Question

A satellite orbits the earth at a height of `3.6xx10^(6)m` from its surface. Compute it’s a kinetic energy, b. potential energy, c. total energy. Mass of the satellite `=500kg` mass of the earth `=6xx10^(24)`kg, radius of the earth `=6.4xx10^(6), G=6.67xx10^(-11)Nm^(2)kg^(-2)`.

Answer 1

Here, `r = R + h = 6.4 xx 10^(6) + 3.6 xx 10^(6) = 10^(7) m`
Orbital velocity of satellite around the earth is given by
`upsilon_(0) = sqrt((GM)/((R + h))) = sqrt((6.67 xx 10^(-11) xx 6 xx 10^(24))/(10^(7))) = sqrt(6.67 xx 6 xx 10^(6)) ms^(-1)`
(a) K.E. of the satellite `= (1)/(2) m upsilon_(0)^(2) = (1)/(2) xx 500 xx 6.67 xx 6 xx 10^(6) = 10^(10) J`
(b) P.E. of the satellite `= - (GMm)/(R + h) = - (6.67 xx 10^(-11)xx (6 xx 10^(24)) xx 500)/(10^(7) = - 2 xx 10^(10) J`
(c ) Total energy `= KE + PE = 10^(10) - 2 xx 10^(10) = - 10^(10)J`