Classical-Mechanics · Unit 11 · Video 2 · Interactive Practice

Gravity Reaches the Moon: Deriving the Lunar Month from Newton's Law

IKey Formulas

Formula	Name	What it says
$\vec{F}_{1,2} = -G\dfrac{m_1 m_2}{r_{1,2}^2}\,\hat{r}_{1,2}$	Universal gravitation	Attractive, radial, falls off as $1/r^2$
$a_r = -\dfrac{4\pi^2 R_{e,m}}{T^2}$	Centripetal acceleration	Inward, set by orbit size and period
$T = \sqrt{\dfrac{4\pi^2 R_{e,m}^3}{G\,m_{\text{earth}}}}$	Orbital period	Gravity supplies the centripetal pull
$T^2 \propto R_{e,m}^3$	Kepler's Third Law	Falls out of the line above

Key Insight: The orbiting body's own mass $m_1$ sits on both sides of $-G\,m_1 m_2 / R^2 = -m_1\,4\pi^2 R / T^2$ , so it cancels — the period depends only on $R_{e,m}$ and the mass of the Earth. With $G = 6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$ , $m_{\text{earth}} = 5.98\times10^{24}\ \mathrm{kg}$ and $R_{e,m} = 3.82\times10^{8}\ \mathrm{m}$ , that formula returns $27.2$ days.

IISetting Gravity Equal to the Centripetal Requirement

Which quantity survives the cancellation, and which one never enters the period at all?

Step 1

IIISidereal Month vs Synodic Month

One full circle against the stars takes $27.2$ days — so why is a full moon $29.5$ days apart?

Elapsed days 0.0

💡 The Sun direction drifts about $\tfrac{1}{12}$ of a turn per month, so the extra sweep costs roughly $T/12 \approx 2.3$ days — the measured synodic month is $29.53$ days.

IVKepler's Third Law from Newton's Law

Move the Moon outward: does the period grow in step with $R$ , or faster?

Orbit radius R (×10⁸ m) 3.82

💡 Kepler extracted $T^2 \propto R^3$ from decades of observation; Newton derived the same law in a few lines from a single force law.

VQuiz Questions

Problem 1 · The Inverse-Square Falloff

Given: $F = G\dfrac{m_1 m_2}{r^2}$ in magnitude. If the Earth–Moon separation were tripled with both masses unchanged, the gravitational pull would be multiplied by:

✅ Correct!

F \propto 1/r^2

, so

r \to 3r

gives

F \to F/3^2 = F/9

❌ Wrong direction. Moving the bodies apart weakens gravity. The factor is

1/3^2 = \tfrac{1}{9}

, not

9

❌ Close. You divided by

r

, not by

r^2

— the falloff is inverse square, giving

\tfrac{1}{9}

❌ Not quite. The separation enters as

r^2

in the denominator: tripling

r

divides

F

9

Show solution

Hold both masses and $G$ fixed, so only the separation matters:

F = G\frac{m_1 m_2}{r^2} \;\Rightarrow\; F \propto \frac{1}{r^2}

Replacing $r$ by $3r$ :

\frac{F_{\text{new}}}{F_{\text{old}}} = \left(\frac{r}{3r}\right)^{2} = \frac{1}{9}

The pull drops to one ninth. (Doubling the separation would drop it to a quarter — the falloff quoted in the video.)

Problem 2 · Does the Orbiting Mass Matter?

Given: A spacecraft with twice the Moon's mass is placed in a circular orbit of exactly the same radius $R_{e,m} = 3.82\times10^{8}\ \mathrm{m}$ . Its orbital period is:

✅ Correct!

m_1

cancelled before the period was ever isolated, so

T

depends only on

R_{e,m}

and

m_{\text{earth}}

❌ Not quite. You kept the orbiting mass in the answer. It appears on both sides of

-G m_1 m_2/R^2 = -m_1 4\pi^2 R/T^2

and divides out completely.

❌ Not quite. Write out

T = \sqrt{4\pi^2 R^3/(G\,m_{\text{earth}})}

— the orbiting body's mass is nowhere in it.

