A satellite orbits the earth at a height of 400 km, above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence ? Mass of the satellite=200 kg, mass of the earth=`6.0xx10^(24)`kg, radius of the earth=`6.4xx10(6)`m, G=`6.67xx10^(-11)Nm^(2)Kg^(-2)`.

Explanation

Total energy of the satellite at a height h from Earth surface = E = K.E. + P.E.
`= (1)/(2) mv^(2) - (GMm)/((R+h))`
`= (1)/(2)m(GM)/(R+h) - (GMm)/(R+h)`
`= - (GMm)/(2(R+h))`
Energy expended to rocket the satellite out of earth.s gravitational field
= - (Total energy)
`= -(-(GMm)/(2(R+h)))`
`= (6.67 xx 10^(11) xx 6 xx 10^(24) xx 200)/(2(64 xx 10^(5) + 4 xx 10^(5)))`
`= (8,004 xx 10^(13))/(136 xx 10^(5))`
`= 58.85 xx 10^(8)J = 5.9 xx 10^(9) J`