A particle is projected from point `A`, that is at a distance `4R` form the centre of the earth, with speed `V_(1)` in a direction making `30^(@)` with the line joining the centre of the earth and point `A`, as shown. Consider gravitational interaction only between thesetwo. (Use `(GM)/R=6.4xx10^(7) m^(2)//s^(2)`). The speed `V_(1)` if particle pasess grazing the surface of the earth is

Options

A. `2sqrt(2)xx10^(3) m//s`
B. `4sqrt(2)xx10^(3) m//s`
C. `4xx10^(3) m//s`
D. `4sqrt(3)xx10^(3) m//s`

Answer

B

Explanation

Coservation of angular momentum
`m.(V_(1)cos60^(@)).4R m.V_(2).R , (V_(2))/(V_(1))=2` conserving energy of the system
`-(GMm)/(4R)+1/2mV_(1)^(2)=-(GMm)/(R)+1/2mV_(2)^(2)`
`1/2mV_(2)^(2)-1/2mV_(1)^(2)=3/4 (GM)/R`
`V_(1)^(2)=1/2 (GM)/R`
or `V_(1)=1/(sqrt(2))sqrt(64+10^(6))=8000/(sqrt(2))m//s`
`=4sqrt(2)xx10^(3)m//s`