Article 3 — Earth mass, the value of g, and its variation with altitude

From Articles 1 and 2, we now have both pieces of the puzzle:

• The law (Newton): F = GMm/r²
• The number (Cavendish): G ≈ 6.67 × 10⁻¹¹ N·m²/kg²

Add two classical measurements (g from Galileo, R_E from Eratosthenes), and everything about Earth's gravity falls out — its mass, its density, the numerical origin of g, and its variation with altitude. This article is the harvest.

Step 1 — the mass of the Earth

For a mass m at Earth's surface (distance R_E from center), the universal law gives:

\[ F = G\,\frac{M_E\, m}{R_E^2} \]

Locally, this is the object's weight:

\[ W = m g \]

where g is the acceleration of free fall. Equating the two:

\[ m g = G\,\frac{M_E\, m}{R_E^2} \]

m cancels (which is itself a deep point: gravitational mass equals inertial mass — the equivalence principle). Solving for M_E:

\[ \boxed{M_E = \frac{g R_E^2}{G}} \]

Numbers:

• g = 9.80 m/s²
• R_E = 6.371 × 10⁶ m
• G = 6.674 × 10⁻¹¹ N·m²/kg²

\[ M_E = \frac{9.80 \cdot (6.371 \times 10^6)^2}{6.674 \times 10^{-11}} = \frac{9.80 \cdot 4.06 \times 10^{13}}{6.674 \times 10^{-11}} \approx 5.97 \times 10^{24}\ \mathrm{kg} \]

Earth's mass ≈ 6 × 10²⁴ kg. Computed — without ever putting the Earth on any scale.

Step 2 — Earth's density and its geological implications

Mass divided by Earth's volume:

\[ \rho_E = \frac{M_E}{(4/3)\pi R_E^3} = \frac{5.97 \times 10^{24}}{(4/3)\pi (6.371 \times 10^6)^3} \]

\[ \rho_E \approx \frac{5.97 \times 10^{24}}{1.083 \times 10^{21}} \approx 5510\ \mathrm{kg/m^3} \]

That is ~5.5× the density of water. Compare:

Material	Density (kg/m³)
Water	1000
Granite (surface rock)	2700
Aluminum	2700
Earth (average)	5500
Iron	7800
Lead	11300

Earth's surface rock has density ~2700 — half the mean. So the Earth's interior must contain much denser material. A fundamental geological insight from a physics experiment: the Earth has a metallic core (today we know: molten iron-nickel outer, solid inner).

Without Cavendish's G, this discovery would have been impossible. Earth science rides on gravity science.

Step 3 — Where does the number g come from?

Now invert. If we know M_E, R_E, and G, what must g be?

\[ g = \frac{G M_E}{R_E^2} \]

Plug in numbers:

\[ g = \frac{6.674 \times 10^{-11} \cdot 5.97 \times 10^{24}}{(6.371 \times 10^6)^2} = \frac{3.98 \times 10^{14}}{4.06 \times 10^{13}} \approx 9.81\ \mathrm{m/s^2} \]

Consistent! This isn't a coincidence, nor circular — because M_E was itself derived from g. But the point is: applied to a different body, the same formula with that body's M and R gives its g. The Moon:

\[ g_\text{moon} = \frac{G M_\text{moon}}{R_\text{moon}^2} = \frac{6.674 \times 10^{-11} \cdot 7.35 \times 10^{22}}{(1.737 \times 10^6)^2} \approx 1.62\ \mathrm{m/s^2} \]

That is ~1/6 of Earth's gravity — which is why Apollo astronauts could bounce.

Mars: g_Mars ≈ 3.7 m/s² (~0.38g).

Jupiter (at "cloud top"): g_J ≈ 24.8 m/s² (~2.5g).

Step 4 — g varies with altitude

g is not a fixed natural constant or a property of an object. It is a local computation — depending on distance from Earth's center. Above the surface, at altitude h:

\[ g(h) = \frac{G M_E}{(R_E + h)^2} \]

We can write this nicely. Define g_0 = GM_E/R_E² (surface gravity):

\[ g(h) = g_0 \cdot \left(\frac{R_E}{R_E + h}\right)^2 = \frac{g_0}{(1 + h/R_E)^2} \]

Examples

Everest summit (h ≈ 8850 m):

\[ h/R_E = 8850 / 6.371 \times 10^6 \approx 1.39 \times 10^{-3} \]

\[ g(\text{Everest}) \approx \frac{9.81}{(1.00139)^2} \approx 9.81 \cdot 0.9972 \approx 9.783\ \mathrm{m/s^2} \]

Difference from sea level: only ~0.3% or roughly 0.03 m/s². A climber at the summit is about 0.3% lighter than at sea level — for an 80 kg person, roughly 240 g. Small but measurable, and relevant to track-and-field records.

International Space Station (h ≈ 400 km):

\[ h/R_E = 4 \times 10^5 / 6.371 \times 10^6 \approx 0.0628 \]

\[ g(\text{ISS}) \approx \frac{9.81}{(1.0628)^2} \approx 8.68\ \mathrm{m/s^2} \]

About 88% of surface gravity! So why do ISS astronauts appear "weightless"? Because they are in free fall — the ISS is constantly "falling" toward Earth, but its horizontal motion is fast enough that it keeps missing the surface. That is an "orbit".

