Question

Suppose there is a planet existed that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

Answer 1

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 .

another 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.

hope it helps

Answer 2

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Refer the attachment above ..

Hope it Helps !!