The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is-

Question

The height at which the acceleration due to gravity becomes $\frac{g}{9}$   (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is-

Accepted Answer

গ. 2R

ব্যাখ্যা: ভূপৃষ্ঠ থেকে $h$ উচ্চতায় অভিকর্ষজ ত্বরণ $g' = g \left(\frac{R}{R+h}\right)^2$। প্রশ্নানুসারে, $g' = \frac{g}{9}$। সুতরাং $\left(\frac{R}{R+h}\right)^2 = \frac{1}{9} \implies \frac{R}{R+h} = \frac{1}{3} \implies R+h = 3R \implies h = 2R$।

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