Newton's Universal Law of Gravitation — Study Notes

Newton extended gravity from Earth to a universal force that acts between all material bodies everywhere in space (universal attraction).
The core puzzle Newton tackled: how must gravity depend on distance to simultaneously (a) reproduce Kepler’s laws for planetary motion, (b) predict Earth’s gravitational acceleration at the surface (as observed by Galileo), and (c) apply to objects falling near Earth.
Newton’s key insight: there is a universal attractive force between any two masses, and its precise form determines the observed motions.
He invented calculus to derive the distance dependence and the mass dependence of the force.
The resulting law of gravitation (two-body version) is one of the most famous equations in science:
F{\text{gravity}} = G \frac{M1 M2}{R^2} where $G$ is the universal gravitational constant, $M1$ and $M_2$ are the interacting masses, and $R$ is their separation.
Implications:
- The force is inversely proportional to the square of the distance: $F \propto 1/R^2$.
- The force is proportional to the product of the masses: $F \propto M1 M2$.
- Gravity is a central force: it acts along the line joining the centers of the two masses.
Consequences for orbits: using $F = ma$ (Newton’s second law) and the inverse-square law, the motion of two gravitating bodies yields the orbits described by Kepler’s laws.

Distance dependence:
- The gravitational force decreases with distance as the inverse square: F \propto \frac{1}{R^2}.
- If distance doubles, the force becomes one quarter as large; if distance triples, the force becomes one ninth as large.
Mass dependence:
- The force is proportional to the masses of the two bodies: F{\text{gravity}} = G \frac{M1 M_2}{R^2}.
Gravitational acceleration on a body of mass $m$ due to a body of mass $M$ at distance $R$ is
a = \frac{F}{m} = G \frac{M}{R^2}.
Surface gravity of Earth:
- The local gravitational acceleration is $g \approx 9.8\ \text{m s}^{-2}.$
Moon’s orbital distance in Earth radii:
- The Moon is about $60$ Earth radii from Earth's center.
- Because gravity scales as $1/R^2$, the Moon’s gravitational acceleration toward Earth is
  g{\text{Moon}} = \frac{gE}{60^2} = \frac{9.8}{3600} \approx 0.00272\ \text{m s}^{-2}.
- This small acceleration explains why the Moon falls toward Earth but keeps missing it, i.e., it orbits.

Mass is an intrinsic property of an object; it does not change with location.
Weight is the force exerted on a mass by gravity:
W = m g.
Because $g$ varies with location, your weight changes on different planets or moons even though your mass stays the same.
If gravity weakens with distance, weight also weakens accordingly.
Earth’s surface gravity is strong enough that objects near the surface accelerate downward at $9.8\ \text{m s}^{-2}$.
Weightlessness in orbit is not zero gravity; it arises because objects in orbit are in free fall, accelerating downward at the same rate as their spacecraft.
- In orbit, astronauts are falling around Earth, not toward Earth, so they feel no contact force Mike- thus no apparent weight.
- This is an example of “weightlessness” due to free fall, not absence of gravity.

Kepler’s laws describe planetary motion; Newton’s law of gravitation provides the physical mechanism behind them.
For a two-body system, Newtonian gravity leads to the relation between orbital period $P$ and semimajor axis $a$:
P^2 = \frac{4\pi^2 a^3}{G\,(M1+M2)}.
Equivalently,
\frac{a^3}{P^2} = \frac{G\,(M1+M2)}{4\pi^2}.
In solar units (using AU for distance, years for period, and solar masses for mass), the relation simplifies to
\frac{a^3}{P^2} = M1 + M2.
Notes:
- $a$ is the semimajor axis of the orbit, $P$ is the orbital period.
- $M1$ and $M2$ are the masses of the two bodies; if one mass greatly exceeds the other, $M1 + M2 \approx M_1$.
- The Moon’s orbit and planetary orbits around the Sun both conform to this framework when gravity is included.
The law also implies we can deduce masses from motions: if we measure $a$ and $P$ for a system, we can infer $M1+M2$.
The Newtonian form generalizes Kepler’s third law by including the masses of both bodies, which is particularly important when the masses are comparable (e.g., binary stars, galaxies).

Scenario: A planet similar to Earth orbits its star at a distance of $a = 1\ \text{AU}$ with an orbital period of $P = 1\ \text{yr}$; the planet’s mass is negligible relative to the star.
Using the two-body relation in solar units:
\frac{a^3}{P^2} = M1 + M2.
With $a = 1$ AU and $P = 1$ yr, this gives $M1 + M2 \approx 1\ M_\odot$ (the star’s mass is about one solar mass).
The textbook passage discusses that the star’s mass in this example is stated as twice the Sun; in standard solar-unit form, $a=1$ AU and $P=1$ yr would imply $M1 \approx 1\ M\odot$ if the planet’s mass is negligible. The discrepancy highlights how the two-body mass term can affect the exact value when not in the limiting case of $M1 \gg M2$.
The general takeaway: the mass of the star (or companion) can be deduced from the observed orbital distance and period using
M1 + M2 = \frac{a^3}{P^2} in solar units (AU, yr, $M_\odot$).
In more general units, the complete form is
P^2 = \frac{4\pi^2 a^3}{G\,(M1+M2)}.
The text emphasizes the power of this reformulation for inferring masses of astronomical objects from their motions across a wide range of systems, from planets to galaxies.
It also notes that in many real cases, you must include both masses (e.g., binary stars, interacting galaxies).

Problem: A star has twice the mass of the Sun, and an Earth-like planet orbits it with $P = 4\ \text{yr}$. What is the semimajor axis $a$ of the planet’s orbit?
Approach (in solar units): using $a^3 / P^2 = M1 + M2$ and $M1 + M2 = 2\ M\odot$, $P = 4\ \text{yr}$, a^3 = (M1+M_2) P^2 = 2 \times 4^2 = 2 \times 16 = 32.
Therefore
a = \sqrt[3]{32} \approx 3.2\ \text{AU}.
This illustrates how a larger stellar mass and a longer orbital period push the planet to a larger orbital distance.

Gravity is a universal, long-range force that acts between all masses, not just on Earth.
The inverse-square nature of gravity explains why gravitational effects weaken with distance and why celestial orbits are stable and predictable under Newton’s laws.
Kepler’s empirical laws are recovered and explained by Newton’s more fundamental gravitational law.
Measuring orbital motions provides a powerful method to determine masses of astronomical objects, from planets to stars to galaxies.
The concept of weightlessness arises from being in free fall, not from the absence of gravity; this is a common source of intuition about spaceflight.

Surface gravity on Earth: g\approx 9.8\ \text{m s}^{-2}.
Moon’s distance from Earth: about R{\text{Moon}} = 60\,RE.
Moon’s surface acceleration due to Earth’s gravity: g{\text{Moon}} = \frac{gE}{60^2} = \frac{9.8}{3600} \approx 0.00272\ \text{m s}^{-2}.
Gravitational force between two bodies:
F{\text{gravity}} = G \frac{M1 M_2}{R^2}.
Orbital relation (two-body; SI form):
P^2 = \frac{4\pi^2 a^3}{G\,(M1+M2)}.
In solar units (AU, yr, $M\odot$): \frac{a^3}{P^2} = M1 + M_2.
For the Check Your Learning example: a = \sqrt[3]{(M1+M2) P^2} = \sqrt[3]{2 \times 4^2} = \sqrt[3]{32} \approx 3.2\ \text{AU}.