The height in terms of radius of the earth (R) , at which the acceleration due to gravity becomes (g)/(9) , where g is acceleration due to gravity on earth's surface, is ______.

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Quantrex Academy · Accepted Answer

Correct answer: C. 2R

The acceleration due to gravity at a height

h

above the surface of the earth is given by

g' = \dfrac{g}{\left(1 + \dfrac{h}{R}\right)^2}

Given that

g' = \dfrac{g}{9}

, we get

\dfrac{g}{9} = \dfrac{g}{\left(1 + \dfrac{h}{R}\right)^2}

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

<br><br>

1 + \dfrac{h}{R} = 3

<br><br>

\dfrac{h}{R} = 2

<br><br>

h = 2R

<br><br>Answer:

2R

The height in terms of radius of the earth (R) , at which the...

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