Final answer:
The period of rotation after the collapse is approximately 9.87 x 10⁷ seconds. The work done by gravity during the collapse is approximately -2.67 x 10⁴⁸ J. The speed of an indestructible person standing on the equator of the collapsed star is approximately 2.10 x 10⁻² m/s.
Step-by-step explanation:
(a) To find the period of rotation after the collapse, we can use the conservation of angular momentum. The angular momentum L of a rotating object is given by the product of its moment of inertia I and angular velocity ω:
L=Iω
For a star undergoing a collapse, the moment of inertia is conserved, so:
I1ω1 = I2ω2
Where I1 and I2 are the moments of inertia before and after the collapse, and ω1 and ω2 are the angular velocities before and after the collapse.
The moment of inertia for a sphere (assumed for simplicity) is given by:
I= 2/5 mr2
where m is the mass and r is the radius.
Before the collapse (initial state), w2 is the final angular velocity.