The critical velocity of a satellite at height h from the surface of the earth is 5 km/s. The same for a satellite around another planet of double the radius and 4 times the mass of the earth and double the height will be

Options

A. `10km//s`
B. `7km//s`
C. `2.50m//s`
D. `3.57km//s`

Answer

b

Explanation

`(V_(C_(2)))/(V_(C_(1)))=sqrt((M_(2))/(M_(1))xx(r_(1))/(r_(2)))=sqrt((M_(2))/(M_(1))xx((R_(1)+h_(1)))/((R_(2)+h_(2))))`
`=sqrt((4xx(R_(1)+h_(1)))/((2R_(1)+2h_(1))))=sqrt((4(R_(1)+h_(1)))/(2(R_(1)+h_(1))))`
`V_(C_(2))=sqrt2xx5`
`=1.141xx5=7.070"km/s."`