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?
Answers
Answer
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.
Answer:
Let Ms =mass of the sun
Me = mass of the earth
r = radius of orbit of the earth around the sun .
r' = radius of planet around the sun.
w = angular speed of the earth
w' = angular speed of the planet
A/c to question,
w' = 2w { planet around the sun twice fast as earth }
Centripital force balance the Gravitational force in orbital motion.
MeV²/r = GMsMe/r²
V² = GMs/r
We know,
V = wr [use this here,
(wr)² = GMs/r
w² = GMs/r³ -----------(1)
Similarly for planet ,
w'² = GMs/r'³ ---------(2)
w²/w'² = (r'/r)³
r'/r = (w/w')^(2/3)
r' = (w/2w)^(2/3) r
= (1/4)⅓ r
= 0.63r hence, size of planet' s orbital is smaller then the orbital size of the earth .
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