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JEE Main 202227 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
27 July
Shift
Evening Shift
Subject
Physics
Chapter
Gravitation

JEE Main 2022 · 27 July · Evening Shift · Physics · Gravitation

A body of mass m is projected with velocity λ v _ e in vertically...

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

Options

  1. AR1+λ2\dfrac{R}{1 + {\lambda}^{2}}
  2. BR1−λ2\dfrac{R}{1 - {\lambda}^{2}}
  3. CR1−λ\dfrac{R}{1 - \lambda}
  4. Dλ2R1−λ2\dfrac{{\lambda}^{2}R}{1 - {\lambda}^{2}}

Correct answer

B. R1−λ2\dfrac{R}{1 - {\lambda}^{2}}

Step-by-step solution

A body of mass m is projected with velocity λ v _ e in vertically upward direction from the surface of the earth into

Escape velocity is given by, ve=2GMeRe{v}_{e} = \sqrt{\dfrac{2G{M}_{e}}{{R}_{e}}}

If the body goes to the height hh from the surface of the earth, using mechanical energy conservation,

⇒12m(λ2GMeRe)2−GMemRe=−GMemRe+h⇒ \dfrac{1}{2}m{\left(\lambda\sqrt{\dfrac{2G{M}_{e}}{{R}_{e}}}\right)}^{2} - \dfrac{G{M}_{e}m}{{R}_{e}} = - \dfrac{G{M}_{e}m}{{R}_{e} + h}

⇒λ2GMemRe−GMemRe=−GMemRe+h⇒ {\lambda}^{2}\dfrac{G{M}_{e}m}{{R}_{e}} - \dfrac{G{M}_{e}m}{{R}_{e}} = - \dfrac{G{M}_{e}m}{{R}_{e} + h}

⇒(Re+h)=Re1−λ2⇒ \left({R}_{e} + h\right) = \dfrac{{R}_{e}}{1 - {\lambda}^{2}}

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