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Question
Suppose there existed a planet that went around the sun twice as fast as the earth.What would be its orbital size as compared to that of the earth?
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Solution 1
Lesser by a factor of 0.63
Time taken by the Earth to complete one revolution around the Sun,
Te = 1 year
Orbital radius of the Earth in its orbit, Re = 1 AU
Time taken by the planet to complete one revolution around the Sun, `T_P = 1/2T_e = 1/2` year
Orbital radius of the planet = Rp
From Kepler’s third law of planetary motion, we can write:
`(R_p/R_e)^3 = (T_p/T_e)^2`
`(R_p/R_e) = (T_p/T_e)^(2/3)`
`=((1/2)/1)^(2/3) = (0.5)^(2/3) = 0.63`
Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.
Solution 2
Here, `T_e = 1 "year"`, `T_p = T_e/2 = 1/2` year
`r_c = 1 A.U.`
Using Kepler's third law,we have `r_p = r_c (T_p/T_e)^(2/3)`
`=>r_p = 1("1/2"/1)^(2/3)` = 0.63 AU
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