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Unit 1 · Lesson 1-4
Orbital Energy and the Vis-Viva Equation
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Orbital Energy and the Vis-Viva Equation

Specific mechanical energy, the vis-viva equation and the sign of ε

An accounting problem
In the last lesson we found that a Molniya satellite moves nearly seven times faster at perigee than at apogee. Stop and consider how strange that is. Nothing pushed it. No engine fired. There is no propellant involved, no external agency, nothing but empty space and an invisible pull. And yet somewhere between apogee and perigee the satellite acquired an enormous amount of speed, then gave every bit of it back on the way out. Twice per orbit, forever, with a precision no machine could match. Something is being traded. Angular momentum told us the exchange rate — rpvp=ravar_p v_p = r_a v_a — but it never told us where the currency comes from. There is a second ledger, and unlike the first one, this one has a sign that tells you your fate.
Watch first · presented by Shoaib Aziz
The transfer orbit in this film is drawn to scale and moves at its true Kepler timing, and every number on screen — 10.07 and 1.62 km/s, ε = −8.1455 three times over, h = 68,198 at both ends — is printed by this lesson's own lab solution.
Kinetic energy is familiar: 12mv2\tfrac{1}{2}mv^2. Dividing by mass as always, the specific kinetic energy is v2/2v^2/2.
Potential energy is where things get interesting. On Earth's surface you learned U=mghU = mgh, measured upward from an arbitrary floor. That works because gg is effectively constant over a few metres. In orbit, gg varies as 1/r21/r^2, so we have to integrate properly — the work done bringing a mass in from infinity:
U(r)=−∫∞rμr′2 dr′=−μrU(r) = -\int_{\infty}^{r} \frac{\mu}{r'^2}\,dr' = -\frac{\mu}{r}

Specific gravitational potential energy. Negative, and approaching zero from below as r → ∞.

⚠Why gravitational potential energy is negative
Students lose more time to this sign than to anything else in the course. There is nothing mysterious about it. Potential energy has no absolute value — only differences matter, so you may put the zero anywhere. On Earth's surface we conventionally put it at the floor, making everything above it positive. In orbital mechanics we put the zero at infinity, because that is the only place that is special for every orbit around every body. And if U=0U = 0 infinitely far away, then anywhere closer must be negative, because you have to do positive work to climb back out. So: U=−μ/rU = -\mu/r. Deepest (most negative) near the body, rising toward zero as you escape. The number is negative not because something is wrong, but because you are in a hole.
It is important to realize that in physics today, we have no knowledge of what energy is.
– Richard Feynman, The Feynman Lectures on Physics, Vol. I, ch. 4, 1963
Figure 1.10
renergy0bound — ellipseε = 0 — parabolic escapeunbound — hyperbolaU(r) = −μ/rNegative total energy means trapped. Zero means just barely free. Positive means gone for good.
The gravitational potential well. Your total energy is a horizontal line: where it cuts the curve is your turning point, the furthest you can get. Below zero you are trapped in the well. At exactly zero you just barely coast to infinity with nothing left. Above zero you leave with speed to spare.
Add the two together and you get the specific mechanical energy, universally written ε\varepsilon (or sometimes E\mathcal{E}):
ε=v22−μr\varepsilon = \frac{v^2}{2} - \frac{\mu}{r}

Specific mechanical energy, in km²/s². Constant along any two-body orbit — this is the second conserved quantity.

Now the remarkable part, and the single most useful equation in the whole subject.
Evaluate ε\varepsilon at any point and you get the same answer. Evaluate it at perigee and at apogee, do a page of algebra, and ε\varepsilon turns out to depend on only one thing: the semi-major axis aa.
ε=−μ2a\varepsilon = -\frac{\mu}{2a}

Energy depends on the size of the orbit and nothing else — not its shape, not its tilt, not where the satellite currently is.

