities tend to keep the objects traveling in a straight line, according to the *Law of Inertia*. If gravitation cases an inward deviation from straight-line travel, the result is an outward centrifugal force. By setting the gravitational force equal to the centrifugal forces, you can derive the required tangential velocities for circular orbits.

The orbit equations can be in simplified forms when the masses of the two objects are the same and when the mass of one object is much greater than that of the other.

Questions you may have include:

- What are the factors involved in the derivation?
- What are the equations for the velocities of the objects?
- What happens when one object is much larger than the other?

This lesson will answer those questions. Useful tool: Units Conversion

## Factors in determining orbital velocities

The linear tangential velocities required for two objects to be in circular orbits around the center of mass (CM) between them is found by comparing their gravitational force of attraction with the outward centrifugal force for each object.

Note: A linear tangential velocity is a straight line velocity is perpendicular to the axis between the two objects. It is tangent to the curved path and is different from rotational velocity.(

See Center of Mass and Tangential Gravitational Motion for more information.)

### Assume no initial radial velocities

When two objects in space are traveling toward the general vicinity of each other, they both have radial and tangential velocities with respect to the center of mass (CM) between them. However, in order to simplify the derivation for circular orbits, we will only look at the case where there are no inward or outward radial velocities and only be concerned about the tangential velocities.

This is similar to the case of Newton's cannonball going into orbit or sending a satellite into orbit around the Earth.

(

See Gravity and Newton's Cannon for more information.)

Since there is no radial motion, separation between the objects remains constant, which is a requirement for circular orbits.

### Gravitational force of attraction

The gravitational force of attraction between two objects is:

F = GMm/R^{2}

where

**F**is the force of attraction between two objects in newtons (N)**G**is the Universal Gravitational Constant = 6.674*10^{−17}N-km^{2}/kg^{2}**M**and**m**are the masses of the two objects in kilograms (kg)**R**is the separation in kilometers (km) between the objects, as measured from their centers of mass

Note: Since force is usually stated in newtons, but motion between astronomical bodies is usually stated in km/s, an adjusted value forGis used, with N-km^{2}/kg^{2}as the unit instead of N-m^{2}/kg^{2}.Gis also sometimes stated as 6.674*10^{−20}km^{3}/kg-s^{2}.(

See Universal Gravitation Equation for more information.)

### Separation of objects

As the objects orbit the CM, their total separation, **R**, remains constant. The individual separations between the objects and CM are also constant and determined by **R** and their masses:

R = R_{M}+ R_{m}

where

**R**is the separation between the center of object_{M}**M**and the CM in km**R**is the separation between the center of object_{m }**m**and the CM in km

The values of **R _{M}** and

**R**are according to the equations:

_{m}

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

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

See Center of Mass Definitions for more information.)

The factors involved can be seen in the illustration below:

Factors in objects orbiting CM

Note: Although the Earth orbits the Sun in a counterclockwise direction, we usually indicate motion in a clockwise direction.(

See Direction Convention for Gravitational Motion for more information.)

### Centrifugal force

The centrifugal inertial force on each object relates to its circle of travel:

F_{M}= Mv_{TM}^{2}/R_{M}

F_{m}= mv_{Tm}^{2}/R_{m}

where

**F**is the centrifugal inertial force on mass_{M}**M****v**is the tangential velocity of mass_{TM}**M****F**is the centrifugal inertial force on mass_{m}**m****v**is the tangential velocity of mass_{Tm}**m**

Note: Centrifugal force is caused by inertia and is not considered a "true" force. It is sometimes called a pseudo- or virtual force.

Substituting **R _{M} = mR/(M + m)** and

**R**in the above equations gives you:

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

F_{M}= Mv_{TM}^{2}(M + m)/mR

F_{m}= mv_{Tm}^{2}(M + m)/MR

## Solve for individual velocities

Since the centrifugal force equals the gravitational force for a circular orbit, you can solve for the velocity.

### Object with mass *m*

In the case of the object with mass **m**:

F_{m}= F

Substitute equations:

mv_{Tm}^{2}(M + m)/MR = GMm/R^{2}

Multiply both sides by **MR** and divide by **m**:

v_{Tm}^{2}(M + m) = GM^{2}/R

Divide both sides by** (M + m)**:

v_{Tm}^{2}= GM^{2}/R(M + m)

Take the square root:

v_{Tm}= ±√[GM^{2}/R(M + m)]

This means the velocity can be in either direction for a circular orbit. Since direction is not relevant here:

vkm/s_{Tm}= √[GM^{2}/R(M + m)]

### Object with mass *M*

Likewise, for the object of mass **M**:

vkm/s_{TM}= √[Gm^{2}/R(M + m)]

## Sizes of objects

The equations for the tangential orbital velocities can be simplified when both objects are the same size, as well as when one object has a much greater mass than the other.

### Objects have same mass

There are situations in space where two stars have close to the same mass and orbit the CM between them. Astronomers call them *double stars*.

Double stars follow same orbit around CM

If the objects are the same mass, then **M = m** and the velocity equation for each becomes:

vkm/s_{TM}= √[Gm^{2}/R(m + m)]

The equation reduces to:

vkm/s_{TM}= √[Gm/2R]

Since both objects or stars have the same orbital velocity and the same separation from the CM, they follow the same orbit around the CM.

### One object much larger than other

Another situation often seen in space is when one object is much larger than the other. In this case, the CM between them is almost at the larger object's geometric center. This results in simplifying the equation for orbital velocity. The small object then seems to orbit the larger object.

For example, the CM between a satellite orbiting the Earth is near the geometric center of the Earth. Likewise, the CM between the Earth and the Sun is near the center of the Sun.

Suppose **M >> m** (**M** is much greater than **m**). Then:

M + m ≈ M

where **≈** means "approximately equal to".

Orbits when one object is much larger than other

Substitute **M + m ≈ M** into the equation for the velocity of the smaller object:

v_{Tm}= √[GM^{2}/R(M + m)]

v_{Tm}= √(GM^{2}/RM)

Reducing the equation results in:

vkm/s_{Tm}= √(GM/R)

This is the same as the standard equation for orbital velocity of one object around another.

(

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

## Summary

When two objects are moving at the correct tangential velocities, they will go in circular orbits around their CM. The velocity equations are determined by setting the gravitational force equal to the outward centrifugal forces caused by their tangential velocities.

The velocity equations are:

vkm/s_{Tm}= √[GM^{2}/R(M + m)]

vkm/s_{TM}= √[Gm^{2}/R(M + m)]

When the mass of each object is the same, the velocity equation is simplified. The same is true when the mass of one object is much greater than that of the other.

Try to simplify your life

## Resources and references

### Websites

**Center of Mass Calculator** - Univ. of Tennessee - Knoxville (Java applet)

**Center of Mass** - Wikipedia

### Books

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

**Top-rated books on Gravitation**

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## Derivation of Circular Orbits Around Center of Mass