Medium MCQ +4 / -1 PYQ · JEE Mains 2022

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 :

(R : radius of earth)

  1. A $\frac{\mathrm{R}}{1+\lambda^{2}}$
  2. B $\frac{R}{1-\lambda^{2}}$ Correct answer
  3. C $\frac{R}{1-\lambda}$
  4. D $\frac{\lambda^{2} \mathrm{R}}{1-\lambda^{2}}$

Solution

<p>Using energy conservation</p> <p>$$ - {{G{M_e}m} \over {{R_e}}} + {1 \over 2}m{\left( {\lambda \sqrt {{{2G{M_e}} \over {{R_e}}}} } \right)^2} = - {{G{M_e}m} \over r}$$</p> <p>$${{G{M_e}m} \over r} = {{G{M_e}m} \over {{R_e}}} - {{G{M_e}m} \over {{R_e}}}{\lambda ^2}$$</p> <p>$r = {{{R_e}} \over {1 - {\lambda ^2}}}$</p>

About this question

Subject: Physics · Chapter: Gravitation · Topic: Escape Velocity

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