💡Energy depends only on the semi-major axis
Two orbits with the same semi-major axis have identical energy, even if one is a perfect circle and the other is a cigar-shaped ellipse grazing the atmosphere. They also have identical periods — which is Kepler's Third Law, and we will derive it properly in Unit 5. This is the counterpart to the previous lesson. Angular momentum knows about shape and plane but not size. Energy knows about size but nothing else. Between them, they pin the orbit down completely.
Set the two expressions for ε\varepsilon equal and rearrange, and you get the equation you will use more than any other:
v2=μ(2r−1a)v^2 = \mu\left(\frac{2}{r} - \frac{1}{a}\right)

The vis-viva equation — Latin for "living force". Give it where you are (r) and how big the orbit is (a), and it gives you your speed. Every manoeuvre calculation in Unit 6 starts here.

Figure 1.11
0.51.01.5246810orbital radius r (Earth radii)v / v₀escape √(2μ/r)circular √(μ/r)a = 3a = 6Every bound orbit lives between the two dashed curves. Touch the top one and you never come back.
Vis-viva plotted for real. Every bound orbit lives in the band between the circular-speed curve and the escape-speed curve. A larger a lifts the whole curve; touch the escape curve anywhere and you never come back.
📐Worked Example: three orbits, one equation
Circular at 400 km. Here r=a=6771r = a = 6771 km: v2=398600(2/6771−1/6771)=398600/6771v^2 = 398600(2/6771 - 1/6771) = 398600/6771, so v=7.67v = \mathbf{7.67} km/s. The circular formula is just vis-viva with r=ar = a. Elliptical, rp=6771r_p = 6771, ra=42,164r_a = 42{,}164 km. First a=(6771+42164)/2=24,468a = (6771+42164)/2 = 24{,}468 km. At perigee: v2=398600(2/6771−1/24468)=101.4v^2 = 398600(2/6771 - 1/24468) = 101.4, so vp=10.07v_p = \mathbf{10.07} km/s. At apogee: v2=398600(2/42164−1/24468)=2.62v^2 = 398600(2/42164 - 1/24468) = 2.62, so va=1.62v_a = \mathbf{1.62} km/s. Check against angular momentum: rpvp=6771×10.07=68,184r_p v_p = 6771 \times 10.07 = 68{,}184 and rava=42164×1.62=68,306r_a v_a = 42164 \times 1.62 = 68{,}306. Equal to rounding. The two conservation laws agree — as they must. Escape from 400 km. Escape means a→∞a \to \infty, so the 1/a1/a term vanishes: v2=2μ/rv^2 = 2\mu/r, giving v=10.85v = \mathbf{10.85} km/s.
Figure 1.12
ε < 00 ≤ e < 1circle / ellipsebound — comes backε = 0e = 1parabolajust barely escapesε > 0e > 1hyperbolagone, with speed to spareε = v²/2 − μ/r · its sign alone decides everything
The sign of one number decides the entire character of the trajectory. This table is the bridge into Unit 2, where we will prove that these four cases are the *only* four cases possible.
🛰The Oberth effect, foreshadowed
Look hard at ε=v2/2−μ/r\varepsilon = v^2/2 - \mu/r. Energy depends on the square of speed. That means adding a fixed Δv\Delta v buys you far more energy when you are already moving fast — and you are fastest at perigee, deep in the well. Burning 1 km/s at perigee of a highly elliptical orbit raises your apogee dramatically more than the same 1 km/s spent at apogee. This is the Oberth effect, and it is why interplanetary missions perform their escape burns as low and as fast as possible. We will come back and use it properly in Unit 6 — but the reason is already sitting in this equation, if you look.
↓But there is a problem
We now have two conserved quantities. Angular momentum h⃗\vec{h} — fixes the plane, sets the sweep rate, ties speed to radius. Energy ε\varepsilon — fixes the size, and its sign decides whether you are trapped or free. Here is the claim Unit 2 rests on: those two numbers determine the orbit completely. Not approximately. Not for special cases. Give me hh and ε\varepsilon and I will tell you the exact shape of the path, and prove that no other shape is possible. That is a very strong claim, and it deserves a real proof — one vector differential equation, and a trick of algebra so elegant it looks like cheating.