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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet?
    [Given : Mass of planet \[=8\times {{10}^{22}}kg;\]
    Radius of planet \[=2\times {{10}^{6}}m,\]
    Gravitational constant
    \[G=6.67\times {{10}^{11}}\]\[\text{ }N{{m}^{2}}/k{{g}^{2}}\]]
    [JEE Main 10-4-2019 Afternoon]

    A) 9                                 

    B) 11

    C) 13                               

    D) 17

    Correct Answer: B

    Solution :

    \[{{F}_{g}}=\frac{m{{\text{v}}^{2}}}{r}\]           \[\frac{\text{GMm}}{{{r}^{2}}}=\frac{m{{\text{v}}^{2}}}{r}\]           \[V=\sqrt{\frac{\text{GM}}{r}}=\sqrt{\frac{(6.67\times {{10}^{-11}})(8\times {{10}^{22}})}{2.02\times {{10}^{6}}}}\]           \[V=1.625\times {{10}^{3}}\]                       \[T=\frac{2\pi r}{V}\]           \[n\times T=24\times 60\times 60\]           \[n\left[ \frac{2\pi (2.02\times {{10}^{6}}}{1.625\times {{10}^{3}}} \right]=24\times 3600\]           \[n=\frac{24\times 3600\times 1.625\times {{10}^{3}}}{2\pi (2.02\times {{10}^{6}})}\] \[n=11\]


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