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?
Solution and Explanation
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, \(T_p = \frac{1}{2} T_e = \frac{1}{2}\) year
Orbital radius of the planet = Rp
From Kepler’s third law of planetary motion, we can write:
\((\frac{R_p}{R_e})^3 = (\frac{T_p }{ T_e})^2\)
\(\frac{R_p}{R_e} = (\frac{T_p}{T_e})^{\frac{2}{3}}\)
\(=(\frac{\frac{1}{2}}{1} )^{\frac{2}{3}}= (0.5)^{\frac{2}{3} }= 0.63\)
Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.
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Concepts Used:
Newtons Law of Gravitation
Gravitational Force
Gravitational force is a central force that depends only on the position of the test mass from the source mass and always acts along the line joining the centers of the two masses.
Newton’s Law of Gravitation:
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- Directly proportional to the product of their masses i.e. F ∝ (M1M2) . . . . (1)
- Inversely proportional to the square of the distance between their center i.e. (F ∝ 1/r2) . . . . (2)
By combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2 [f(r)is a variable, Non-contact, and conservative force]