Question

(a) At what height above the earth is the acceleration due to gravity 62% of its value at the surface?

(b) What is the speed of a satellite orbiting at that height?

I got 1210.49km for question a by using the formula, but it is wrong.

Answer 1

F = GMm / r²

F is the force of gravity between M and m, and also represents the weight of the object, or mg. The radius r for an object at some altitude will include the earth’s radius. The above may be written as:

mg = GMm / r²

The small mass, m, divides out, and solved for r, you get:

r = √[GM / g]

At some altitude h, this is:

r + h = √[GM / g]
h = √[GM / g] - r
   = √[(6.67 x 10⁻¹¹N∙m²/kg²)(5.97 x 10^24kg) / 0.620(9.81m/s²)] - 6.38 x 10^6m
h = 1.711x 10^6m (1,711 km)

The speed is:

v = √[GM / r + h]
= √[(6.67 x 10⁻¹¹N∙m²/kg²)(5.97 x 10^24kg) / (6.38 x 10^6m + 1.711 x 10^6m)]
= 7015 m/s (7.015 km/s)