A satellite orbits the earth at a height of 500km from its surface.
Mass of the earth `=6xx10^(24)kg`
Radius of the earth `=6.4xx10^(6)m`
Mass of the satellite = 300kg
`G=6.67xx10^(-11)Nm^(2)kg^(-2)`
Calculate (i) kinetic energy (ii) potential energy and (iii) total energy of the satellite.

Explanation

Height of the satellite `h=500xx10^(3)m`
Mass of the satelliite `m=300kg`
Mass of the earth `M=6.0xx10^(24)kg`
Radius of the earth `R=6.4xx10^(6)m`
`G=6.67xx10^(-11)Nm^(2)kg^(-2)`
Orbital velocity of a satellite,
`v=sqrt((GM)/(R+h))`
(i) Kinetic energy `=(1)/(2)mv^(2)`
`=(1)/(2)mxx((GM)/(R+h))`
`K.E=(1xx300xx6.67xx10^(-11)xx6.0xx10^(24))/(2xx(6.4xx10^(6)+0.5xx10^(6)))`
`K.E=(150xx6.67xx6xx10^(13))/(6.9xx10^(6))`
`=8.7xx10^(9)J`
(ii) Potential energy `=-(GMm)/(R+h)`
`P.E=-(6.67xx10^(-11)xx6xx10^(24)xx300)/((6.4xx10^(6)+0.5xx10^(6)))`
`=(-6.67xx1800xx10^(13))/(6.9xx10^(6))`
`=-17.4xx10^(9)J`
(iii) Total energy
`T.E=K.E+P.E`
`=8.7xx10^(9)J-17.4xx10^(9)J`
`=-8.7xx10^(9)J`