I've seen this several related questions to this, but they tend to focus on complex maneuvers to minimize the delta-V. If I wanted to mass-drive, railgun or tether-spin a mass so that it would eventually crash-dive into the sun, what is that value?

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    $\begingroup$ Have you got some evidence you really really need to get rid of? Like Musk and the body he flung into space? $\endgroup$
    – Richard
    Commented Sep 7, 2019 at 19:18
  • $\begingroup$ You're not far off. I'm working out the actual details to the silly old ideal about hiring nuclear waste into the sun, mostly just to illustrate falling launch costs. Of course, you wouldn't have to ACTUALLY get it into the sun...a lower energy orbit than the Earth would do the trick. $\endgroup$ Commented Sep 8, 2019 at 1:51

2 Answers 2


Actually the answer is a bit more complex than "Earth orbits at 30 m/s, so you have to stop that velocity and drop in. Thus the delta-V is 30 km/s." The question states that you start from Earth orbit, and that makes a big difference.

Let's assume an almost ideal situation: the object to be sent to the sun is in a 200-km (Low!!) LEO, whose orbit plane is parallel to the ecliptic plane, so there's no out-of-plane component to the required delta-V. At that altitude, the object orbits Earth at about 7.78 km/s.

Also, the question says you need only to get to the sun, you don't have to dive straight-line into it. That will save some more delta-V.

It makes a little difference if you start from Earth at its perihelion, at its aphelion, or somewhere in between. I'll treat the two extreme cases.

If Earth is at perihelion, its heliocentric velocity is ~30.29 km/s. At that distance from the sun, a very eccentric orbit with perihelion at the sun's photosphere (guaranteed complete entry!) would have an aphelion velocity of ~2.91 km/s. If the object were in free space (not orbiting Earth, but moving around the sun at Earth's orbital velocity), the delta-V would be the difference of those two, 27.38 km/s. And that's a useful number: it tells you that when you've escaped Earth's gravity, you have to be receding from Earth at 27.38 km/s with respect to Earth's barycenter, in the direction opposite to Earth's velocity vector. That relative speed after escaping the gravity field is called the "V-infinity", i.e. the velocity the object would have at essentially infinite distance from the primary.

But the object is not in free space, it's in LEO. Assuming an impulsive delta-V (instantaneous, so no losses due to increasing altitude as a finite-duration burn progresses)(but boy oh boy, what a kick that would be!), the delta-V needed at 200 km altitude to yield a V-infinity of 27.38 km/s is ~21.72 km/s. Immediately after the delta-V the object would be going 29.50 km/s relative to Earth's surface, parallel to the surface. As it ascends through the gravity field it would decelerate, winding up moving essentially radially away from Earth at 27.38 km/s. If the delta-V is timed properly, that V-infinity is directed "backward", so the object's heliocentric velocity would be 2.91 km/s, and it falls into the sun's photosphere.

So in that scenario, the required delta-V is ~21.72 km/s.

But if Earth is at aphelion, its orbit velocity is down to ~29.29 km/s. From there the eccentric orbit that grazes the sun's photosphere has an aphelion velocity of ~2.82 km/s, so the Earth-centered V-infinity needed is the difference of those two, ~26.47 km/s. The delta-V required to get that v-infinity from the 200 km orbit is ~20.89 km/s.

Initiating "the plunge" from Earth's aphelion saves ~0.83 km/s of delta-V over initiating from perihelion.

If, for some reason, you want to do the straight-line plunge into the sun, the delta-V goes up. Now you need a V-infinity equal to Earth's orbit velocity. From the 200 km orbit at Earth's perihelion that delta-V is ~24.44 km/s, and from aphelion it is ~23.51 km/s.

Note that all these are quite different from the 30 km/s result if you do the simple free-space stop'n'drop.

No references, I did all these calculations myself using introductory orbital mechanics formulae. I assumed only Earth's GM, I didn't include the moon. The result would vary somewhat with the position of the moon when you do this mad dash.

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    $\begingroup$ Very nice answer! So I've just asked How much less delta-v would it take to reach the Sun using Venus and Earth flyby's compared to direct? $\endgroup$
    – uhoh
    Commented Sep 7, 2019 at 0:26
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    $\begingroup$ @uhoh Thanks again, @uhoh! Actually, it's faster to do a few Venus and Earth gravity assists to get out to Jupiter with a high V-infinity of approach, then use a Jupiter flyby for the stop'n'drop. $\endgroup$ Commented Sep 7, 2019 at 2:26
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    $\begingroup$ I'd love to see anything in LEO that can impart 20.89km/s of delta V to anything. $\endgroup$
    – Joshua
    Commented Sep 7, 2019 at 3:26
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    $\begingroup$ @Joshua Orion could do it. Detonating that many nukes in LEO would devastate Earth with the EMP, though. $\endgroup$ Commented Sep 7, 2019 at 13:16
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    $\begingroup$ @Joshua I suspect an earth-grazing asteroid would do the job. Although I do also suspect that your probe wouldn't be able to send any telemetry after that boost. $\endgroup$ Commented Sep 8, 2019 at 13:00

It costs approximately 30km/s to transfer from LEO to the Sun according to this delta-v map.

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    $\begingroup$ This is why that delta V map is a pile of stuff. $\endgroup$ Commented Sep 7, 2019 at 5:30
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    $\begingroup$ @DavidHammen You're right on the money. Delta-V is not linear, so you can't do what the map implies. Plus it doesn't specify the list of assumptions that go into each of the listed delta-Vs. $\endgroup$ Commented Sep 7, 2019 at 17:48
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    $\begingroup$ If you are looking for a better dv map, this one gives 21.29 km/s to low sun orbit, which is quite close to the answer of @TomSpilker $\endgroup$
    – Polygnome
    Commented Sep 8, 2019 at 10:02

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