Question

A body weighs 900N on earth's surface. At a distance of twice earth radius above the earth's surface , the weight of the body is: *100N *200N *300N *450?

Answer 1

100 N.

Explanation:

The body's mass will aways be the same. However, the acceleration of gravity, `g`, varies according to the altitude. According to Newton's gravitational law:

`F = (GmM)/d^2`,

where `G = 6.67 * 10^-11 m³kg^(-1)s^(-2)` is the gravitational constant, `m` and `M` are the masses that attract each other and `d` is the distance between them.

On the Earth's surface, `d = R_E`, where `R_E` is the Earth's radius. In this case, `F = 900 N` (the body's weight). At a distance of twice `R_E` above the Earth's surface (which equals a distance of `3R_E`), then:

`F' = (GmM)/(3R_E)^2`;

`F' = (GMm)/(9R_E^2)`.

Since `F = (GMm)/(R_E^2)` (the body's weight on the Earth's surface), then:

`F' = F/9`.

Since F = 900 N:

`F' = 900/9`;

`F' = 100 N.`