Question

an astronaut weighs 8.00x10^2 on the surface of earth. what is the weight of the astronaut 6.37x10^6 meters above the surface of the earth?

Answer 1

To solve this problem, we use the Law of Universal Gravitation:

F = Gm1m2/d^2

where m1 and m2 are two objects. In this case, earth and man. d is the distance between the objects. Lastly, G is the gravitational constant. Since the mass of the earth and man are constant, this is lumped up with G into k. The equation would be:

F = k/d^2
k = Fd^2
$F_{1} d_{1} ^{2} =F_{2} d_{2} ^{2}$

The radius of earth, d1, is equal to 6.371E+6 m. Thus, d2 = 2d1

(8E+2)(d1)^2 = F2(2d1)^2

(8E+2)(d1)^2 = 4F2(d1)^2

(8E+2)=4F2
F2 = 200 Newtons