The Timpson Lyceum · University Physics I (2025)

Gravitation: Newton's Law of Gravity – Lecture 13 (University Physics I)

Overview

This lecture replaces the surface-only picture of gravity ($F = mg$, $U = mgy$) with Newton's law of universal gravitation, a single law that works everywhere in the universe. Every particle attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers. From this one law we recover surface gravity, learn how $G$ and the mass of the Earth were measured, build a new gravitational potential energy valid at astronomical scale, and derive escape speed. We then apply it to orbits — finding orbital speed, period, and energy for circular orbits — and connect everything to Kepler's three laws of planetary motion, which turn out to be theorems that follow from Newton's law. This material is foundational: later topics, especially electricity and magnetism, lean on the same inverse-square reasoning.

Key Concepts

Newton's law of universal gravitation

• Every particle attracts every other particle with the force $F = \frac{G\,m_1 m_2}{r^2}$
• $r$ is the distance between the two masses' centers; the force is attractive, directed along the line joining them.
• $G = 6.67\times10^{-11}\ \mathrm{N\,m^2/kg^2}$ is the universal gravitational constant — a true constant, the same everywhere in the universe. Its odd units exist only to make the right-hand side come out in newtons; the value depends on the units chosen to measure mass and distance.
• This law is a starting assumption (a law), observed by induction, not derived from anything deeper.

Weight and gravitational field

• Weight is the sum of the gravitational forces acting on an object; in practice the nearest large body dominates and all other bodies are negligible.
• At a planet's surface, weight is $w = \dfrac{G\,m\,M}{R^2}$, where $M$ and $R$ are the planet's mass and radius.
• Comparing with $w = mg$ gives the surface gravitational acceleration: $g = \frac{G M}{R^2}$
• For Earth ($M \approx 5.97\times10^{24}\ \mathrm{kg}$, $R \approx 6.37\times10^{6}\ \mathrm{m}$) this yields $g \approx 9.8\ \mathrm{m/s^2}$ — the number Galileo measured directly.

Measuring $G$ and the mass of the Earth

• $G$ is found with a torsion balance: two small masses on a fine fiber, with a mirror that deflects a laser. Bringing large masses nearby produces a tiny torque; the laser amplifies the resulting twist, letting you back out the force and solve for $G$.
• Once $G$ is known, the mass of the Earth follows from the measurable surface gravity: $M = gR^2/G \approx 6\times10^{24}\ \mathrm{kg}$.
• The Earth's radius has been measurable since antiquity (e.g., shadow/sun-angle methods), but its mass had no good measurement before Newton's law made $M = gR^2/G$ possible. Dividing mass by volume gives an average density of about $5.5\ \mathrm{g/cm^3}$ (water is $1\ \mathrm{g/cm^3}$).

Gravitational potential energy at astronomical scale

• $U = mgy$ is only valid near a surface where $g$ is constant. The general form is $U = -\frac{G\,m\,M}{r}$
• The zero of $U$ is chosen at $r \to \infty$ (infinitely far away), so $U$ is negative everywhere finite. Only changes in $U$ are physical, so the negative sign and reference choice cause no trouble.
• This is obtained by integrating the work done by gravity, $W = \int_{r_1}^{r_2} -\dfrac{G m M}{r^2}\,dr$, which gives $W = U_1 - U_2$, consistent with the work–energy theorem $W = K_2 - K_1$.
• Energy conservation (gravity only) still reads $K_1 + U_1 = K_2 + U_2$.

Escape speed

• The minimum launch speed for which an object never returns: set $U_1 + K_1 = U_2 + K_2$ with $U_2 = 0$ and $K_2 = 0$ at infinity.
• The launch mass cancels, leaving $v_{esc} = \sqrt{\frac{2 G M}{R}}$
• For Earth this is $v_{esc} \approx 1.12\times10^{4}\ \mathrm{m/s}$ (about $11{,}000\ \mathrm{m/s}$). Even though gravity acts at all distances, its inverse-square falloff means the total "braking" is finite, so a finite speed suffices.

