Energy in Orbits

Positive energy from orbital motion

Fire a thruster on the Hubble Space Telescope (535 km orbit, launched 1990) to add energy. The orbit expands — Hubble moves to a higher altitude. At that higher altitude, the orbital speed is lower ($v = \sqrt{GM/r}$ decreases as $r$ increases). You added energy, yet the satellite slowed down. This counter-intuitive result follows directly from calculating the kinetic energy of an orbiting satellite.

From the circular orbit condition $\frac{GMm}{r^2} = \frac{mv^2}{r}$, we get $v^2 = \frac{GM}{r}$. Substituting into $KE = \frac{1}{2}mv^2$:

Energy quantities for a circular orbit as a function of radius $r$. Note that $|U| = 2KE$ and $E_\text{total} = -KE$ for all circular orbits.

Orbital KE: $KE = GMm/(2r)$ (always positive). Derived from $v^2 = GM/r$. $KE \propto 1/r$: higher orbit → lower speed → less KE. Do not confuse $v = \sqrt{GM/r}$ (orbital) with $v_e = \sqrt{2GM/r}$ (escape).

Pause — copy the highlighted orbital KE formula and the $1/r$ trend into your book before moving on.

Always negative — zero at infinity

We just saw that orbital KE is always positive. That raises a question: what about gravitational potential energy — why is the standard formula $U = -GMm/r$ negative? This card answers it → the reference is set at infinity ($U = 0$); all finite orbits are below that, so $U < 0$.

In orbital mechanics, potential energy is defined with zero at infinite separation. This is the only convention that makes energy conservation work cleanly. All finite orbits therefore have negative potential energy.

The negative sign reflects gravity being an attractive force. To bring a satellite from distance $r$ to infinity, you must do positive work on it — raising $U$ from a negative value up to zero.

More negative = deeper in the gravitational well = closer to the central mass. A satellite at $r = 7.0 \times 10^6$ m has more negative $U$ than one at $r = 10.0 \times 10^6$ m. It is harder to move to infinity from a lower orbit.

Gravitational PE: $U = -GMm/r$ — always negative at finite $r$; zero at infinity. Lower orbit → more negative $U$ (deeper in gravitational well). Cannot use $U = mgh$ for orbital problems — different reference.

Add the highlighted GPE formula and the reference-point rule to your notes before the check below.

The sum that determines whether an object remains bound

We just saw that KE is positive and $U$ is negative. That raises a question: when you add them together, what does the sign of the total energy tell you? This card answers it → $E_\text{total} = -GMm/(2r)$ (always negative = bound); binding energy $= |E_\text{total}|$.

Adding KE and U for a circular orbit gives a result that is both simple and physically profound. The total energy is always negative — which tells us everything about whether the satellite is trapped or free.

The total energy is always negative for any bound orbit. This is the defining characteristic: the satellite does not have enough energy to escape to infinity (where $E = 0$).

Total energy: $E_\text{total} = -GMm/(2r)$ (always negative for bound orbits). Virial theorem: $E_\text{total} = -KE = U/2$. Binding energy $= |E_\text{total}|$. More negative $E$ → lower orbit → more tightly bound → more energy needed to escape.

Pause — copy the highlighted total energy formula and the Virial theorem into your book before moving on.

The minimum speed to break free from gravity

We just saw that $E_\text{total} < 0$ for any bound orbit. That raises a question: what initial speed is required to make $E_\text{total} = 0$ and just barely escape? This card answers it → set $\tfrac{1}{2}mv_e^2 - GMm/r = 0$, giving $v_e = \sqrt{2GM/r}$.

Escape velocity is the minimum initial speed needed for an object to reach infinity with zero kinetic energy remaining. Setting total energy to zero at the launch point gives us the condition for escape.

For Earth at the surface ($r = R_E = 6.37 \times 10^6$ m):

Escape velocity: $v_e = \sqrt{2GM/r}$ (set $E_\text{total} = 0$). $v_e = \sqrt{2}\,v_\text{orbital}$ at same $r$ (NOT $2\times$). Mass $m$ cancels — escape velocity is independent of the object's mass. Earth's surface: $v_e \approx 11.2$ km/s.

Add the highlighted escape velocity formula and the factor-of-$\sqrt{2}$ rule to your notes before the check below.

Applying orbital energy to deep-space travel

We just saw the key misconceptions corrected. That raises a question: how do spacecraft engineers apply orbital energy principles to plan real missions beyond the solar system? This card answers it → gravity assists transfer energy from planets; Oberth effect maximises engine efficiency at periapsis.

Voyager 1 escaped the solar system using gravity assists (slingshot manoeuvres) from Jupiter and Saturn. These increased its total mechanical energy by transferring orbital energy from the planets — the planets lost an infinitesimal amount, the spacecraft gained a decisive boost.

The Oberth effect explains why burning engines at periapsis (closest approach) is most fuel-efficient. At closest approach, orbital speed is highest. The same $\Delta v$ produces a larger change in kinetic energy $\Delta E = \frac{1}{2}m(v + \Delta v)^2 - \frac{1}{2}mv^2 \approx mv\,\Delta v$ when $v$ is large. This is why spacecraft fire their engines at the closest point during flybys — the same fuel burns more effectively there.

Gravity assists transfer orbital energy from a planet to a spacecraft. Oberth effect: burn at periapsis (highest $v$) → largest energy gain because $\Delta E \approx mv\,\Delta v$. Both follow directly from $E_\text{total} = KE + U$.

Add the highlighted Oberth effect rule to your notes before the activities.