Question

At what distance above the surface of the earth is the acceleration due to the earth’s gravity one hundred times smaller than the one at the surface of Earth? Express your result in terms of the radius of Earth R_E. SHOW YOUR WORK.

Formulas to use:

W = mg

F = GmM/r^2

F = ma

Answer 1

Gravitational force on a mass m which situated at surface of earth i.e. distance from it and center of earth = Re by law of gravitation is

F = GMm/(Re^2) where G = gravitational constant and M = mass of earth

Now by Newton's law 2nd law, F = ma

Here a = acceleration due to gravity = g

So, F = ma = mg = G Mm/ (Re)^2

Or, g = GM/(Re)^2

Now let at distance r from center of earth the acceleration due to gravity,

g' = one hundredth times smaller than g = (1/100)g

g'= g/100 = Gm/(r^2)

Or, Gm/(r^2) = {Gm/(Re)^2}/100

Or, 1/r^2 = 1/{(Re)^2×100}

Or, r^2 = (Re)^2×100

Or, r = Re × √100 = 10 ×Re = 10Re

So, distance from surface of earth = r -Re = 10Re -Re = 9Re ( 9 times of earth's radius).