# Homework Help: Meteor Gravitational Field Question

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1. Nov 3, 2014

### Ess_Elle

1. The problem statement, all variables and given/known data
A 12 kg meteor experiences an acceleration of 7.2 m/s^2 when falling towards the earth.
a) How high above the earth's surface is the meteor?
b) What force will a 30 kg meteor experience at the same altitude?

2. Relevant equations

F
g= Gm1m2/r2 m1m2

r2 = Gm1m2/Fg

r = √Gm1m2/Fg

3. The attempt at a solution

Please help, I have been stuck on this question for hours trying to figure out if what I am doing is right. I am really confused right now, but here is my attempt at the solution:

Fg= Gm1m2/r2 m1m2

r2 = Gm1m2/Fg

r = √Gm1m2/Fg

r = √(6.67x10^11)(5.98x10^24)/7.2 m/s

r = 7442987.005 = 7.44 x 106 m

Paraphrase: The altitude above the earth’s surface is 7.44 x 106 m - 6.38 x 106 m = 1060000 = 1.06 x 106 m.

Now, I am confused because in an example in my lesson, they divide the entire sq. root equation by the force of gravity found, rather than by acceleration (which in this case is 7.2 m/s). I am unsure if I am supposed to divide by Fg rather than a? See below what I mean:

Fg = mg
Fg = (12 kg)(9.8 N/kg)
Fg = 117.6 = 118 N

which would make the equation, r = sq.rt| (6.67 x10^-11)(5.98x10^24)/118 N?

This again confuses me even more, because in this question, I think that Fg = ma = Gm1m2/r^2. So maybe if I am to be dividing by force of gravity, that means, rather than multiplying mg, I would have to multiply ma? See below:

Fg = ma
Fg = (12 kg)(7.2 m/s)

Fg = 86.4 N

If the above equation is correct, then would I have to divide r = sq.rt| (6.67 x10^-11)(5.98x10^24)/86.4 N?

2. Nov 3, 2014

### Ess_Elle

Oh, I forgot to post my solution to part b). I think I have done part b) correctly. See below:

F = ma

F = (30 kg)(7.2 m/s)

F = 216 N

Paraphrase: Therefore, a 30 kg meteor at the same altitude will experience a force of 216 N.