First slide

Acceleration due to gravity and its variation

Question

The height at which the value of acceleration due to gravity becomes 50% of that at the surface of the earth.(Radius of the earth =6400 km ) is

Easy

A

2650
B

2430
C

2250
D

2350

Solution

$A c c e l e r a t i o n d u e t o g r a v i y d e c r e a s e s b y 50 % m e a n s g' = g / 2 g^{'} = \frac{G M}{{(R + h)}^{2}} \Rightarrow g^{'} = \frac{g R^{2}}{{(R + h)}^{2}} = \frac{g}{{(1 + \frac{h}{R})}^{2}} \Rightarrow \frac{g}{2} = \frac{g}{{(1 + \frac{h}{R})}^{2}} \Rightarrow 2 = {(1 + \frac{h}{R})}^{2} \Rightarrow \sqrt{2} = (1 + \frac{h}{R}) \Rightarrow \frac{h}{R} = \sqrt{2} - 1 = 1.414 - 1 = 0.414 \Rightarrow h = 0.414 \times 6400 k m = 2649.6 k m$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App