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 –

• A. 2R
• B. $\frac { R } { \sqrt { 2 } }$
• C. R / 2
• D. $\sqrt { 2 } R$

Answer 1

Answer: (A)

2R

Answer 2

g = $\frac { \mathrm { GH } } { ( \mathrm { R } + \mathrm { h } ) ^ { 2 } }$

$\therefore$ $\frac { G M } { 9 R ^ { 2 } }$ = $\frac { \mathrm { GM } } { ( \mathrm { R } + \mathrm { h } ) ^ { 2 } }$

Ž 3R = R + h

h = 2R

So option (1) is correct.