Hohmann transfer — derivation, minimum energy transfer: Question bank

Every item below leans on three tools you must already trust: the Vis-viva equation ( $v^{2} = μ (2/ r - 1/ a)$ ), Kepler's Third Law (period depends only on $a$ ), and the link between Orbital energy & semi-major axis ( $ε = - μ /2 a$ ). If any of those symbols feel unearned, read those notes first — this page uses them, it doesn't rebuild them.

True or false — justify

Cover the answers. Speak the justification, not just the verdict.

At apoapsis of the transfer ellipse you are moving faster than the outer circular speed.

False. At apoapsis you are at the slow end of the ellipse and are below the circular speed for that radius (

v_{a} < v_{c 2}

); that gap is exactly why a second prograde burn is required.

Raising your orbit means you must slow down because you are "climbing."

False. You raise an orbit by adding energy, i.e. burning prograde to speed up; the coasting slow-down happens on its own as you climb, but every intentional burn in a raise is an acceleration.

Both Hohmann burns are prograde when going from a smaller to a larger orbit.

True. The periapsis burn adds energy to stretch the circle into the ellipse, and the apoapsis burn adds energy again to circularize higher — both point along velocity.

The Hohmann transfer is the cheapest possible way to move between any two circular orbits.

False in general. It is the cheapest two-impulse tangential transfer, and it wins for radius ratios up to

r_{2} / r_{1} \approx 11.94

; beyond that a Bi-elliptic transfer can cost less.

On the transfer ellipse you should plug

a = r_{1}

into vis-viva at periapsis.

False. On the ellipse

a = a_{t} = (r_{1} + r_{2}) /2

everywhere; only the radius

r

changes between

r_{1}

and

r_{2}

. Using

a = r

silently turns your ellipse back into a circle.

A Hohmann transfer works between two orbits that are in different planes.

False as stated. The classic derivation assumes coplanar orbits; a plane change needs an extra out-of-plane

Δ v

that adds to the Delta-v budget.

Because velocities are tangent at both burns, you may subtract speeds as plain numbers.

True. Tangency means the old and new velocity vectors are collinear, so

Δ v = ∣ v_{new} ∣ - ∣ v_{old} ∣

is exact — the very reason tangent burns are chosen.

The transfer time depends on the launch site or the mass of the spacecraft.

False. Half-period is

π a_{t}^{3} / μ

: it depends only on

a_{t}

and the central body's

μ

, not on the vehicle's mass (Two-body problem cancels the small mass).

A Hohmann transfer can also lower an orbit.

True. Run it in reverse: burn retrograde at the outer circle to drop into an ellipse whose apoapsis is

r_{2}

and periapsis

r_{1}

, then retrograde again at periapsis to circularize lower. Both burns are then retrograde.

Spot the error

Each line contains a flawed statement; the reveal names the flaw.

"You brake at apoapsis to slow into the bigger circle."

The error is the word brake. At apoapsis

v_{a} < v_{c 2}

, so you must fire prograde to speed up; braking would drop you back toward periapsis.

"The transfer ellipse has semi-major axis

r_{2}

since apoapsis reaches

r_{2}

Confuses apoapsis radius with semi-major axis. The full major axis is

r_{1} + r_{2}

, so

a_{t} = (r_{1} + r_{2}) /2

, always less than

r_{2}

"Fire the first burn anywhere on the inner circle; the geometry is symmetric."

The burn point becomes periapsis, so the ellipse's low point is fixed there. That's fine — but its high point (apoapsis) then lands

18 0^{\circ}

away, so timing the departure to make apoapsis coincide with the target does matter (phasing).

"A single burn from the inner circle can put you straight onto the outer circle."

One burn only produces an ellipse tangent to the inner circle; without a second burn at

r_{2}

you'd swing right back down to

r_{1}

. Circles need matched speed and radius.

"Since kinetic energy is

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

, a burn deep in the well and one far out give the same energy gain for the same

Δ v

Ignores the Oberth effect. The gain is

d (\frac{1}{2} v^{2}) = v d v

, which grows with the current speed

v

; the same

Δ v

buys more energy where you're already fast (low, at periapsis).

