What would the set up be for this question?

Question

When a falling meteor is at a distance above the earth’s surface of 3 times the earth’s radius what is its free fall acceleration due to the earth’s gravitational force exerted on it?

Answer 1

So the meteor is 4 times the Earth's radius away from the Earth's centre.

Forumula: Fg=G[(Mm)/r^2]
G=6.673x10^-11 (N*m^2/kg^2)
M=5.98x10^24 (kg)
m=unknown(just say it is 500kg)
r=4(6.38x10^6)=2.552x10^7 m

Fg=6.673x10^-11[5.98x10^24(500)/(2.552x10^7)^2]
=306N


Suppose there are no other forces, so Fnet=Fg+Fn and Fnet=Fg
Fg=Fnet=ma
a=Fg/m
a=306 (N) /500kg)
a=0.612 (m/s^2) toward the earth

*notice that the mass of the meteor might be different from yours.

Answer 2

your meteor is three times above the distance of earth's radius. At such a large distance there will be no gravitational force exerted on it and meteor will simply revolve round the earth. also the formula for calculating gravitational force is: F=G(m1*m2)/r^2 where G is gravitational constant, m1 is mass of the 1st object, m2 is mass of the 2nd object and r is the distance between the two objects.

Answer 3

As altitude increases, the attraction of gravity decreases as the square of the distance of the body to the center of the earth. If it's 3R above the surface, it's 4R from the center of the earth. At 1R, g=9.8 m/S^2

At 4R, the acceleration due to gravity is 9.8 / 4^2 m/s^2 = 0.61 m/s^2.