<p>This example has less to do with scaling relations, and more to do with dimensional analysis - theoretical physicists like to express equations in terms of dimensionless ratios, because then it is really easy to plug in numbers and get a sense of how the equation behaves for certain parameter values.</p>
<p>In this case, you have <a href="http://astro.physics.uiowa.edu/ITU/glossary/keplers-third-law/" rel="nofollow noreferrer">Kepler's third law</a>, which is a well-known result of the force of gravity and the laws of motion. To derive this form of Kepler's third law in dimensionless ratios, try to write the expression in terms of ratios that you want given the units that are relevant for the regime that you're interested in.</p>
<p>Kepler's third law states that the total mass of a binary system is</p>
<p><span class="math-container">$$M = m_1 + m_2 = \frac{4\pi}{G} \frac{a^3}{T^2} $$</span></p>
<p>where <span class="math-container">$G$</span> is the gravitational constant, <span class="math-container">$a$</span> is the semimajor axis and <span class="math-container">$T$</span> is the period. Since the RHS of the above equation is equal to a quantity of mass, it must have dimensions of mass (otherwise we did algebra incorrectly), but we don't have units yet as these are all variables and known constants. To get units into the equation, multiply by 1 a couple times, and rearrange:</p>
<p><span class="math-container">$$M = m_1 + m_2 = \frac{4\pi}{G} \bigg(\frac{a(1~AU)}{1~AU}\bigg)^3 \bigg(\frac{1yr}{T(1yr)}\bigg)^2$$</span></p>
<p><span class="math-container">$$M = m_1 + m_2 = \frac{4\pi}{G} \bigg(\frac{1~AU^3}{1yr^2}\bigg) \bigg(\frac{a}{1~AU}\bigg)^3 \bigg(\frac{1yr}{T}\bigg)^2$$</span></p>
<p>Now we need to finish simplifying by substituting for <span class="math-container">$G$</span> in terms of the units that we already chose. The dimension of <span class="math-container">$G$</span> is: <span class="math-container">$[G] = [length]^3/[mass][time]^2$</span>. You'll notice that <span class="math-container">$G$</span>, by definition, has the exact dimensions that will cancel out the lengths and times to leave behind mass (this is a good way to check that you've done things correctly, especially when doing this for more complicated formulas). To put <span class="math-container">$G$</span> in terms of <span class="math-container">$~AU$</span> and <span class="math-container">$yrs$</span>, carefully <a href="https://www.wolframalpha.com/input/?i=%286.6e-11%29*%282e30%29*%28pi%20*10%5E7%29%5E2%2F%281.5e11%29%5E3" rel="nofollow noreferrer">do the same thing to get</a>:</p>
<p><span class="math-container">$$G = 6.6\times10^{-11} \bigg( \frac{m^3}{kg~s^2} \bigg) \bigg(\frac{1~AU}{1.5\times10^{11}m}\bigg)^3 \bigg(\frac{2\times10^{30} kg}{1M_{\odot}}\bigg) \bigg(\frac{\pi\times10^7 s}{1 yr}\bigg)^2 = 38.6 \frac{~AU^3}{M_{\odot}yr^2}$$</span></p>
<p>Thus,</p>
<p><span class="math-container">$$M = m_1 + m_2 = \frac{4\pi}{G} \bigg(\frac{1~AU^3}{1yr^2}\bigg) \bigg(\frac{a}{1~AU}\bigg)^3 \bigg(\frac{1yr}{T}\bigg)^2 \approx \frac{1}{3} \bigg(\frac{a}{1~AU}\bigg)^3 \bigg(\frac{1yr}{T}\bigg)^2 M_{\odot}$$</span></p>
<p>since <span class="math-container">$4 \pi/38.6 \approx 1/3$</span>.</p>
<p>So, the equation in your image should be an approximately equal to sign, not an equal sign, since they've approximated the value of the numerical prefactor as i did, and especially since they just dropped the numerical prefactor as it is of order unity.</p>
<p>Once you get the hang of this process, it's kind of fun and is very helpful. :)</p>

CC BY-SA 4.0
2021-12-25T22:26:34.927Z
2021-12-25T22:32:17.763Z
2021-12-25T22:32:17.763Z
Daddy Kropotkin
<p>In this picture , Keplers third law is written in a scaling relationship with solar mass.</p>
<p><span class="math-container">$$a^3=\frac{G(m_1+m_2)}{4\pi^2}P^2 \rightarrow m_1+m_2 = \frac{4\pi^2a^3}{GP^2}$$</span></p>
<p><span class="math-container">$$m_1+m_2=(1 M_{\odot})\, \left( \frac{a}{1\ \text{AU}}\right)^3\,
\left(\frac{P}{1\ \text{year}}\right)^{-2}$$</span></p>
<p><a href="https://i.sstatic.net/DRWph.png" rel="nofollow noreferrer">original screenshot</a></p>
<p>I'm struggling in these scaling relationships.How that worked?</p>

