A man can jump vertically to a height of 1.5m on the earth. Calculate the radius of a planet of the same mean density as that of the earth from whose gravitational field he could escape by jumping. Radius of earth is 6.41×106m.
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Solution
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h=u22ge ∴u=√2geh (i) For the asked planet this u sbould be equal to the escape velocity from its surface. √2geh=√2gpRp or geh=gpRp GMeR2e.h=GMpR2pRp or (43πR3e)ρhR2e=(43πR3p)ρRpR2p or Rp=√Reh =√(6.41×106)(1.5) =3.1×103 m
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