Classical-Mechanics · Unit 7 · Video 2 · Interactive Practice

Gravity: The First Fundamental Force

IKey Formulas

Formula	Name	What you need
$\vec{F}_{1,2} = -\,G\,\dfrac{m_1 m_2}{r_{1,2}^{\,2}}\,\hat{r}_{1,2}$	Universal Law of Gravitation	Two masses, separation $r$
$G = 6.67\times10^{-11}\ \text{N}\cdot\text{m}^2\cdot\text{kg}^{-2}$	Gravitational constant	Same for every pair of masses
$\|\vec{F}\| = m g$	Near-surface gravity	Local $g$ (free-fall acceleration)
$g = 9.80616 - 0.025928\cos 2\varphi + 0.000069\cos^2 2\varphi - 3.086\times10^{-4} h$	Helmert's equation	Latitude $\varphi$ , elevation $h$

Key Insight: Gravitational mass (how strongly you respond to gravity) equals inertial mass (how strongly you resist acceleration). This Principle of Equivalence lets us write both as a single $m$ — and it is why every object falls with the same $g$ .

IIInverse-Square Falloff

The pull weakens as the inverse square of separation — how fast does moving the masses apart shrink it?

💡 Gravity has infinite range: even at $r = 10$ the pull is still $F_0/100$ — small, but never zero.

IIIEquivalence: Do Heavy Things Fall Faster?

A heavy ball feels more gravitational force — so does it fall faster than a light one in vacuum?

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💡 The heavy ball feels $4\times$ the force but has $4\times$ the inertia — $a = F/m = g$ cancels the mass. That equality of gravitational and inertial mass is the Principle of Equivalence.

IVHow $g$ Varies with Latitude

The value of $g$ is not one number — it climbs from the equator to the pole.

💡 Two effects make the pole stronger: the equatorial bulge sits about $21\ \text{km}$ farther from Earth's center, and rotation flings equatorial objects outward — both weakening $g$ where the planet is widest.

VQuiz Questions

Problem 1 · Inverse-Square Reasoning

Given: Two masses feel a gravitational force $F$ at separation $r$ . If the separation triples to $3r$ , what is the new force?

✅ Correct! Tripling

r

multiplies the force by

(1/3)^2 = 1/9

❌ Close, but square it. You divided by

3

, but the law is inverse-square — divide by

3^2 = 9

❌ Not quite. Force scales as

1/r^2

: multiply by

\left(r_{\text{old}}/r_{\text{new}}\right)^2

Show solution

The magnitude of the gravitational force is $F = G\dfrac{m_1 m_2}{r^2}$ . Replacing $r$ with $3r$ :

F' = G\frac{m_1 m_2}{(3r)^2} = G\frac{m_1 m_2}{9r^2} = \frac{F}{9}

Distance enters as $r^2$ , so tripling the distance cuts the force to one ninth, not one third.

Problem 2 · The Equivalence Pitfall

Given: An $8\ \text{kg}$ ball and a $2\ \text{kg}$ ball are released together in a vacuum near Earth's surface. Which one accelerates faster?

✅ Correct!

F = mg

is larger for the heavy ball, but

a = F/m = g

is identical for both.

❌ Close, but think about inertia. It feels

4\times

the force, yet also has

4\times

the inertia — so

a = F/m = g

for both.

❌ Not quite. In vacuum

a = F/m = (mg)/m = g

; the mass cancels.

Show solution

The gravitational force on a mass $m$ near the surface is $F = mg$ . Newton's second law gives the acceleration:

a = \frac{F}{m} = \frac{m g}{m} = g

The mass cancels, so acceleration is $g$ for every object. This only works because gravitational mass (in $F = mg$ ) equals inertial mass (in $F = ma$ ) — the Principle of Equivalence. Both balls stay level and land together.

Problem 3 · Reading Helmert's Equation

Given: In $g = 9.80616 - 0.025928\cos 2\varphi + \dots$ , the value $\cos 2\varphi = 1$ at the equator ( $\varphi = 0\degree$ ) and $\cos 2\varphi = -1$ at the pole ( $\varphi = 90\degree$ ). Where is $g$ larger, and by about how much?

✅ Correct! The

\cos 2\varphi

term flips from

-0.025928

+0.025928

; the gap is

2(0.025928) \approx 0.05

❌ Check the sign. The

\cos 2\varphi

term is subtracted, so

\cos 2\varphi = -1

at the pole makes

g

larger there.

❌ Not quite. Compare the term at both ends:

2 \times 0.025928 \approx 0.052

, pole minus equator.

Show solution

Keep only the $\cos 2\varphi$ term (the $\cos^2 2\varphi$ term is equal at both, since $\cos^2 = 1$ there):

g_{\text{pole}} - g_{\text{eq}} = \big[-0.025928(-1)\big] - \big[-0.025928(1)\big] = 2(0.025928) \approx 0.052\ \text{m/s}^2

Numerically $g_{\text{eq}} \approx 9.780$ and $g_{\text{pole}} \approx 9.832$ . The pole is stronger by about $0.05\ \text{m/s}^2$ , because subtracting a negative adds.

Problem 4 · From the Universal Law to $g$

Given: Model Earth as $M = 6.0\times10^{24}\ \text{kg}$ and a $1\ \text{kg}$ mass on its surface at $r = 6.4\times10^{6}\ \text{m}$ , with $G = 6.67\times10^{-11}$ . What gravitational force does Earth exert on the mass?

✅ Correct!

\dfrac{GMm}{r^2} \approx 9.8\ \text{N}

— exactly the weight

mg

of a

1\ \text{kg}

mass.

❌ Close, but square the distance. You divided by

r

, not

r^2

(6.4\times10^{6})^2 = 4.1\times10^{13}

❌ Not quite. Don't forget the

r^2

in the denominator;

GMm

alone is

4.0\times10^{14}

Show solution

Apply the Universal Law of Gravitation:

F = \frac{GMm}{r^2} = \frac{(6.67\times10^{-11})(6.0\times10^{24})(1)}{(6.4\times10^{6})^2}

= \frac{4.0\times10^{14}}{4.1\times10^{13}} \approx 9.8\ \text{N}

This matches the near-surface weight $mg = (1)(9.8) = 9.8\ \text{N}$ : the universal law reproduces the constant $g$ we use at Earth's surface.

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Gravity: The First Fundamental Force

IKey Formulas

IIInverse-Square Falloff

IIIEquivalence: Do Heavy Things Fall Faster?

IVHow ggg Varies with Latitude

VQuiz Questions

IVHow $g$ Varies with Latitude