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According to the following diagram it would require ~648.69 km/s to do all of the intermediate transfers that would land you at or around the sun's surface at perigee. Is this a real number? How would they have calculated this number? Is the sun's mass and other quantities known well enough for this to be absolutely accurate? It seems like a massive number; am I reading this wrong?

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I understand partially the vis-viva equations, but not in terms of the sun. It seems odd to use it around the body you're currently orbiting. Do I have to think of it in terms of a different frame of reference? I was thinking this is the delta-v for a non-circularized orbit with a lowest point being at the suns surface and the highest at Low Sun Orbit (that's my thought on the 440 number)... but I'm not sure.

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    $\begingroup$ Seems plausible. The delta v for landing is mostly cancellation of orbital speed, and solar orbit speed at Earth’s altitude is 30km/s — from low solar orbit it would be much higher. WP gives the sun’s mass to four decimal places; I don’t know if it’s better known than that. $\endgroup$ – Russell Borogove Aug 8 at 15:23
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    $\begingroup$ If you had the spacecraft that could tolerate the heat of the surface, you sould be able to aerobrake, so that chart should have a red arrow, reducing the delta v requirement by as much aerobrake as you can tolerate. $\endgroup$ – Joshua Aug 8 at 23:22
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    $\begingroup$ @Joshua Aerobraking changes how much delta-V you need from your rocket, but not how much dV you actually need to change orbits. It's just that part of the dV can come from atmospheric drag instead of your horribly inefficient rocket engine. The problem is that this is not a nice number that works for every spacecraft, unlike the total dV - it depends the aerodynamics and size of your craft, the mission parameters etc.; the blue number is actually the minimal dV you need, regardless of how you achieve that dV. $\endgroup$ – Luaan Aug 9 at 6:05
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    $\begingroup$ @Joshua: "Aerobraking in the photosphere" is one of the more hair-raising concepts I've heard of in space travel, but I suppose it'd technically be possible. We'll have to get right on that if we need a harder challenge after we colonize Venus. $\endgroup$ – Michael Seifert Aug 9 at 17:21
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    $\begingroup$ "It seems like a massive number; am I reading this wrong?" The sun is massive - about 99% of the mass of the Solar System. $\endgroup$ – Jacob Krall Aug 9 at 22:46
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Addressing

is the sun's mass and other quantities known well enough for this to be absolutely accurate?

Well, the key to this is the vis-viva equation in your question. It's not actually important for us to know the mass of the Sun precisely, so long as we know the product $GM$ (another answer makes mention of this).

And that product is, of course, the same for any body orbiting the Sun, the key insight behind Kepler's third law. So if we have an accurate knowledge of the period and semimajor axis of any body orbiting the Sun (including Earth) we have an accurate picture of $GM$. Well, we know Earth's orbital period to better than a millisecond per year (10^-12 fractional accuracy), and Earth's semimajor axis to better than 4km in 1 AU (10^-7 fractional accuracy) so we should know that GM figure to about 1 part in 10 million (all the various exponents conveniently cancel out). Seems pretty good to me.

In fact, $G$ is the one that we have the least handle on (we know it to a little better than 4 digits, according to CODATA), and that is what limits our estimate of the mass of the Sun to the same precision (in fact, Wikipedia's info box says the sun is $1.9885 \times 10^{30}$ kg and then links to an archived older version of the same page that says $1.9891 \times 10^{30}$ kg). but A) that's not a terrible estimate anyway, and B) as already pointed out, it's immaterial because we know $GM$ better than we know $G$ or $M$.

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The Vis-viva equation is

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

The $GM$ product for the Sun is 1.327E+20 m^3/s^2.

If 1 AU is 150E+09 meters, then when you are in a circular ($r=a$) Heliocentric orbit at Earth escape/capture point your velocity is 29.7 km/s.

If you then change to an ellipse with aphelion still at 1 AU and perihelion at the surface of the Sun, then $a = (1 AU +695700 km)/2 = 75.35 $ million km and your velocity at aphelion ($r= 1AU$) is now only 2.86 km/s, and at perihelion (grazing the Sun) it's 616.2 km/s. You've had to lose $\Delta v$ of 26.9 km/s to get into that orbit, and you'll have to lose another 616.2 km/s to stop on the Sun the next time you graze it.

