Somewhat inspired by this question and its answers, does it require less delta-v for an object to fall into the Sun the further away it is from the Sun?

It makes sense that an object has to shed its orbital velocity before falling into the Sun, but counterintuitive that it is harder to reach the Sun from Earth (or even Mercury) than it is from Pluto.

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    $\begingroup$ To collide with the sun? Yes, it takes less delta-v from a more distant circular orbit. If you want a circular orbit close to the sun, it'll take more delta-v, though. $\endgroup$
    – Ghedipunk
    Sep 25, 2019 at 22:12
  • $\begingroup$ @Ghedipunk I remember running the numbers several years ago, when I was more active in KSP, and I recall thatthe break-even delta-v circular orbit radius between deorbit and escape occurred at a specific value between 4 and 5 radio of the object being orbited, regardless of mass. Unfortunately, I don't recall any more significant figures. $\endgroup$
    – notovny
    Sep 27, 2019 at 10:44

1 Answer 1



1st scenario: A spacecraft orbiting the Sun at Earth distance vs. Pluto distance, shedding its orbital velocity

The orbital velocity decreases with distance, according to the following formula, where $r$ is the orbital radius, and $\mu$ is the mass parameter (it's just a shorthand we use)

$$v_{circular} = \sqrt{\frac{\mu}{r}}$$

The orbital velocity at Earth distance is 30 km/s, at Pluto it's 4.7 km/s. Shedding this velocity to fall straight down is clearly easier at Pluto distance.

(we don't quite need to fall straight down, we can still have some horizontal velocity as the Sun is not a point, but it doesn't change the qualitative answer)

2nd scenario: A spacecraft orbiting the Sun at Earth distance vs. Pluto distance, shedding its orbital velocity, but this time a little smarter

If falling into the Sun was easier when farther away, why don't we first try to go farther out then? Turns out that's a little bit more efficient.

You can't get farther away than escaping the Solar system. If you do that, and you "at infinity" so to speak is crawling away at basically 0 velocity, you can just make a tiny burn with the rocket engine to turn around and fall back into the Sun.

So how much does it cost to escape?

Escape velocity can be calculated in the following way:

$$v_{escape} = \sqrt{\frac{2\mu}{r}} = \sqrt{2}\cdot v_{circular}$$

We are already travelling at circular velocity, so the extra velocity change needed is $v_e - v_c \approx 0.41 v_c$

Crashing into the Sun from Earth distance now costs only 12 km/s, and from Pluto distance 1.9 km/s. Which one is cheaper didn't change, since we multiplied by the same constant (0.41)

3rd scenario: But what if the Earth and Pluto are still there?

If we start from the surface of (or an orbit around) these locations, the calculation has an extra step, as we have to escape their gravitational field first.

After escaping Pluto, we would want to have solar system escape velocity, and this we know is 1.9 km/s larger than the speed Pluto is travelling at. We want this at a sufficiently large distance from Pluto, so let's call it $v_{\infty}$

The following equation is what we use to reach a target velocity after escaping:

$$v^2 = v_{\infty}^2 + v_e^2$$


$$v = \sqrt{v_{\infty}^2 + v_e^2}$$

Escape velocity at the surface of Pluto is 1.2 km/s, so calculating the $v$ above gives us a required burn of 2.3 km/s.

2.3 km/s is not enough to even get into low Earth orbit, and it's not even enough to escape Earth when starting from orbit.

It's thus easier to reach the Sun from Pluto than from Earth.

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    $\begingroup$ How do you dare calling yourself Hohmannfan and then advertise bi-elliptical transfers over hohmann transfers :) $\endgroup$
    – Philipp
    Sep 26, 2019 at 13:04
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    $\begingroup$ I'm a Hohmannfan up to 11.94 :) $\endgroup$ Sep 26, 2019 at 13:15
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    $\begingroup$ Just a general comment - this becomes less surprising if you consider how long the trip takes. It is very cheap to give up Pluto's orbital velocity, but then it will be a long, slow fall. If you set a short time limit, it is cheaper to get to the Sun from our orbit than Pluto's. $\endgroup$ Sep 27, 2019 at 5:10
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    $\begingroup$ Of course, this assumes you can actually hit the Sun from such a distance. Not a big deal for a spacecraft that can do corrections along the way but, OP's mention of "falling" into the Sun implies "dropping" something without propulsion to me. The optimal path would still need less energy, but actually hitting that path, with all of the gravitational influences along the way... most likely, you'd end up on an extremely eccentric orbit around the Sun, passing relatively nearby at ridiculous velocities. $\endgroup$
    – Luaan
    Sep 27, 2019 at 8:42
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    $\begingroup$ @Hohmannfan There's a big difference between having a rocket and being shot out of a cannon :) $\endgroup$
    – Luaan
    Sep 27, 2019 at 8:51

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