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:

## Expert Answer

\[ F _ G = F _ C \]

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

\[ v = \frac { 2 \pi R } { T } \]

**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**.*