Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2011

  • question_answer
      The escape velocity of an object from the earth depends upon the mass of the earth \[\text{(}{{\text{M}}_{\text{e}}}\text{),}\] its mean density \[\text{( }\!\!\rho\!\!\text{ ),}\] its radius \[\text{(}{{\text{R}}_{e}}\text{),}\] and the gravitational constant \[\text{(G),}\] Thus, the formula for escape velocity is

    A) \[{{v}_{es}}={{R}_{e}}\sqrt{\frac{8\pi }{3}\text{G }\!\!\rho\!\!\text{ }}\]              

    B) \[{{v}_{es}}=M\sqrt{\frac{8\pi }{3}\text{G}{{\text{R}}_{e}}}\]

    C) \[{{v}_{es}}=\sqrt{2GM{{R}_{e}}}\]                        

    D) \[{{v}_{es}}=\sqrt{\frac{2GM}{R_{e}^{2}}}\]

    Correct Answer: A

    Solution :

    Escape velocity \[{{v}_{es}}=\sqrt{2g{{R}_{e}}}\] \[=\sqrt{2\frac{G{{M}_{e}}}{R_{e}^{2}}\times {{R}_{e}}}\]                           \[\left[ g=\frac{G{{M}_{e}}}{R_{e}^{2}} \right]\] \[=\sqrt{2\frac{G{{M}_{e}}}{{{R}_{e}}}}\][mass = volume \[\times \] density] \[=\sqrt{\frac{2G}{{{R}_{e}}}V\rho }\] \[=\sqrt{\frac{2G\times \frac{4}{3}\pi R_{e}^{3}\rho }{{{R}_{e}}}}=\sqrt{\frac{8}{3}\pi GR_{e}^{2}\rho }={{R}_{e}}\sqrt{\frac{8\pi }{3}G\rho }\]


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