NEET-XII-Physics
11: Gravitation
- #37A particle is fired vertically upward from earth’s surface and it goes up to a maximum height of 6400 km. Find the initial speed of particle.Ans : The particle attained a maximum height 6400 km, which is equal to the radius of the Earth.
Total energy of the particle on the Earth's surface is given by
`` {E}_{e}=\frac{1}{2}M{V}^{2}+\left(\frac{-gmM}{r}\right)....\left(1\right)``
Now, total energy of the particle at the maximum height is given by
`` {E}_{3}=\left(\frac{-GMm}{R\mathit{+}h}\right)+0``
`` \Rightarrow {E}_{3}=\left(\frac{\mathit{-}GMm}{\mathit{2}R}\right)...\left(2\right)\left(\therefore g=R\right)``
From equations (1) and (2), we have:
`` -\frac{GMm}{R}+\frac{1}{2}m{v}^{2}=-\frac{GMm}{2R}``
`` \Rightarrow \left(\frac{1}{2}\right)m{v}^{2}=GMm\left(-\frac{1}{2R}+\frac{1}{R}\right)``
`` \Rightarrow {v}^{2}=\frac{GM}{R}``
`` =\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}}{6400\times {10}^{3}}``
`` =\frac{40.02+{10}^{13}}{6.4\times {10}^{6}}``
`` =6.2\times {10}^{7}=0.62\times {10}^{8}``
`` \therefore v=\sqrt{0.62\times {10}^{8}}``
`` =0.79\times {10}^{4}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
`` =79\,\mathrm{\,km\,}/\,\mathrm{\,s\,}``
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