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The square of time period of revolution around the Sun is directly proportional to the _______ of the planet from the Sun. |
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Answer» Let M and R `-=` mass and radius of the Earth, G `-=` gravitional constant, `h -=` altitude of an orbit, `r = R + h -=` radius of the circular orbit, `v_(c )` and `T -=` critical orbitical speed and period of the satellite. `v_(c)=sqrt((GM)/(r))" "...(1)` In one period (T), the satellite travels a distance equal to the circumference of its circular orbit. `therefore T =("circumference")/("critical orbital speed")=(2pir)/(v_(c))` Substituting for `v_(c)` from Eq. (1), `T = (2pir)/(sqrt(GM//r))=2pisqrt((r^(3))/(GM))" "...(2)` `therefore T^(2)=((4pi^(2))/(GM))^(3)" "...(3)` Since `4pi^(2)//GM ` is constant for a givn planet, `T^(2) prop r^(3)" "...(4)` THus, the square of the period or revolution of a satellite is proportional to the cube of its orbital radius. Equation (2) shows that `T prop (1)/(sqrt(M))`. |
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