Gravitational field strength (ratios)

[HerroFish]

Homework Statement

A 6.2*10^2-kg satellite above Earth's surface experiences a gravitational field strength of magnitude 4.5-N/kg. Knowing the gravitational field strength at Earth's surface and Earth's radius, how far above Earth's surface is the satellite? (Use ratio and proportion.)

Homework Equations

Fg = (GMm)/r²
g = GM/r²
g_E = 9.8-N/Kg
r_E = 6.37*10^6-m

The Attempt at a Solution

g_s/g_E = r_s²/r_E²
r_s= √((r_E²g_s)/g_E)

and my answer comes out to be 4.3*10^6-m. But the answer at the back of the book is 3.0*10^6-m, what did I do wrong? Thanks

 

[cepheid]

The field strength is INVERSELY proportional to the square of the radius. So it should be

g_s/g_E = r_E²/r_s²

which is the reciprocal of what you had on the right hand side.

Think about it: for g_s ,the r_s is on the bottom, and for 1/g_E, the r_E is on top. It also makes intuitive sense. We know g_s is smaller than g_E, so the ratio should have the smaller distance on top.

 

[cepheid]

Don't forget, also that r_s is the distance between the satellite and the centre of its orbit, which is not quite what the problem is asking for.

 

[HerroFish]

ah that make sense, thanks for everything!