It has been mentioned several times on this site that it is "easier" (less delta-v, and hence less fuel) to reach the escape velocity of the solar system, than to reach the planet Mercury or the surface of the Sun:

However, none of these questions answer why it is so. Is there an explanation (preferably conceptual rather than calculations) why it is easier to escape the solar system than get to Mercury or the Sun?

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    $\begingroup$ It's all delta-Vee . What is your question? $\endgroup$ Commented Jul 29, 2020 at 12:14
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    $\begingroup$ @CarlWitthoft: Can you write an explain-xkcd style answer that is suitable for the general public? $\endgroup$
    – DrSheldon
    Commented Jul 29, 2020 at 13:34
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    $\begingroup$ I think Speedphoenix did pretty well there :-) $\endgroup$ Commented Jul 29, 2020 at 17:18
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    $\begingroup$ The surprising corollary to this question is that if you are in orbit around a black hole, it is easier to escape the black hole than to fall into it. $\endgroup$
    – DrSheldon
    Commented Jul 29, 2020 at 20:56
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    $\begingroup$ @Cireo Orbiting at distance of 4.8 * 0 = 0 over an object of radius 0 does tend to break the equation, yes. $\endgroup$
    – notovny
    Commented Jul 29, 2020 at 23:25

5 Answers 5


Because the earth goes very fast around the sun.

If you want to get to the sun, you need to slow down almost completely so that your speed relative to the sun becomes almost zero.
If you don't slow down (almost) completely, your probe will miss the sun when you 'drop' it, so it will eventually come back and you'll end up in an elliptical orbit.
Kind of like how if you threw a marble in a kitchen bowl, without dropping it perfectly still, it will go very fast near the center of the bowl but miss it, turn around on the other side, and probably not hit the center coming back too.

The earth orbits the sun at a speed of about 29.78 km/s (107,208 km/h; 66,616 mph). This means you'll need to accelerate by 29.78 km/s behind the earth to go to the sun.

According to the Escape Velocity wikipedia page the speed required to escape the solar system if you were at the earth's distance from the sun is 42.1 km/s, but the actual escape velocity for something in the earth's system is 16.6 km/s, this is because the earth goes fast, so you get a boost by having that speed to begin with.

This means that you need about twice the raw speed to go to the sun than to leave the solar system.
(This does not take into account gravity assists from planets, nor the earth's gravity well)

If you just want to get to mercury that's actually much easier, because mercury's orbit is much wider than the sun, so you don't need to aim for something $1.4*10^6$ km in diameter (the sun), but rather $1.2 * 10^8$ km (mercury's orbit). You need to accelerate backwards quite less to reach it (though you do need to get the timing right).

Problem is, you'll be going very fast because the sun will have pulled you in quite a bit, so you'll either crash into mercury, or pass it by very quickly without stopping. So if you want to hang out on or around mercury you would need to slow down a lot, not from the earth's orbital velocity, but to counteract the speed you reached from the sun's pull.

A good way to slow down is to use gravity assists on the various planets. This is what the ESA/JAXA probe BepiColombo uses. The spacecraft does nine gravity assists on the earth, venus and mercury overall.
It has currently completed the first assist with earth. The next one will be with Venus on October 15 2020.

Here is a good animation showing it.

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    $\begingroup$ I'd have thought the additional velocity required would just be the difference between escape velocity and orbital velocity, which is 42.1 - 29.8 = 12.3 km/s. Is the additional 4.3 km/s required to escape the gravity of Earth itself? I notice that's the escape velocity from "the Earth's system", not just something in Earth orbit that ignores the Earth entirely. $\endgroup$ Commented Jul 29, 2020 at 16:05
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    $\begingroup$ @NuclearWang Yes, it is to escape Earth's gravity. In order to have speed 12.3 km/s relative to Earth far away from Earth, you need to have speed $\sqrt{12.3^2 + 11.2^2} = 16.6$ km/s near Earth's surface. $\endgroup$
    – Litho
    Commented Jul 29, 2020 at 17:50
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    $\begingroup$ Note: If you're willing to take thousands of years to get to the sun, it's about the same delta v as escaping the solar system -- raise the aphelion to not-quite-infinity, wait until you reach aphelion, and then it's very, very cheap to lower the perihelion. $\endgroup$ Commented Jul 29, 2020 at 20:49
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    $\begingroup$ @PaŭloEbermann the "direct" way would achieve exactly the same result $\endgroup$
    – IMil
    Commented Jul 30, 2020 at 0:11
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    $\begingroup$ @d-b: en.wikipedia.org/wiki/Bi-elliptic_transfer $\endgroup$ Commented Jul 30, 2020 at 12:59

