# Delta-V chart mathematics

Could someone explain me the mathematics behind the Delta-V charts one can find on internet such as: Delta-Vs for inner Solar System (Image converted from source Wikipedia SVG)

How are they created?

Say I would like to make an estimation of the delta-v required for putting an object on low earth orbit, how should I proceed in order to calculate that amount regardless of the rocket specifications or the payload? Is it even possible?

• At first glance I thought this question was overly broad, but I can see where answers to it would be helpful to the site. I also see an attempt to provide a concise answer, so while it may seem overly broad, it appears that potential for concise answers exists. – James Jenkins Sep 20 '13 at 15:18
• The more I look at this, the deeper explanation develops. Tricky part is that you must imagine everything on either side of Mars transfer orbit to be burns done at either LEO or LMO locations. That's how they can have 5 different points all in a row - they're all different degrees of eccentricity, up to and beyond hyperbolic orbits. That's Oberth effect on the solar system scale. – AlanSE Sep 22 '13 at 15:36
• 30km/s delta V... to crash into the surface of the sun? – Magic Octopus Urn Jul 5 '18 at 14:02
• This chart is rather lousy with regard to getting to and from the Moon. Low Earth orbit to low Lunar orbit is about 4 km/s, 3.1 for translunar injection and 0.9 for lunar orbit insertion; the chart suggests 4.8 (4.1+0.7). Besides, nobody goes to L4/L5 to get to the Moon. From low lunar orbit to the surface or the reverse is about 1 km/s; the chart says 1.6. – David Hammen Sep 26 '18 at 8:19
• Mars is similarly bad. Delta V is not additive, particularly when one uses transfer orbits. Getting to Mars' surface by first entering Mars orbit at Deimos altitude, then transferring to Phobos altitude, and then transferring to low Mars orbit is dumb. – David Hammen Sep 26 '18 at 8:19

This is a large question, but we can certainly boil it down. You need several levels of requisite knowledge. I'll break it down as so:

1. Relevance of Delta v for propellent budget
2. Conversion between gravitational potential and its corresponding velocity
3. The basic physics of Hohmann transfers
4. Non-ideal factors going from surface to orbit

Not all of these are fully necessary, so other people will take a different route to explaining it. This sequence of points just reflects my own intuition.

Propellent budget

The idea of a Delta-v map would be useless if it was not additive. Think about it. That's the entire point of a map. If I can't add segments to calculate the total distance, then it's not a "map". But there is an interesting criticism of this point - that the "distance" on this map doesn't scale linearly with fuel needed.

We tend to think of fuel as proportional to distance driven. But this is actually incorrect. As your gas tank is full it cases more rolling friction with the road, so your car is more efficient with a nearly empty gas tank. You and I neglect this because our energetic budget is small compared to the weight of the car in gas. In rocketry, it matters to the extreme. BUT, the calculation for the delta-v is still linear. In that way, it is a strong mathematical analogy to distance traveled by car.

Rocket equation:

$$\Delta v = v_\text{e} \ln \frac {m_0} {m_1}$$

In the delta-v maps, you are counting $$\Delta v$$, and they faithfully add linearly.

Gravitational potential

From physics you should be familiar with the concept of "potential" as $$GM/r$$. This has units of $$m^2/s^2$$. In this form, you can apply energy balance. If you think about energy conservation manifested in an equation, divide that equation by the mass of your test mass. This is then the governing energy balance in a stationary reference frame.

If we had a perfect reference frame, then we would apply all the energy balance equations with the naive gravitational potential above. In other words, that quantity would be additive. But it's not because we care much more about the rocket's reference frame than the Earth's or the sun's.

In the rocket's frame, units of $$m/s$$ make a heck of a lot more sense. Consider the situation:

A rocket lies at rest relative to the Earth in the upper atmosphere. It fires its rockets toward the center of Earth, the burn finishes quickly, and imparts just enough momentum to get out of Earth's gravity well.

For this case, the problem is answerable with relative ease.

energy balance:

$$\frac{1}{2} m v^2 = \frac{GM_E m}{R_E+(200 km)}$$

divide by m.

$$\frac{1}{2} v^2 = \frac{GM_E }{R_E+(200 km)}$$

Here is the actual calculation

$$\Delta v = \sqrt{\frac{2 GM_E}{R_E+(200 km)}} = 10,925 \frac{m}{s}$$

That is an escape velocity calculation, and it is a very simple example of a delta v segment.

