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Question: If a body is projected vertically upwards with velocity $v = Nv_e$, where N<1 and $v_e$ is escape ve...

If a body is projected vertically upwards with velocity v=Nvev = Nv_e, where N<1 and vev_e is escape velocity from the surface of the Earth, then maximum altitude h attained by the body is given as (R: Radius of the Earth)

A

R1−N2\frac{R}{1-N^2}

B

RN21−N2\frac{RN^2}{1-N^2}

C

RN1−N\frac{RN}{1-N}

D

R1−N\frac{R}{1-N}

Answer

RN21−N2\frac{RN^2}{1-N^2}

Explanation

Solution

Apply conservation of mechanical energy. The initial energy of the body at the surface of the Earth is Ei=12mv2−GMmRE_i = \frac{1}{2}mv^2 - \frac{GMm}{R}, where v=Nvev = Nv_e and ve=2GMRv_e = \sqrt{\frac{2GM}{R}}. Thus, v2=N22GMRv^2 = N^2 \frac{2GM}{R}. The energy at maximum altitude hh is Ef=−GMmR+hE_f = -\frac{GMm}{R+h}, where the velocity is zero. Equating initial and final energies: 12m(N22GMR)−GMmR=−GMmR+h\frac{1}{2}m \left(N^2 \frac{2GM}{R}\right) - \frac{GMm}{R} = -\frac{GMm}{R+h} mN2GMR−GMmR=−GMmR+h\frac{mN^2GM}{R} - \frac{GMm}{R} = -\frac{GMm}{R+h} Dividing by GMmGMm: N2R−1R=−1R+h\frac{N^2}{R} - \frac{1}{R} = -\frac{1}{R+h} N2−1R=−1R+h\frac{N^2-1}{R} = -\frac{1}{R+h} 1−N2=RR+h1-N^2 = \frac{R}{R+h} R+h=R1−N2R+h = \frac{R}{1-N^2} h=R1−N2−R=R−R(1−N2)1−N2=RN21−N2h = \frac{R}{1-N^2} - R = \frac{R - R(1-N^2)}{1-N^2} = \frac{RN^2}{1-N^2}