1. 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 RE. SHOW YOUR WORK.

heres a few things if it helps

Formulas:

Weight: w = mg

Gravitational force: F = G(mM/r2)

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.640 m/s^2 if

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.640 m/s^2 if the acceleration due to gravity at the surface has magnitude 9.80m/s^2

Take the radius of the Earth to be 6.38 * 10^6 m

Inverse square law is the key of it.

(9.8 m/s^2)/(2^x) = 0.640 m/s^2

You're setting up a proportion here, earth's surface constant (9.8m/s^2) divided by 2(since it's square law) to the what power will make the given acceleration due to gravity at another distance.

0.640(2)^x = 9.8m/s^2

Simplify

15.3125 = 2^x

Simplify

log15.3125/log2 = x

Solve for x

x = 3.94

Since x is the amount of times the radius the unknown distance was, subtract the radius from the distance since we're measuring from the surface.

3.94(6.38E6) - 6.38E6 = 1.85E7 m

= 1.85 x 10^7 m.