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

Answer

D

Explanation

Conserving angular momentum `m.(v_1 cos 60^@)AR=m.v_2R,v_2/v_1=2`
Conserving energy of the system `-(GMm)/(4R)+1/2mv_1^2 =-(GMm)/R+1/2mv_2^2`
`1/2v_2^2-1/2v_1^2=3/4(GM)/R` or `v_1^2=1/2(GM)/2`
`v_1=1/sqrt2sqrt(64xx10^6)=8000/sqrt2` m/s = `4000sqrt2` m/s
P=400