Weightlessness in the ISS does not mean "no gravity" — it means "gravity is balanced by centrifugal effect in the rotating frame", or equivalently: both astronaut and station fall with the same acceleration. (Equivalence principle again!)

Moon's distance (r ≈ 384000 km):

\[ h/R_E \approx 60.3\quad\Rightarrow\quad g_\text{at moon} \approx \frac{9.81}{60.3^2 + \ldots} \approx 0.0027\ \mathrm{m/s^2} \]

(Same number from Article 1's "Moon test" — same physics from a different angle.)

Small-altitude approximation

For h ≪ R_E, use the binomial expansion:

\[ (1 + h/R_E)^{-2} \approx 1 - 2h/R_E \]

\[ g(h) \approx g_0 \left(1 - \frac{2h}{R_E}\right) \]

With g_0 = 9.81, R_E = 6.371 × 10⁶ m:

\[ \Delta g = g_0 \cdot \frac{2h}{R_E} \approx 3.08 \times 10^{-6}\ h\ (\text{h in m}) \]

So g decreases by about 3 × 10⁻⁶ m/s² per meter of altitude. For the 30th floor of a building (h ≈ 100 m), the change is about 3 × 10⁻⁴ m/s² — around 30 ppm. Precision balances can detect this.

Step 5 — Gravitational potential energy

At Earth's surface with g approximately constant, high school gave us:

\[ U = m g h \]

But this only works for small altitudes (where g is nearly constant). What about satellites, or escape from Earth? We need the full formula. Using the definition of work (see The definite integral for physicists if integration is unfamiliar):

\[ U(r) - U(\infty) = -\int_\infty^r F\, dr' = -\int_\infty^r \left(-\frac{G M m}{r'^2}\right) dr' = -G M m \left[\frac{1}{r'}\right]_\infty^r \]

Taking U(∞) = 0 (the conventional zero — no gravity at infinity):

\[ \boxed{U(r) = -\frac{G M m}{r}} \]

Three notes:

1. The minus sign means the object is generally "trapped" by Earth's gravity — to escape to infinity, you need to inject positive energy.
2. The zero reference is a choice. We took U(∞) = 0. For near-surface problems, U(R_E) ≈ -GMm/R_E — a large negative number, but only differences matter.
3. Relation to mgh: for small h ≪ R, U(R+h) - U(R) = GMm(1/R - 1/(R+h)) ≈ GMmh/R² = mgh. So mgh is the near-surface approximation of U = -GMm/r.

Step 6 — Escape velocity

Question: for an object to "escape" from Earth's surface (reach infinity and not return), what minimum speed is needed?

By energy conservation. Initially:

• Kinetic energy: (1/2)mv²
• Potential energy: -GMm/R

At infinity (minimum energy to just "reach"):

• Kinetic energy: 0 (barely)
• Potential energy: 0 (by convention)

Conservation:

\[ \frac{1}{2} m v_\text{esc}^2 - \frac{G M m}{R} = 0 \]

\[ \boxed{v_\text{esc} = \sqrt{\frac{2 G M}{R}}} \]

For Earth:

\[ v_\text{esc} = \sqrt{\frac{2 \cdot 6.674 \times 10^{-11} \cdot 5.97 \times 10^{24}}{6.371 \times 10^6}} \approx \sqrt{1.25 \times 10^8} \approx 11.2\ \mathrm{km/s} \]

11.2 km/s — any object launched from Earth's surface at this speed (or higher) will never come back. Apollo lunar rockets went slower, since they only needed to reach lunar orbit, not full escape.

For the Moon: v_esc ≈ 2.4 km/s (small M) For Mars: v_esc ≈ 5.0 km/s For Jupiter: v_esc ≈ 59.5 km/s (six times Earth's!)

What you should be able to do

After this article, you should be able to:

• Compute Earth's mass with M_E = gR²/G from three accessible measurements
• Evaluate Earth's density and infer the presence of a metallic core from the discrepancy with surface rock
• Derive g ≈ 9.8 from the universal law of gravitation and apply the same to any other body
• Write g(h) = g₀·R²/(R+h)² and compute values at mountain tops, the ISS, Moon's distance
• Explain why ISS astronauts appear "weightless" despite gravity being ~88% of surface
• Derive U(r) = -GMm/r by integrating the force and show why mgh only works at low altitude
• Derive escape velocity from energy conservation and compute it for different bodies

Wrap-up

The three articles you just read completed a three-century story:

1. Newton gave the form of the law
2. Cavendish gave the number G
3. The rest of us got everything else with one multiplication and one division

The path from F = GMm/r² to escape velocity was walked without leaving Earth. A beautiful example of the power of physics: small laws, enormous consequences.

📚 Primary sources: Isaac Newton (1687) + Henry Cavendish (1798) — both in Principia and Philosophical Transactions. 📖 Open reference: OpenStax University Physics Vol 1 — §13.2: Gravitation Near Earth's Surface and §13.4: Gravitational Potential Energy and Total Energy. 📖 Feynman Lectures Vol I — Ch 7: The Theory of Gravitation.

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