### Orbital Velocity of Satellite-derivation of 2 expressions

##### October 21, 2017

# Orbital Velocity of Satellite

Definition: **Orbital Velocity of a satellite** is the minimum velocity it has to maintain to continue its circular motion in its orbit. We know for a circular motion to continue, a force called **centripetal force** is required to work upon the object in circular motion. And this force is a real and acts towards the centre of the circle. As this is a real force it’s to be supplied by a real mechanism or system. A satellite moves in a circular track and the centripetal force acting on it is supplied by the gravitational force acting between the earth and the satellite. With the help of this concept or information we can easily go for the **derivation of Orbital Velocity**.

## Derivation of Orbital Velocity

###### Deriving the expression or formula #1

The **Gravitational Force **between the earth and the satellite = **F _{g} = (G.M.m)/r^{2} ……………… (1)**

The **centripetal force** acting on the satellite =** F _{c} = mV^{2}/r ……………………….. (2)**

Here, M is the mass of earth and m is the mass of the satellite which is having a uniform circular motion in a circular track of radius r around the earth. V is the linear velocity of the satellite at a point on its circular track.

Now this r is the sum of the radius of the earth(R) and the height(h) of the satellite from the surface of the earth.

r = R + h

Now equating, equation 1 and 2 we get,

**F _{g }= F_{c}**

**=> (G.M.m)/r ^{2} = mV^{2}/r **

** V = [(GM)/r]**^{1/2} ……………………………….. (3)

^{1/2}

**This is the first equation or expression of Orbital Velocity of a satellite. Here r = R +h**

## Another expression or formula of Orbital velocity of a satellite

###### Deriving the expression or formula #2

For a mass of m on earth’s surface, the following is true:

**mg = (GMm)/R ^{2} ………………………. (4) **Note, on earth surface h=0 and r = R. And gravitational force on a mass is equal to its weight on the surface.

From equation 4 we get this equation, **GM = g. R ^{2} …………………….. (5)**

Substituting this expression of GM in equation 3 (Orbital velocity), we get,

**V = [(gR ^{2})/r]^{1/2} **

** V = R . (g/r)**^{1/2} ……………………. (6)

^{1/2}……………………. (6)

**This is the second expression of Orbital Velocity** of a satellite. **Here, as said earlier, r = R +h**

## Summary & Conclusions

**2 sets of Orbital Velocity Equations or expressions are as follows-**

**V**_{orbital} = [(GM)/r]^{1/2}

_{orbital}= [(GM)/r]

^{1/2}

**V**_{orbital} = R . (g/r)^{1/2}

_{orbital}= R . (g/r)

^{1/2}

Here are some important pointers, which are important to know.

###### Orbital Velocity and Radius of the orbit

From both the equations it’s evident that as the radius of the circular path increases, the minimum velocity requirement of the satellite (orbital velocity) to maintain its circular track decreases.

In other words, if the satellite rotates closer to the earth, it has to move faster. If the radius is more then this orbital velocity is less.

###### Any Dependency on mass?

This orbital velocity doesn’t depend on the mass of the satellite.

###### Nearby Orbit and Orbital velocity

Let’s consider an orbit which is pretty close to the earth. Now if the height of the satellite (h) from the surface of the earth is negligible with respect to the Radius of the earth, then we can write r=R+h = R (as h is negligible).

So** from equation 6** we get another form of equation for orbital speed (when h is negligible)

**V _{orbital} = R . (g/r)^{1/2} = R . (g/R)^{1/2} = (gR)^{1/2}**

** ****Kepler’s third law:**

We can use the equations of Orbital velocity to derive this famous law. Read the Post on Kepler’s 3rd Law.

### Related study- highly suggested for you:

###### Escape Velocity

###### Centripetal Force

Law of Gravitation

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