At what height from the surface of Earth the gravitational force will be reduced by 10% if the radius of Earth is 6400 km?

Explanation

Gravitational force on the Earth surface
`F_(s)=(GMm)/(R_(E)^(2))`
Gravitational force at height h
`F_(h)=(GMm)/((R_(E)+h)^(2))`
`therefore(F_s)/(F_h)=((R_(E)+h)^(2))/(R_(E)^(2))=(10)/(9)`
`3(R_(E)+h)=3.162R_(E)`
`3h=(3.162-3)R_(E)`
`=0.162R_(E)`
`(R_(E)+h)/(R_E)=(3.162)/(3)`
`therefore` Height h `=(0.162xx6400)/(3)`
(or) h = 345.6 km