Question

In a science fiction story, a mysterious star has mass M and a period of rotation T. Suddenly, the diameter shrinks by a fraction f. What would the period be after this change?

M = 1.76x10³¹ kg

T = 35.20 days

f = 0.30

Answer 1

The period of rotation of the star after its diameter shrinks by a fraction f is inversely proportional to the square of the diameter. Given that the diameter shrinks by 30%, the new period of rotation would be approximately 71.84 days.

Step-by-step explanation:

We are tasked with determining the new period of rotation of a mysterious star that shrinks in diameter by a fraction f. We know that the period of rotation is inversely proportional to the square of the diameter (D2), given that the mass M remains constant. The initial period T is 35.20 days, and the star shrinks in diameter by a fraction f of 0.30.

Using the relationship Pfinal = Pinitial × (Dinitial / Dfinal)2 and since the diameter shrinks by a fraction f, we can say Dfinal = (1 - f) × Dinitial. Substituting the values, we get Pfinal = T × (1 / (1 - f))2. Calculating with the given numbers, Pfinal = 35.20 days × (1 / (1 - 0.30))2.

After performing the calculation:

1. 1 - f = 1 - 0.30 = 0.70
2. (1 / 0.70)2 = (1 / 0.49)
3. Pfinal = 35.20 days × 1 / 0.49
4. Pfinal ≈ 71.84 days

So, after the diameter shrinks, the new period of the star would be approximately 71.84 days.