Universal Gravitation and Orbits

🌍 Newton's Law of Universal Gravitation

Part 1 of 7 — The Gravitational Force

The Law

$F = G\frac{m_1 m_2}{r^2}$

Gravitational Field

$\vec{g} = -\frac{GM}{r^2}\hat{r}$

At Earth's surface: $g \approx 9.8 \text{ m/s}^2$

🔑 Gravity is always attractive. The force is along the line connecting the two masses.

📝 Worked Example — Deriving Surface Gravity

Earth has mass $M = 5.97 \times 10^{24} \text{ kg}$ and radius $R = 6.37 \times 10^6 \text{ m}$. Show how $g$ at the surface follows from Newton's law, then evaluate it.

Step 1 — Force on a test mass $m$. At the surface, $r = R$, so

$F = \frac{GMm}{R^2}$

Step 2 — Identify the field. The gravitational field is the force per unit mass, $g = F/m$, so the test mass cancels:

$g = \frac{GM}{R^2}$

Step 3 — Evaluate.

$g = \frac{(6.674 \times 10^{-11})(5.97 \times 10^{24})}{(6.37 \times 10^6)^2} \approx 9.8 \text{ m/s}^2$

🔑 Surface gravity depends on a planet's mass and radius — not on the falling object's mass, which is why all objects fall at the same rate.

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🛸 Gravitational Potential Energy & Orbits

Part 2 of 7 — Energy in Gravitational Systems

Gravitational Potential Energy

$U = -\frac{GMm}{r}$

Note the negative sign — $U = 0$ at $r = \infty$.

Circular Orbits

For a satellite in circular orbit:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

$v_{\text{orbit}} = \sqrt{\frac{GM}{r}}$

$T = 2\pi\sqrt{\frac{r^3}{GM}}$ (Kepler's Third Law)

Escape Velocity

$v_{\text{escape}} = \sqrt{\frac{2GM}{r}} = \sqrt{2} \cdot v_{\text{orbit}}$

🔑 Escape velocity is $\sqrt{2}$ times orbital velocity at the same radius.

📝 Worked Example — Where Does $U = -\frac{GMm}{r}$ Come From?

The potential energy is defined as the work done against gravity to bring a mass from infinity to radius $r$. Derive it by integrating the force.

Step 1 — Set up the work integral. Potential energy equals minus the work done by gravity moving the mass in from $\infty$ to $r$:

$U(r) = -\int_{\infty}^{r} \vec{F}\cdot d\vec{r} = -\int_{\infty}^{r} \left(-\frac{GMm}{r'^2}\right) dr'$

Step 2 — Evaluate the integral. The antiderivative of $r'^{-2}$ is $-r'^{-1}$:

$U(r) = GMm \int_{\infty}^{r} \frac{dr'}{r'^2} = GMm\left[-\frac{1}{r'}\right]_{\infty}^{r} = GMm\left(-\frac{1}{r} + 0\right)$

Step 3 — Result.

$U(r) = -\frac{GMm}{r}$

🔑 The negative sign and the $r = \infty$ reference both fall directly out of the integration — bound systems have negative potential energy.

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🪐 Kepler's Laws

Part 3 of 7 — Planetary Motion

Kepler's Three Laws

Kepler's Third Law (Detailed)

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

Example: Earth orbits at 1 AU with $T = 1$ year. For Mars at 1.52 AU:

$T_{\text{Mars}} = \left(\frac{1.52}{1}\right)^{3/2} \approx 1.87 \text{ years}$

📝 Worked Example — Deriving Kepler's Third Law

Derive $T^2 = \frac{4\pi^2}{GM}r^3$ for a circular orbit from Newton's law of gravitation.

Step 1 — Balance gravity and centripetal force. Using $a_c = \frac{4\pi^2 r}{T^2}$ for uniform circular motion,

$\frac{GMm}{r^2} = m\,\frac{4\pi^2 r}{T^2}$

Step 2 — Cancel $m$ and isolate $T^2$. Cross-multiplying,

$T^2 = \frac{4\pi^2 r^3}{GM}$

Step 3 — Apply it. For a geostationary satellite ($T = 86400 \text{ s}$) around Earth ($GM = 3.99 \times 10^{14} \text{ m}^3/\text{s}^2$):

$r = \left(\frac{GM\,T^2}{4\pi^2}\right)^{1/3} = \left(\frac{(3.99 \times 10^{14})(86400)^2}{4\pi^2}\right)^{1/3} \approx 4.2 \times 10^7 \text{ m}$

🔑 The constant $\frac{4\pi^2}{GM}$ is the same for every satellite of the same central body — that is the heart of Kepler's Third Law.

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⚡ Gravitational Potential

Part 4 of 7 — Potential and Field

Gravitational Potential (per unit mass)

$V = -\frac{GM}{r}$

Relationship to field:

$\vec{g} = -\nabla V = -\frac{dV}{dr}\hat{r}$

Shell Theorem

🔑 Only the mass at radii smaller than your position matters (for spherical symmetry).

📝 Worked Example — Field from the Gradient of Potential

Given the gravitational potential $V(r) = -\frac{GM}{r}$, recover the field $\vec{g}$ by differentiation, and find $g$ at $r = 2R_E$ above Earth's center.

