**Orbital Velocity **is the minimum velocity an object or particle has to maintain to continue its circular motion in its orbit or in other words it is the minimum velocity of the object in circular motion that generates enough centripetal force to keep it in the circular orbit of given radius. We will discuss this in details as well as we will focus on the **derivation of Orbital Velocity formula.**

A satellite maintains its own orbital speed (*we use this term as well*) when it rotates around a planet or a planet does maintain its own requred orbital speed while rotating around the sun.

This orbital speed for every rotating object depends on some parameters. This will be clear as we go through the its formula. Also we will see how this centripetal force is supplied.

## Orbital Velocity

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 formula**.

We will derive the formula(2 sets) and later will find out the ** Orbital Velocity formula** for a nearby orbit.

**(so you get total 3 sets of equations)**

## orbital velocity formula – flash card

## Derivation of Orbital Velocity formula

### Orbital Velocity formula – #1

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

**G is the Gravitational Constant.**

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)

**This is the first equation or formula of Orbital Velocity of a satellite.**

**Here r = R +h**

## Orbital velocity formula of a satellite

### Orbital Velocity 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, we get,

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

** V = R . (g/r) ^{1/2} ……. (6)**

**This is the second formula**.

**Here, as said earlier,**

**r = R +h**

Next we will derive the 3rd equation and that is for a **NEARBY ORBIT,** i.e. for an orbit which has **negligible height above the earth’s surface**.

## Orbital Vel for *Near orbit*

### formula #3

**Let’s consider an orbit which is pretty close to the earth. (formula #3)**

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)**.

**From equation 3 (the fundamental form of orbital velocity equation), we get an equation of nearby orbit’s Orb. velocity:

V = [(GM)/R]^{1/2}………….(3.1)

**And **from equation 6** we get another form of equation for *orbital speed (when h is negligible) at near earth orbit:*

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

**= R . (g/R)**

^{1/2}**V**=

_{orbital}**(gR)**

^{1/2}……….(6.1)here, R = r

## Orbital Velocity Formula

**Here you get a set of Orbital Velocity formula as follows-**

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

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

**And for Nearby Orbit**

**V _{orbital} **

**= [(GM)/R]**

^{1/2}** V _{orbital} = (gR)^{1/2}**

## Orbital velocity of a satellite – Numerical problem

**Assume that a satellite orbits Earth 225 km above its surface. Given that the mass of Earth is 5.97 x 10**^{24}** kg and the radius of Earth is 6.38×10**^{6}** m, what is the satellite’s orbital speed?**

** Solution:**** **h =2.25×10^{5} m (height of the satellite’s orbit)

r_{E} =6.38×10^{6} m (Earth’s radius)

m_{E} =5.97×10^{24} kg (Earth’s mass)

G = 6.67×10^{-11} Nm^2/kg^2

Orbital velocity = V =?

We get the orbital radius r by adding the height of the satellite’s orbit to Earth’s radius.

**r** =h +r_{E} =2.25×10^{5} m + 6.38×10^{6} m = 6.61×10^{6} m

So, Orbital speed.

V = √[(G m_{E})/r] = [G m_{E})/r ]^{1/2 } = 7.76 x 10^{3} m/s

Here are some 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 velocity in orbit is less.

#### Any Dependency on mass?

This velocity in a orbit doesn’t depend on the mass of the satellite.

Tips to remember:

If radius of satellite orbit is made N times of the radius of the earth then its orbital velocity would be (1/N)^{(1/2)} times of the near earth orbit orbital velocity.

If 2 satellites with radii r1 and r2 are orbiting in circular orbits, then the ratio of their velocities is v1/v2 = (r2/r1)^{(1/2)}, where v1 and v2 are orbital velocities.

**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.**

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