Question

A weight watcher who normally weighs 400 N stands on top of a very tall ladder so she is one Earth radius above the Earth's surface. How much would she weigh there

Answer 1

Her weight at the height of R will be 100 N

Step-by-step explanation:

The weight of the body on the surface of earth can be found by the relation

W₁ =
`(GMm)/(R^2)` I

Here G is gravitational constant and M is the mass of earth .

m is the mass of the body

and R is the radius of earth or the distance of body from the center of the earth .

Now she moves to height equal to R . Thus the distance from center of earth becomes 2 R .

Thus Weight now is W₂ =
`(GMm)/((2R)^2)` II

Dividing II by I , we have

`(W_2)/(W_1)` =
`(1)/(4)`

But W₁ = 400 N

Thus W₂ =
`(400)/(4)` = 100 N