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AMU Medical AMU Solved Paper-2015

  • question_answer
    An asteroid, whose mass is \[2.0\times {{10}^{-4}}\] times the mass of Earth, revolves in a circular orbit around the sun at a distance that is twice the Earths distance form the sun. The period of revolution of asteroid in years is (mass of the sun \[=1.99\times {{10}^{30}}kg\], radius of the Earths orbit \[=1.5\times {{10}^{11}}m\],\[G=6.67\times {{10}^{-11}}{{m}^{3}}/{{s}^{2}}\,kg\])

    A)  0.28 yr                 

    B)  2.8 yr

    C)  10yr                                     

    D)  28 yr

    Correct Answer: B

    Solution :

                    Centripetal force is obtained from gravitational force of attraction                 \[\frac{GMm}{{{r}^{2}}}=m\left( \frac{{{v}^{2}}}{r} \right)\]                 \[\frac{GMm}{{{r}^{2}}}=\frac{m\times 4{{\pi }^{2}}\times r}{{{T}^{2}}}\]                 \[T=\sqrt{\frac{4{{\pi }^{2}}{{r}^{3}}}{GM}}\]                                 \[=\sqrt{\frac{4\times 3.14\times 1.5\times {{10}^{11}}}{6.67\times {{10}^{-11}}\times 1.99}}=2.8\,yr\]


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