Tools of the Trade

Section 2 of 5

The Ratio Method

2.2 Tool 2: The Ratio Method

Escaping the “Big Number” Trap

The mass of the Sun is $M_\odot = 2 \times 10^{33}$ g; the mass of Earth is $M_\oplus = 6 \times 10^{27}$ g. What can we do with two numbers like these?

Subtraction fails. $M_\odot - M_\oplus = 1.999\ldots \times 10^{33}$ g — still just a giant number that conveys no intuition.

Division works.

$\frac{M_\odot}{M_\oplus} = \frac{2 \times 10^{33}}{6 \times 10^{27}} = \frac{2}{6} \times 10^{6} \approx 333{,}000$

The Sun is about 333,000 times more massive than Earth. That tells you something.

Ratio method

Comparing quantities by division rather than subtraction, so that shared constants cancel and only the physically meaningful scaling survives.

The Cancellation Trick

Most physical laws are proportionalities, $A = k\,B^n$ , where $k$ is some constant — often hiding a $\pi$ , a $G$ , or both. Compare two systems and the constant disappears:

$\frac{A_2}{A_1} = \frac{k\,B_2^{\,n}}{k\,B_1^{\,n}} = \left(\frac{B_2}{B_1}\right)^{n}$

The ratio method means you never need $G$ , $\pi$ , or any other constant numerically. You only need the scaling — the exponent $n$ .

Worked Example 3How Many Earths Fit Inside the Sun?

Problem

The Sun’s radius is 109 times Earth’s, $R_\odot / R_\oplus = 109$ . Volume goes as $V = \tfrac{4}{3}\pi R^3$ . How many Earths fit inside the Sun?

StepTake the ratio

Volume scales as $R^3$ , so the $\tfrac{4}{3}\pi$ cancels: $V_\odot / V_\oplus = (R_\odot / R_\oplus)^3 = 109^3$ .

Dimensional check

A ratio of like quantities is dimensionless: $[R^3]/[R^3] = [L^3]/[L^3] = 1$ . The answer is a pure count, as it must be ✓.

Result

$109^3 \approx 1.3 \times 10^6$ — over a million Earths. We never computed a single volume in cm³; the ratio was enough.

Scaling Intuition

The exponent in a scaling law is the physical story:

Scaling	Physical Meaning	Example
$A \propto B$	Linear	Schwarzschild radius vs. mass
$A \propto B^2$	Area scaling	Surface area, cross-section
$A \propto B^3$	Volume scaling	Mass (at fixed density)
$A \propto B^{-2}$	Inverse-square	Gravity, light intensity
$A \propto B^{3/2}$	Kepler scaling	Orbital period vs. radius

Scaling relationship

A proportionality showing how one quantity depends on another, $A \propto B^n$ . The exponent $n$ carries the physics: when $n > 1$ , small changes in the input drive large changes in the output.

When $n > 1$ , small changes in input produce big changes in output — the heart of why a scaling relationship is so predictive.

Applying Ratios: Mars’s Orbital Period

Kepler’s Third Law, written as a ratio against Earth as the reference, sheds every constant:

$\left(\frac{P}{1\,\text{yr}}\right)^2 = \left(\frac{a}{1\,\text{AU}}\right)^3$

Mars orbits at $a = 1.52$ AU. Substituting, $(P/1\,\text{yr})^2 = (1.52)^3 \approx 3.5$ , so $P/1\,\text{yr} = \sqrt{3.5} \approx 1.87$ . We predict $P_\text{Mars} \approx 1.9$ years; the measured value is 1.88 years — excellent agreement, with no $G$ and no heavy calculation.