A particle of mass 'm' is projected with a velocity v = kVe (k < 1) from the surface of the earth. (Ve = escape velocity) The maximum height above the surface reached by the particle is :
- Rk21-k2
- R ( k1-k2 )2
- R ( k1+k2 )2
- R2k1+k
Answer
(1) Rk21-k2
Solution
Energy conservation between the surface and the highest point (height h):
12mv2 - GMmR = - GMmR+h
v = kVe and Ve2 = 2GMR
12m k2 2GMR = GMmR - GMmR+h
k2R = 1R - 1R+h
1R+h = 1-k2R
R + h = R1-k2
h = R1-k2 - R = Rk21-k2
Answer: (1)