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The ratio of accelerations due to gravity g_(1) : g_(2) on the surface of two planets is 5 : 2 and the ratio of their respective average densities rho_(1) : rho_(2) is 2 : 1. What is the ratio of respective escape velocities v_(1) : v_(2) from the surface of the planets ? |
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Answer» `5: 2` `v_(E )=sqrt((2GM)/(R ))=sqrt(2gR)""` ….(i) when g is GRAVITY, R is the radius of the planet, G is the gravitational constant and M is the mass of the planet. As mass`="volume" xx" density or " M=(4)/(3)piR^(3)rho` `:. M_(1)=(4)/(3)piR_(1)^(3)rho_(1)" and " M_(2)=(4)/(3)piR_(2)^(3)rho_(2)` Let `M_(1)` and `M_(2)` be the masses of the planets and `rho_(1)` and `rho_(2)` be the densities of the planets. As we know, `g=(GM)/(R^(2))` `:. (g_(1))/(g_(2))=(M_(1)R_(2)^(2))/(M_(2)R_(1)^(2))=(5)/(2) rArr ((4)/(3)piR_(1)^(3)xxR_(2)^(2)rho_(1))/((4)/(3)piR_(2)^(3)xxR_(1)^(2)rho_(2))=(5)/(2)` `[:. g_(1) : g_(2)=5 : 2]` `" or (R_(1))/(R_(2))=(5)/(2)xx(rho_(2))/(rho_(1))=(5)/(2)xx(1)/(2)=(5)/(4)[`:.` rho_(1) : rho_(2)=2 : 1]` Now, the ratio of respective escape velocities `v_(e_(1)) : v_(e_(2))=sqrt(g_(1)R_(1)):sqrt(g_(2)R_(2))=(5)/(2sqrt(2))""`[Using eqn.(i)] |
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