The height in terms of radius of the earth , at which the acceleration due to gravity becomes , where is acceleration due to

Question

The height in terms of radius of the earth $(R)$ , at which the acceleration due to gravity becomes $\frac{g}{9}$ , where $g$ is acceleration due to gravity on earth's surface, is _____ . [2026]

Smart Achievers · Accepted Answer

(3)

$g_{h} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

$\frac{g}{9} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

$1 + \frac{h}{R} = 3 \Rightarrow h = 2 R$

Solution

Q.
The height in terms of radius of the earth $(R)$ , at which the acceleration due to gravity becomes $\frac{g}{9}$ , where $g$ is acceleration due to gravity on earth's surface, is _____ . [2026]

1 $\sqrt{3} R$

2 $2 \sqrt{2} R$

3 $2 R$

4 $\frac{4 R}{9}$

Ans.
(3)

$g_{h} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

$\frac{g}{9} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

$1 + \frac{h}{R} = 3 \Rightarrow h = 2 R$

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Solution

Q. The height in terms of radius of the earth (R), at which the acceleration due to gravity becomes g9, where g is acceleration due to gravity on earth's surface, is _____ . [2026]

1 3R

2 22R

3 2R

4 4R9

Ans. (3) gh=g(1+hR)2 g9=g(1+hR)2 1+hR=3⇒h=2R