Since the diameter of the star suddenly shrinks to 0.350 times its present size, the new period of the star is 4.2 days
To find the new period, we need to know Kepler's third law of planetary motion
What is Kepler's third law of planetary motion?
Kepler's third law of planetary motion states that the square of the period of orbit of a planet, T equals the cube of its radius, R.
So, T² ∝ R³
and T₂²/T₁² = R₂³/R₁³ where
- T₁ = initial period of star = 20.30 days,
- T₂ = final period of star after shrinking,
- R₁ = initial radius of star and
- R₂ = final radius of star after shrinking
Given that the diameter suddenly shrinks to 0.350 times its present size and its radius is proportional to its diameter. So, its radius shrinks 0.350 times its initial size. Also, since there is a uniform mass distribution before and after, the mass remains the same.
So, R₂ = 0.350R₁
The new period of the star
Making T₂ subject of the formula, we have
T₂ = [√(R₂/R₁)³]T₁
Substituting the values of the variables into the equation, we have
T₂ = [√(R₂/R₁)³]T₁
T₂ = [√(0.350R₁/R₁)³]20.30 days
T₂ = [√(0.350)³]20.30 days
T₂ = [√0.042875]20.30 days
T₂ = [0.2071]20.30 days
T₂ = 4.2 days
So, the new period of the star is 4.2 days
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