If a spaceship orbits the earth at a height of 500 km from its surface, then determine its (i) kinetic energy, (ii) potential energy, and (iii) total energy (iv) binding energy. Mass of the satellite = 300 kg, Mass of the earth `= 6xx10^(24) kg`, radius of the earth `= 6.4 xx 10^(6) m, G=6.67 xx 10^(-11) "N-m"^(2) kg^(-2)`. Will your answer alter if the earth were to shrink suddenly to half its size ?

Explanation

The orbital velocity of a satellite at a height h is `v = sqrt(( GM)/( R + h))`
K.E. `= ( 1)/( 2) mv^(2) = (1)/( 2) mv^(2) = (1)/( 2) m ((GM)/( R + h))`
Given `G = 6.67 xx 10^(-1) Nm^(2) kg^(-2)`
`R = 6.4 xx 10^(6) m`
`M = 6 xx 10^(24) kg`
`h = 500 xx 10^(3) m,m = 300 kg`
K.E. `= (( 1)/( 2) xx 300 xx 6.67 xx 10^(-11) xx 6 xx 10^(24))/( 6.4 xx 10^(6) + 500 xx 10^(3))`
`= ( 300 xx 6.67 xx 10^(-11) xx 6 xx 10^(24))/( 2 xx 6.9 xx 10^(6))`
`= 8.7 xx 10^(9) J`
P.E. `= ( - GMm)/( R + h)`
`= ( - 6.67 xx 10^(-11) xx 6 xx 10^(24) xx 300)/( 6.4 xx 10^(6) + 500 xx 10^(3))`
`= ( - 6.67 xx 10^(-11) xx 6 xx 10^(24) xx 300)/( 6.9 xx 10^(6))`
`= - 1.74 xx 10^(9) J`
Total energy `= K.E. + P.E. = 8.7 xx 10^(9) - 17.4 xx 10^(9)`
`= ( 8.7 - 17.4 ) 10^(9)`
`= - 8.7 xx 10^(9) J`
When the earth shrinks to half the size R will become `( 6.4 xx 10^(6))/( 2) = 3.2 xx 10^(6) m`. Hence, K.E., P.E. and T.E. will increase.