Question

A 1400 kg spacecraft is 3 Earth radii above the Earth’s surface. What is the acceleration due to gravity at this distance above the earth

Answer 1

using Newton's second law we get the acceleration of gravity at the planet's surface:
a = F/m = {F = G(m1m2)/(r2)}/m2 = G(m1)/(r2)

so the acceleration of gravity at the surface is:
a = G(m1)/(r2)
I will use 6.38x106m as the radius of the earth, so this planet has a radius of 3*(6.38x106m) = 1.914x107m
Earth has a mass of 5.98x1024kg, and so let's just plug this into our equation:
a = (6.67x10-11 Nm2/kg2)(5.98x1024 kg)/(1.914x107m)2 = 1.088 m/s2

a = 1.1 m/s2