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24 March, 16:02

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

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  1. 24 March, 16:19
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    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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