How high above the the earth's surface is the meteor?

[fa08ti]

a 12 kg meteor experiences an acceleration if 7.2 m/s^2. when falling towards the earth

a: how high above the the Earth's surface is the meteor?
b: what force will a 30 kg meteor experience at the same altitude?

attempt:

i'm, not sure which equation to use..would it be
v= sqrt(Gm/r)?

 

[Pi-Bond]

You need a formula which relates acceleration due to the Earth's gravity and the distance from it's center. Do you know such a formula?

 

[fa08ti]

would this be correct?

gh = GM / (R + h )2

 

[Pi-Bond]

If R is the radius of the earth, and h above the surface of the earth, yes.

g=\frac{GM}{(R+h)^{2}}

 

[fa08ti]

i used the formula and got a negative value?

 

[Dick]

i used the formula and got a negative value?

How did you get a negative value? g should come out to 9.8m/s^2 if h=0. Do you know why? To get 7.2m/s^2, h should certainly be positive.

 

[fa08ti]

i used the mass of the meteor and i think i should have used the mass of earth?

 

[Dick]

i used the mass of the meteor and i think i should have used the mass of earth?

Very correct.

 

[fa08ti]

thanks..i just want someone to clarify something..does the mass of the object not matter? if so, why not?

 

[gneill]

thanks..i just want someone to clarify something..does the mass of the object not matter? if so, why not?

The mass of the object does not matter. Every mass falls with the same acceleration (in a given gravitational field).

F = G\frac{M m}{r^2}
but F = m a so that a = F/m Thus

a = \frac{F}{m} = G\frac{M}{r^2}

Only the mass of the Earth, M, matters for the acceleration of mass m in its field (that's the acceleration with respect to the Earth, of course).

So, why should this be so? It is so because inertial mass happens to be equal to gravitational mass for any object with mass (Look up "equivalence principle").

 

[fa08ti]

thanks everyone sooo much