Vis-viva equation v² = GM(2 - r − 1 - a) — derivation: Question bank

This is a child of the vis-viva derivation. That parent page proves the key fact we lean on here: $ε = \frac{1}{2} v^{2} - \frac{GM}{r} = - \frac{GM}{2 a}$ The middle piece is total specific energy; the right piece is its constant value. We do not re-derive it — but the figures on this page show why it comes out depending only on $a$ .

Before we start, the cast of symbols (so nothing is used unexplained):

$r$ = your instantaneous distance from the central body's centre (changes every second).
$a$ = the semi-major axis, a fixed size-label for the whole orbit.
$v$ = your current speed. $GM$ = strength of the central body's gravity (the "standard gravitational parameter").
$ε = \frac{1}{2} v^{2} - \frac{GM}{r}$ = specific (per-kilogram) total energy — the same number everywhere on one orbit.
$e$ = eccentricity (how squashed the ellipse is), $r_{p} = a (1 - e)$ = closest point, $r_{a} = a (1 + e)$ = farthest point.

True or false — justify

Each line: claim ::: verdict + one-sentence reason.

For a circular orbit

r

and

a

are the same number.

True — a circle's "semi-major axis" is just its radius, so

r = a

everywhere and vis-viva collapses to

v = GM / r

The specific energy

ε

changes as the satellite moves from perigee to apogee.

False —

ε = - GM / (2 a)

depends only on

a

, a fixed orbit property, so it is identical at every point; only the split between kinetic and potential shifts (the flat line in figure s01).

Two orbits with the same

a

but different eccentricity have the same total energy.

True — energy depends only on

a

, not on shape, so a near-circular and a very elongated orbit of equal

a

carry equal

ε

(exactly the three curves in figure s02).

A more eccentric orbit is always a higher-energy orbit.

False — eccentricity does not appear in

ε = - GM / (2 a)

; only

a

sets the energy, so eccentricity alone tells you nothing about energy.

Doubling your speed at a point doubles your orbital energy.

False — energy has a

\frac{1}{2} v^{2}

kinetic term, so doubling

v

quadruples the kinetic part, and the potential term is unchanged, so energy does not simply double.

On an elliptical orbit the satellite moves fastest at perigee.

True — perigee has the smallest

r

, so

2/ r

is largest, making

v^{2}

largest there.

For a bound orbit the specific energy

ε

is negative.

True — bound means you cannot reach infinity, so the total energy sits below the zero-energy escape threshold;

ε = - GM / (2 a) < 0

Escape from a body needs a speed exactly twice the local circular speed.

False — it needs

2

times the circular speed, since

v_{esc}^{2} = 2 GM / r

while

v_{c i r c}^{2} = GM / r

A parabolic escape trajectory has

ε = 0

True — it just barely reaches infinity with zero leftover speed, so kinetic plus potential energy sums to exactly the boundary value zero.

For a hyperbolic orbit the semi-major axis

a

is negative.

True — by convention

a < 0

for unbound orbits, which makes

- 1/ a > 0

so that

v^{2} > 2 GM / r

, i.e. speed above escape.

Spot the error

Each line names a flawed line of reasoning; the answer states what breaks.

"Since

r = a

works for a circle, I can always use

r

in place of

a

The equality

r = a

holds everywhere only for a circle; on an ellipse

r

sweeps from

r_{p}

r_{a}

while

a

stays fixed, so it equals

a

at exactly two points (where the orbit crosses radius

a

) and differs everywhere else — substituting one for the other is wrong in general.

"At apogee the orbit is slow, so its energy is lower there than at perigee."

Energy is conserved and identical at both apsides; at apogee kinetic energy is low but potential energy is correspondingly high, so the sum is unchanged.

a

is a length, and lengths are positive, so

a > 0

always."

For hyperbolic (unbound) orbits

a

is defined to be negative; treating it as positive would wrongly predict

v^{2} < 2 GM / r

for an escaping body.

"I'll use

ε = \frac{1}{2} v^{2} - GM / r

with the full potential energy

- GM m / r

in it."

Those are inconsistent —

ε

is specific (per unit mass), so its potential term is

- GM / r

(no

m

); mixing per-mass and total quantities gives units off by a factor of

m

"The derivation uses only energy conservation."

It uses two conservation laws — energy and angular momentum (

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

); with only energy you cannot solve for the two apsis speeds separately.

"Escape speed depends on which direction you launch."

