# Help with kinetic energy/potential energy ratio

1. Apr 20, 2010

### Princess19

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

What is the ratio of kinetic energy to potential energy, of a satellite orbiting a mass M around a radius R.

2. Relevant equations

ke = 1/2 (mv^2)/r, pe = -Gmm/r^2 (im not really sure of these, i dont have my book with me)

3. The attempt at a solution

this was a problem on my test, and i dont know if i got it right, so i would like to know what the answer is.

i had no idea how to do this so i just did

2ke = pe

ke/pe = 1/2.

what is the answer? i know i got it wrong hehe

any input is appreciated! thanks!

2. Apr 20, 2010

### kuruman

Start from
KE = (1/2)mv2
PE = -GMm/r

Write Newton's 2nd Law for a circular orbit and solve for the speed v.
Replace v in the KE equation.
Take the ratio and see what you get.

3. Apr 20, 2010

### Princess19

all i want to know is if i have it right. were not gonna have a final exam so i wont see this stuff ever again.

anyways ill try it cuz im bored...

(1/2)mv2/-GMm/r

thats its i have no idea what else

4. Apr 20, 2010

### kuruman

You don't have it right.

5. Apr 20, 2010

### Princess19

haha i knew it

so how do u do this problem? all i can get is to that first part i just did

6. Apr 20, 2010

### kuruman

In my posting #2 I gave you three steps. You skipped to the third without doing the first two. OK, once more

Step 1
Write Newton's 2nd Law for a circular orbit and solve for the speed v.

7. Apr 20, 2010

### Princess19

i still dont know how to do it

8. Apr 20, 2010

### Princess19

can someone help me?

9. Apr 20, 2010

### zachzach

Maybe you should check your sign???

10. Apr 20, 2010

### Princess19

im so lost, wat sign?

im up to this part

(1/2)mv^2/GMm/r

11. Apr 20, 2010

### kuruman

Newton's 2nd Law says F = ma. For a satellite of mass m orbiting the Earth in a circular orbit, what is F and what is a?

12. Apr 20, 2010

### Princess19

F = m.v^2 / r ??

13. Apr 20, 2010

### kuruman

Correct for the ma part. Now what is the force F for the gravitational attraction?

14. Apr 20, 2010

### Princess19

mg ?

15. Apr 20, 2010

### kuruman

No. It is the Law of Universal gravitational force that has upper case G not lower case g in it.

16. Apr 20, 2010

### Princess19

F = GMm/r^2 ?

17. Apr 20, 2010

### kuruman

That's the one. Now let's go to step 2. You have F (posting # 16) and you have ma (posting #12). Set the two equal and solve for the quantity mv2.

18. Apr 20, 2010

### Princess19

ok so

F = ma
GMm/r^2 = m . v^2 /r

GMm/r = mv^2

is that it? what next?

19. Apr 20, 2010

### kuruman

Step 3. You know that KE = (1/2)mv2. What is KE in terms of GMm/r?

20. Apr 20, 2010

### Princess19

i dont understand lol

is it GMm/2r?