Question

At what hieght from the surface of the earth the accelertaion due to gravity is the same as at a depth of 160km

Answer 1

Answer:

Given:

An object is taken to a depth of 160km.

To find:

Height above the earth surface at which the acceleration due to gravity will be same.

Formulas used:

Let depth be "d", height be "h" and radius of Earth be r

So at a depth d :

$g" = g(1 - \frac{d}{r} )$

At a height h:

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

Calculation:

The height has been considered much smaller than the radius of Earth such that h << r.

Equating the 2 Equations:

$g(1 - \frac{160}{r} ) = \frac{g}{ ({1 + \frac{h}{r} )}^{2} }$

For h <<< r , we can write:

$= > g(1 - \frac{160}{r} ) = g(1 - \frac{2h}{r} )$

$= > \frac{160}{r} = \frac{2h}{r}$

$= > h = \frac{160}{2}$

$= > h = 80 \: km$

So final answer is 80 km

Answer 2

QUESTION :-

At what hieght from the surface of the earth the accelertaion due to gravity is the same as at a depth of 160km.

ANSWER:-

[ NOTE]:-

DEPTH BE DENOTE AS 'D'.

HEIGHT BE DENOTE AS 'H'

RADIUS BE DENOTE AS 'R'

[FORMULA'S ]:-

FORMULA OF DEPTH:-

$g' = g(1 - \frac{d}{r} ).........(1)$

FORMULA OF H.

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

EQUATING THE 2 EQUATION.

$g = (1 - \frac{160}{r} ) = \frac{g}{(1 + \frac{h}{r}) {}^{2} }$

FOR HEIGHT & RADIUS:-

$\cancel{g}(1 - \frac{160}{r} ) = \cancel{g}(1 - \frac{2h}{r} ) \\ \frac{160}{r} = \frac{2h}{r} \\ .°. \frac{160}{\cancel{r}} = \frac{\cancel{r}}{2h} \\ .°.h = \frac{\cancel{160}}{\cancel{2}} \\ .°. height = 80km.$

THANKS.