The height at which the weight of a body&nbsp | vedclass

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Question

Accleration due to gravity at a height $h$ from the suface of earth is 

$g' = \frac{g}{{{{\left( {1 + \

Vedclass · Accepted Answer

$3R$

Vedclass · Answer

Accleration due to gravity at a height $h$ from the suface of earth is 

$g' = \frac{g}{{{{\left( {1 + \frac{h}{R}} ight)}^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)$

Where g is the acceleration due to gravity at the surface of earth and $R$ is the radius of earth.

Multiplying by $m\,\, (mass\,\, of\,\ the\,\, body)$ on both sides in $(i)$, we get

 $mg' = \frac{{mg}}{{{{\left( {1 + \frac{h}{R}} ight)}^2}}}$

$	herefore $ Weight of body at height $h,\, w'\, =\, mg'$

Weight of body at surface of earth, $W\,=\,mg$

According to question, $W' = \frac{1}{{16}}W$

$	herefore \frac{1}{{16}} = \frac{1}{{{{\left( {1 + \frac{h}{R}} ight)}^2}}}$

${\left( {1 + \frac{h}{R}} ight)^2} = 16\,\,or\,\,1 + \frac{h}{R} = 4$

$or\,\frac{h}{R} = 3\,\,or\,\,h = 3R$.

The height at which the weight of a body becomes ${\frac{1}{16}}^{th}$ , its weight on the surface of earth (radius $R$), is

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