# Kepler’s Third Law

As we have derived the equations of **Orbital velocity** of *satellite* in our last post, this is the right time to discuss on an important Law of Physics, which talks on the *period of its revolution* and *how* the** period of revolution of a satellite depends on the radius of its orbit**.

This is known as

**Kepler’s Third Law**or 3

^{rd}Law of Kepler.

This law states that

**square of the Orbital Period of Revolution is directly proportional to the cube of the radius of the orbit**.

We will derive the equation for

**Kepler’s 3rd Law**using the concept of

**Period of Revolution**and the equation of orbital velocity. . In this process, the

**equation of Time Period Of Revolution of earth satellite**would be derived as well. We’ll also solve sample numerical problem here using this law. So let’s start!

## Time Period of Revolution of the Satellite about the earth and Kepler’s Third Law derivation

Say, Period of revolution is T and the radius of the orbit is r.

And **r = R + h** = Radius of the earth + height of the satellite from the surface of the earth.

### Time Period of revolution of earth satellite

So **period of revolution = T = (2 π r) /V, **where V is the Orbital Velocity of the satellite. ……………………….. (1)

Now from the previous post on Orbital Velocity of Satellite, we know that orbital velocity **V = R √(g/r) = R (g/r)^(1/2)** .

Here g is the acceleration due to gravity and R is the radius of the earth** **…………….(2)****[ ready Reference: Orbital Velocity ]**

From equation 1 and 2, we get,

T = (2 **π** r)/V = [(2 **π** r)] /[R (g/r)^(1/2)] =[ 2**π** / R]. √[r ^3/g]

Here we get an important equation for **Period of Revolution of earth satellite**:

**T = [ 2π / R]. √[r ^3/g] ……………………………(3)** here R is the radius of the earth and r = R + h

In the next section we will use the above equation to **derive Kepler’s Third Law**

### Derivation of Kepler’s Third Law

Now if we square both side of equation 3 we get the following:

**T^2** =[ (4 . **π**^2)/(R^2)]. **r^3**/g ……………………………(4)

Here, **(4. π^2)/(R^2)** and **g** are constant as the values of **π**** (**Pi), g and R are not changing with time.

So we can say, **T^2 ∝ r ^3**. …………….. (5) .[**Kepler’s Third Law equation]**

**Here this Kepler’s Third Law equation says that square of the Orbital Period of Revolution is directly proportional to the cube of the radius of the orbit.**

If the orbit is not circular in the truest sense and rather elliptical, then this law states like this,**square of the Orbital Period of Revolution varies with the cube of the semi-major axis of the orbit. (ref:Wiki)**

This is known as **Kepler’s third Law. **

From this equation of **Kepler’s Third Law** it comes out clearly that the mass of the object in revolution has no effect on the Period of Revolution.

**And this law is applicable for the revolution of any planet and satellite.**

**Now let’s solve a numerical problem using this formula, in the next paragraph.**

### Kepler’s Third Law – Sample Numerical Problem using Kepler’s 3rd law:

**Two satellites Y and Z are rotating around a planet in a circular orbit. The radii of the orbits for Y and Z are 4R and R respectively. Now if the speed of Y satellite is 3v then what is the speed of satellite Z?**

**Solution:**

**We would use the Kepler’s third law here.**

**Say, radius of the orbit of Y and Z are Ry and Rz respectively.As given data, Ry = 4R and Rz = R**

From the 3rd Law of Kepler,

Ty^2/Tz^2 =Ry^3/Rz^3 = (4R)^3/R^3 = 64. So, Ty/Tz = 8 ……………. (1)

Continuing with this equation (equation 1) and bringing in the orbital velocity in it we get:

(2 **π **Ry/Vy) / (2 **π** Rz/Vz) = 8

=> (Ry Vz) /(Rz Vy) = 8

=> (Ry/Rz).(Vz/Vy) = 8

=> (4R/R) .(Vz/Vy) = 8

=> (Vz/Vy) = 8/4 = 2

=> **Vz = 2. Vy = 2. 3v = 6v**

**So the speed of satellite Z is 6v (answer)**

### Period of revolution of earth satellite – Numerical Problem

**Q1) What is the value of Period of Revolution of earth satellite? ignore the height of satellite above the surface of earth.**

**Given: gravitational acceleration (g) = 10 m/s^2. Radius of earth = R = 6400 km. Take π as 3.14**

**Solution:**

**Period of revolution = T = [ 2π / R]. √[r ^3/g] **

Here r = R + h. As h is to be ignored according to the question, so r =R here.

So T = 2 **π √(R/g) ……………. (1)**

**Given: R= 6400 km = 6400 X 10^3 meter. g = 10 m/s^2**Putting the values given, we get,

T = 2 X 3.14 X **√ [(6400X10^3)/10] = 2 X 3.14 X 800 = 5024 seconds = 83.73 minutes.**