Question

92. A man weight ‘W’ on the surface of earth and his weight at a height ‘R’ from surface of earth is (R
is Radius of earth)
a)
4
W
b)
2
W
c) W d) 4W

Answer 1

Given:

Weight of man on the surface of Earth= W

Radius of Earth= R

To Find:

Weight of man at a height 'R' from the surface of the Earth

Solution:

We know that,

• Weight= m×g

where,

m is the mass of the body

g is the acceleration due to gravity

• The value of acceleration due to gravity at height h above the earth's surface is given by

$\pink{\boxed{\bf{g'=\dfrac{g}{\bigg(1+\dfrac{h}{R_{e}}\bigg)^{2}}}}}$

where,

g is acceleration due to gravity at the surface of Earth

Re is the radius of the Earth

Now,

$\longrightarrow\rm{Weight\:of\:Man\:on\:Earth=m\times g}$

$\longrightarrow\rm{W=m\times g}-----(1)$

Let the acceleration due to gravity at height R be g'

So,

$\longrightarrow\rm{g'=\dfrac{g}{\bigg(1+\dfrac{h}{R_{e}}\bigg)^{2}}}$

$\longrightarrow\rm{g'=\dfrac{g}{\bigg(1+\dfrac{\cancel{R}}{\cancel{R}}\bigg)^{2}}}$

$\longrightarrow\rm{g'=\dfrac{g}{(1+1)^{2}}}$

$\longrightarrow\rm{g'=\dfrac{g}{(2)^{2}}}$

$\longrightarrow\rm{g'=\dfrac{g}{4}}$

Now,

Let the weight of man at a height 'R' from the surface of Earth be W'

$\longrightarrow\rm{Weight\:of\:Man\:on\:height\:R=m\times g'}$

$\longrightarrow\rm{W'=m\times \dfrac{g}{4}}$

$\longrightarrow\rm{W'=\dfrac{m\times g}{4}}$

From (1), we get

$\longrightarrow\rm\green{W'=\dfrac{W}{4}}$

Hence, the weight of man at a height 'R' from the surface of Earth is W/4.