A satellite of mass `3 xx 10^(5)` kg is orbiting in a circular orbit around the earth at a height of `3.6 xx 10^(6)m`. Calculate (i) orbital velocity of the satellite (ii) Potential energy of satellite. Given, mass of earth `=6 xx 10^(24)kg`.

Explanation

We are given: `M = 6 xx 10^(24)kg, m =3 xx 10^(5)kg, h= 3.6 xx 10^(6)m`
We are to calculate (i) orbital velocity of the satellite (ii) Potential energy of the satellite.
We know that, `R= 6400km= 6.4 xx 10^(6)m`
Using relations (i) orbital velocity `v_(o)= ((GM)/(R+h))^(1//2)` (ii) Potential energy `= - (GMm)/(r )`
We know that,
(i) Orbital velocity, `v_(o)= ((GM)/(R+h))^(1//2)`
`therefore v_(o)= sqrt((6.67 xx 10^(-11) xx 6 xx 10^(24))/((6.4 xx 10^(6) + 3.6 xx 10^(6))))= [(6.67 xx 6 xx 10^(13))/(10.0 xx 10^(6))]^(1//2)`
`=6.32 xx 10^(3) m s^(-1) = 6.32 km s^(-1)`
(ii) Potential energy `= - (GMm)/(r )= - (GMm)/((R+h))`
`therefore P.E. = - (6.67 xx 10^(-11) xx 6 xx 10^(24) xx 3 xx 10^(5))/(10 xx 10^(6)) = 12 xx 10^(12)J` [Ignoring- ve sign]