What will be the value of the acceleration due to gravity at a point 3 km above the earth's surface ? Diameter of the earth =12800 km and `g=980cm*s^(-2)` on the surface of the earth.

Explanation

Radius of the earth `R=(12800)/2=6400 km` .
As the height of the point (h=3 km) is negligibly small in comparison to R, the acceleration due to gravity at that height,
`g^.=g(1-(2h)/R)=980(1-(2xx3)/(6400))=980xx(6394)/(6400)= 979.08 cm*s^(-2)`.