BHU PMT BHU PMT (Screening) Solved Paper-2005

  • question_answer
    Two planets are revolving around the earth with velocities\[{{v}_{1}}\]and\[{{v}_{2}}\]and in radii\[{{r}_{1}}\]and\[{{r}_{2}}({{r}_{1}}>{{r}_{2}})\] respectively. Then:

    A)  \[{{v}_{1}}={{v}_{2}}\]                 

    B) \[{{v}_{1}}>{{v}_{2}}\]

    C)   \[{{v}_{1}}<{{v}_{2}}\]                

    D)  \[\frac{{{v}_{1}}}{{{r}_{1}}}=\frac{{{v}_{2}}}{{{r}_{2}}}\]

    Correct Answer: C

    Solution :

                      Key Idea: Gravitational force provides the required centripetal force. The gravitational force exerted on the planet is                 \[{{F}_{1}}=\frac{GMm}{r_{1}^{2}}=\frac{g{{R}^{2}}m}{r_{1}^{2}}\] where                   \[GM=g{{R}^{2}}\] Similarly               \[{{F}_{2}}=\frac{GMm}{r_{2}^{2}}=\frac{g{{R}^{2}}m}{r_{2}^{2}}\] This gravitational force provides the necessary centripetal force, hence we have                 \[\frac{g{{R}^{2}}m}{r_{1}^{2}}=\frac{mv_{1}^{2}}{{{r}_{1}}}\] \[\Rightarrow \]               \[v_{1}^{2}=\frac{g{{R}^{2}}}{{{r}_{1}}}\] \[\Rightarrow \]               \[{{v}_{1}}=\sqrt{\frac{g{{R}^{2}}}{{{r}_{2}}}}\] Similarly                 \[{{v}_{2}}=\sqrt{\frac{g{{R}^{2}}}{{{r}_{2}}}}\] \[\therefore \]  \[\frac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\frac{{{r}_{2}}}{{{r}_{1}}}}\] Given, \[{{r}_{1}}>{{r}_{2}}\] \[{{v}_{2}}>{{v}_{1}}\] Note: Greater the distance of planet above earth's surface, smaller is the speed of planet. Because speed of planet does not depend upon the mass of planet.


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