Test Index

Gravitation Part 1

Section: Physics

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Question : 83 of 100

Marks: +1, -0

The height at which the acceleration due to gravity becomes

\frac{g}{9}

g=

the acceleration due to gravity on the surface of the earth) in terms of

R

, the radius of the earth, is:

$\frac{R}{\sqrt{2}}$
$\frac{R}{2}$
$\sqrt{2}R$
$2R$

Solution:

Given that, at height

h

from ground the acceleration due to gravity becomes

\frac{g}{9}

.

We know acceleration at earth surface due to gravity

g=\frac{G M}{R^2}

and acceleration at height

h

due to gravity

g'=\frac{G M}{(R+h)^2}

So

\frac{g}{9}=\frac{G M}{(R+h)^2}

\Rightarrow \frac{g}{9}=\frac{G M}{R^2} \cdot \frac{R^2}{(R+h)^2}

=g \cdot \left(\frac{R}{R+h}\right)^2

\Rightarrow \frac{1}{9}= \left(\frac{R}{R+h}\right)^2

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

\Rightarrow 3R=R+h

\therefore h=2R

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