Question

At what height from the centre of the earth surface the acceleration due to gravity will be 1/4 of its value on the surface of the earth?

Answer 1

Answer:

To find:

At what height the Gravitational acceleration will be ¼th as that of the surface of Earth.

Formulas used:

At a height h from earth surface, radius of Earth be "r" and new gravity be g2

$g2 = \dfrac{g}{ {(1 + \frac{h}{r}) }^{2} }$

Calculation:

As per the question, g2 = ¼g

$= > \dfrac{g}{4} = \dfrac{g}{ {(1 + \frac{h}{r}) }^{2} }$

Cancelling g on both sides:

$= > \dfrac{1}{4} = \dfrac{1}{ {(1 + \frac{h}{r}) }^{2} }$

$= > ( {1 + \dfrac{h}{r} )}^{2} = 4$

$= > 1 + \dfrac{h}{r} = 2$

$= > \dfrac{h}{r} = 1$

$= > h = r$

So at a height equal to Radius of the Earth, the gravitational acceleration will be ¼ as the surface of the Earth .

So final answer :

$\boxed{ \red{height \: = radius \: of \: earth}}$

Answer 2

Answer:

Let the required height be h.

Radius of earth be R

So the question says that acceleration at height h is

g/4.

Therefore :

$\frac{g}{4} = \frac{g}{ {(1 + \frac{h}{r}) }^{2} }$

=> 4 = (1 +h/R)²

=> 1 + h/R = 2

=> h/R = 2 - 1 = 1

=> h = R .

At a height of R , the value of gravitation acceleration will become g/4.