Irodov 1.204: Scaling the Solar System Keeps Time – Amit & Deepali Sinha Foundation

Problem Irodov 1.204

A model of the Solar System is built at a fixed linear scale factor $\eta$ (every length, including every body’s own size and every orbital distance, shrinks by the factor $\eta$ ), using the same materials, and therefore the same density, as the real Sun and each real planet. Determine how the orbital period of a planet in the model compares with the orbital period of the real planet it represents.

The actual system (black) and a model scaled down by η, built from the same density (gray). Every length shrinks by η; the question is what that does to T. — The actual system (black) and a model scaled down by $\eta$ , built from the same density (gray). Every length shrinks by $\eta$ ; the question is what that does to $T$ .

Given

Linear scale factor $\eta$
density unchanged for every body

Find

How does the orbital period

T

change?

Solution

Step 1Write Kepler’s third law for the two-body system

For a planet of mass $m$ orbiting a star of mass $M$ with semi-major axis $a$ , the exact two-body form of Kepler’s third law is

T^2 = \frac{4\pi^2 a^3}{G(M+m)}. \tag{1}

Step 2Find how mass scales under the model’s construction

Every linear dimension of the model, including the size of each body and every orbital distance, is $\eta$ times the actual one; since the density is kept the same, a body’s mass, proportional to density times volume, scales as the cube of its linear size,

M' = \eta^3 M, \qquad m' = \eta^3 m, \qquad a' = \eta a. \tag{2}

Step 3Substitute the scaled quantities into Kepler’s law

\begin{align*} T'^2 &= \frac{4\pi^2 (a')^3}{G(M'+m')} = \frac{4\pi^2 (\eta a)^3}{G\,\eta^3(M+m)} \\ &= \frac{\eta^3}{\eta^3}\cdot\frac{4\pi^2 a^3}{G(M+m)} = T^2. \tag{3}\end{align*}

T' = T

(4)

Check

The same cancellation shows up in the orbital speed: since $v = \sqrt{G(M+m)/a}$ , the model’s speed is $v' = \sqrt{G\eta^3(M+m)/(\eta a)} = \eta\, v$ , a factor $\eta$ slower. The model’s orbit is also a factor $\eta$ shorter, so the ratio of path length to speed, the period, is unchanged either way it is computed.

Remark

The period depends on the body only through its mean density, not through its size: doubling every length while keeping the density fixed leaves $T$ untouched. A model built from a lighter material, rather than a smaller one, would change the period; a model merely scaled down in size, at the same density, would not.