Suppose there is 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?
Answers
Answer:
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,
Tp = 1/2Te = 1/2 year
Orbital radius of the planet = Rp
From Kepler’s third law of planetary motion, we can write:
(Rp /Re)^3 = (Tp/Te)^2
Rp /Re = (Tp/Te)^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.
Answer:
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, TP = ½Te = ½ year
Orbital radius of the planet = Rp
From Kepler’s third law of planetary motion, we can write:
(Rp / Re)3 = (Tp / Te)2
(Rp / Re) = (Tp / Te)2/3
= (½ / 1)2/3 = 0.52/3 = 0.63
Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.
Explanation:
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