The height of the point verrtically aboce the earth 's surface at which acceleration due to gravity become 1% if its value at the surface is (Radius of the earth = R)

Options

A. 8R
B. 9R
C. 10R
D. 20R

Answer

B

Explanation

`g=(GM/R^2)`
`g_h =(GM)/(R+h)^2`
therefore `g_h = 1%of g = (1xxg)/100 = (g)/100`
Rightarrow `(g/100)=(GM)/(R+h)^2`
Rightarrow `(g)/(100/g)=(GM)/(R^2/((GM)/(R+h)^2`
`100 = ((R+h)^2))/R^2`
`100R^2 = (R+h)^2`
`10R = R+h`
`Rightarrow h= 10R-R
therefore h= 9R`