Gravitational potential energy

[Calpalned]

Homework Statement

For a satellite of mass $m_s$ in a circular orbit of radius $r_s$ around the Earth, determine its kinetic energy K.

Homework Equations

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

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?

 

[robphy]

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

 

[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.

 

[Calpalned]

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

No velocity isn't given

 

[Calpalned]

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

 

[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 $\frac{G M_T m}{(R_T + h)^2} = \frac{m v^2}{(R_T + h)}$

 

[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.