The acceleration due to gravity at a height $2 km$ above the earth surface is same as that of acceleration due to gravity at depth $^{'} d^{'}$ below the surface of earth. Then d is equal to

4 km

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 km

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 km

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8 km

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $4 km$
$g_{h} = g (1 - \frac{2 h}{R})$ for $h << R_{e}$

$g_{d} = g (1 - \frac{d}{R})$

$g (1 - \frac{2 h}{R}) = g (1 - \frac{d}{R})$

$\frac{2 h}{R} = \frac{d}{R}$

$d = 2 h = 4 k m$

Suggest Corrections

Similar questions

Q. The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:

Q. At which height from the surface of the Earth, The acceleration due to gravity is the same as at a depth 160 km below the surface of Earth. Radius of Earth is 6400 km

Q. Calculate the acceleration due to gravity at a point

64 k m

above and

32 k m

below the surface of earth.
Given Radius of Earth

= 6400 k m

. Acceleration due to gravity at the surface of earth

= 9.8 m s^{- 2}

Q. Derive an expression for acceleration due to gravity at a depth 'd' below the earth's earth surface.

The ratio of acceleration due to gravity at a depth h below the surface of earth and at a height h above the surface of earth for h << radius of earth

Join BYJU'S Learning Program

The acceleration due to gravity at a height 2 km above the earth surface is same as that of acceleration due to gravity at depth ′d′ below the surface of earth. Then d is equal to

The correct option is A 4 km gh=g(1−2hR) for h<<Re gd=g(1−dR) g(1−2hR)=g(1−dR) 2hR=dR d=2h=4 km

The acceleration due to gravity at a height $2 km$ above the earth surface is same as that of acceleration due to gravity at depth $^{'} d^{'}$ below the surface of earth. Then d is equal to

The correct option is A $4 km$
$g_{h} = g (1 - \frac{2 h}{R})$ for $h << R_{e}$

$g_{d} = g (1 - \frac{d}{R})$

$g (1 - \frac{2 h}{R}) = g (1 - \frac{d}{R})$

$\frac{2 h}{R} = \frac{d}{R}$

$d = 2 h = 4 k m$