A body weight W Newton the surface of the earth- Its weight a height equal to half the radius of the earth will be -dfrac-W-2-dfrac-2W-3-dfrac-4W-9-dfrac-W-4-

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Toppr Admin · Accepted Answer

Given-xA0-R-6400km-w-mg-x21D2-g-wmNow the weight of the body at the height is 12Rg-x2032-g-1-x2212-dR-So the weight of the body at the 12RW-x2032-W-1-x2212-R2R-WR2R-W2

A body weight W Newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be :
$\frac{W}{2}$
$\frac{2 W}{3}$
$\frac{4 W}{9}$
$\frac{W}{4}$

Given, $R = 6400 k m, w = m g \Rightarrow g = \frac{w}{m}$

Now the weight of the body at the height is $\frac{1}{2} R$

$g^{'} = g (1 - \frac{d}{R})$

So the weight of the body at the $\frac{1}{2} R$

$W^{'} = W (1 - \frac{R}{2 R}) = W \frac{R}{2 R}) = \frac{W}{2}$

A body weight W Newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be :W22W34W9W4

Given, R=6400km,w=mg⇒g=wmNow the weight of the body at the height is 12Rg′=g(1−dR)So the weight of the body at the 12RW′=W(1−R2R)=WR2R)=W2

A body weight W Newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be :
$\frac{W}{2}$
$\frac{2 W}{3}$
$\frac{4 W}{9}$
$\frac{W}{4}$

Given, $R = 6400 k m, w = m g \Rightarrow g = \frac{w}{m}$

Now the weight of the body at the height is $\frac{1}{2} R$

$g^{'} = g (1 - \frac{d}{R})$

So the weight of the body at the $\frac{1}{2} R$

$W^{'} = W (1 - \frac{R}{2 R}) = W \frac{R}{2 R}) = \frac{W}{2}$