Question

The height at which weight of a body becomes 1/9th its weight on surface of earth

Answer 1

Let height of body is h from the earth's surface.

using formula,

acceleration to due gravity at height h from the earth's surface is given by,

$g'=\frac{g_0}{\left(1+\frac{h}{R}\right)^2}$

where, $g_0$ is acceleration due to gravity on earth's surface

we also know, weight of body is directly proportional to acceleration due to gravity [ as you know, W = mg ].

so, $W'=\frac{W_0}{\left(1+\frac{h}{R}\right)^2}$

weight of body at height h = weight of body on earth's surface.

W' = $\frac{W_0}{9}$

$\frac{W_0}{\left(1+\frac{h}{R}\right)^2}=\frac{W_0}{9}$

or, $\left(1+\frac{h}{R}\right)^2=9$

or, 1 + h/R = 3

or, h = 2R

hence, height of body is twice the radius of the earth.

Answer 2

Answer:

2R is the correct answer

Explanation:

weight on earth's surface is twice