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 these two. (Use `(GM)/R=6.4xx10^(7) m^(2)//s^(2)`). The speed `V_(1)` if particle pasess grazing the surface of the earth is

Answer

8000

Explanation

Conserving angular momentum `m(V_(1) cos 60^(@)).4R =m.V_(2) R , (V_(2))/(V_(1))=2`
Conserving energy of the system
`4R =m.V_(2) R , (V_(2))/(V_(1))=2`
`-(GMm)/(4R)+(1)/(2)mV_(1)^(2)=-(GMm)/(R)+(1)/(2)mV_(2)^(2)`
`(1)/(2)V_(2)^(2)-(1)/(2)V_(1)^(2)=(3)/(4)(GM)/(R)`
or `V_(1)^(2)=(1)/(2)(GM)/(R)`
`V_(1)=(1)/(sqrt(2))sqrt(64xx10^(6))=(8000)/(sqrt(2))m//s`.