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KVPY Sample Paper KVPY Stream-SX Model Paper-8

  • question_answer
    A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is \[3\text{ }\times {{10}^{5}}\] times Heavier than the Earth and is at a distance \[2.5\times {{10}^{4}}\] times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is \[{{v}_{e}}=11.2km{{s}^{-1}}\]. The minimum initial velocity \[({{v}_{s}})\] required for the rocket to be able to leave the Sun-Earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet)

    A) \[{{v}_{s}}=22km\,{{s}^{-1}}\]          

    B) \[{{v}_{s}}=42km\,{{s}^{-1}}\]

    C) \[{{v}_{s}}=62km\,{{s}^{-1}}\]          

    D) \[{{v}_{s}}=72km\,{{s}^{-1}}\]

    Correct Answer: B

    Solution :

    \[\frac{1}{2}mV_{e}^{2}-\frac{G{{M}_{e}}m}{{{R}_{e}}}-\frac{G{{M}_{e}}m\times 3\times {{10}^{5}}}{2.5\times {{10}^{4}}{{R}_{e}}}=0\] \[\frac{V_{e}^{2}}{2}=\frac{G{{M}_{e}}}{{{R}_{e}}}\left[ 1+\frac{3\times {{10}^{5}}}{2.5\times {{10}^{4}}} \right]\]\[{{v}_{e}}=\sqrt{13\left( \frac{2G{{M}_{e}}}{{{R}_{e}}} \right)}=\sqrt{13}\times 11.2\approx 42\]


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