Step 1 — Differentiate the potential. The radial field is the negative gradient:

$g = -\frac{dV}{dr} = -\frac{d}{dr}\left(-\frac{GM}{r}\right)$

Step 2 — Carry out the derivative. Since $\frac{d}{dr}(r^{-1}) = -r^{-2}$:

$g = -\left(\frac{GM}{r^2}\right) \quad\text{(magnitude } \frac{GM}{r^2}, \text{ directed inward)}$

Step 3 — Evaluate at $r = 2R_E$. Because $g_{\text{surface}} = \frac{GM}{R_E^2}$,

$g(2R_E) = \frac{GM}{(2R_E)^2} = \frac{1}{4}\,g_{\text{surface}} \approx \frac{9.8}{4} \approx 2.5 \text{ m/s}^2$

🔑 The field is the slope of the potential curve; potential is a scalar, which often makes energy problems easier than vector force problems.

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🛰️ Energy of Orbiting Bodies

Part 5 of 7 — Total Energy in Orbits

Energy Summary for Circular Orbits

$KE = \frac{1}{2}mv^2 = \frac{GMm}{2r}$

$PE = -\frac{GMm}{r}$

$E_{\text{total}} = KE + PE = -\frac{GMm}{2r}$

🔑 Total energy is negative for bound orbits. $E = 0$ at the boundary (parabolic trajectory = escape).

📝 Worked Example — Total Energy and the Virial Relation

Show that a circular orbit has $E_{\text{total}} = -\frac{GMm}{2r}$, then find the energy needed to raise a satellite from radius $r_i$ to $r_f$.

Step 1 — Kinetic energy from the orbit condition. With $v^2 = \frac{GM}{r}$,

$KE = \frac{1}{2}mv^2 = \frac{1}{2}m\,\frac{GM}{r} = \frac{GMm}{2r}$

Step 2 — Add potential energy. Since $PE = -\frac{GMm}{r}$,

$E_{\text{total}} = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$

Step 3 — Energy to change orbit. The work needed equals $\Delta E$:

$\Delta E = E_f - E_i = -\frac{GMm}{2r_f} + \frac{GMm}{2r_i} = \frac{GMm}{2}\left(\frac{1}{r_i} - \frac{1}{r_f}\right)$

For $r_f > r_i$ this is positive — you must add energy to climb to a higher orbit.

🔑 Notice $E_{\text{total}} = -KE = \tfrac{1}{2}PE$: the virial relation for an inverse-square bound orbit.

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🛠️ Gravitation Workshop

Part 6 of 7 — Strategies and Practice

Common Problem Types

📝 Worked Example — Escape Speed via Energy Conservation

A projectile is launched straight up from a planet's surface (radius $R$, mass $M$) and just barely escapes. Derive the escape speed from energy conservation, then evaluate for Earth.

Step 1 — Set up energy conservation. "Just barely escapes" means $KE = 0$ and $U = 0$ at $r = \infty$:

$\frac{1}{2}mv_{esc}^2 + \left(-\frac{GMm}{R}\right) = 0 + 0$

Step 2 — Solve for $v_{esc}$. The mass $m$ cancels:

$\frac{1}{2}v_{esc}^2 = \frac{GM}{R} \quad\Rightarrow\quad v_{esc} = \sqrt{\frac{2GM}{R}}$

Step 3 — Evaluate for Earth. With $GM = 3.99 \times 10^{14} \text{ m}^3/\text{s}^2$ and $R = 6.37 \times 10^6 \text{ m}$:

$v_{esc} = \sqrt{\frac{2(3.99 \times 10^{14})}{6.37 \times 10^6}} \approx 1.12 \times 10^4 \text{ m/s} = 11.2 \text{ km/s}$

🔑 Escape speed comes straight from "total energy = 0"; it is independent of launch direction (ignoring air drag and rotation).

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📋 Gravitation Review

Part 7 of 7 — Master Summary

Essential Formulas

📝 Worked Example — Synthesis: Speed, Energy, and "Weighing" a Planet

A satellite orbits a planet in a circular orbit of radius $r = 8.0 \times 10^6 \text{ m}$ with period $T = 7200 \text{ s}$. Find the orbital speed, the planet's mass, and the satellite's total-energy sign.

Step 1 — Orbital speed from geometry. The satellite covers one circumference per period:

$v = \frac{2\pi r}{T} = \frac{2\pi (8.0 \times 10^6)}{7200} \approx 6.98 \times 10^3 \text{ m/s}$

Step 2 — "Weigh" the planet with Kepler's Third Law. Solving $T^2 = \frac{4\pi^2 r^3}{GM}$ for $M$:

$M = \frac{4\pi^2 r^3}{G\,T^2} = \frac{4\pi^2 (8.0 \times 10^6)^3}{(6.674 \times 10^{-11})(7200)^2} \approx 5.8 \times 10^{24} \text{ kg}$

Step 3 — Energy sign. Because $E_{\text{total}} = -\frac{GMm}{2r} < 0$, the orbit is bound — consistent with a closed circular path.

🔑 Measuring a satellite's $r$ and $T$ lets you compute the central body's mass — the same method used to weigh the Sun, Jupiter, and black holes.

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