A person goes to height equal to $2 R$ from surface of earth (where $R$ is radius of earth), weight of the person at this height is $W_{1}$ and weight of person at surface of earth is $W_{2}$ , then $\frac{W_{1}}{W_{2}}$ will be

1 : 1

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1 : 2

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1 : 9

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1 : 3

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Solution

The correct option is C $1 : 9$
Weight of the person can be written using Newton's law of gravitation as:
$W = \frac{G M m}{r^{2}}$
Where, $r =$ distance from the center of earth.
Here,
$W_{1} =$ Weight at $r_{1} = 3 R$
$W_{2} =$ Weight at $r_{2} = R$

$\frac{W_{1}}{W_{2}} = \frac{\frac{G M m}{r_{1}^{2}}}{\frac{G M m}{r_{2}^{2}}} = \frac{r_{2}^{2}}{r_{1}^{2}} = \frac{R^{2}}{(3 R)^{2}} = \frac{1}{9}$

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A person goes to height equal to 2R from surface of earth (where R is radius of earth), weight of the person at this height is W1 and weight of person at surface of earth is W2, then W1W2 will be

The correct option is C 1:9Weight of the person can be written using Newton's law of gravitation as: W=GMmr2 Where, r= distance from the center of earth. Here, W1= Weight at r1=3R W2= Weight at r2=R W1W2=GMmr21GMmr22=r22r21=R2(3R)2=19

A person goes to height equal to $2 R$ from surface of earth (where $R$ is radius of earth), weight of the person at this height is $W_{1}$ and weight of person at surface of earth is $W_{2}$ , then $\frac{W_{1}}{W_{2}}$ will be