# Escape speed Mechanical Energy

1. Jul 29, 2012

### jamesbiomed

1. The problem statement, all variables and given/known data

Ok, this really IS my last question here.

Again, I have a fairly close answer that isn't right.

A 1630-kg communications satellite is released from a space shuttle to initially orbit the Earth at a radius of 9. 106 m. After being deployed, the satellite's rockets are fired to put it into a higher altitude orbit of radius 3. 107 m. What is the minimum mechanical energy supplied by the rockets to effect this change in orbit?

2. Relevant equations
Minimum Escape Speed:
V=sqrt(2GM/R)
G=6.67*10^-11
M=5.97*10^24

R should be the radius in whatever case. (It's the radius of the orbit so it is indeed the distance to earth's center)

KE=1/2 m v^2

3. The attempt at a solution

I plugged both radius's into two seperate eq'ns

And got

Vi=sqrt(2GM/Ri)=9406.84
Vf=sqrt(2GM/Rf)=5152.339

Then I put on my thinking cap and decided (Vi+Vf)/2 would give me the average velocity needed.

Then plugged into 1/2 mv^2 and got 4.30 e10 J

the correct answer is 2.53 e 10 J so I think my scope/basic methods are correct.

Any ideas?

Thanks a lot
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 29, 2012

### TSny

You've forgotten an important type of energy.

3. Jul 29, 2012

### TSny

Check this formula: V=sqrt(2GM/R)

4. Jul 29, 2012

### HallsofIvy

Staff Emeritus
You titled this "Escape speed" and mention escape speed in the text but this problem has nothing to do with escape speed because the object stays in orbit. There is, of course, a increase in potential energt but, if I am not mistaken, the speed necessary to orbit at a greater height is less so there will be a decrease in kinetic energy. (Ah, while I was typing this, TSny gave the formula for orbital speed.)

5. Jul 29, 2012

### jamesbiomed

You mean as in E=1/2 m v^2- GMm/Re+h?

I thought kinetic meant mechanical-my fault.

So I'm going to try the average kinetic energy minus the difference in potential energy

6. Jul 29, 2012

### jamesbiomed

Sorry, didn't see either of these before I posted most recently===

TSny was telling me to check that formula, I think, because it is the one for escape speed.

Is orbital speed sqrt (GM/R)?

My book honestly doesn't have an orbital formula.

7. Jul 29, 2012

### TSny

Yes, now you have the correct formula for the speed while in a circular orbit. Note that the speed will decrease in going to a larger orbit. So, the kinetic energy will decrease. But you'll also need to consider the potential energy. You don't need to write the radius as Re + h since they give you the radius of the orbit directly.

8. Jul 29, 2012

### TSny

You can easily derive this formula from F = ma for circular motion where F is the force of gravity and a = v2/R for centripetal acceleration.

9. Jul 29, 2012

### jamesbiomed

Oh I see, thank you

10. Jul 29, 2012

### jamesbiomed

The value I'm getting for KE is very, very small. Subtracting the values for PE and dividing by two got my the correct answer on a practice

11. Jul 29, 2012

### jamesbiomed

Basically, that's given me a lucky guess and it doesn't hold up. But the values I'm getting are normally about twice where they should be

12. Jul 29, 2012

### TSny

Total orbital energy E = KE + PE. What is the change in total orbital energy as you go from the smaller orbit to the larger orbit? Where does the gain in energy come from?