Specific orbital energy ε = −GM - 2a: Visual walkthrough

Step 0 — The cast of characters (earn every symbol)

WHAT. Before any equation, let us name the pieces on a drawing.

WHY. The contract of this whole page: no letter appears until you can point to it in a figure.

PICTURE.

Figure — Specific orbital energy ε = −GM - 2a

Look at the drawing:

The big dot is the planet, mass $M$ (a number of kilograms). $M$ = "how strong the puller is."
The small dot is the satellite. Its own mass is $m$ , but watch — $m$ will cancel out and vanish, which is the whole point of the word "specific."
The straight chalk-blue line from planet-centre to satellite has length $r$ — the distance, measured in metres. Not from the surface; from the centre.
The pink arrow is the velocity $v$ : how fast (its length) and which way (its direction) the satellite moves right now.
$G$ is the gravitational constant, a fixed number of nature ( $6.674 \times 1 0^{- 11}$ in SI units) that sets the strength of gravity everywhere.

Step 1 — The energy account never changes

WHAT. We claim $ε$ is the same number at every point of the orbit.

WHY. Gravity is a conservative force: the work it does depends only on where you start and end, never the wiggly path between. That is exactly the condition for "kinetic + potential = constant." So as the satellite swings around, the two pieces trade back and forth but their sum is frozen. This is the gravitational PE doing its bookkeeping.

PICTURE.

Watch the two bars. Near the planet (left) the pink motion bar is tall and the blue position bar is deep — the satellite is fast and low. Far away (right) the motion bar shrinks and the position bar rises toward zero — slow and high. The white line marked $ε$ — the sum — stays flat across the whole trip. That flat line is the conserved quantity.

Step 2 — Two special points where the arrow is "sideways"

WHAT. Pick the two turning points of the ellipse: perigee (closest, distance $r_{p}$ ) and apogee (farthest, distance $r_{a}$ ).

WHY. At these two points only, the velocity arrow is exactly perpendicular to the distance line. That makes them the cleanest places to write energy — no awkward angle between $v$ and $r$ . We equate $ε$ at both because Step 1 says it is the same value.

PICTURE.

The ellipse: planet at the pale-yellow focus. Left tip = perigee, the pink arrow $v_{p}$ points straight up (perpendicular to the short blue line $r_{p}$ ). Right tip = apogee, arrow $v_{a}$ points straight down (perpendicular to the long blue line $r_{a}$ ). Writing $ε$ at each and setting them equal:

$fast, close \frac{v _{p}^{2}}{2} - deep well \frac{GM}{r _{p}} = slow, far \frac{v _{a}^{2}}{2} - shallow well \frac{GM}{r _{a}}$

Each term labelled: left side is the account at perigee, right side at apogee — equal because the balance never changes.

Step 3 — The second law: angular momentum ties the two speeds

WHAT. We get a second equation linking $v_{p}$ and $v_{a}$ : $r_{p} v_{p} = r_{a} v_{a}$ .

WHY. Gravity always points along the line to the planet — straight at the centre. A force through the centre exerts zero twist (torque). No twist means the "spin quantity" — angular momentum per kilogram — is conserved. It equals distance times the sideways speed. At perigee and apogee the arrow is already perpendicular, so the whole speed is sideways, and that spin quantity is simply $r_{p} v_{p}$ at one tip and $r_{a} v_{a}$ at the other. Equal spin gives $r_{p} v_{p} = r_{a} v_{a}$ .

PICTURE.

Two shaded "sweep" wedges of equal area — one thin-and-long at perigee, one fat-and-short at apogee. Equal areas in equal time is Kepler's flavour of the same law. Rearranged:

$v_{a} = v_{p} \cdot small / big < 1 \frac{r _{p}}{r _{a}}$

The label shows why $v_{a}$ is smaller: at apogee the lever arm $r_{a}$ is long, so the speed must be short to keep the product fixed.

Step 4 — Solve for the perigee speed (pure algebra, shown honestly)

WHAT. Put $v_{a} = v_{p} r_{p} / r_{a}$ into the energy equation of Step 2 and isolate $v_{p}^{2}$ .

WHY. We have two equations and two unknown speeds. Eliminating $v_{a}$ leaves $v_{p}^{2}$ written in geometry alone ( $r_{p}$ , $r_{a}$ , $GM$ ) — no speeds left to chase.

PICTURE.

The figure is a "flow" of the cancellation. Substituting gives $\frac{v _{p}^{2}}{2} (1 - \frac{r _{p}^{2}}{r _{a}^{2}}) = GM (\frac{1}{r _{p}} - \frac{1}{r _{a}}) .$ The left bracket factors as $\frac{( r _{a} - r _{p} ) ( r _{a} + r _{p} )}{r _{a}^{2}}$ and the right as $\frac{r _{a} - r _{p}}{r _{p} r _{a}}$ . The red slash in the figure kills the common factor $(r_{a} - r_{p})$ on both sides, leaving

$v_{p}^{2} = \frac{2 GM r _{a}}{r _{p} ( r _{a} + r _{p} )}$

Every symbol earned: $2 GM$ = strength, $r_{a}$ on top / $r_{p} (r_{a} + r_{p})$ below = the geometry of the ellipse.

