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triden
Dec10-04, 02:35 PM
http://upload.cybermart.ca/2004/hepmesad.gif


I can't decide what to do with this question. I am studying for a final. I converted 24 hours to seconds and tried using the Fg = -gmm/r but I dont think thats going to work...

any leads? Thanks guys
theFuture
Dec10-04, 02:43 PM
You're on the right track. Think about circular motion, too.
triden
Dec10-04, 02:47 PM
You're on the right track. Think about circular motion, too.

ok I think im getting it.

r*(omega)^2 = GM/r^2

then i can cancel the R's to get

Omega^2 = GM/r ?
dextercioby
Dec10-04, 03:12 PM
ok I think im getting it.

r*(omega)^2 = GM/r^2

then i can cancel the R's to get

Omega^2 = GM/r ?

Yes,it's correct.To check your answer,though,u should be gettin round about 35000km.
Gamma
Dec11-04, 12:10 AM
you mean omega^2 = GM/r^3
frankR
Dec11-04, 01:57 AM
That's one of my favorite problems.

I think the answer is about 24,000mi. Don't forget to subtract the radius of the earth. I always forget to do that!
saltrock
Dec11-04, 02:39 AM
i Think radius should be (R+r) and use F=GMm/(R+r)^2
dextercioby
Dec11-04, 08:20 AM
i Think radius should be (R+r) and use F=GMm/(R+r)^2

Why complicate?Use "r" as your length variable (the radius of the trajectory)and then,once u got the result,subtract the mean radius of the Earth which is round about 6371km.