Manipal Medical Manipal Medical Solved Paper-2006

  • question_answer
    The eccentricity of earths orbit is 0.0167. The ratio of its maximum speed in its orbit to its minimum speed is:

    A)  2.507           

    B)  1.0339

    C)  8.324           

    D)  1.000

    Correct Answer: B

    Solution :

     Eccentricity, \[e=\frac{{{r}_{\max }}-{{r}_{\min }}}{{{r}_{\max }}-{{r}_{\min }}}=0.0167\] Therefore, \[\frac{{{r}_{\max }}}{{{r}_{\min }}}=\frac{1+e}{1-e}\] \[=\frac{1-0.0167}{1+0.0167}\] As angular momentum remains constant, hence \[{{r}_{\max }}\times v{{v}_{\min }}={{r}_{\min }}\times {{v}_{\max }}\] \[\therefore \] \[\frac{{{v}_{\max }}}{{{v}_{\min }}}=\frac{{{r}_{\max }}}{{{r}_{\min }}}=\frac{1+0.0167}{1-0.0167}\] \[=1.0339\]


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