A small satellite revolves around a planet of average density `10 g*cm^(-3)`. The radius of the orbit of the satellite is slightly more than the radius of the planet. Find the time period of rotation of the satellite. `G=6.68xx10^(-8)` CGS unit.

Explanation

Let mass of the planet =M, radius =R and density `=rho`.
`therefore M=4/3 piR^3rho`
As per the question , radius of the orbit of the satellite =R.if its mass=m and orbital speed =v.
centripetal force `=(mv^2)/R`.
Mutual force of gravitation between the planet and satellite provides this centripetal force.
`therefore (mv^2)/R =(GMm)/(R^2)`
or, `v^2=(GM)/R=G/R4/3 pi R^3rho =4/3piGrhoR^2`
or, `v=R sqrt(4/3piGrho)`
Time period = `("circumference of orbit")/("orbital speed")=(2piR)/(v)`
`=(2pi)/(sqrt(4/3piGrho))=sqrt((3pi)/(Grho))=sqrt((3xxpi)/(6.68xx10^(-8)xx10)) =3756 s =1h 2 min 36 s`.