At what height above the earth's surface would the value of acceleration due to gravity be half of what is on the surface? Calculate.

Let us say the distance at which the acceleration due to gravity wil be half that distance from the centre of the earth is R'. Now at that point gravitaional force of attraction must be

G*M*m/R'*R' = m*g/2

 

or, G*M/R'*R' = G*M/R*R*2

 

or, R' = 1.414*R, where R is the radius of the earth. Hence from the surface of the earth that

point is at a distance of 1.414-1 = 0.414*R

 

Now put the value of R = 6400km and get the value of 0.414*R. that is the answer.

 

Regards,