Problems

#02

The amount of angular momentum $l$ depends on the product of mass $m$, angular speed $v$, and distance $r$, so $l=mvr$. The idea here is that the cloud mass does not change as it collapses. The angular momentum will also not change. So if the size $r$ gets smaller, then the angular speed $v$ will go up.

But we need to find the rotation period $P$. The rotation speed and period are related by $v=2\pi r/P$. The problem is about $r$ and $P$. We do not know anything about $v$, so we can get rid of $v$ in favor of $P$. The angular momentum $l$ becomes

$$l = mvr = m \frac{2\pi r^2}{P}$$

The left-hand side $l$ does not change, nor the mass $m$. The relation between size and period is such that $P$ will go up or down as the square of the size $r$.

Initially we have that $r=0.2$ LY, and $P=10^8$ yrs.
(a) If the new size is 100 AU, what is the new $P$? One LY has $9.5\times 10^{17}$ cm and 1 AU is $1.5\times 10^13$ cm, so $0.2 {\rm LY} = 0.2\times 9.5\times 10^{17}/1.5\times 10^13 1.3\times 10^4$ AU. Shrinking to 100 AU is a drop by 130. The change in period depends on the square of this, or about 17000, so the period drops (because the cloud spins up) to become about $P= 6000$ yrs.
(b) Another drop of 50 times in size compared to case (a), so a drop of $50^2=2500$ in period (because the period depends on the square of the size), becoming about $P=2.4$ yrs.

#12

This requires Kepler's 3rd Law. For AU, years, and solar masses, $a^3/P^2 = M$. We want the period, so $P=\sqrt{a^3/M}$. Using the values given, we have that $P=\sqrt{0.042^3/1.06} = 0.0084$ yrs. If the date were 2004 Dec 1, then it has been 6 yrs since the planet's discovery. The number of orbits is $6/0.0084=710$ times.