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Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    The orbital angular momentum of a satellite revolving at a distance r from the centre is L. If the distance is increased to 16r, then the new angular momentum will be

    A)  16 L                      

    B)         64 L

    C)  \[\frac{L}{4}\]                  

    D)         4 L

    Correct Answer: D

    Solution :

    \[L=mvr=m\sqrt{\frac{GM}{r}}r=m\sqrt{GMr}\] \[\therefore \]  \[L\propto \sqrt{r}\] \[\Rightarrow \]               \[\frac{{{L}_{2}}}{{{L}_{1}}}=\sqrt{\frac{{{r}_{2}}}{{{r}_{1}}}}\]                 \[=\sqrt{\frac{16r}{r}}=4\] \[\Rightarrow \]               \[{{L}_{2}}=4L\]


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