Show solution

Step 1 — Newton's second law, radial component. With $m_1$ the orbiting mass and $m_2 = m_{\text{earth}}$ :

-G\frac{m_1 m_2}{R^2} = m_1\left(-\frac{4\pi^2 R}{T^2}\right)

Step 2 — Cancel $m_1$ . It multiplies both sides, so it divides out:

G\frac{m_2}{R^2} = \frac{4\pi^2 R}{T^2} \;\Longrightarrow\; T = \sqrt{\frac{4\pi^2 R^3}{G\,m_2}}

Only $R$ and $m_{\text{earth}}$ survive. Doubling the spacecraft's mass doubles the gravitational pull and doubles the inertia it must overcome — the two effects exactly offset, so the period stays 27.2 days.

Problem 3 · Scaling the Orbit Up

Given: A hypothetical moon circles the Earth at four times the Earth–Moon distance, $R' = 4R_{e,m}$ . Use $T^2 = \dfrac{4\pi^2 R^3}{G\,m_{\text{earth}}}$ and the known period $T = 27.2$ days.

By what factor is $T^2$ multiplied?

So the new period is about

✅ Correct!

T'^2 = 4^3 T^2 = 64\,T^2

, so

T' = 8T = 8(27.2) \approx 218

days.

❌ Check the cube.

T^2 \propto R^3

, so scaling

R

4

scales

T^2

4^3 = 64

— not by

4

4^2

❌ You stopped at $T^2$ . The factor

64

applies to

T^2

; take the square root,

\sqrt{64} = 8

, before multiplying

27.2

days.

❌ Recompute.

T' = \sqrt{64}\,T = 8 \times 27.2

days.

Show solution

Step 1 — Only $R$ changes, since $4\pi^2$ , $G$ and $m_{\text{earth}}$ are all fixed:

\frac{T'^2}{T^2} = \frac{R'^3}{R^3} = \left(\frac{4R}{R}\right)^{3} = 4^3 = 64

Step 2 — Take the square root:

T' = \sqrt{64}\;T = 8T = 8 \times 27.2\ \text{days} = 217.6 \approx 218\ \text{days}

Common slips: using $4$ or $4^2$ for the cube ( $109$ or $435$ days), or reporting $64 \times 27.2 \approx 1740$ days by forgetting the square root.

Problem 4 · A Faster Year (Transfer)

Given: Imagine the Earth raced around the Sun in a little under half the time it actually does — a year of exactly 6 sidereal months ( $163.2$ days) — while the Moon's orbit stayed the same ( $T = 27.2$ days). The time from one full moon to the next would be about:

✅ Correct! The Moon has to gain a full turn on a Sun direction that now drifts

\tfrac{1}{6}

of a turn per month, so the rates subtract:

\tfrac{1}{T_{\text{syn}}} = \tfrac{1}{27.2} - \tfrac{1}{163.2}

, giving

32.6

days.

❌ That is the sidereal month. After one full circle against the stars the Sun has moved on, so the Moon must sweep further to sit opposite it again.

❌ The gap depends on the Earth's year.

29.5

days is the real synodic month, set by the real

365

-day year. A shorter year makes the Sun direction drift faster, so the gap widens.

❌ Wrong direction. The synodic month is always longer than the sidereal one, because the Moon must chase a Sun direction that has moved ahead.

❌ Not quite. Full moon recurs when the Moon gains one whole turn on the Sun direction, so the angular rates subtract:

1/T_{\text{syn}} = 1/T - 1/T_{\text{year}}

Show solution

Step 1 — How far does the Sun direction drift in one sidereal month? The year is exactly 6 sidereal months, so in one month the direction to the Sun rotates by $\tfrac{1}{6}$ of a turn.

Step 2 — Full moon recurs when the Moon gains a whole turn on that direction. Angular rates subtract, so the reciprocals do too:

\frac{1}{T_{\text{syn}}} = \frac{1}{T} - \frac{1}{T_{\text{year}}} = \frac{1}{27.2} - \frac{1}{163.2}

Step 3 — Solve:

T_{\text{syn}} = \frac{T\,T_{\text{year}}}{T_{\text{year}} - T} = \frac{27.2 \times 163.2}{163.2 - 27.2} = \frac{4439.04}{136} = 32.64\ \text{days}

The quick estimate $T + T/6 \approx 31.7$ days undershoots: the Sun keeps drifting during the extra $4.5$ days of chase, so the Moon must sweep further still. For the real Earth the same exact relation gives $29.5$ days — a faster year always widens the sidereal–synodic gap.

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