Two satelites of a planet have period 32 days and 256 days. If the radius of orbit of former is R, find the orbital radius of the latter.

Options

A. radius of the orbit of the second is `4R`
B. radius of the orbit of the second `8R`
C. total mechanical energy of the second is greater than that of the first
D. kinetic energy of the second in more than that of the first

Answer

A, C

Explanation

Using kepler's third law,
`R_(2) = R_(1) ((T_(2))/(T_(1)))^(2//3) = R ((256)/(32))^(2//3)`
`K.E.` of satellite `E_(K) = (1)/(2)m upsilon_(2)^(2) = (1)/(2)m(GM)/(r ) = (GMm)/(2r)`
As `E_(K) prop (1)/(r )`
so `(E_(K_2))/(E_(K_1)) = (r_(1))/(r_(2)) = (R )/(4R) = (1)/(4)` or `E_(K_2) lt E_(K_1)`
Total mechanical energy, `E = PE + KE`
`= - (GMm)/(r ) + (GMm)/(2r) = - (GMm)/(2r)`
Here -ve sign shown that as `r` increases `E` increases.
So total mechancial energy of the second is greater than of the first satellite.