Abstract
Fuel is taken to buy a fixed return. This paper shows that the same fuel yields energies differing by $\mathbf{3.90}$ depending on where it is burned—the fuel’s own equation contains no place. No new theorem or law is claimed.
First, a $\Delta v$ of $1$ km/s requires a fuel fraction of $\mathbf{0.283469}$, matching $1-e^{-\Delta v/v_{e}}$ (Section 2).
Second, this is the core of the paper. With the same fuel, the specific energy gained differs by $\mathbf{3.000000}$ between $v=3$ km/s and $v=10$ km/s (Section 2).
Third, that ratio matches the closed form $(v_{2}+\Delta v/2)/(v_{1}+\Delta v/2)$ (Section 2).
Fourth, burning from rest gains only $\Delta v^{2}/2$, which is the floor (Section 2).
Fifth, on an orbit of eccentricity $0.6$ the gap is $\mathbf{3.9025}$—closing to $\mathbf{1.019865}$ at eccentricity $0.01$ (Section 3).
Sixth, the separator is whether velocity or energy is being added (Section 4).
Keywords: Oberth effect, Tsiolkovsky rocket equation, spaceflight, orbital mechanics, propellant.
Scope (what is not claimed)
No new theorem or law is claimed—the Oberth effect and the Tsiolkovsky equation are both standard. No mission is planned—real trajectory design, gravity assists and finite burn times are not the subject; only impulsive burns. Energy conservation is not doubted—including the energy carried off by the propellant, the total is conserved. What is counted here is only what remains with the vehicle. One exhaust velocity—$v_{e}=3$ km/s throughout; the choice of specific impulse is not treated. Relativity is excluded—speeds are taken small compared with light. Relation to earlier papers: Paper 483 showed that a radius cancels when two forces share an exponent—this is the converse: a place that does not cancel. Paper 505 showed that the sphere of influence is three spheres—in the same field, here one quantity depends on place. Paper 300 showed that a shared root can be decided—an increment of velocity and an increment of energy are distinct roots. Paper 481 showed that without separators a refutation has no destination—this paper writes one of them. What is added is giving the fuel fraction as $0.283469$ against its closed form, showing a factor of $3.000000$ from the same fuel, confirming the ratio against its closed form, giving $3.9025$ between perihelion and aphelion, and showing the gap closing to $1.019865$.
1. Notation and the Question
Setting: A vehicle moving at speed $v$ receives an impulsive $\Delta v$. The gain in energy per unit mass is
$$\Delta E=\frac{1}{2}(v+\Delta v)^{2}-\frac{1}{2}v^{2} =v\,\Delta v+\frac{\Delta v^{2}}{2}$$while the fuel required follows Tsiolkovsky’s $\Delta v=v_{e}\ln(m_{0}/m_{1})$, which contains no $v$. Below, $\Delta v=1$ km/s and $v_{e}=3$ km/s.
Observation 1.1 Fuel is taken to buy a fixed return. The same fuel, the same gain—one ledger, runs the picture.
Observation 1.2 But the expression above contains $v$. The term $v \Delta v$—absent from the fuel equation, present in the energy equation.
Question: By how much does the same fuel differ in return.
2. The Fuel Is the Same and the Return Is Not
Computation 2.1: Apply $\Delta v=1$ km/s at three speeds.
| Speed $v$ at the burn | Gain in specific energy |
|---|---|
| $3$ km/s | $3.500000\times10^{6}$ J/kg |
| $10$ km/s | $\mathbf{1.050000\times10^{7}}$ J/kg |
| $30$ km/s | $3.050000\times10^{7}$ J/kg |
Observation 2.2 The fuel required is the same in all three. A fraction of $\mathbf{0.283469}$—matching $1-e^{-\Delta v/v_{e}}=0.283468689426$, with $v$ appearing nowhere.
Observation 2.3 Yet the return differs by $3.000000$. $1.050000\times10^{7}$ against $3.500000\times10^{6}$—the same fuel, burned in a different place.
Observation 2.4 The ratio matches its closed form. $3.000000000$ against $(v_{2}+\Delta v/2)/(v_{1}+\Delta v/2)=3.000000000$—an independent check.
Observation 2.5 Burning from rest gains only $\Delta v^{2}/2$. $5.000000\times10^{5}$ J/kg—the floor: even at $v=0$ the gain is not zero.
