Why is Wikipedia's delta-v from the Moon to Mars transfer so much lower than mine?

Question

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

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?

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Source: Delta-Vs for inner Solar System

Answer 1

@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??