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

Explanation

Let mass of earth =M
radius of earth =R
acceleration due to gravity on the surface of earth =g
`:.g=(GM)/(R^(2))`……….1
Let required height at which the value of acceleration due to gravity is half of its value on the surface of earth =h=?
`:.` using `g.=(GM)/((R+h)^(2))` we get
`g//2=(GM)/((R+h)^(2))`
Divinding 1 by 2 we get
`g/(g//2)=(GM)/(R^(2))xx((R+h)^(2))/(GM)`
or `2=((R+h)/R)^(2)`
or `sqrt(2)=(R+h)/R`
ro `R+h=sqrt(2)R`
or `h=sqrt(2)R-R`
`=R(sqrt(2)R--1)`
`=R(1.414-1)`
i.e. h=0.414R