The decrease in the value of g wrt earth-s surface on going to a height R2 above the earth-s surface will be-

Acceleration due to gravity at a height from the earth's surface is given by, g' = g(1 + hR)^2 At nbsp; height, h = R2 g' = g(1 + 12) = 4g9 So, decrease i ...

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Standard XII

Physics

Application of Projectile to a Height

The decrease ...

Question

The decrease in the value of $g$ (wrt earth's surface) on going to a height $\frac{R}{2}$ above the earth’s surface will be?

\frac{g}{2}

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\frac{5 g}{9}

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\frac{4 g}{9}

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\frac{g}{3}

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Solution

The correct option is B $\frac{5 g}{9}$
Acceleration due to gravity at a height from the earth's surface is given by,

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

At height, $h = \frac{R}{2}$

$g^{'} = \frac{g}{(1 + \frac{1}{2})} = \frac{4 g}{9}$

So, decrease in the value of acceleration due to gravity $= g - g^{'}$

i.e. $= g - \frac{4 g}{9} = \frac{5 g}{9}$

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The decrease in the value of g (wrt earth's surface) on going to a height R2 above the earth’s surface will be?

The correct option is B 5g9Acceleration due to gravity at a height from the earth's surface is given by, g′=g(1+hR)2 At height, h=R2 g′=g(1+12)=4g9 So, decrease in the value of acceleration due to gravity =g−g′ i.e. =g−4g9=5g9

The decrease in the value of $g$ (wrt earth's surface) on going to a height $\frac{R}{2}$ above the earth’s surface will be?