<mathematics><keplers-laws>
Scaling Relations
Particle king
# Scaling Relations

(Asked by: username_0 on 2021-12-25)

In this picture , Keplers third law is written in a scaling relationship with solar mass.

$$a^3=\frac{G(m\_1+m\_2)}{4\pi^2}P^2 \rightarrow m\_1+m\_2 = \frac{4\pi^2a^3}{GP^2}$$

$$m\_1+m\_2=(1 M\_{\odot})\, \left( \frac{a}{1\ \text{AU}}\right)^3\,
\left(\frac{P}{1\ \text{year}}\right)^{-2}$$

[original screenshot](https://i.sstatic.net/DRWph.png)

I'm struggling in these scaling relationships.How that worked?

**Tags:** mathematics, keplers-laws


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## 1 answers

### Answer by username_1 on 2021-12-25. Score: 4 | Accepted answer

This example has less to do with scaling relations, and more to do with dimensional analysis - theoretical physicists like to express equations in terms of dimensionless ratios, because then it is really easy to plug in numbers and get a sense of how the equation behaves for certain parameter values.

In this case, you have [Kepler's third law](http://astro.physics.uiowa.edu/ITU/glossary/keplers-third-law/), which is a well-known result of the force of gravity and the laws of motion. To derive this form of Kepler's third law in dimensionless ratios, try to write the expression in terms of ratios that you want given the units that are relevant for the regime that you're interested in.

Kepler's third law states that the total mass of a binary system is

$$M = m\_1 + m\_2 = \frac{4\pi}{G} \frac{a^3}{T^2} $$

where $G$ is the gravitational constant, $a$ is the semimajor axis and $T$ is the period. Since the RHS of the above equation is equal to a quantity of mass, it must have dimensions of mass (otherwise we did algebra incorrectly), but we don't have units yet as these are all variables and known constants. To get units into the equation, multiply by 1 a couple times, and rearrange:

$$M = m\_1 + m\_2 = \frac{4\pi}{G} \bigg(\frac{a(1~AU)}{1~AU}\bigg)^3 \bigg(\frac{1yr}{T(1yr)}\bigg)^2$$

$$M = m\_1 + m\_2 = \frac{4\pi}{G} \bigg(\frac{1~AU^3}{1yr^2}\bigg) \bigg(\frac{a}{1~AU}\bigg)^3 \bigg(\frac{1yr}{T}\bigg)^2$$

Now we need to finish simplifying by substituting for $G$ in terms of the units that we already chose. The dimension of $G$ is: $[G] = [length]^3/[mass][time]^2$. You'll notice that $G$, by definition, has the exact dimensions that will cancel out the lengths and times to leave behind mass (this is a good way to check that you've done things correctly, especially when doing this for more complicated formulas). To put $G$ in terms of $~AU$ and $yrs$, carefully [do the same thing to get](https://www.wolframalpha.com/input/?i=%286.6e-11%29*%282e30%29*%28pi%20*10%5E7%29%5E2%2F%281.5e11%29%5E3):

$$G = 6.6\times10^{-11} \bigg( \frac{m^3}{kg~s^2} \bigg) \bigg(\frac{1~AU}{1.5\times10^{11}m}\bigg)^3 \bigg(\frac{2\times10^{30} kg}{1M\_{\odot}}\bigg) \bigg(\frac{\pi\times10^7 s}{1 yr}\bigg)^2 = 38.6 \frac{~AU^3}{M\_{\odot}yr^2}$$

Thus,

$$M = m\_1 + m\_2 = \frac{4\pi}{G} \bigg(\frac{1~AU^3}{1yr^2}\bigg) \bigg(\frac{a}{1~AU}\bigg)^3 \bigg(\frac{1yr}{T}\bigg)^2 \approx \frac{1}{3} \bigg(\frac{a}{1~AU}\bigg)^3 \bigg(\frac{1yr}{T}\bigg)^2 M\_{\odot}$$

since $4 \pi/38.6 \approx 1/3$.

So, the equation in your image should be an approximately equal to sign, not an equal sign, since they've approximated the value of the numerical prefactor as i did, and especially since they just dropped the numerical prefactor as it is of order unity.

Once you get the hang of this process, it's kind of fun and is very helpful. :)


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