Step 5 — Feed it back and watch the speed vanish

WHAT. Substitute that $v_{p}^{2}$ into $ε = \frac{v _{p}^{2}}{2} - \frac{GM}{r _{p}}$ .

WHY. $ε$ is what we actually want, and Step 4 removed $v$ . Plugging in must collapse everything to pure distances.

PICTURE.

$ε = v_{p}^{2} /2 \frac{GM r _{a}}{r _{p} ( r _{a} + r _{p} )} - PE at perigee \frac{GM}{r _{p}} = \frac{GM}{r _{p}} \cdot \frac{r _{a} - ( r _{a} + r _{p} )}{r _{a} + r _{p}} = \frac{GM}{r _{p}} \cdot \frac{- r _{p}}{r _{a} + r _{p}} .$

The $r_{p}$ cancels (shown crossed out in chalk), leaving the clean intermediate result:

$ε = - \frac{GM}{r _{a} + r _{p}}$

Read the labels: numerator over the sum of the two extreme distances. Almost there.

Step 6 — The ellipse's own geometry finishes it

WHAT. Replace $r_{a} + r_{p}$ by $2 a$ .

WHY. Look at the ellipse end-to-end: the perigee distance plus the apogee distance is the full major axis, which by definition is $2 a$ (twice the semi-major axis $a$ ). This is pure shape, no physics.

PICTURE.

The chalk ruler runs tip to tip: the closest gap $r_{p}$ and the far gap $r_{a}$ butt together to span the whole long axis $= 2 a$ . Substituting $r_{a} + r_{p} = 2 a$ :

$ε = - \frac{GM}{2 a} ■$

The result the parent promised — energy depends on $a$ alone (times $GM$ ). Shape, speed, and the satellite's mass all cancelled on the way.

Step 7 — Every case, no gaps

WHAT. Check the sign of $ε$ across all orbit shapes, including the degenerate ones.

WHY. A picture of only ellipses would leave the reader stranded at escape or fly-by. We cover all of the conics.

PICTURE.

Ellipse (bound): $a > 0 \Rightarrow ε = - \frac{GM}{2 a} < 0$ . Nested, comes back. Circle is the special case $r_{p} = r_{a} = a$ .
Parabola (escape edge): $a \to \infty \Rightarrow ε \to 0$ . Just enough to coast to infinity arriving with zero speed. Set $ε = 0$ in vis-viva to read off escape speed $v_{esc} = 2 GM / r$ .
Hyperbola (unbound fly-by): $ε > 0 \Rightarrow a = - GM /2 ε < 0$ . Negative $a$ is the honest signature of leaving forever — not a mistake.

The one-picture summary

One frame chains it all: (1) flat conserved line $ε$ → (2) equate at perigee & apogee → (3) angular momentum links the speeds → (4–5) algebra erases $v$ giving $- GM / (r_{a} + r_{p})$ → (6) ellipse geometry $r_{a} + r_{p} = 2 a$ → (7) the box $ε = - GM /2 a$ , with the sign strip showing ellipse/parabola/hyperbola. See Kepler's Third Law and Hohmann transfer orbit for where this single number does real work.

Recall Feynman retelling of the whole walkthrough

A satellite has a bank balance made of speed-money and height-money. Gravity is an honest banker — it never adds or removes money, only lets the two kinds swap, so the total is a flat line for the whole trip. I peek at the balance at the two easy spots, the nearest and farthest points, where the velocity arrow points cleanly sideways, and I set the two balances equal. A second rule — spin can't leak because gravity pulls dead-centre — tells me the far point must move slower in exact proportion to how far it is. With those two facts I can chase the speed out of the equations entirely; it cancels, and I'm left with just the two extreme distances. Then I notice those two distances, laid end to end, are the full length of the loop, $2 a$ . So the balance is simply $- GM /2 a$ : bigger loop, richer (less negative) account. If the account ever climbs to zero, the satellite waves goodbye and never returns.

Active-recall

At which two points is

v ⊥ r

, letting us write energy cleanly?

Perigee and apogee (the turning points of the ellipse).

Why does the satellite's own mass

m

disappear?

We use energy per kilogram ("specific"), so

m

cancels.

Which conservation law gives

r_{p} v_{p} = r_{a} v_{a}

Conservation of angular momentum (central force ⇒ zero torque).

What geometric fact turns

r_{a} + r_{p}

into

2 a

The closest and farthest distances together span the full major axis

= 2 a

What does negative

a

signal?

An unbound hyperbolic trajectory, since

a = - GM /2 ε

with

ε > 0