Question

An asteroid exerts a 360-N gravitational force on a nearby spacecraft. If the spacecraft moves to a position three times as far from the center of the asteroid, the force will be

zero
40N
120N
360N
1080N

Answer 1

Answer:

40 N

Explanation:

The gravitational force between the asteroid and the spaceship is given by:

$F=G\frac{mM}{R^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$M$ is the mass of the asteroid

$m$ is the mass of the spaceship

$R$ is the distance between the asteroid and the spaceship

The initial force is equal to:

$F=G\frac{mM}{R^2}=360 N$

Later, the spaceship moves to a position 3 times as far from the center of the asteroid, so R' = 3R. Therefore, the new force will be

$F'=G\frac{mM}{R'^2}=G\frac{mM}{(3R)^2}=G\frac{mM}{9R^2}=\frac{1}{9}F$

so, the force is decreased by a factor 9. Since the initial force was F=360 N, the new force will be

$F'=\frac{360 N}{9}=40 N$