For the energy threshold only the speed matters, not direction, because kinetic energy

\frac{1}{2} v^{2}

ignores direction;

v_{esc}

is the same for any launch angle (ignoring atmosphere/ground).

"Angular momentum

h

also appears in the final vis-viva formula, so I need

e

to use it."

No — the whole point is that

h

(and hence

e

) cancels in the energy step, leaving

v^{2} = GM (2/ r - 1/ a)

with only

r

and

a

; you never need

e

to apply vis-viva.

"At apogee the velocity has a radial part, so

h = r_{a} v_{a}

is wrong."

At the apsides the radial velocity is exactly zero (distance is momentarily at a turning point), so

v ⊥ r

and

h = r_{a} v_{a}

holds cleanly there.

Why questions

Why does the total orbital energy depend only on

a

and not on eccentricity?

In the derivation the apogee speed is fixed by angular momentum, and when you substitute it into the energy expression the eccentricity-carrying factor

(r_{a} - r_{p})

cancels top-and-bottom (a difference-of-squares cancellation), leaving pure

- GM / (2 a)

— visually, figure s02 shows the flat energy line staying put as

e

changes.

Why is gravitational potential energy taken as negative?

It is measured relative to zero at infinity; since you must add energy to climb out to infinity, a bound body sits in an energy "well" below zero, hence negative (the bowl in figure s01).

Why does the derivation evaluate energy specifically at perigee and apogee?

At those two points the velocity is purely tangential, killing the awkward radial component, so

v

appears cleanly and angular momentum reads simply as

r v

Why does setting

a \to \infty

give the escape condition?

An infinite semi-major axis makes

- 1/ a \to 0

ε \to 0

, exactly the boundary between bound (negative energy) and free (positive energy) motion — the rim of the well.

Why is escape speed

2

times circular speed rather than some other factor?

Circular energy is

ε = - GM / (2 r)

while escape needs

ε = 0

; raising kinetic energy from

GM / (2 r)

GM / r

doubles

v^{2}

, and

2

is the doubling of speed.

Why can vis-viva be called "conservation of energy in disguise"?

It is literally

\frac{1}{2} v^{2} - GM / r = - GM / (2 a)

rearranged; the left side is total specific energy and the right side is its constant value

ε

Why do a slow apogee pass and a fast perigee pass carry the same energy?

Kinetic energy lost as the satellite climbs (rising

r

) is exactly gained back as potential energy, keeping the conserved sum constant.

Edge cases

At which points on an ellipse does

r

actually equal

a

At exactly two points — where the orbit crosses the circle of radius

a

(the ends of the semi-minor axis); there

v^{2} = GM (2/ a - 1/ a) = GM / a

, the same speed a circular orbit of radius

a

would have.

What happens to vis-viva at the exact parabolic boundary

a \to \infty

The

1/ a

term vanishes, leaving

v^{2} = 2 GM / r

— the escape-speed profile, valid at every

r

along the parabola.

For a hyperbola, what is the speed far from the body (

r \to \infty

2/ r \to 0

, so

v^{2} \to GM (- 1/ a) = - GM / a

; since

a < 0

this is positive, the leftover "hyperbolic excess speed" the body keeps at infinity.

Can

v^{2}

come out negative from vis-viva, and what would that mean?

Yes if you plug an

r

larger than the orbit's apogee (or beyond a bound orbit's reach); a negative

v^{2}

signals a physically forbidden point — the body never actually reaches that

r

on that orbit.

What is the degenerate

e = 1

radial (straight-line drop) case, and does vis-viva still work?

When

e \to 1

with finite

a

the ellipse collapses to a line through the focus (

r_{p} \to 0

), and the body falls straight in and back out; energy conservation

\frac{1}{2} v^{2} - GM / r = - GM / (2 a)

still gives the correct speed at each

r

, but angular momentum is zero so

h = r_{p} v_{p}

degenerates — the orbit shape breaks down even though the vis-viva speed stays valid.

At perigee versus apogee, which term of

ε

dominates?

At perigee the kinetic term

\frac{1}{2} v^{2}

is large and potential is deeply negative; at apogee kinetic is small and potential is shallower — their sum stays fixed at

- GM / (2 a)

Is there an orbit where

r

never changes?

Yes — the circle, where

r = a

is constant, so

v

is constant too; it is the one case where "instantaneous distance" and "orbit size" coincide permanently.

What is the specific energy of a low circular orbit compared to a high one?

The low orbit has smaller

a

, so

ε = - GM / (2 a)

is more negative (deeper well); higher orbits are less negative, i.e. closer to escape.