(a) Find the acceleration due to gravity in a mine of depth `640 km` if the value at the surface is `9.800 m//s^(2)`. The radius of the earth is `6400 km`
(b) Find the height over the earth's surface at which the weight of a body becomes half of its value at the surface.

Explanation

(a) `g'=g(1-h/(R_(e)))=g(1-640/6400)=g(1-1/10)=(9g)/10`
`=0.9xx9.8=8.82 m//s^(2)`
(b) `g'=g/((1+h/(R_(e)))^(2)) implies g/2=g/((1+h/(R_(e)))^(2))`
`(1+h/(R_(e)))^(2)=2 implies 1+h/(R_(e))=sqrt(2)`
`h/(R_(e))=(sqrt(2)-1)`
`h=(sqrt(2)-1)R_(e)=(1.4-1)xx6400`
`=2560 km`