Question

The height at which the weight of a body becomes 1/16 th its weight on the surface of the earth is?

Answer 1

at the height of moon.....

Answer 2

Answer:

3R ( R = Radius of the Earth)

Explanation:

Weight, w = mg     [m = mass (constant)]  

$g = \frac{GM}{R^{2}}$  where M is mass of the Earth and R is its Radius.

So,

$\frac{w_1}{w_2} = \frac{mg_1}{mg_2} = \frac{g_1}{g_2}$ ____ (1)   where $w_1$ is weight on Earth's surface and $w_2$ is weight at a height from the surface of the Earth.

$g_1 = \frac{GM}{R^2}$

$g_2 = \frac{GM}{(R+h)^2}$ because we are trying to find g at a certain height.

So, eqn (1) gives,

$\frac{w_1}{w_2} = \frac{g_1}{g_2}$

= $\frac{\frac{GM}{R^2} }{\frac{GM}{(R+h)^2}} = \frac{(R+h)^2}{R^2}$ _____ (2)

According to question, $w_1 = 16 w_2 = \frac{w_1}{w_2} = 16$

So, eqn (2) becomes

16 = $\frac{(R+h)^2}{R^2}$

Taking square roots,

4 = $\frac{R+h}{R}$

4R = R+h

h = 3R

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