Q. 34.0( 41 Votes )
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?
Now planet revolves in orbit around sun due to attractive gravitational force between planet and sun, from Kepler’s third law of planetary motion we know that square of time period one complete rotation of planet T around the sun is proportional to the cube of mean distance (average distance) between the planet and sun R
Le the Time taken by the earth for one complete revolution be Te
Te = 1 Year
Let mean distance of the earth from the sun or orbital radius be Re
Since the planet is moving twice as fast as earth the time taken by the planet to complete one complete revolution is half of that taken by earth
Let the Time period of revolution of the planet be Tp, then we have
TP = 1/2Te = 1/2 year
Let, the orbital radius of this planet be RP
Now, according to the Kepler’s third law of planetary motion, we have
Using both equations we get the relation
Or we can say
Simplifying we get the relation for the orbital radius of the planet as
Since planet took half as much time as earth so we have
Tp/Te = 1/2
i.e. the radius of the planet is
Therefore, the radius of the orbit of this planet is 0.63 times the radius of the orbit of the Earth.
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