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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Physics
Chapter
Gravitation

BITSAT 2019 · Physics · Gravitation

The escape velocity from the surface of earth is v _ e . A body is...

The escape velocity from the surface of earth is ve.{v}_{e}. A body is projected with velocity 3ve.3{v}_{e}. With what constant velocity will it move in the inter planetary space?

Options

  1. Ave{v}_{e}
  2. B3ve3{v}_{e}
  3. C2ve\sqrt{2}{v}_{e}
  4. D22ve2\sqrt{2}{v}_{e}

Correct answer

D. 22ve2\sqrt{2}{v}_{e}

Step-by-step solution

& If vv be the velocity in the inter planetary space and Ui,Uf{U}_{i,}{U}_{f} initial and final PE

Then, by the law of conservation of energy,

Ui+Ki=Uf+Kf{U}_{i} + {K}_{i} = {U}_{f} + {K}_{f}

−GMemRe+12m(3ve)2=0+12mv2\dfrac{- G{M}_{e}m}{Re} + \dfrac{1}{2}m\left(3{v}_{e}\right)^{2} = 0 + \dfrac{1}{2}mv^{2}

−GMeRe+92ve2=12v2\dfrac{- G{M}_{e}}{Re} + \dfrac{9}{2}{v}_e^{2} = \dfrac{1}{2}v^{2}

⇒−2GMeRe+9Ve2=v2⇒ \dfrac{- 2G{M}_{e}}{Re} + 9{V}_e^{2} = v^{2}

⇒−2gRe2Re+9ve2=v2[∵GMe=gRe2]⇒ \dfrac{- 2g{R}_e^{2}}{Re} + 9{v}_e^{2} = v^{2} \left[∵ G{M}_{e} = g{R}_e^{2}\right]

−2gRe+9×2gRe=v2[∵ve2=2gRe]- 2g{R}_{e} + 9 \times2g{R}_{e} = v^{2} \left[∵ {v}_e^{2} = 2g{R}_{e}\right]

16gRe=v216g{R}_{e} = v^{2}
⇒v=8⋅2gRe===22ve</p>⇒ v = \sqrt{8} \cdot \sqrt{2g{R}_{e}} ===2\sqrt{2}{v}_{e}</p>

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