# Homework Help: Energy for circular orbit

1. Jan 21, 2008

### bray d

[SOLVED] energy for circular orbit

This one should be easy for you guys, I've been workin on it for a while and need pointed in the right direction. For starters, here's the question:

Neglecting Earth's rotation, show that the energy needed to launch a satellite of mass m into circular orbit at altitude h is [(GMm)/R]*[(R+2h)/(2*(R+h))]
where G is 6.67x10^-11, M=mass of the earth, R=radius of the earth

I believe the problem has to do with conservation of energy, so I found the change in potential energy between the surface of the earth and height h. I think I need to find the change in kinetic energy from the surface and height h then add U and K to find the total energy. I did this and came up with an incorrect answer. Where am I going wrong?

P.S. nice site!

2. Jan 21, 2008

### hage567

Last edited: Jan 21, 2008
3. Jan 21, 2008

### bray d

hmmm, welp that link confirms my thought process...right? anyways here's what I've been getting:

first I found the change in potential energy:

U= -GMm(1/r1-1/r2)
= -GMm(1/R-1/(R+h))
= -GMm/R+GMm/(R+h)

Then I tried to find the change in kinetic energy:

to find v^2 in K=(1/2)mv^2 I equated the gravitational force to rotational acceleration since the gravity is the force that causes the acceleration:

GMm/r^2=mv^2/r

solving for v^2 I got:

v^2 = GM/r

Plugging that into the kinetic energy equation I got:

K = (1/2)GMm/r

therefor the change in kinetic energy would be the kinetic energy at the surface minus the kinetic energy at height h:

K = GMm/2R - GMm/2(R+h)

then to find total energy I added U and K:

E = U + K
= GMm/(R+h) - GMm/R + GMm/2R - GMm/2(R+h)

I then simplified by creating common denominators and adding like terms:

E = (2GMm-GMm)/2(R+h)+(-2GMm+GMm)/2R
= GMm/2(R+h) - GMm/2R

wow, that was not fun to type lol

4. Jan 21, 2008

### Dick

1) There is NO initial kinetic energy at the surface. It's not in orbit at the surface, it's just sitting there. 2) Check the signs on your changes in kinetic and potential energy. Both should be positive. Other than that, well done really.

Last edited: Jan 22, 2008
5. Jan 22, 2008

### bray d

that's what's confusing me. I initially thought there would be no initial kinetic energy, but then though about how it got to orbit. in order to get into orbit there must be some kind of kinetic energy right? so I figured there would have to be more initial kinetic energy then there was at orbit.

yep, dunno where that negative came from in the delta U equation. I'll rework it and give an update. thanks for the reply!

6. Jan 22, 2008

### Dick

You aren't told how it would get to orbit. Presumably a rocket or something. The energy provided by that unknown thing is what you want to compute. Update fast, I'm fading. It worked ok for me though. I got the answer you were looking for.

7. Jan 22, 2008

### bray d

ok, I understand. Yep, I corrected the sign in the delta U equation and took K = 0 at the surface. this gave me the correct answer after some algebraic manipulation. thanks again for the reply, it's greatly appreciated. have a nice night

8. Jan 22, 2008

### Dick

You too. Nice work.