Hohmann transfers

In real life, of course we need to take the most efficient route possible. It looks like this: 1. You start in a circular orbit
2. You burn to go into a elliptical transfer orbit
3. Once you reach the desired orbit, you burn again to make your path circular

It never makes sense to "stop along the way", and doing so will always cost you more propellant. That might seem a little confusing to you considering the diagram, with many stop-points. But those are basically transfer orbits, as in the above diagram, and then moving from one scale to another.

Given that a Hohlmann transfer has two burns, you have to expressions for the Delta v. Here they are with notation consistent with the above image.

$$\Delta v = \sqrt{\frac{GM}{R}} \left( \sqrt{\frac{2 R'}{R+R'}} - 1 \right),$$

$$\Delta v' = \sqrt{\frac{GM}{R'}} \left( 1 - \sqrt{\frac{2 R}{R+R'}}\,\! \right) ,$$

These account for a large fraction of the numbers you see in the image. These add together to get the ultimate requirement on your rocketry, so it makes sense to put them in a "map". For increase orbit from LEO to GEO, for instance, you could have 3 dots, and two segments, where first dot is LEO, second is the transfer orbit, and 3rd is GEO. I think they didn't put this on the map (although they could have) because no one cares too much about LEO-to-GEO transfer orbit.

Non-ideal factors

Going from the surface of Earth to low Earth orbit includes other factors:

• gravity drag
• air resistance
• some additional minimum increase in elevation to avoid quick orbit decay

These explain why going to LEO is about 10 km/s, instead of the literal speed of LEO which is more like 7.9 km/s. Gravity drag and elevation increase each contribute on the order of 1 km/s, so the final answer isn't surprising. Not all bodies will have these same factors. This is just an example of special considerations for that map.

I also realize this answer isn't comprehensive. It explains maybe half of the graph.

• Thanks a lot for having taken the time to answer me! Still there is one point that cant quite figure out regarding the escape velocity calculation. If I understood correctly, the 10.925m/s we get as a result of the escape velocity calculation would be the Delta-V amount required for a LEO transfer from the rocket launch pad to 200km above the ground ? – Duom Sep 21 '13 at 18:50
• @Duom No, that's quite different. Escape is firing directly up, then going so far you never fall back down. IRL, this leaves you floating helplessly about the inner solar system, but this is academic and we're not concerned with what happens after. The figure of 200km is frequently used for LEO. I calculated escape velocity at 200km up at rest relative to Earth. There was no particular reason for that. I just wanted to apply the formula once. It's an example that demonstrates the basic physics because it's not complicated by reference frame transformations of kinetic energy. – AlanSE Sep 21 '13 at 21:07
• Okay, I think I got it, I edited my question with an example, hopefully it's correct! – Duom Sep 22 '13 at 14:04
• Escape velocity will lead to escape regardless of your direction. – Erik Jun 19 '15 at 13:44
• "The idea of a Delta-v map would be useless if it was not additive." Actually that is how this map is most often misused. For example LEO to L4 is 4.1 km/s. L4 to Lunar orbit is .7. The naive user will add 4.1 to .7 to conclude LEO to lunar orbit is 4.8 km/s. But a direct route from LEO to lunar orbit takes more like 4 km/s. – HopDavid Jun 19 '15 at 16:58

I have been crunching some numbers and, with @AlanSE's answer, I think I’m on the right path. I’ll take the example of a Delta-V map calculation for the launch a rocket to LEO, followed by a Hohmann transfer to the moon. Correct me if I'm wrong: We have this equation for the delta-V required to lift off and to reach an altitude of 200km:

\begin{align} \Delta v_{0} & = \sqrt{\frac{\mu}{R}} \\ & = \sqrt{\frac{6.673\times 10^{-11} * 5.97219 \times 10^{24}}{6371000}} \\ & = 7909.034~\text{m}/\text{s} \end{align}
\begin{align} v & = \sqrt{2 * Gh} \\ & = \sqrt{ 2 * 9.8 * 200000 } \\ & = 1979.899~\text{m}/\text{s} \end{align}
$$\text{Where } \mu = GR_{earth}$$

So we have ~9888.9 m/s Δv to reach 200KM above the ground.

