Question

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 1

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 .

Step-by-step 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 .