Hohmann Δv calculation — both maneuvers: Exercises

Throughout, we reuse the toolkit:

Circular speed (from Circular orbital velocity): $v_{circ} = μ / r$ .
Vis-viva equation: $v^{2} = μ (\frac{2}{r} - \frac{1}{a})$ .
Transfer ellipse: $a_{t} = \frac{r _{1} + r _{2}}{2}$ .
$μ$ is the Standard gravitational parameter $= GM$ of the central body.

Unless a problem says otherwise, use $μ_{\oplus} = 3.986 \times 1 0^{5} km^{3} / s^{2}$ for Earth.

Figure — Hohmann Δv calculation — both maneuvers

That picture is the map for every problem: an inner circle (radius $r_{1}$ , chalk blue), an outer circle (radius $r_{2}$ , pale yellow), and the pink transfer ellipse touching both. Burn 1 happens at the pink dot on the inner circle; burn 2 at the pink dot on the outer circle.

Level 1 — Recognition

L1.1

Which two speeds does $Δ v_{1}$ compare, and at which radius?

Recall Solution

WHAT: $Δ v_{1} = v_{p} - v_{c 1}$ , both evaluated at radius $r_{1}$ . WHY: Burn 1 is fired on the inner circle. Before the burn you move at the circular speed $v_{c 1} = μ / r_{1}$ ; after the burn you are on the ellipse, so your speed is the ellipse's periapsis speed $v_{p}$ . Both are "at $r_{1}$ ", so the two velocity vectors are parallel (tangential) and the vector subtraction collapses to a subtraction of speeds. Answer: it compares the circular speed $v_{c 1}$ and the transfer periapsis speed $v_{p}$ , both at $r_{1}$ .

L1.2

For $r_{2} > r_{1}$ , is the transfer ellipse faster or slower than the outer circle at apoapsis? Therefore, is burn 2 prograde (speed up) or retrograde (slow down)?

Recall Solution

WHAT: Compare $v_{a}$ (ellipse at apoapsis) with $v_{c 2}$ (outer circle). WHY: At apoapsis the spacecraft is at its slowest point on the ellipse (farthest from the focus, energy traded into height). The circle at that same radius needs more speed to stay circular. So $v_{a} < v_{c 2}$ . Answer: the ellipse is slower at apoapsis; burn 2 must speed you up (prograde), giving $Δ v_{2} = v_{c 2} - v_{a} > 0$ .

Level 2 — Application

L2.1

Earth transfer with $r_{1} = 7000$ km, $r_{2} = 20000$ km. Compute $a_{t}$ , $v_{c 1}$ , $v_{p}$ , and $Δ v_{1}$ .

Recall Solution

Step 1 — $a_{t}$ : $a_{t} = \frac{7000 + 20000}{2} = 13500$ km. (midpoint of periapsis + apoapsis) Step 2 — $v_{c 1}$ : $v_{c 1} = 398600/7000 = 7.546$ km/s. Step 3 — $v_{p}$ (vis-viva at $r = r_{1}$ ): $v_{p} = 398600 (\frac{2}{7000} - \frac{1}{13500}) = 9.187 km/s .$ Step 4 — $Δ v_{1}$ : $9.187 - 7.546 = 1.641$ km/s.

L2.2

Same orbits as L2.1. Compute $v_{c 2}$ , $v_{a}$ , and $Δ v_{2}$ , then the total.

Recall Solution

$v_{c 2}$ : $398600/20000 = 4.465$ km/s. $v_{a}$ (vis-viva at $r = r_{2}$ ): $398600 (\frac{2}{20000} - \frac{1}{13500}) = 3.213$ km/s. $Δ v_{2}$ : $4.465 - 3.213 = 1.252$ km/s. Total: $1.641 + 1.252 = 2.893$ km/s.

L2.3

Use the factored forms to confirm $Δ v_{1}$ from L2.1 without computing $v_{p}$ separately: $Δ v_{1} = \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) .$

Recall Solution

WHAT: substitute directly. WHY: the factored form is algebraically identical to $v_{p} - v_{c 1}$ ; it just hides the vis-viva step. $Δ v_{1} = 7.546 (\frac{2 \cdot 20000}{27000} - 1) = 7.546 (1.2172 - 1) = 1.641 km/s .$ Matches L2.1 exactly. ✔

Level 3 — Analysis

L3.1

For LEO→GEO ( $r_{1} = 6678$ km, $r_{2} = 42164$ km) the parent found $Δ v_{1} \approx 2.42$ , $Δ v_{2} \approx 1.47$ km/s. Explain in energy terms why the first burn is the larger one, and state which physical effect this connects to.