Observation 2.6 The faster the burn, the smaller that floor is in relative terms. $0.016393$ at $v=30$ km/s—only $1.6$ per cent, the rest being $v \Delta v$.
Observation 2.7 The fuel side responds to $\Delta v$. Doubling $\Delta v$ takes the fraction to $0.486583$—it does not respond to $v$ but does to $\Delta v$, so the equation is doing its work.
3. Measuring Inside an Orbit
Computation 3.1: Apply the same $\Delta v$ at perihelion and aphelion of an orbit of eccentricity $0.6$.
| Position | Speed | Gain in specific energy |
|---|---|---|
| Perihelion | $59.5694$ km/s | $\mathbf{6.006938\times10^{7}}$ J/kg |
| Aphelion | $14.8923$ km/s | $\mathbf{1.539235\times10^{7}}$ J/kg |
| Ratio | $4.0000$ | $\mathbf{3.9025}$ |
Observation 3.2 Burning at perihelion is $3.9025$ times better. The same fuel—only the timing differs.
Observation 3.3 The energy ratio is nearer to one than the speed ratio. $3.9025$ against $4.0000$—the $\Delta v^{2}/2$ term sits equally on both sides and dilutes the ratio.
Computation 3.4: Lower the eccentricity.
| Eccentricity | Perihelion | Aphelion | Ratio |
|---|---|---|---|
| $0.60$ | $59.5694$ km/s | $14.8923$ km/s | $3.902549$ |
| $0.30$ | $40.5897$ km/s | $21.8560$ km/s | $1.837973$ |
| $0.10$ | $32.9282$ km/s | $26.9413$ km/s | $1.218173$ |
| $0.01$ | $30.0840$ km/s | $29.4883$ km/s | $\mathbf{1.019865}$ |
Observation 3.5 Lowering the eccentricity moves the ratio monotonically to one. $3.902549\to1.837973\to1.218173\to1.019865$—with no step rising.
Observation 3.6 At $0.01$ it is still not one. A difference of $1.987\times10^{-2}$—the other side of the boundary: approaching a circle does not extinguish the effect.
Observation 3.7 So the size of the effect is set by the shape of the orbit. The eccentricity—not the fuel and not the vehicle.
4. The Separator
The separator: is velocity being added, or energy.
Observation 4.1 Velocity, and the place is irrelevant. $\Delta v$—Tsiolkovsky’s equation contains no $v$.
Observation 4.2 Energy, and the place matters. $v \Delta v+\Delta v^{2}/2$—$v$ is in it.
Observation 4.3 The two are distinct roots by the criterion of Paper 300. Both may be called “what the burn bought”—one an increment of velocity, one of energy, differing even in units.
Observation 4.4 Conflate them and energy appears from nowhere. The same fuel buying $3.9025$ times more—in fact the propellant carries off a different amount depending on where the burn happens.
Observation 4.5 Paper 483 runs the other way. There, matching exponents made the radius cancel—here, the $v$ one might wish to cancel does not. What cancels is decided by where it sits in the expression.
Observation 4.6 This is one of the separators of Paper 481. When “the same fuel gave a different result” is reported—whether the measurement is poor or the quantities compared were different sends the repair elsewhere.
Observation 4.7 To be honest—only what remains with the vehicle is counted. The energy carried off by the propellant is not in the ledger—including it, the total is conserved. The “gain” belongs to the vehicle alone, and no energy appears from nowhere.
5. Roll Call and Closing
- Fuel fraction $0.283469$ (Observation 2.2): containing no $v$.
- A factor of $3.000000$ from the same fuel (Observation 2.3): the core.
- The ratio matches its closed form (Observation 2.4): an independent check.
- $5.000000\times10^{5}$ J/kg even from rest (Observation 2.5): the floor.
- The floor is $1.6$ per cent at $30$ km/s (Observation 2.6): the rest is $v\Delta v$.
- $3.9025$ between perihelion and aphelion (Computation 3.1): eccentricity $0.6$.
- Nearer to one than the speed ratio (Observation 3.3): diluted by $\Delta v^{2}/2$.
- $1.019865$ at eccentricity $0.01$ (Computation 3.4): monotone approach.
- The separator is velocity or energy (Observation 4.3): distinct roots by Paper 300.
- Only the vehicle’s side was counted (Observation 4.7): the total is conserved.
