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

40 N

Step-by-step explanation:

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

`F=G(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(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(mM)/(R'^2)=G(mM)/((3R)^2)=G(mM)/(9R^2)=(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'=(360 N)/(9)=40 N`