Recall Solution

WHAT: $Δ v_{1} > Δ v_{2}$ even though the outer orbit is slower. WHY: Specific orbital energy is $ε = - μ /2 a$ (see Specific orbital energy). The transfer ellipse has a much larger $a_{t}$ than circle 1, so burn 1 must inject a large energy jump. Because you are deep in the gravity well at $r_{1}$ (speeds large), a given $Δ v$ there buys a lot of energy — but you also need a lot of energy, so the required $Δ v$ is big. Burn 2 only tweaks an already-high orbit up to circular, a smaller energy change. Connection: this is exactly the reasoning behind the Oberth effect — burns deep in the well, where speed is high, are the energy-efficient ones, and also the ones that dominate the budget when raising an orbit.

L3.2

Show algebraically that $Δ v_{1} = v_{c 1} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1)$ is always positive for any $r_{2} > r_{1}$ , and that it equals zero when $r_{2} = r_{1}$ .

Recall Solution

WHAT: analyse the bracket $B = \frac{2 r _{2}}{r _{1} + r _{2}} - 1$ . WHY $B > 0$ : $B > 0 ⟺ \frac{2 r _{2}}{r _{1} + r _{2}} > 1 ⟺ 2 r_{2} > r_{1} + r_{2} ⟺ r_{2} > r_{1}$ . True by assumption, and $v_{c 1} > 0$ , so $Δ v_{1} > 0$ . ✔ Degenerate case $r_{2} = r_{1}$ : then $\frac{2 r _{2}}{r _{1} + r _{2}} = \frac{2 r _{1}}{2 r _{1}} = 1$ , so $B = 0$ and $Δ v_{1} = 0$ . Physical meaning: no transfer needed — you are already on the target circle. The math self-cancels correctly.

L3.3

Does raising to a larger $r_{2}$ always increase $Δ v_{1}$ ? What is the limit of $Δ v_{1}$ as $r_{2} \to \infty$ ?

Recall Solution

WHAT: behaviour of $Δ v_{1} (r_{2})$ . Monotonic? The factor $\frac{2 r _{2}}{r _{1} + r _{2}} = \frac{2}{r _{1} / r _{2} + 1}$ increases as $r_{2}$ grows (denominator shrinks), so $Δ v_{1}$ increases monotonically with $r_{2}$ . Yes — always more. Limit: as $r_{2} \to \infty$ , $\frac{2 r _{2}}{r _{1} + r _{2}} \to 2$ , so $Δ v_{1} \to v_{c 1} (2 - 1) .$ Physical meaning: this is the escape-like periapsis kick — pushing to infinity from the inner circle costs the escape increment $(2 - 1) v_{c 1}$ , since escape speed is $2 v_{c 1}$ . Beyond that, $Δ v_{1}$ never grows further; the extra cost of an ever-larger orbit is bounded.

Level 4 — Synthesis

L4.1

A Mars orbiter: $μ_{Mars} = 4.283 \times 1 0^{4} km^{3} / s^{2}$ , $r_{1} = 3690$ km (low Mars orbit), $r_{2} = 20428$ km (areostationary). Find the total Hohmann Δv.

Recall Solution

$a_{t}$ : $(3690 + 20428) /2 = 12059$ km. $v_{c 1}$ : $42830/3690 = 3.407$ km/s. $v_{p}$ : $42830 (\frac{2}{3690} - \frac{1}{12059}) = 4.439$ km/s. $Δ v_{1}$ : $4.439 - 3.407 = 1.032$ km/s. $v_{c 2}$ : $42830/20428 = 1.448$ km/s. $v_{a}$ : $42830 (\frac{2}{20428} - \frac{1}{12059}) = 0.802$ km/s. $Δ v_{2}$ : $1.448 - 0.802 = 0.646$ km/s. Total: $1.678$ km/s. (Notice again $Δ v_{1} > Δ v_{2}$ — same deep-well logic as L3.1, now around Mars.)

L4.2

Lowering an orbit: reverse L2.1 — go from $r_{1} = 20000$ km down to $r_{2} = 7000$ km around Earth. Compute both burns. Are they prograde or retrograde? Compare the total to L2.2.

