JEE Main & Advanced Sample Paper JEE Main Sample Paper-25

  • question_answer
    Suppose, the gravitational force varies inversely as the \[{{n}^{th}}\] power of distance. Then, the time period of a planet in circular orbit of radius R around the sun will be proportional to:

    A)  \[{{R}^{\left( \frac{n+1}{2} \right)}}\]                         

    B)  \[{{R}^{\left( \frac{n-1}{2} \right)}}\]

    C)  \[{{R}^{n}}\]                                     

    D)  \[{{R}^{\left( \frac{n-2}{2} \right)}}\]

    Correct Answer: A

    Solution :

    The necessary centripetal force required for a planet to move round the sun = gravitational force exerted on it i.e., \[\frac{m{{v}^{2}}}{R}=\frac{G{{M}_{e}}m}{{{R}^{n}}}\,\Rightarrow \,v={{\left( \frac{G{{M}_{e}}}{{{R}^{n-1}}} \right)}^{1/2}}\] now \[T=\frac{2\pi R}{v}=2\pi R\times {{\left( \frac{{{R}^{n-1}}}{G{{M}_{e}}} \right)}^{1/2}}\] \[=2\pi {{\left( \frac{{{R}^{2}}\times {{R}^{n-1}}}{G{{M}_{e}}} \right)}^{1/2}}\,=2\pi \left( \frac{{{R}^{(n+1)/2}}}{{{(G{{M}_{e}})}^{1/2}}} \right)\]\[\Rightarrow \,T\propto {{R}^{\left( \frac{n+1}{2} \right)}}\]


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