Squiggle Math III, The Time Conundrum

We are bounded in a nutshell of Infinite Space: Week 11: Worksheet #20: Problem #2: Squiggle Math III, The Time Conundrum

2.

a. Time for some more squiggle math!! So from the previous problem on this worksheet relating temperature, luminosity, and the distance between the star and the object, we can rewrite it in terms of its variables as: $$T \sim L_\star^{1/4} a^{-1/2},$$ but since the temperature we can consider it to be constant for the purposes of this problem, we find that:   $$a^{1/2} \sim L_\star^{1/4}$$

b. We can take the previous result relating the temperature to the distance, and find the relationship is also: $$a^2 \sim L .$$ Also, we know from a previous worksheet we did a couple of months ago that luminosity and mass are inherently linked in a star, which can be described with its proportion as:   $$L \sim\ M^4,$$ and by plugging these in, we find that:   $$a^2 \sim M^4$$ $$a \sim M^2,$$ which means $$a_{HZ} \sim M_\star^2 ,$$ where the looked for variable becomes: $$\alpha = 2$$

c.  Furthermore, we can use these proportions to establish how Kepler’s third law works according to squiggle math: $$P^2 = \frac{4\pi^2 a^3}{G M_\star},$$ becomes $$P^2 \sim \frac{a^3}{M_\star} ,$$ and according to our previous part, we can substitute in and find that: $$P^2 \sim \frac{M_\star^{2^{3}} }{M_\star},$$ which means $$P^2 \sim M_\star^5$$ $$P \sim M^{5/2}.$$

Now, we want ot find the period of the Earth if the Sun were half its current mass, such that: $$P \sim \left(\frac{1}{2} M_\star \right)^{5/2} ,$$ turns into   $$P \sim {\frac{1}{2}}^\frac{5}{2} M_\star^{5/2}$$ $$\sqrt{32} P \sim M_\star^{5/2},$$ which means the new Earth period is related to the original period by:   $$\frac{1}{\sqrt{32}} P_\oplus = P_{\frac{1}{2} M_\star },$$ and thus, the Period of rotation becomes $$P_{\frac{1}{2} M_\star} = \frac{365}{\sqrt{32} } = 64 ~days,$$ our new definition of a year.