Recall Solution

WHAT: now the target is inside. The transfer ellipse has apoapsis at the starting (outer) radius and periapsis at the target (inner) radius, so $a_{t}$ is unchanged: $13500$ km. Burn 1 (at outer radius $20000$ km): you must drop onto the ellipse, whose speed there ( $v_{a} = 3.213$ km/s) is less than the circular speed there ( $4.465$ km/s). So you slow down: $Δ v_{1} = 4.465 - 3.213 = 1.252$ km/s — retrograde. Burn 2 (at inner radius $7000$ km): you arrive at the ellipse's periapsis ( $v_{p} = 9.187$ km/s), faster than the inner circle ( $7.546$ km/s), so you slow down again: $Δ v_{2} = 9.187 - 7.546 = 1.641$ km/s — retrograde. Total: $2.893$ km/s — identical to L2.2. WHY identical: the total Δv budget of a Hohmann transfer is symmetric — lowering costs exactly what raising cost, just with both burns retrograde instead of prograde.

Level 5 — Mastery

L5.1

Prove the symmetry result from L4.2 in general: the total Δv to go $r_{1} \to r_{2}$ equals the total to go $r_{2} \to r_{1}$ .

Recall Solution

WHAT: show $Δ v_{tot}$ is unchanged when you swap $r_{1} \leftrightarrow r_{2}$ . WHY it works: for raising, the two burns compare {circle, ellipse} at $r_{1}$ and at $r_{2}$ . For lowering, the same ellipse (same $a_{t}$ , since $a_{t}$ is symmetric in $r_{1}, r_{2}$ ) is ridden, and the two burns again compare {circle, ellipse} at $r_{2}$ and at $r_{1}$ — the same two speed gaps, just executed retrograde. Formally the raise total is $∣ v_{p} - v_{c 1} ∣ + ∣ v_{c 2} - v_{a} ∣,$ and the lower total is $∣ v_{a} - v_{c 2} ∣ + ∣ v_{c 1} - v_{p} ∣,$ which are term-by-term equal because $∣ x ∣ = ∣ - x ∣$ . Hence equal totals. ∎

L5.2

A student proposes skipping burn 2: fire only burn 1 to reach apoapsis at $r_{2}$ , then just "let it be an ellipse." Using $r_{1} = 6678$ , $r_{2} = 42164$ km, explain what orbit results, and compute the Δv you'd still need later to circularize.

Recall Solution

WHAT happens with only burn 1: the craft is on the transfer ellipse — periapsis $r_{1}$ , apoapsis $r_{2}$ — forever coasting between them (an Impulsive maneuver approximation treats each burn as instantaneous, so with no second burn the ellipse simply repeats). It is not on GEO. At apoapsis its speed is $v_{a} = 398600 (2/42164 - 1/24421) = 1.609$ km/s, but a GEO circle needs $v_{c 2} = 398600/42164 = 3.075$ km/s. Δv still owed: $Δ v_{2} = 3.075 - 1.609 = 1.467$ km/s — precisely the second burn. Skipping it doesn't save fuel; it just leaves you on an ellipse, and you owe the same $1.467$ km/s whenever you finally circularize.

L5.3

Limiting sanity check. For LEO ( $r_{1} = 6678$ km, $v_{c 1} = 7.726$ km/s), verify numerically that $Δ v_{1} \to v_{c 1} (2 - 1)$ as $r_{2} \to \infty$ , and interpret.

Recall Solution

Target value: $v_{c 1} (2 - 1) = 7.726 \times 0.41421 = 3.200$ km/s. Check with a huge $r_{2}$ , say $r_{2} = 1 0^{10}$ km: $Δ v_{1} = 7.726 (\frac{2 \cdot 1 0 ^{10}}{6678 + 1 0 ^{10}} - 1) \approx 7.726 (2 - 1) = 3.200 km/s .$ Interpretation: escape speed from LEO is $2 v_{c 1} = 10.926$ km/s, and $Δ v_{1}$ from circular is $(2 - 1) v_{c 1} = 3.200$ km/s. Reaching infinitely far costs no more first-burn Δv than barely escaping — a beautiful bound that motivates why, past some radius ratio, the Bi-elliptic transfer (which flings out past $r_{2}$ and back) can undercut Hohmann.

Recall Self-test summary (reveal after finishing)

Every burn here was a speed match between the two orbits that actually meet at a radius. Raising means both burns are prograde. ::: Lowering means both burns are retrograde, with the same total. Which radius sits in the numerator of $2 r / (r_{1} + r_{2})$ ? ::: The radius you are reaching toward — near burn uses $r_{2}$ on top, far burn uses $r_{1}$ on top. What does $Δ v_{1}$ approach as $r_{2} \to \infty$ ? ::: $v_{c 1} (2 - 1)$ , the escape-like periapsis kick.