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 .
