JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Question Bank Self Evaluation Test - Gravitation

  • question_answer
    The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If \[{{t}_{1}}\] is the time for the planet to move from C to D and \[{{t}_{2}}\] is the time to move from A to B then

    A) \[{{t}_{i}}=4{{t}_{2}}\]

    B) \[{{t}_{i}}=2{{t}_{2}}\]

    C) \[{{t}_{i}}={{t}_{2}}\]

    D) \[{{t}_{i}}>{{t}_{2}}\]

    Correct Answer: B

    Solution :

    [b] According to Kepler's law, the areal velocity of a planet around the sun always remains constant. \[\operatorname{SCD} : {{A}_{1}}-{{t}_{1}}\](areal velocity constant) \[\operatorname{SAB} : {{A}_{2}}-{{t}_{2}}\] \[\frac{{{A}_{1}}}{{{t}_{1}}}=\frac{{{A}_{2}}}{{{t}_{2}}},\,\,{{t}_{1}}={{t}_{2}}.\frac{{{A}_{1}}}{{{A}_{2}}},\left( given\,{{A}_{1}}=2{{\operatorname{A}}_{2}} \right)\] \[={{t}_{2}}.\frac{2{{\operatorname{A}}_{2}}}{{{A}_{2}}}\therefore {{t}_{1}}=2{{t}_{2}}\]


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