What is the weight of this spacecraft at the following distances from Earth's surface?

Question

The radius of Earth is about 6.38x10^3 km. A 7.30x10^3 N spacecraft travels away from Earth. What is the weight of the spacecraft at the following distances from Earth's surface?

(a) 6.38x10^3 km

(b) 1.35x10^4 km

Answer 1

According to newtons law of gravitation the weight force acting on the spacecraft is inversely proportional to the distance to the center of earth:

F = G · M · m / r²

(G gravitation constant, M *** of earth, m mass of spacecraft,)

Let F₀ be the weight on earth surface (r =R) :

F₀ = G·M·m/R²

The weight at distance d to earth surface (r=R+d) is

F = G·M·m/(R+d)²

So the ratio of the weight at two different distances is

F/F₀ = (R /(R+d))²

<=>

F = F₀ / (1 + (d/R) )²

(a)

F = 7.30×10³ N / (1 + (6.38×10³km / 6.38×10³ km) )²

= 1.825×10³ N

(b)

F = 7.30×10³ N / (1 + (1.35×10⁴km / 6.38×10³ km) )²

= 7.52×10² N

Answer 2

a) 7.3e3 *6.38e3/(6.38e3+6.38e3)^2

b) 7.3e3 *6.38e3/(6.38e3+1.35e4)^2