# Gravitational Field Question

1. Jun 3, 2012

### EE123

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

Fg = ma , Fg =Gm1m2 / r^2

3. The attempt at a solution

a = Fg / m, thus:

Fg / m = Gm1m2 / r^2

then:

r = √ (Fg / Gm1m2(m))

I don't know how to solve the question. The altitude is the radius from the earth? and since a = Fg / m, :S I equated this with the Fg = Gm1m2 / r^2, :|.

2. Jun 3, 2012

### HallsofIvy

Staff Emeritus
Do you understand what "Gm1m2/r^2" means? What is G? What are m1 and m2?
A really major error is setting "a= Fg/m= Gm1m2/r^2". "Gm1m2/r^2" is the force due to gravity, not the acceleration. What you want is Fg= ma= Gm1m2/r^2.

Now, again, what are m1 and m2?

That wolyld be correct but you didn't! You equated it to a, not Fg.

3. Jun 3, 2012

### EE123

Yes I do understand what the equation: Fg = Gm1m2 / r^2 this is what my notes say (taking an online physics course, they provide notes)

G = gravitational field constant, m1 and m2 are the two masses, Fg = the force of gravity acting on each object,

r = the distance between the centres of the two masses.

this is the newtons law of universal gravitation in the above equation form.

So ma = Gm1m2 / r^2 ??

which means r = √ (ma / Gm1m2)?

m1 would be the earths mass and m2 would be the meteor mass? but the mass, m2, would cancel out right?

4. Jun 3, 2012

### SammyS

Staff Emeritus
It looks as if you have three different masses, m1, m2 and m. There are only two objects, so there should only be two different masses.

5. Jun 3, 2012

### Villyer

The top equation is right, but you solved for r wrong.
And you are right about m2, it does cancel.