The height at which g will be 1ex1- -1ex4- th of its value at the earth-s surface is

Given quantities:g^'=g/4Formula Used:g/g^'=(R+h)^2/R^20ex0exg^'→ acceleration due to gravity at height0ex0exg→ acceleration due to gravity at surface0ex0exR→ ra ...

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Question

The height at which g will be $\frac{1}{4}$ th of its value at the earth's surface is

$h = R$

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$h = \frac{R}{2}$

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$h = 2 R$

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none of these

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Solution

The correct option is A
$h = R$

Given quantities:
$g^{'} = \frac{g}{4}$
Formula Used:
$\frac{g}{g^{'}} = \frac{{(R + h)}^{2}}{R^{2}} g^{'} \to a c c e l e r a t i o n d u e t o g r a v i t y a t h e i g h t g \to a c c e l e r a t i o n d u e t o g r a v i t y a t s u r f a c e R \to r a d i u s h \to h e i g h t$
The calculation for the height:
$\frac{g}{g^{'}} = {(\frac{R + h}{R})}^{2} 4 = {(\frac{R + h}{R})}^{2} (\frac{R + h}{R}) = \sqrt{4} = 2 R + h = 2 R h = R$
Hence, option A is correct.

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The height at which g will be 14 th of its value at the earth's surface is

The correct option is A h=RGiven quantities:g'=g4Formula Used:gg'=R+h2R2g'→accelerationduetogravityatheightg→accelerationduetogravityatsurfaceR→radiush→heightThe calculation for the height:gg'=R+hR24=R+hR2R+hR=4=2R+h=2Rh=RHence, option A is correct.

The height at which g will be $\frac{1}{4}$ th of its value at the earth's surface is