# NEET Physics Gravitation Questions Solved $T=\frac{2\mathrm{\pi R}}{v}=\frac{2\mathrm{\pi R}}{\sqrt{\frac{GM}{R}}}=\frac{2{\mathrm{\pi R}}^{3}{2}}}{\sqrt{GM}}\phantom{\rule{0ex}{0ex}}T=\frac{2\pi {R}^{3}{2}}}{\sqrt{G\frac{4}{3}{\mathrm{\pi R}}^{3}\mathrm{\rho }}}=\frac{2{\mathrm{\pi R}}^{3}{2}}}{2\sqrt{\frac{\mathrm{\pi }}{3}\rho G}{R}^{3}{2}}}=\sqrt{\frac{3\mathrm{\pi }}{\rho G}}$

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Crack NEET with Online Course - Free Trial (Offer Valid Till August 24, 2019) $\left(2\right)\phantom{\rule{0ex}{0ex}}{K}_{i}+{U}_{i}={K}_{f}+{U}_{f}\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{\left(k{v}_{e}\right)}^{2}-\frac{GMm}{R}=0-\frac{GMm}{r}\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{\left(k\sqrt{\frac{2GM}{R}}\right)}^{2}-\frac{GMm}{R}=0-\frac{GMm}{r}\phantom{\rule{0ex}{0ex}}\frac{{k}^{2}}{R}-\frac{1}{R}=-\frac{1}{r}\phantom{\rule{0ex}{0ex}}r=\frac{R}{1-{k}^{2}}$

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