# Space Shuttle- Gravitation

1. Jan 15, 2007

### Gyroscope

1. The problem statement, all variables and given/known data
A space shuttle of 68000 kg is placed in orbit at an altitude of 1000 km from the surface of the Earth. What energy must we provide?

3. The attempt at a solution

I will need to provide energy in order to decrease the potential energy and an additional energy for the kinetic energy of the rotation.

$$\Delta E_p=-G\frac{Mm}{R+h} + G\frac{Mm}{R}$$

$$E_c=\frac{1}{2}mv^2=\frac{1}{2}mGM/(R+h)$$

So,
$$E=-G\frac{Mm}{R+h}+ G\frac{Mm}{R}+\frac{1}{2}mGM/(R+h)$$

Am I right?

Last edited by a moderator: Jan 15, 2007
2. Jan 15, 2007

### Hootenanny

Staff Emeritus
You are almost correct, but you have made one 'small' mistake in the first line which has carried through.

3. Jan 15, 2007

### Gyroscope

I have just corrected my mistake. My doubt was whether to consider or not the kinetic energy of rotation. Because when the space shuttle reaches an height of 1000 km relative to the surface of the Earth it will need an extra impulse from the fuel engines.

4. Jan 15, 2007

### Kurdt

Staff Emeritus
The potential energy equation looks fine except you do not square the distance and you have the correct overall idea. The kinetic energy term looks fine as you have used circular motion to derive the velocity.

EDIT: Beaten by Hoot and the coffee machine again!

5. Jan 15, 2007

### Gyroscope

Kurdt, thank you anyway for your kind help!

6. Jan 15, 2007

### D H

Staff Emeritus
Gyroscope:

The Shuttle has kinetic energy prior to launch. You have solved for the final kinetic energy, not the change in kinetic energy. The distinction is important: It is why the Shuttle is launched to the East.

7. Jan 15, 2007