# Find an expression for the square of the orbital period.

This question aims to find the expression for the square of the orbital period and expression in terms of G, M, and R.

The distance between two objects of masses M and m is represented by R. The potential energy between these masses having a distance R is given by:

$U = \frac { – G M m } { R }$

Here, U is the potential energy which is the energy of an object at rest.

Many forces are acting on the planet. One of them is gravitational pull which holds the planet in its orbit. It is a force acting on the center of mass of any object which pulls it downwards. Centripetal force helps to keep an object moving in orbit without falling. Gravitational force balances out the centripetal force acting on the planet. It is written as:

$F _ G = F _ C$

$\frac { G M m } { R ^ 2 } = \frac { m v ^ 2 } { R } ….. 1$

$v = \frac { 2 \pi R } { T }$

v is the angular velocity of the satellite.

By substituting the equation of velocity in the 1:

$\frac { G M m } { R ^ 2 } = \frac { m (\frac { 2 \pi R} { T } ) ^ 2 } { R }$

Rearranging the above equation to find the time period:

$\frac { G M m } { R ^ 2 } = \frac { \frac { 4 m \pi ^ 2 R ^ 2} { T ^ 2} } { R }$

$\frac { G M } { R ^ 2 } = \frac { 4 \pi ^ 2 R } { T ^ 2 }$

$T ^ 2 = \frac { 4 \pi ^ 2 R } { G M }$

The potential energy U is:

$U = \frac { – G M m } { R }$

## Numerical Solution

The potential energy of the object is $\frac { – G M m } { R }$ and the expression for the square of orbital period is $\frac { 4 \pi ^ 2 R } { G M }$.

## Example

We can also find the kinetic energy K of the satellite which is the energy of an object in motion in terms of potential energy.

The gravitational force balance out the centripetal force acting on the planet:

$F _ G = F _ C$

$\frac { G M m } { R ^ 2 } = \frac { m v ^ 2 } { R }$

$v ^ 2 = \frac { G M } { R }$

The Kinetic energy of the satellite is calculated by putting the expression of velocity in the formula of kinetic energy:

$K = \frac { 1 } { 2 } m v ^ 2$

$K = \frac { 1 } { 2 } m ( \frac { G M } { R } )$

$K = \frac { GmM}{2R}$

$K = \frac { -1 } { 2} U$

Image/Mathematical drawings are created in Geogebra.