That's 643.1 km/s, plus the 12.6 km/s to get from Earth's surface to Earth escape/capture, so the total I get is 655.7 km/s.

That's pretty close to your number. I've treated the Earth's orbit as a circle at 150 million km exactly, which it isn't.

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    $\begingroup$ Wow, when explained like that it's much easier to understand the math behind those charts. I was definitely overthinking the application of that equation... Thanks for expanding on the actual calculations. $\endgroup$ – Magic Octopus Urn Aug 8 at 18:13
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    $\begingroup$ @MagicOctopusUrn: Of course, you could eliminate over 90% of the $\Delta v$ requirement by aerobraking on the Sun. Surviving that maneuver is left as an engineering exercise, but then again, if you're going to land on the Sun you presumably have some way of dealing with it, anyway. $\endgroup$ – Ilmari Karonen Aug 10 at 0:49
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    $\begingroup$ @IlmariKaronen delta-v is just a measure of absolute value of acceleration, independent of how it is produced. You aren't eliminating it, you are just naming your methods. $\endgroup$ – uhoh Aug 10 at 1:04
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It's a "real number", a delta-v chart isn't really concerned with the realism of various missions. It's quite simply a table of velocity changes, how you would go about achieving these velocity changes is outside its scope.

As to how to calculate it, the parts of the journey in space can be calculated like any other part of the diagram, using the patched conic approximation to split it up into two body systems, and from then the Vis-viva equation (written on the chart). The other lading/ascension values on the chart appears to consider atmospheric drag, but they skipped it for the Sun since any realistic drag model for a spacecraft there doesn't exist. It's just the speed you would crash into the Sun with from low orbit. (or more accurately, the way you have described it in your question, as a small elliptical transfer). Using the Vis-viva equation for the body you are currently orbiting isn't odd at all, it's the normal use case.

Are the relevant quantities for the Sun known well enough to give a reliable number? Yes and no. The Sun's mass is known very accurately due to accurate measurements of the planets orbiting it, but the other relevant number, the radius is more so-so. The Sun doesn't really have a solid surface, so the surface radius is more a question of definition than accuracy of measurements.

Lastly, it seems like a massive number, can it be right? Going from the surface of a body, and then navigating some in space, you would expect spending about the same delta-v as the escape velocity of the body. Given the Sun's escape velocity of 620km/s, this seems correct.

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  • $\begingroup$ Just want to clarify, that the 440 delta-v is not for a "circular orbit at the surface of the sun" (I know that doesn't make sense) but an orbit at which Apogee is LSO and Perigee is Sun Surface? However, the 178 delta-v is a transfer from Earth orbit to a circular Low Sun Orbit both probably approximated with 0 eccentricity? +1 anyway, for the links and the factoids. $\endgroup$ – Magic Octopus Urn Aug 8 at 15:52
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    $\begingroup$ 1. Yes, that's correct. I can edit that in to better address that part of the question. 2. Not quite. The node at 178 distance is the Earth-LSO transfer, at which point you are already in orbit around the Sun in an ellipse touching LSO and Earth's orbit. $\endgroup$ – Hohmannfan Aug 8 at 15:54
  • $\begingroup$ Ah, so literally none of these are transfers to circular orbits, just the delta-v for the first burn (assumed instantaneous) at the apogee (being LEO) to lower from LEO to LSO on only one side of the orbit. Thanks! $\endgroup$ – Magic Octopus Urn Aug 8 at 18:11
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    $\begingroup$ @MagicOctopusUrn Some of the numbers in the diagram are to or from circular orbits. The numbers are for the conic element of the flight plan. Example: Earth<->Moon. 9.4km/s from ground to 250x250km orbit, 2.44+0.68 for TLI, 0.14 to capture into a highly elliptical orbit around the moon, 2.94 to go from elliptical into 400x400km circular, and then 1.73 to the surface. On the way back you need 1.73km surface -> 400x400km circular, then 0.68+0.14 to get back on a trajectory that intersects the earths atmosphere, and the other numbers are void due to aerobraking. $\endgroup$ – Polygnome Aug 9 at 8:20

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