Determine the gravitational potential on the surface of earth, given that radius of the earth is `6.4 xx 10^(6) m`: its mean density is `5.5 xx 10^(3)kg m^(-3) , G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`.

Options

A. `-6.297 xx 10^(7) J kg^(-1)`
B. `3.11 xx 10^(11) J kg^(-1)`
C. `4 xx 10^(8) J kg^(-1)`
D. `-2 xx 10^(9) J kg^(-1)`

Answer

A

Explanation

Here, `R = 6.4 xx 10^(6) m`,
`rho = 5.5 xx 10^(3) kg m^(-3)`,
Mass of the earth, `M = (4)/(3) pi R^(3) rho`.
Gracitational potential on the surface of earth is
`V = (-GM)/(R ) = (-G)/(R ) xx (4)/(3) pi R^(3) rho`
`= (-4)/(3) pi G R^(2) rho`
`= (-4)/(3) xx (22)/(7) xx (6.67 xx 10^(-11))`
`xx (6.4 xx 10^(6))^(2) xx (5.5 xx 10^(3))`
`= - 6.297 xx 10^(7) J kg^(-1)`