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  • A body of mass  is projected with velocity <img alt="\lambda v_{\mathrm{e}}" src="https://en

A body of mass \mathrm{m} is projected with velocity \lambda v_{\mathrm{e}} in vertically upward direction from the surface of the earth into space. It is given that v_{\mathrm{e}} is escape velocity and \lambda<1. If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be :
( \mathrm{R} : radius of earth)
 

Option: 1

\mathrm{\frac{R}{1+\lambda^{2}}}


Option: 2

\mathrm{\frac{R}{1-\lambda^{2}}}


Option: 3

\frac{\mathrm{R}}{1-\lambda}


Option: 4

\mathrm{\frac{\lambda^{2} R}{1-\lambda^{2}}}


Answers (1)

best_answer

Applying energy consevation between pt.(1) and (2)

\mathrm{T E_1 =T E_2 }

\mathrm{\frac{-G M m}{R} +\frac{1}{2} m\left(\lambda V_e\right)^2=\frac{-G M m}{(R+h)}+0 }

\mathrm{\because V_e =\sqrt{\frac{2 G M}{R}} }

\mathrm{\frac{-G M m}{R} +\lambda^2 \frac{G M m}{R}=\frac{-G M m}{R+h}}

The height from the centre of earth is \mathrm{\left (R+h \right )}

\mathrm{\frac{1}{R+h}=\frac{1}{R}-\frac{\lambda ^{2}}{R}}

\mathrm{\frac{1}{R+h}=\frac{1-\lambda ^{2}}{R}}

\mathrm{\Rightarrow \frac{1}{R+h}=\frac{R}{1-\lambda ^{2}}}

Hence 2 is correct option.

Posted by

Sayak

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