"Vis-viva says

v

depends on

r

only, so speed is fixed once you pick a radius."

v^{2} = μ (2/ r - 1/ a)

also depends on

a

. At the same

r_{1}

, the circular orbit (

a = r_{1}

) and the transfer ellipse (

a = a_{t}

) have different speeds — that difference is

Δ v_{1}

Why questions

Why must a Hohmann use two impulses rather than one?

One burn changes energy but leaves you on an ellipse that recrosses

r_{1}

; a second burn at

r_{2}

is needed to raise periapsis up to

r_{2}

and make the orbit circular again.

Why place burns tangent to velocity instead of at some angle?

A tangential burn spends all its

Δ v

changing speed (hence energy) and none on redirecting the velocity, so no fuel is wasted turning — this is what makes each burn minimal.

Why is the first burn done at the inner (low) orbit rather than after coasting outward?

The Oberth effect: you are moving fastest deep in the gravity well, so a given

Δ v

there yields the largest gain in orbital energy

ε

Why does the transfer time not depend on

r_{1}

and

r_{2}

separately, only on their sum?

Kepler's Third Law ties period to

a^{3}

, and

a_{t} = (r_{1} + r_{2}) /2

, so half the period

π a_{t}^{3} / μ

sees only the combination

r_{1} + r_{2}

Why is

Δ v_{2}

(the circularizing burn) usually smaller than

Δ v_{1}

for a large raise?

Far out at

r_{2}

every speed is small, so the difference

v_{c 2} - v_{a}

between two small speeds is small; near

r_{1}

the speeds are large so their difference

v_{p} - v_{c 1}

is larger.

Why does bi-elliptic sometimes beat Hohmann only at huge ratios?

Going out far first lets the plane-raising/lowering burns happen where speeds are tiny (extreme Oberth savings), but you pay extra to reach that far point; only when

r_{2} / r_{1} ≳ 11.94

does the saving outweigh the cost.

Why can we treat the spacecraft's mass as irrelevant in every speed formula?

In the Two-body problem the small mass cancels; the equations use specific quantities (energy and

μ = GM

per unit spacecraft mass), so trajectories are mass-independent.

Edge cases

What happens to

Δ v_{total}

and to the ellipse as

r_{2} \to r_{1}

The ellipse degenerates to the original circle,

a_{t} \to r_{1}

, and both burns shrink to zero — you're already there, no maneuver needed.

What does the transfer look like as

r_{2} \to \infty

(escape-like)?

Periapsis speed

v_{p} \to 2 μ / r_{1}

, the local escape speed, so

Δ v_{1}

approaches the escape burn; the ellipse becomes an ever-flatter, ever-longer path and transfer time

\to \infty

r_{2} < r_{1}

(target is inside the start orbit), is it still a Hohmann?

Yes — it's the same maneuver reversed. Periapsis is now the outer burn point's opposite; both burns are retrograde and you lose energy to fall inward.

Is a Hohmann defined between two elliptical (non-circular) orbits?

The classic Hohmann is between two circular coplanar orbits. Transferring between ellipses is a more general problem; the clean two-tangent-burn result assumes circular endpoints.

At exactly

r_{2} / r_{1} = 11.94

, which transfer is cheaper?

They are essentially tied — that ratio is the crossover where Hohmann and the (limiting) bi-elliptic cost the same; below it Hohmann wins, above it bi-elliptic can.

If your engine can only burn for a long time (not an instant), does the ideal Hohmann still apply?

Not exactly. The derivation assumes impulsive (instantaneous) burns; a long finite burn spreads thrust over changing positions, incurring gravity losses that raise the real Delta-v budget above the ideal.

Recall One-line self-test before you leave

Cover this: name the single reason tangent burns let you use scalar (not vector) subtraction. ::: The old and new velocity vectors are collinear at each burn, so only their magnitudes differ.