Calculate the height at which the value of acceleration due to gravity becomes 50% of that at the surface of the earth. (Radius of the earth =6400 km )

Explanation

g at height , `g_(h)=g/((1+h/R)^2)`
In this problem , `g_(h)=50/100g = 0.5 g`
`0.5g =g/((1+h/R)^2)implies(1+h/R)^2=g/(0.5g)=2`
`1+h/R=sqrt2,h/R=sqrt2-1implies` the height ,
`h = 0.414 R = 0.414 xx 6400 = 2650` km.