A staellite in a circular orbit of raidus R has a period of 4h another satellite with orbital radius 3R around the same planet will have a period (in h)

Options

A. `16`
B. `4`
C. `4sqrt(27)`
D. `4sqrt(8)`

Answer

c

Explanation

According to kepler s third law
`T^(2) prop R^(3) rarr (R_(2))/(R_(1))^(3//2)`
`therefore (T_(2))/(T_(1))=(3R)/(R )^(3//2)=sqrt(27)`
`therefore T_(2)=sqrt(27)T_(1)=sqrt(27)xx=s4sqrt(27)h`