# Gravitational potential energy

1. Mar 27, 2015

### Calpalned

1. The problem statement, all variables and given/known data
For a satellite of mass $m_s$ in a circular orbit of radius $r_s$ around the Earth, determine its kinetic energy K.

2. Relevant equations
$K = \frac {1}{2}mv^2$
Gravitational potential energy $U(r) = - \frac {GmM_E}{r}$

3. The attempt at a solution
My answer is $K = \frac {1}{2}m_sv^2$ My textbook indicates that this is wrong and avoids using $K = \frac {1}{2}mv^2$ without saying why. Perhaps I can't use that equation in this question, but if that's the case, what's the reason? However my textbook states that the total energy for objects far from Earth's surface is the combination of both kinetic energy and gravitational potential energy
$\frac {1}{2}mv_1^2 - G \frac {mM_E}{r_1} = \frac {1}{2}mv_2^2 - G \frac {mM_E}{r_2}$
Clearly, $K = \frac {1}{2}mv^2$ can be used in the same situation as gravitational potential energy. So why can't $K = \frac {1}{2}mv^2$ be used to solve the question, but it can be used for the total energy?

2. Mar 27, 2015

### robphy

Was v given? (You didn't list it.)

3. Mar 27, 2015

### TSny

The kinetic energy is definitely given by $\frac{1}{2}m_sv^2$. However, it could be that you are asked to express this kinetic energy in terms of the radius of the orbit. That will require you to express the speed of the satellite in terms of the radius of the orbit.

4. Mar 27, 2015

### Calpalned

No velocity isn't given

5. Mar 27, 2015

### Calpalned

Thanks everyone! I see what I did wrong. Velocity isn't given so I cannot use $K = \frac {1}{2} mv^2$

6. Mar 27, 2015

### mattt

You must use $K=\frac{1}{2}m v^2$ but first you have to calculate $v$ in terms of $G,M_T, R_T+h$ using $\frac{G M_T m}{(R_T + h)^2} = \frac{m v^2}{(R_T + h)}$

7. Mar 27, 2015

### BvU

To follow a circular trajectory of radius $r_s$, what centripetal force is required ? What is it that exercises this force ? You have expressions for both, so you can equate them to find $v$, and then the kinetic energy is $K={\tfrac {1} {2}} mv^2$, expressed in the desired variables.