Question

numerical question:

Radius of the earth is 6.4 x 10^3 km and value of acceleration due to gravity on its surface is 9.8 m/s^2. find the value of acceleration due to gravity produced on a meteor at a distance of 9850 m from the earth surface.

Answer 1

Answer :

Acceleration due to gravity at height 9850m is 9.8 m/s²

Step-by-step Explanation:

Given : Radius of Earth = 6.4 × 10³ km = 6400 km

Acceleration due to gravity on earth surface = 9.8 m/s²

Height from the earth surface = 9850m = 9.85 km

To find : acceleration due to gravity at height = ?

We know that,

g' = g / ( 1 + h/Re ) ²

Where,

g' = acceleration due to gravity at height

g = acceleration due to gravity on earth surface

h = height

Re = Radius of Earth

Substituting the given value in above formula we get,

g' = 9.8 / ( 1 + 9.85/ 6400) ²

g' = 9.8 / ( 6409.85 /6400 )²

g' = 9.8 / ( 1 )²

= 9.8

g' = 9.8m/s²