Question 64 of 223

A body weight $W$, is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be:

A$\dfrac{W}{91}$B$\dfrac{W}{100}$C$\dfrac{W}{9}$D$\dfrac{W}{3}$

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Answer: (B) $\dfrac{W}{100}$

Final distance from the centre of earth wil be $R+9R=10R$.
Now acceleration due to gravity at any height(distance from centre r) is given by,
$g=\dfrac{GM}{{r}^{2}}$.
Therefore,
${g}^{'}=\dfrac{GM}{{\left(10R\right)}^{2}}$
Now, value of $g$ at surface will be, $\dfrac{GM}{{R}^{2}}$.
Hence,
${g}^{'}=\left(\dfrac{g}{100}\right)$.
Then, ${W}^{'}=m{g}^{'}=\dfrac{mg}{100}=\dfrac{W}{100}$