How high does a rocket have to go above Earth's surface until its weight is half what it would be on Earth?

Answer 1

Gravitational pull provides weight to each object. This pull is from the center of the earth which is 6378 km from the surface. To decrease your weight by half, one needs to travel another 2641.9 km from the surface.

Answer 2

The mass of an object at any point will remain the same( unless its velocity approaches that of light, of course )So, for the weight to reduce to half its value, the acceleration due to gravity has to reduce to half its value on earth.

now, the relation between acceleration due to gravity and distance from the center of the earth is given by

g = 1/R2

where R is the radius of the earth which is 6400 kms (approx)

let 'h' be the height at which the weight of the rocket will be half its initial value.

so, gh / g = R2 / (R+h)2

but gh = g/ 2

substituting it in the above expression yields h=2649.6 kms (approx)

so, at this height, the weight of the rocket will be half its initial weight.

Answer 3

Because of the inverse-square law, you would have to be at a distance, from Earth's center, that is square root of 2 (about 1.414) times the distance you would be on the surface. So, 1.414 - 1 = 0.414 times Earth's radius from the SURFACE.

Answer 4

For a g/2 (half gravity) effect on an object, a ball park figure is that it would have to be about 2500 km or about 1550 miles up.

Answer 5

[ sqrt(2) - 1 ] Earth radius = 1,640 miles (rounded) above the surface.