A particle of mass 'mmm' is projected with a velocity v=kvev = kv_ev=kve (k<1k < 1k<1) from the surface of the earth. (vev_eve = escape velocity). The maximum height above the surface reached by the particle is
R(k1−k)2R\left(\frac{k}{1-k}\right)^2R(1−kk)2
R(k1+k)2R\left(\frac{k}{1+k}\right)^2R(1+kk)2
R2k1+k\frac{R^2k}{1+k}1+kR2k
Rk21−k2\frac{Rk^2}{1-k^2}1−k2Rk2
Given v=kvev = kv_ev=kve, where k<1k < 1k<1. Thus v<vev < v_ev<ve. From conservation of mechanical energy, 12mv2−GMmR=−GMm(R+h)\frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{(R+h)}21mv2−RGMm=−(R+h)GMm. ⇒v22=GMR−GM(R+h)=GMhR(R+h)\Rightarrow \frac{v^2}{2} = \frac{GM}{R} - \frac{GM}{(R+h)} = \frac{GMh}{R(R+h)}⇒2v2=RGM−(R+h)GM=R(R+h)GMh. Since ve2=2GMRv_e^2 = \frac{2GM}{R}ve2=R2GM, we have 12k2ve2=GMhR(R+h)⇒12k2(2GMR)=GMhR(R+h)⇒k2=hR+h⇒k2(R+h)=h⇒k2R=h(1−k2)⇒h=Rk21−k2\frac{1}{2}k^2v_e^2 = \frac{GMh}{R(R+h)} \Rightarrow \frac{1}{2}k^2(\frac{2GM}{R}) = \frac{GMh}{R(R+h)} \Rightarrow k^2 = \frac{h}{R+h} \Rightarrow k^2(R+h) = h \Rightarrow k^2R = h(1-k^2) \Rightarrow h = \frac{Rk^2}{1-k^2}21k2ve2=R(R+h)GMh⇒21k2(R2GM)=R(R+h)GMh⇒k2=R+hh⇒k2(R+h)=h⇒k2R=h(1−k2)⇒h=1−k2Rk2.
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