The height at which the weight of a body becomes 1/16th its weight on the surface of earth (Radius R ) is:

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15R

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Solution

The correct option is D 3R
$\frac{G M m}{{(R + h)}^{2}} = \frac{1}{16} \frac{G M m}{R^{2}} \frac{1}{{(R + h)}^{2}} = \frac{1}{16 R^{2}} o r, {(\frac{R}{R + h})}^{2} = \frac{1}{16} \Rightarrow \frac{R}{R + h} = \frac{1}{4} . ∴ R + h = 4 R \Rightarrow h = 3 R$

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The height at which the weight of a body becomes 1/16th its weight on the surface of earth (Radius R ) is:

The correct option is D 3RGMm(R+h)2=116GMmR21(R+h)2=116R2or,(RR+h)2=116⇒RR+h=14.∴R+h=4R⇒h=3R

The correct option is D 3R
$\frac{G M m}{{(R + h)}^{2}} = \frac{1}{16} \frac{G M m}{R^{2}} \frac{1}{{(R + h)}^{2}} = \frac{1}{16 R^{2}} o r, {(\frac{R}{R + h})}^{2} = \frac{1}{16} \Rightarrow \frac{R}{R + h} = \frac{1}{4} . ∴ R + h = 4 R \Rightarrow h = 3 R$