suppose there exists a planet that went around the sun twice as fast as the earth.what would be its orbital size as compare to earth
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When you say orbital size I'm assuming the area of that orbit assuming it's almost circular
I shall also give you the radius
Since you haven't mentioned anything about the mass of the planet I'm going to assume it's same as earth's
from the above pic we can see that the radius of that planet is 1/4th of earth's radius and it's orbits area is 1/16th of earth's initial radius
I shall also give you the radius
Since you haven't mentioned anything about the mass of the planet I'm going to assume it's same as earth's
from the above pic we can see that the radius of that planet is 1/4th of earth's radius and it's orbits area is 1/16th of earth's initial radius
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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.
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