Question

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 ?

Answer 1

Given height, `h=500 " km" = 500 xx 10^(3) "m"`
Mass of spaceship, m = 300 kg, Mass of the earth, `M = 6 xx 10^(24) kg`
Radius of the earth, `R = 6.4 xx 10^(6) "m",G=6.67 xx 10^(-11) "N-m"^(-2) kg^(-2)`
(i) Kinetic energy `= (1)/(2)mv^(2)=(1)/(2)m.(GM)/(R+h)" "(becausev=sqrt((GM)/(R+h)))`
`=(1)/(2)xx(300xx6.67xx10^(-11)xx6xx10^(34))/(6.4xx10^(6)+500xx10^(3))=8.7 xx10^(9) J`
(ii) Potential energy
`=-(GMm)/((R+h))=(6.67xx10^(-11)xx6xx10^(24)xx300)/(6.4xx10^(6)+500xx10^(3))`
`=-17.4xx10^(9)J`
(iii) Total energy `= KE +PE`
`=8.7xx10^(9)-17.4xx10^(9)=-8.7 xx10^(9)J`
(iv) Binding energy = -(Total energy) `= 8.7 xx 10^(9) J`.