Closing: Fuel is taken to buy a fixed return—the same fuel, the same gain, one ledger. But $\Delta E=v \Delta v+\Delta v^{2}/2$ contains $v$, which is absent from the fuel equation and present in the energy equation. The fuel is the same and the return is not—applying $\Delta v=1$ km/s gives $3.500000\times10^{6}$ J/kg at $v=3$ km/s, $\mathbf{1.050000\times10^{7}}$ at $10$ km/s and $3.050000\times10^{7}$ at $30$ km/s. The fuel required is the same throughout, a fraction of $\mathbf{0.283469}$ matching $1-e^{-\Delta v/v_{e}}=0.283468689426$, with $v$ appearing nowhere. Yet the return differs by $\mathbf{3.000000}$, and the ratio matches $(v_{2}+\Delta v/2)/(v_{1}+\Delta v/2)$. Burning from rest gains only $\Delta v^{2}/2=5.000000\times10^{5}$ J/kg, the floor. The faster the burn, the smaller that floor in relative terms—only $1.6$ per cent at $30$ km/s. The fuel fraction does not respond to $v$ but does to $\Delta v$, moving to $0.486583$ when $\Delta v$ doubles. Measuring inside an orbit—on an orbit of eccentricity $0.6$, perihelion at $59.5694$ km/s gains $\mathbf{6.006938\times10^{7}}$ J/kg and aphelion at $14.8923$ km/s gains $\mathbf{1.539235\times10^{7}}$, a ratio of $\mathbf{3.9025}$. The same fuel; only the timing differs. The energy ratio $3.9025$ is nearer to one than the speed ratio $4.0000$, the $\Delta v^{2}/2$ term sitting equally on both sides and diluting it. Lowering the eccentricity through $0.60$, $0.30$, $0.10$ and $0.01$ gives $3.902549$, $1.837973$, $1.218173$ and $\mathbf{1.019865}$, monotonically approaching one. At $0.01$ a difference of $1.987\times10^{-2}$ remains, so approaching a circle does not extinguish the effect. The size of the effect is set by the shape of the orbit, not by the fuel or the vehicle. One thing separates them—is velocity being added, or energy. Velocity, and the place is irrelevant: Tsiolkovsky’s equation contains no $v$. Energy, and the place matters: $v \Delta v+\Delta v^{2}/2$ has $v$ in it. The two are distinct roots by the criterion of Paper 300—both may be called “what the burn bought”, but one is an increment of velocity and one of energy, differing even in units. Conflate them and the same fuel buying $3.9025$ times more looks like energy from nowhere, when in fact the propellant carries off a different amount depending on where the burn happens. Paper 483 runs the other way: there matching exponents made the radius cancel, while here the $v$ does not. What cancels is decided by where it sits in the expression. This is one of the separators of Paper 481: when “the same fuel gave a different result” is reported, whether the measurement is poor or the quantities compared were different sends the repair elsewhere. To be honest—only what remains with the vehicle is counted. The energy carried off by the propellant is not in the ledger, and including it the total is conserved. The “gain” belongs to the vehicle alone, and no energy appears from nowhere.
References
- Battin, R. H. (1999). An Introduction to the Mathematics and Methods of Astrodynamics (rev. ed.). AIAA.
- Oberth, H. (1929). Wege zur Raumschiffahrt. R. Oldenbourg, München.
- Tsiolkovsky, K. E. (1903). Исследование мировых пространств реактивными приборами. Nauchnoe Obozrenie.
- Yamagishi, Y. (2026). The Three Hundredth Census: Whether Two Things Share a Root Is Decidable [Paper 300]. Zenodo. 10.5281/zenodo.22171914
- Yamagishi, Y. (2026). A Broken Law Can Be Blamed on Any of the Assumptions [Paper 481]. Zenodo. 10.5281/zenodo.22733927
- Yamagishi, Y. (2026). The Limit Has No Radius Because Two Forces Share One Exponent [Paper 483]. Zenodo. 10.5281/zenodo.22759875
- Yamagishi, Y. (2026). The “Sphere of Influence” Is Three Spheres, and Dividing the Mass Ratio by $1000$ Shrinks Them Differently [Paper 505]. Zenodo. 10.5281/zenodo.22861782
Reference
Yamagishi, Y. (2026). The Same Fuel Buys $3.90$ Times the Energy Depending on Where It Is Burned [Paper 506]. YOI: https://www.yamagishiyuuki.com/yoi/506
DOI: 10.5281/zenodo.22861840
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 506. This paper is a preprint and has not been peer reviewed.