3.2.20 · D5Orbital Mechanics & Astrodynamics
Question bank — Hohmann Δv calculation — both maneuvers
Vocabulary reminder (all built in the parent parent note):
- = magnitude of the velocity change a burn supplies.
- = circular speeds on the inner/outer circle.
- = transfer-ellipse speeds at periapsis (near) and apoapsis (far), from the vis-viva equation.
- = transfer semi-major axis; = standard gravitational parameter.
True or false — justify
A Hohmann transfer works between any two orbits.
False. It is defined only for two coplanar, circular orbits; different planes or eccentric orbits need extra maneuvers or a modified transfer.
Both Hohmann burns are prograde (speed increases) when raising an orbit.
True. At periapsis you go from up to the faster , and at apoapsis from the slow up to — both are positive speed additions for a raise.
The total Δv equals the difference of the two circular speeds, .
False. You never ride between the two circles directly; you ride an ellipse, so the real cost is circle↔ellipse at each end: and .
For raising an orbit, the first (periapsis) burn is always the larger of the two.
True for the classic LEO→GEO-scale jump, because it injects the energy to reach a distant apoapsis deep in the gravity well; it can flip for modest ratios, so it is a tendency, not a law.
The Hohmann transfer is the cheapest possible two-impulse transfer between circular orbits for all radius ratios.
False. For very large ratios () a bi-elliptic transfer with three burns can beat it in total Δv.
Making the target orbit ever larger makes the total Δv grow without bound.
False. As , (an escape-like kick) and , so the total approaches a finite limit near the escape budget.
The transfer ellipse is tangent to both circles.
True. Tangency (velocity vectors parallel at each end) is exactly what lets each burn be a pure scalar speed change with no wasted sideways thrust.
A Hohmann transfer takes exactly half the period of the transfer ellipse.
True. You coast from periapsis to apoapsis — one half-orbit — so the transfer time is .
Δv is a vector, so you cannot just subtract speeds.
False here. Δv is a vector change in general, but because Hohmann burns are tangential the before/after velocities are parallel, so its magnitude collapses to a plain speed subtraction.
Spot the error
"At apoapsis I'm arriving on the transfer, so ."
The sign is backwards: at apoapsis the ellipse is slower than the outer circle, so you must add speed — .
"I'll compute using because the burn is at ."
Wrong : at periapsis but the orbit is the ellipse, so vis-viva uses , not .
"Δv depends on the spacecraft's mass, so a heavier ship needs more Δv."
No: Δv is a per-unit-mass velocity budget; cancels out of every orbital-speed formula. Mass affects the fuel (via the rocket equation), not the Δv.
", so I'll plug in ."
— the product of and the central body's mass. Using alone omits the planet entirely and gives nonsense.
"The transfer's semi-major axis is ."
That's not even the axis length. The major axis spans periapsis+apoapsis from the focus: , so .
"For lowering an orbit (), the formulas break and both burns become negative."
They don't break; the geometry just reverses — both burns become retrograde (speed decreases), so you take the magnitudes and the total Δv is the same as the reverse raise.
"A burn takes many minutes, so I should model gravity acting during the burn."
For textbook Hohmann we use the impulsive maneuver approximation: the burn is treated as instantaneous at one point, valid when burn time orbital period.
Why questions
Why must there be exactly two burns, no more, no less?
One burn at periapsis raises apoapsis onto ; a second at apoapsis raises periapsis onto to circularize. One burn can only fix one of the two circle conditions.
Why are tangential (prograde) burns the most efficient here?
Because velocity is already parallel to the desired thrust, every bit of Δv adds directly to speed — none is spent turning the velocity vector sideways.
Why is the deep (periapsis) burn the energy-expensive one?
Deep in the well the orbit is fastest, and adding speed where you already move fastest buys the most energy (the Oberth effect) — but reaching a distant apoapsis simply demands a large energy injection there.
Why does the outer (apoapsis) burn tend to be smaller?
The outer circle is slow, and the transfer already coasts up nearly to that speed, so only a small top-up is needed to circularize.
Why does specific orbital energy depend only on , not on where you are in the orbit?
Kinetic and potential energy trade off along the ellipse but their sum is conserved; that constant equals , so it is fixed by size alone (see Specific orbital energy).
Why does vis-viva reduce to the circular-speed formula when ?
Set : , exactly — a circle is just an ellipse whose semi-major axis equals its radius.
Why can't you simply "jump" straight from the inner circle to the outer one?
A single instantaneous position-jump violates orbital mechanics; you must follow a continuous conic path (the ellipse) that physically connects the two radii.
Edge cases
What happens if ?
The transfer ellipse degenerates to the same circle, , and both Δv's are zero — no maneuver needed since you're already on the target orbit.
What if the target is lower, ?
Same machinery, reversed: both burns are retrograde (you slow down at each end). The Δv magnitudes match the corresponding raise, so lowering costs the same as raising between the same two radii.
As , what does approach and what does that represent?
, the extra speed to just barely escape from — the periapsis kick becomes an escape-like burn while .
If is only slightly larger than , roughly how does total Δv behave?
It shrinks toward zero smoothly, since the transfer ellipse is nearly the starting circle; tiny radius changes cost tiny, near-linear Δv.
What if you tried a Hohmann between two circles in different planes?
The tangency and pure-scalar-subtraction reasoning fails; you'd need an added plane-change Δv (a vector addition), making it no longer a simple two-scalar-burn problem.
Does the transfer time change if you use the same radii but a heavier planet (larger )?
Yes: , so a larger shortens the coast time even though the geometry () is unchanged.
Recall One-line self-test
The single fact that fixes every sign error ::: In a raise, both burns speed you up ( at periapsis, at apoapsis), so both Δv's are positive prograde additions.