Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital says as a compared to that of Earth.
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see the image you will understand
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Hii dear,
◆ Answer-
R' = 0.63 R
◆ Explaination-
Consider T and T' be the period of revolution earth and other planet respectively. Also R and R' be their respective orbital radius.
By Kepler's law of periods,
(T'/T)^2 = (R'/R)^3
We are given 2T' = T
0.5^2 = (R'/R)^3
R'/R = (0.25)^(1/3)
R'/R = 0.63
R' = 0.63 R
Thus we can say that orbital radius of the planet will be 0.63 times that of orbital radius of earth.
Hope that is useful...
◆ Answer-
R' = 0.63 R
◆ Explaination-
Consider T and T' be the period of revolution earth and other planet respectively. Also R and R' be their respective orbital radius.
By Kepler's law of periods,
(T'/T)^2 = (R'/R)^3
We are given 2T' = T
0.5^2 = (R'/R)^3
R'/R = (0.25)^(1/3)
R'/R = 0.63
R' = 0.63 R
Thus we can say that orbital radius of the planet will be 0.63 times that of orbital radius of earth.
Hope that is useful...
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