An orbit is a free fall that misses the ground

Picture a cannon on top of a mountain that pokes above the atmosphere, aimed sideways. Fire slowly: the cannonball lands close. Fire harder: it lands farther away. The ball is in free fall the whole time, but the Earth curves away underneath. At some critical speed, the ball falls toward the Earth at the same rate the Earth curves away. It keeps falling forever. That is an orbit.

Newton put this thought experiment in the Principia, and it still kills the most common orbit misconception. A satellite in a stable circular orbit has one force on it: gravity, pulling toward the central body. No engine. No sideways push. The forward motion is just inertia, the same thing that keeps a hockey puck sliding on ice. Gravity bends that straight-line tendency into a circle.

Standard circular-motion bookkeeping works here. A particle in a circle of radius $r$ at speed $v$ has centripetal acceleration $a_c = v^2 / r$ pointing inward. Whatever supplies that acceleration is the net force. For an orbit, that "whatever" is gravity, and only gravity. Setting gravity equal to $m v^2 / r$ is the whole game.

Start with Newton's second law applied radially. The force of gravity on a satellite of mass $m$ at distance $r$ from a central body of mass $M$ is $F_\text{grav} = G M m / r^2$, directed radially inward. The centripetal requirement for circular motion is $F_c = m v^2 / r$, also directed radially inward. Setting them equal:

$$\dfrac{G M m}{r^2} = \dfrac{m v^2}{r}$$

The satellite mass $m$ cancels. Solving for $v$:

$$v = \sqrt{\dfrac{G M}{r}}$$

Orbital speed depends on the central mass and the radius, and nothing else. Heavier central body, faster orbit at the same $r$. Larger orbit, slower speed. Doubling $r$ does not halve $v$; it cuts $v$ by $\sqrt{2}$.

The period $T$ is the time to complete one orbit. The satellite travels a distance $2 \pi r$ at speed $v$, so:

$$T = \dfrac{2 \pi r}{v} = 2 \pi \sqrt{\dfrac{r^3}{G M}}$$

This is Kepler's third law in disguise: $T^2 \propto r^3$ at fixed $M$. Doubling $r$ multiplies $T$ by $2^{3/2} = 2 \sqrt{2} \approx 2.83$.

Two forms show up in AP problems. The first plugs in $G$ and $M$ directly. The second uses the surface trick: at radius $R$, $g = G M / R^2$, so $G M = g R^2$. Substituting:

$$v = \sqrt{\dfrac{g R^2}{r}}\qquad T = 2 \pi \sqrt{\dfrac{r^3}{g R^2}}$$

Either form is fine. Use whichever matches the numbers the problem gives you.

Worked example: a satellite at twice Earth's radius

Find the orbital speed for a circle of radius $r = 2 R_E$ around Earth, using the $g R^2$ shortcut.

Same answer either way. The $g R_E^2$ shortcut just saves you from carrying $G = 6.67 \times 10^{-11}$ N·m$^2$/kg$^2$ and $M_E = 5.97 \times 10^{24}$ kg through the algebra.

The §2 equations describe motion, not energy. To say what "bound" means, you need three quantities: kinetic energy $K$, gravitational potential energy $U_g$, and total mechanical energy $E = K + U_g$.

The kinetic energy is the usual one:

$$K = \dfrac{1}{2} m v^2 = \dfrac{G M m}{2 r}$$

where the second form substitutes $v^2 = GM/r$. So $K$ is positive, and at a circular orbit it is half of $GMm/r$.

The potential energy is the tricky one. The familiar $U_g = m g h$ form is a flat-Earth approximation; it breaks for large altitude changes. For orbits, set $U_g = 0$ at infinity, which forces:

$$U_g = -\dfrac{G M m}{r}$$

The minus sign is forced by the reference, not chosen. With zero at infinity and gravity pulling inward, every finite separation sits below zero. Closer in, more negative. Farther out, less negative. At infinity, zero. If you insist $U_g$ must be positive, you are using $mgh$ where it no longer applies.

Add $K$ and $U_g$ for the total mechanical energy of a circular orbit:

$$E = K + U_g = \dfrac{G M m}{2 r} - \dfrac{G M m}{r} = -\dfrac{G M m}{2 r}$$

$E$ is negative, with magnitude equal to $K$. A bound orbit has negative total energy. The energy you would have to add to push the satellite out to infinity (arriving with $K = 0$) is exactly $|E| = GMm/(2r)$. That is the binding energy.

Three relations are worth memorizing for circular orbits:

• $K = -E$: kinetic energy and total energy are equal in magnitude, opposite in sign.
• $U_g = 2 E$: potential energy is twice the total energy (and twice as deep as $|E|$).
• $|U_g| = 2 K$: the gravitational well is twice as deep as the kinetic energy is tall.

How does $E$ change as $r$ grows? Since $E = -GMm/(2r)$, larger $r$ means smaller $|E|$, so $E$ rises toward zero from below. It becomes less negative, not more. At $r \to \infty$, $E \to 0$. A higher orbit has more total energy than a lower one, even though its $K$ and $v$ are smaller. The extra energy lives in $U_g$ (less negative) as you climb. Higher orbit, less negative $E$, lower $v$.

Endnote: elliptical orbits

Real orbits are usually elliptical. The formulas still apply, but $r$ varies along the orbit, and $K$ trades against $U_g$ between the closest point (perihelion) and the farthest (aphelion). At perihelion: small $r$, very negative $U_g$, large $K$, large $v$. At aphelion: large $r$, less negative $U_g$, small $K$, small $v$. Total $E$ stays put through both.

Angular momentum $L = m v r_\perp$ also stays put: gravity is a central force, so its torque about the central body is zero. But $L$ conserved is not the same as angular speed $\omega = v / r$ conserved. Both $v$ and $r$ change between perihelion and aphelion, and $\omega$ drops sharply. Gravity preserves $L$, not $\omega$.

Three misconceptions dominate Topic 6.6. Each one is the natural extension of an everyday intuition into a regime where that intuition stops working.

Ten short scenarios. Each one targets one of the misconceptions from §4. Pick chips, hit Check answer. Wrong answers explain where the trap reasoning fails. Progress saves automatically.