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I tried to write a nice answer to this first question by a new user, but I've crashed and burned.

from the Moon to Mars transfer

I calculated the escape from the Moon to be about 2,376 m/s which is close to 2.3 km/s shown in the plot below using

$$v_{Esc} = \sqrt{\frac{2GM_{Moon}}{R_{Moon}}}.$$

Adding Earth's orbital velocity around the Sun of 29,700 m/s to the Moon's orbital velocity around Earth of 1,000 m/s gives me a heliocentric orbital velocity of 30,700 m/s at 150 million km.

If I want an aphelion of 208 million km, I'd need a semimajor axis of 179 million km, which means I'd want a velocity at 1 AU of 32,100 m/s calculated using the vis-viva equation:

$$v = \sqrt{GM_{Sun}\left(\frac{2}{r} - \frac{1}{a} \right) }$$

where $r$=150 million km and $a$=179 million km.

That means that after escaping the Earth-Moon system I'd need an additional 32,100-30,700 = 1,400 m/s.

But the plot in Wikipedia reaches Mars transfer orbit with only an additional 600 m/s bump.

Question: What's the source of the disparity between my calculation and that one?


Delta-Vs for inner Solar System

Source: Delta-Vs for inner Solar System

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@DavidHammen and I agree: that delta-V plot is at best misleading, and at worst — well, Dave is a smart fellow, and he wisely didn't want to use the word here on SESE. It turns out the delta-Vs for all these transfers depend tremendously on how you do them.

An example: as you point out, the plot says the delta-V to go from "Earth C3=0" to "Mars transfer" is 0.6 km/s. I won't go into the differences among the various combinations of departing Earth at perihelion or aphelion or somewhere between, and arriving at Mars with Mars at perihelion or aphelion or somewhere between; for now I'll just assume those orbits are circular, but I'll assume that we get to Mars at a heliocentric distance of 208 million km, as you did; the precise distance won't make a qualitative difference in the result. Let's look at two ways of doing the transfer orbit injection.

The first is to actually get out to where C3=0 takes you, escaped from Earth. You're orbiting the sun now at the same speed as Earth, and with the circular-orbit approximation that's 29.78 km/s. You need to burn to the transfer orbit, which has a perihelion velocity of 32.12 (plus a bit) km/s. So you need to speed up by 2.34 km/s, and of course that's the delta-V to get onto that transfer orbit.

The second is to assume you're in a C3=0 orbit but you're at perigee (I'll assume 200 km altitude), and you do your TMI (trans-Mars injection) burn there. When you escape from Earth you'll need your V-infinity to be the same 2.34 km/s we saw above. But now you're down deep in Earth's gravity well, so the Oberth effect comes to your aid. At 200km altitude, a C3=0 orbit has a velocity of 11.01 km/s. An Earth escape orbit with a V-infinity of 2.34 km/s has a perigee velocity (assuming the same 200 km altitude) of 11.25 (plus a bit) km/s. So getting assistance from Hermann Oberth, you could actually go from C3=0 to TMI for only 0.24 (plus a bit) km/s! That's about one tenth of the free-space delta-V!

That Wikipedia chart does not consider this at all! And that is why Dave and I take exception to it. Delta-V is not linear!! Any chart that implies it is, is ... well ... I won't use the word either, Dave.

Exercise for the student: assuming an impulsive delta-V precisely aligned with the velocity vector, and assuming the orbits as I described above, at what orbit altitude (or geocentric radius, if you prefer) would you have to perform the TMI burn to make the needed burn magnitude 0.600 km/s??

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  • $\begingroup$ So is the tl;dr of your answer that I forgot to swing past Earth and do a burn at perigee, using the Oberth effect? The 0.6 km/s there would be the equivalent of my 1.4 km/s far from cis-lunar? $\endgroup$ – uhoh Sep 10 '19 at 5:38
  • $\begingroup$ @uhoh No. As far as I can see you failed to include the delta-v from your burn needed to escape the moon into your transfer orbit. If you escape the moon prograde (in relation to the sun), then your apohelion is already half-way to mars. This ofc assumes you leave when the moon is behind the earth (viewed from the sun), not in front. This is actually one of those moments where KSP gives really nice visualizations :D I'll try to snap a few images over the course of the day. $\endgroup$ – Polygnome Sep 10 '19 at 7:06
  • $\begingroup$ @Polygnome My question says "I calculated the escape from the Moon to be about 2,376 m/s which is close to 2.3 km/s shown in the plot below using $$v_{Esc} = \sqrt{\frac{2GM_{Moon}}{R_{Moon}}}."$$ That gets me to 30,700 m/s heliocentric with an aphelion less than half-way to Mars. The last 1,400 m/s puts my aphelion at Mars. What exactly did I fail to do? Also, I'm pretty sure the answer to my comment is "Yes, by using the Oberth effect swinging past Earth, you'll only need the 600 m/s shown in the graphic, not the 1,400 m/s you calculate by launching directly to transfer" rather than No. $\endgroup$ – uhoh Sep 10 '19 at 8:10
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    $\begingroup$ @uhoh To answer the question in your comment: No, the point is that by choosing the altitude where you do the TMI injection burn, you can make the TMI delta-V anything between 240 m/s and 2.34 km/s. To answer my "Exercise for the student": to get the 0.6 km/s from a normal moon-to-Earth transfer orbit (the way Apollo did it), you need to do the (impulsive) burn at 7023 km geocentric radius. That whole range assumes a start from a C3=0 orbit, so the plot's spec of 0.6 km/s is essentially arbitrary. Why choose 7023 km? There's no reason, it's arbitrary. $\endgroup$ – Tom Spilker Sep 10 '19 at 15:56
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    $\begingroup$ I always get in trouble when I use, hmmm, descriptive terms, regardless of how appropriate said descriptive terms might seem to be. $\endgroup$ – David Hammen Sep 11 '19 at 1:47

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