A 620 kg satellite above the earth's surface experiences a gravitational field strength of 4.5 N/kg.

Question

Knowing the gravitational field strength at Earth’s sur-

face and Earth’s radius, how far above Earth’s surface

is the satellite? (Use ratio and proportion.)

I'm using this gs=GMs/Rs^2

ge=GMe/Re^2

but I'm not sure how to completely finish the equation?

Answer 1

I am not sure what you mean by "Ms" and "Me".

Just to make sure, let me remind you that mass is an intrinsic property of

matter, independent of gravity.

Let

G be the gravitational constant,

M be the mass of the Earth,

R_e be the radius of the Earth,

R_s be the distance of the satellite from the center of the Earth,

g_e = 9.81 N/m be gravitational field strength on the Earth,

g_s = 4.5 N/m be gravitational field strength in the satellite's orbit.

We have the equations for gravitational field strength

g_s = GM/R_s²

g_e = GM/R_e²

g_s/g_e = R_e²/R_s²

R_s = R_e√(g_e/g_s)

R_s - Re = Re(√(ge/gs) - 1)

Now we need to calculate the radius of the Earth:

The authors of the international units defined 1 meter to be the way it is so that

the equator is exactly 40,000 km.

So

40000km = 2πR_e

R_e = 20000/π km

Substituting actual numbers

R_s - R_e = 20000/π×(√(9.81/4.5) - 1) = 3033 km

So that is how far the satellite is from the Earth's surface.

Notice that it is independent of the mass of the satellite; I do not know why you were given that.