RAJASTHAN PMT Rajasthan - PMT Solved Paper-1995

  • question_answer
    A satellite is revolving close to surface of planet of radius R, its time period is T- Time period of satellite for a planet of radius 3R will be:

    A) \[\text{3}\sqrt{\text{3}}\text{T}\]          

    B)                        \[\frac{\sqrt{\text{3}}}{2}\text{T}\]        

    C)        \[\frac{1}{3\sqrt{\text{3}}}\text{T}\]     

    D)        \[\frac{1}{2}\text{T}\]

    Correct Answer: A

    Solution :

    According to Keplers law, \[{{T}^{2}}\propto {{R}^{3}}\] \[\therefore \] \[T\propto {{R}^{3/2}}\]                 \[\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3/2}}={{\left( \frac{R}{3R} \right)}^{3/2}}=\frac{1}{3\sqrt{3}}\]                 \[{{T}_{2}}=3\sqrt{3}\,\,{{T}_{1}}=3\sqrt{3}T\]


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