A body weighs 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

At the surfaceW-GMmR2-I-W-x2032-GMm-3R2-2-II-WW-x2032-GMmR2GMm9R2-4-x21D2-WW-x2032-94-x21D2-W-x2032-4W9

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

$At the surface$
$W = \frac{G M m}{R^{2}} . . . . . . (I)$
$W^{'} = \frac{G M m}{(\frac{3 R}{2})^{2}} . . . . . . . (I I)$
$\frac{W}{W^{'}} = \frac{\frac{G M m}{R^{2}}}{\frac{G M m}{9 R^{2} / 4}}$
$\Rightarrow \frac{W}{W^{'}} = \frac{9}{4}$
$\Rightarrow W^{'} = \frac{4 W}{9}$

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

At the surfaceW=GMmR2......(I)W′=GMm(3R2)2.......(II)WW′=GMmR2GMm9R2/4⇒WW′=94⇒W′=4W9

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

$At the surface$
$W = \frac{G M m}{R^{2}} . . . . . . (I)$
$W^{'} = \frac{G M m}{(\frac{3 R}{2})^{2}} . . . . . . . (I I)$
$\frac{W}{W^{'}} = \frac{\frac{G M m}{R^{2}}}{\frac{G M m}{9 R^{2} / 4}}$
$\Rightarrow \frac{W}{W^{'}} = \frac{9}{4}$
$\Rightarrow W^{'} = \frac{4 W}{9}$