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JEE Main & Advanced Sample Paper JEE Main Sample Paper-41

  • question_answer
    The gravitational potential energy of a body of mass m at the earth's surface \[-mg{{\operatorname{R}}_{e}}.\]Its gravitational potential energy at a height \[{{\operatorname{R}}_{e}}\]from earth's surface will be (Here, \[{{\operatorname{R}}_{e}}\] is the radius of the earth)

    A)  \[-2mg{{\operatorname{R}}_{e}}\]            

    B)  \[2mg{{\operatorname{R}}_{e}}\]

    C)  \[\frac{1}{2}mg{{\operatorname{R}}_{e}}\]                     

    D)  \[-\frac{1}{2}mg{{\operatorname{R}}_{e}}\]

    Correct Answer: D

    Solution :

    \[\Delta U={{U}_{2}}-{{U}_{1}}\] \[=\frac{mgh}{1+\frac{h}{{{\operatorname{R}}_{e}}}}=\frac{mg{{\operatorname{R}}_{e}}}{1+\frac{{{\operatorname{R}}_{e}}}{{{\operatorname{R}}_{e}}}}=\frac{mg{{\operatorname{R}}_{e}}}{2}\] \[\therefore \]\[{{U}_{2}}=\Delta U+{{U}_{1}}\]             \[{{U}_{2}}=\frac{mg{{\operatorname{R}}_{e}}}{2}-mg{{\operatorname{R}}_{e}}\] \[\Rightarrow \]            \[{{U}_{2}}=-\frac{1}{2}mg{{\operatorname{R}}_{e}}\]


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