Circular orbits

• An orbit is continual free-fall: the body's sideways motion carries it past the planet as fast as it falls inward. Setting gravity equal to the centripetal requirement, $\dfrac{G M}{r^2} = \dfrac{v^2}{r}$, gives the orbital speed $v = \sqrt{\frac{G M}{r}}$
• Higher orbits require less speed. Here $r$ is measured from the planet's center.
• Period (time for one revolution): from $v = \dfrac{2\pi r}{T}$, $T = 2\pi\sqrt{\frac{r^3}{G M}}$
• Total orbital energy ($E = K + U$ with the orbital speed substituted): $E = -\frac{G\,m\,M}{2r}$

Kepler's laws (theorems provable from Newton's law)

• First law: each planet moves in an ellipse with the Sun at one focus. (A circle is the special case where the two foci coincide.) An ellipse satisfies $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$; $a$ is the semi-major axis.
• Second law: the line from the Sun to a planet sweeps out equal areas in equal times, i.e. $\dfrac{dA}{dt}$ is constant. This is exactly conservation of angular momentum: since gravity is central, the torque $\tau = rF\sin(0^\circ) = 0$, so $L = mvr\sin\phi$ is conserved, and $\dfrac{dA}{dt} = \dfrac{L}{2m}$.
• Third law: $T = 2\pi\sqrt{\dfrac{a^3}{G M}}$, so $T \propto a^{3/2}$ with the same proportionality constant for every body orbiting the same primary.
• Eccentricity measures how off-center the focus sits: $e = \dfrac{a - r_{\text{min}}}{a}$, where $0$ is a perfect circle. Halley's comet has $e \approx 0.97$.

Spherical mass distribution

• A key justification used throughout: a spherically symmetric body attracts an external point mass as though all its mass were concentrated at its center. This is proven by summing the potential energy of thin spherical shells (each shell summed over rings, using $dm/m_{\text{shell}} = dA/A_{\text{shell}}$), which integrates to $U = -\dfrac{G m M}{r}$.

Worked Example

Weight of a robot on Mars. A robot weighs $3{,}430\ \mathrm{N}$ on Earth. Find its weight on Mars, given Mars' mass $M_{\text{Mars}} = 6.42\times10^{23}\ \mathrm{kg}$ and radius $R_{\text{Mars}} = 3.44\times10^{6}\ \mathrm{m}$.

First recover the robot's mass from its Earth weight, then apply Newton's law at Mars' surface:

Mars has about one-tenth Earth's mass, yet the robot weighs roughly one-third of its Earth weight rather than one-tenth — because Mars' much smaller radius places the surface closer to its center, and the inverse-square dependence on $R$ partly offsets the smaller mass.

Speed of a low-Earth-orbit satellite. A $1{,}000\ \mathrm{kg}$ satellite orbits $300\ \mathrm{km}$ above Earth's surface. Find its orbital speed (use $M = 5.97\times10^{24}\ \mathrm{kg}$, $R = 6.37\times10^{6}\ \mathrm{m}$). The orbital radius is $r = R + 3\times10^{5}\ \mathrm{m}$:

The corresponding period is about $90\ \mathrm{minutes}$ — roughly an hour and a half per orbit.

Summary

• Newton's law $F = \dfrac{G m_1 m_2}{r^2}$ governs gravity everywhere; surface gravity $g = \dfrac{G M}{R^2}$ is just its value at a planet's surface, and weight is the dominant gravitational force on a body.
• $G$ is a universal constant (torsion-balance measured); once known, it lets us weigh the Earth via $M = gR^2/G \approx 6\times10^{24}\ \mathrm{kg}$.
• The general gravitational potential energy is $U = -\dfrac{G m M}{r}$ with the zero set at infinity; energy conservation $K_1 + U_1 = K_2 + U_2$ still applies.
• Escape speed $v_{esc} = \sqrt{\dfrac{2GM}{R}} \approx 11{,}000\ \mathrm{m/s}$ for Earth, and it is independent of the escaping object's mass.
• For circular orbits, $v = \sqrt{\dfrac{GM}{r}}$, $T = 2\pi\sqrt{\dfrac{r^3}{GM}}$, and $E = -\dfrac{G m M}{2r}$.
• Kepler's three laws (ellipses, equal areas in equal times, $T \propto a^{3/2}$) are consequences of Newton's law; the second law is conservation of angular momentum, and a spherical body acts gravitationally as if all its mass sat at its center.