The force of gravity on a spacecraft some distance from the earth is 800N What would that force be if its distance to the earth's centre were one half as great?

Answer 1

gravitational attraction between two bodies conforms to an inverse square law, so the force applied to the spacecraft by the earth is proportional to one over the displacement squared

in this case, one quarter squared is one sixteenth, so the force would be multiplied by one over one sixteenth, which is sixteen

the force would therefore be 800 x 16 = 12800N

Answer 2

If the distance between two objects is reduced by a factor of 1/2, then the gravitational

attraction between them increases by a factor of (2)2 = 4. The new weight of the

spaceship at the reduced distance would be 3,200 N.

Answer 3

If the distance between two objects is reduced by a factor of 1/3, then the gravitational

attraction between them increases by a factor of (3)2 = 9. The new weight of the

spaceship at the reduced distance would be 7,200 N.

Answer 4

If the distance between two objects is reduced by a factor of 1/10, then the gravitational

attraction between them increases by a factor of (10)2 = 100. The new weight of the

spaceship at the reduced distance would be 80,000 N.

Answer 5

If the distance to the Earth's center is halved, the force of gravity would increase by a factor of 4. This is because gravitational force is inversely proportional to the square of the distance between the objects. Therefore, if the new force is 800N, when the distance is halved, it would become 3200N.