# An artificial satellite moving in a circular orbit around the Earth has total (kinetic +...

## Question:

An artificial satellite moving in a circular orbit around the Earth has total (kinetic + potential) energy equal to {eq}E_0 {/eq}. Then its potential energy is

{eq}(a) E_0 \\ (b) 1.5 E_0 \\ (c) 2E_0 \\ (d) E_0 {/eq}

## Mechanical Energy

The classical mechanical energy of an object is the sum of its kinetic and potential energies. This means that the total mechanical energy of an object can be written as:

{eq}\displaystyle E = K+U {/eq}

Here E is the total mechanical energy, K is the kinetic energy of the object and U is the potential energy of the object. In the case of an object orbiting another much larger object the only two energies are the kinetic energy and gravitational potential energy, so the mechanical energy can be written as:

{eq}\displaystyle E = \frac{1}{2} mv^2 - \frac{GmM}{r} {/eq}

Here m and M are the mass of the smaller and larger objects respectively, v is the smaller object's velocity, G is the gravitational constant, and r is the distance between the objects.

In order for an object to stay in a circular orbit around another object the force of gravity must be equal to the centrifugal force so that there is no linear acceleration on the object. This means that:

{eq}\displaystyle F_g = F_c \: \rightarrow \: \frac{GmM}{r^2} = \frac{mv^2}{r} \: \rightarrow \: v = \sqrt{\frac{GM}{r}} {/eq}

So the orbiting object needs a tangential (circular) velocity of {eq}\displaystyle \sqrt{\frac{GM}{r}} {/eq} in order to maintain a circular orbit. We are told that the total energy of the satellite is {eq}\displaystyle E_0 {/eq} which represents the total mechanical energy of the satellite. This can be written as:

{eq}\displaystyle E_0 = \frac{1}{2}mv^2- \frac{GmM}{r} {/eq}

Plugging in the velocity we know the satellite needs to maintain this orbit, we get:

{eq}\displaystyle E_0 = \frac{1}{2}m \left( \sqrt{\frac{GM}{r}} \right)^2 - \frac{GmM}{r} = \frac{GmM}{2r} - \frac{GmM}{r} = -\frac{GmM}{2r} {/eq}

So we have found the value of {eq}\displaystyle E_0 {/eq} to be {eq}\displaystyle -\frac{GmM}{2r} {/eq}. We can also write this as:

{eq}\displaystyle E_0 = \frac{1}{2} \left( - \frac{GmM}{r} \right) = \frac{1}{2} U \: \rightarrow \: U = 2 E_0 {/eq}

Thus the potential energy is equal to {eq}\displaystyle 2E_0 {/eq} making (c) the correct answer.