Hohmann transfer to the moon

\begin{align} r_{1} & = 6571000~\text{m}~\text{(Earth radius plus 200km)} \\ r_{2} & = 362600000~\text{m}~\text{(Moon average position at perigee)} \\ \end{align}
\begin{align} \mu_{earth} & = 3.98524 \times 10^{14}~\text{m}^3/\text{s}^2 \\ \mu_{moon} & = 4.903120210000 \times 10^{12}~\text{m}^3/\text{s}^2 \end{align}
\begin{align} \Delta v_{1} & = \sqrt{\frac{\mu}{r_{1}}} * \left(\sqrt{ \frac{2*r_{2}}{r_{1}+r_{2}} } - 1\right) \\ & = 3127.331~\text{m}/\text{s} \end{align}
\begin{align} \Delta v_{2} & = \sqrt{\frac{\mu}{r_{2}}} * \left(1 - \sqrt{ \frac{2*r_{1}}{r_{1}+r_{2}} }\right) \\ & = 94.345~\text{m/s} \end{align}

• Just a small correction - your red arrow at V2 to enter lunar orbit would be more intuitive if it pointed in the reverse direction - i.e. you must lose velocity relative to the moon in order to settle into orbit. – user10525 Jun 18 '15 at 10:37

Most of these delta V maps are based on Hohmann orbits and patched conics.

I'll use the example of an earth to Mars Hohmann: Earth's moving 30 km/s, Transfer orbit at perihelion is moving 33 km/s so 3 km/s to match velocities.

Transfer at aphelion is 21.5 km/s and Mars is 24 km/s. So 2.5 km/s to match velocities.

So where do the 33, 30, 21.5 and 24 come from?

The vis viva equation.

$V=\sqrt{\mu_{sun}(2/r - 1/a)}$

$\mu_{sun}$ = Gravitational Constant * Mass of Sun

a = semi major axis of ellipse

r = distance from sun

In the case of a circular orbit, circular orbit's radius is the same as semi-major axis so r = a. Substituting r for a reduces the vis viva equation to:

$V=\sqrt{\mu_{sun}/r}$

The speed of a circular orbit.

For earth orbit r = ~150,000,000 kilometers. For Mars orbit r = ~225,000,000 kilometers

For the transfer orbit a = (150,000,000 + 225,000,000)/2 kilometers.

At perihelion of transfer orbit r = 150,000,000 At aphelion of transfer orbit r = 225,000,000

Plugging those numbers into the vis viva should give you numbers in the neighborhood of 33 km/s, 30 km/s, 21.5 km/s and 24 km/s.

I've rounded a lot for simplicity and also earth and Mars orbits aren't perfectly circular and coplanar. The circular, coplanar model isn't exact but it can give you ball park numbers.

Once inside of a planet's sphere of influence, the paths are no longer well modeled as ellipses about the sun but as hyperbolas about a planet.

A hyperbola's speed:

$V=\sqrt{V_{inf}^2+V_{esc}^2}$

Doesn't that remind you of the Pythagorean theorem? I use this memory device to remember a hyperbola's speed: What's $V_{inf}$? In the case of earth departure it's 3 km/s. (Remember the transfer's perihelion velocity of 33 km/s and earth's 30 km/s?)

What's $V_{esc}$? Escape velocity in the earth's neighborhood is

$V_{esc} = \sqrt{2\mu_{earth}/r}$

One way to think of a parabolic escape orbit is as an ellipse with infinite semi-major axis. In which case 1/a would be zero. So the vis viva equation would reduce to the Vesc equation above.

$\mu_{earth}$ = Gravitational constant * mass of earth.

You might notice that escape velocity is circular velocity times square of two.

For our example, let's set r to be 300 km above earth's surface. r = 6678 km.

Plugging that r into the equation for escape velocity above you should get about 10.9 km/s.

So the hyperbola for Trans Mars Insertion (TMI) is $\sqrt{10.9^2+3^2}$ which comes to 11.3 km/s.

Setting r = 6678 km, a low, circular earth orbit has speed 7.7 km/s.

11.3 - 7.7 = 3.6. So 3.6 km/s for TMI.

On arrival at Mars you'd use a similar process. But $\mu_{Mars}$ would be gravitational constant times mass of Mras. Vinf would be 2.5 km/s.

I've made a Hohmann spreadsheet that uses the equations I've given. It can give you delta Vs for various scenarios. It interesting to change the apoapsis and periapsis of parking orbits about a planet.

I used this same spreadsheet to calculate most the numbers is my own delta V map.

Using Google or Wikipedia it's easy to find planetary radii, orbital radii and masses for various bodies.

If you remember the vis viva equation as well as the pythagorean memory device for hyperbolic orbits, it's not hard to patch conics.