@TomSpilker showed that the minimum delta-v to go from Earth orbit into the Sun is about 20.89 km/s. That's going from 200 km LEO to "perihelion at the sun's photosphere (guaranteed complete entry!)" starting from Earth's aphelion (lowest orbital velocity).

But that's with a direct ellipse, not using Venus or Earth flybys.

Is it possible to estimate, perhaps by looking at Parker Solar Probe's several flyby's and its delta-v budget how much less it would take to reach the Sun's photosphere cleverly using flybys of Venus and Earth?

  • $\begingroup$ Actually, you can get the required delta-V down a bit more by starting from a very eccentric Earth orbit instead of a circular orbit. You have to make sure the line of apsides is in the right direction for the perigee burn to send you in the right direction. $\endgroup$ Commented Sep 7, 2019 at 2:31
  • $\begingroup$ I got curious and did the calculation: doing the eccentric Earth orbit approach with apogee around the moon's orbit (before the mad dash), you can get the perigee delta-V down to ~18.71 km/s. $\endgroup$ Commented Sep 7, 2019 at 2:41
  • $\begingroup$ @TomSpilker isn't that just robbing Peter to pay Paul? The 200 km LEO is kind-of a standard starting point; you need to use delta-v to get into that eccentric orbit, is there any net-savings if you start from 200 km circular and go via the eccentric orbit versus going directly (except of course if you use the Moon)? fyi I've just asked What is the highest velocity a spacecraft has left Earth? $\endgroup$
    – uhoh
    Commented Sep 7, 2019 at 4:16
  • 1
    $\begingroup$ Depends on where you start. If that object started from lunar orbit (maybe it was manufactured there!), a relatively small delta-V gets you to the 200-km perigee. $\endgroup$ Commented Sep 7, 2019 at 4:21
  • $\begingroup$ Your best bet is Venus and Earth flybys until you have enough to reach Jupiter. It can sling you into the sun. $\endgroup$ Commented Sep 7, 2019 at 13:17

1 Answer 1


The potential of fly-by ping pong is pretty much unlimited, provided you have enough time to your disposal.

Given an initial transfer with a perihelion slightly lower than the orbit of Venus, a Venus flyby can increase the aphelion to a bit further out than the Earth's orbit.
On the following Earth flyby, the perihelion can be lowered, and you can just repeat this pattern until the eccentricity is high enough that the perihelion is inside the Sun.

In general: Given two planetary masses, repeated flybys can provide arbitrary eccentricity adjustments (in practice, you would likely want to speed up this scheme by involving Jupiter)

Cost: slightly more than 1 Venus transfer, ~3.5 km/s of $\Delta v$

As you probably know, just repeated Earth flybys will not have the same degree of eccentricity adjustment power, as the $v_{\infty}$ stays the same. However, the Earth system is not a single body, it's a two-body system including the Moon. If you instead of going for a full Venus transfer escape Earth with a small but non-zero $v_{\infty}$, repeated flybys of the Earth-Moon system can slowly increase this velocity, until you can reach Venus and execute the above scheme. This takes a very long time.
While going down the rabbit hole, an Earth escape isn't quite necessary either, as multiple Moon flybys will eventually give you an escape with some $v_{\infty}$.

Cost: a Moon transfer, ~3.12 km/s of $\Delta v$

For the mere 0.4 km/s of savings, one could question of the centuries of added flight time is worth it. On the other hand, this is the absolute minimum.

  • $\begingroup$ So to the question as asked; "How much less delta-v..." what would be the numerical answer? I'm sure the information is here but I'm not sure how to extract it. Thanks! $\endgroup$
    – uhoh
    Commented Sep 8, 2019 at 2:58
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    $\begingroup$ That would be a matter of subtraction. 20.89 minus either 3.50 or 3.12 $\endgroup$ Commented Sep 8, 2019 at 6:12

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