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?

Open in App

Solution

Answer: Lesser by a factor of 0.63

Time taken by the Earth to complete one revolution around the Sun,

T_e = 1 year

Orbital radius of the Earth in its orbit, R_e= 1 AU

Time taken by the planet to complete one revolution around the Sun,

Orbital radius of the planet = R_p

From Kepler’s third law of planetary motion, we can write:

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.

Suggest Corrections

Similar questions

Q. 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 ?

Q. Planet X has twice the mass as the earth and orbits its star as the same orbital distance that the earth orbits the sun. If Planet X takes 365 and

1 / 4

days to complete one orbit around its star, how does the angular momentum of Planet X compare to the angular momentum of the earth?
It can be assumed that both orbits are circular.

Q. If a new planet is discovered rotating around the sun with its orbital radius double that of earth, then what will be its time period?
(In earth days)

Q. A planet has twice the radius but the mean density is

\frac{1}{4} t h

as compared to earth. What is the ratio of escape velocity from to that from the planet?

Suppose a small planet is discovered that is 14 times far from the sun as the Earth. The distance of earth from the sun $(1.5 \times 10^{11} m)$ . Use Kepler's law of harmonies to predict the orbital period of such a planet. GIVEN: $\frac{T^{2}}{R^{3}} = 2.97 \times 10^{- 19} s^{2} / m^{3}$

Join BYJU'S Learning Program