An asteroid is in circular orbit around the Sun. Its distance from the Sun is 7.3 times the average distance of Earth from the Sun. The period of this asteroid is A : 3.8 Earth years B : 144 Earth years C : 20 Earth years D : 7.3 Earth years E : 53 Earth years

Answer :

Answer:

An asteroid is in circular orbit around the Sun. Its distance from the Sun is 7.3 times the average distance of Earth from the Sun. The period of this asteroid is  C : 20 Earth years .

Explanation:

Kepler's law

T² ∝ R³

T = orbital period of the planet

R = semi major axis of the orbit.

Now

( T₁ / T₂ )² =  ( R₁ / R₂ )³

T₁ =  Earth time period = 1 year

T₂ = Asteroid time period

R₁ =  The distance between Earth and Sun

R₂ = The distance between Asteroid and Sun

From the question ,

R₂ = 7.3 R₁

applying on the above equation ,

( T₁ / T₂ )² =  ( R₁ / R₂ )³

( 1 / T₂ )² = ( R₁ / 7.3 R₁ )³

( 1 / T₂ )² = ( 1 / 7.3 )³

T₂ = 19.7 year

Hence ,

approximately T₂ = 20 years.

The period of this asteroid is  C : 20 Earth years .

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