Changing orbits requires delta-v. To reach the Sun, you need to subtract delta-v such that your velocity relative to the Sun is near zero, which allows you to "fall straight down" into the Sun - your required delta-v is nearly equal to your orbital speed. To escape the solar system, you need to add sufficient delta-v in order to reach escape velocity - due to the relationship between gravitational potential energy and kinetic energy, it works out that escape velocity at a particular (circular) orbital height/speed is equal to sqrt(2) of the orbital speed.

In other words, no matter what circular orbit around the Sun you're starting from, you can decrease your velocity by 100% to go directly into the Sun, or you can increase your velocity by 41% to escape the solar system.

I originally expected the answer to have something to do with the locaton of Earth with respect to the Sun, but it turns out it doesn't matter how far away you are, as the ratio is always the same. An object in Neptune's orbit has a relatively low orbital velocity but doesn't have as far to go to exit the solar system, while an object in Mercury's orbit has a relatively high orbital velocity but much further to go escape. But either way, the ratio of delta-v to escape the solar system vs. reach the Sun is always the same - it is always cheaper to escape than to hit the center directly!

You'll also notice I said hit the center directly, as in the straight-line, most direct path. As pointed out in the comments, you can also get to the Sun for as little as it costs to escape, as long as you're willing to take a much longer path. To do so, add 41% to your velocity and escape the solar system, coasting an arbitrarily large distance from the Sun as your velocity approaches zero. From here, you cancel 100% of your near-zero velocity, and fall all the way back into the Sun. As one is willing to take a longer and longer trip, one can approach the minium energy route to the Sun, costing just 41% of the delta-v. Practical orbits will fall somewhere between these extremes of minimum-time/maximum-energy and minimum-energy/maximum-time.

This answer ignores influence of other bodies which might perturb orbits or provide gravitational assists, and effectively treats the Sun as a 0-radius body that must be hit in the center. In actuality, you could reduce your speed by slightly less than 100% and still hit the outer edge of the Sun, but it's a close approximation at these distance scales.

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    $\begingroup$ this equivalence is the true meaning of the common phrase "LEO is halfway to anywhere." In fact, being in ANY orbit is halfway to anywhere. $\endgroup$ Commented Jul 29, 2020 at 16:01
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    $\begingroup$ Is that ration (41%) valid for any system? (Like moons around planets) or is it slightly different? $\endgroup$ Commented Jul 29, 2020 at 16:57
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    $\begingroup$ Because of this it's often cheaper to 'go out' to 'go in', a bi-elliptic transfer. At the limit, if you raise your aphelion to infinity, then it takes 0 m/s to null your orbital velocity of 0 m/s and drop something directly into the Sun (but it would take an infinite amount of time), so it would only take solar escape dV total. In practice, you'd raise your aphelion high (as high as you're willing to wait) before nulling your now much smaller orbital velocity once you get to aphelion to drop your perihelion into the Sun. This is cheaper than directly cancelling your initial orbital velocity. $\endgroup$
    – tylisirn
    Commented Jul 29, 2020 at 16:58
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    $\begingroup$ @Speedphoenix It should be valid for any system, since it's derived only from the balance of gravitational potential and kinetic energy, which will be the same anywhere. It checks out for objects sharing the Moon's orbit, which will have an orbital velocity around the Earth of 1km/s, and an Earth escape velocity of 1.4km/s. $\endgroup$ Commented Jul 29, 2020 at 17:59
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    $\begingroup$ @Joshua I'm not assuming any gravity assist here, but an efficient mission plan would almost certainly use one or more. I think I follow your point that if you can get to Jupiter for some fixed delta-v, you can use a gravity assist to either head into the Sun or leave the solar system "for free", so either maneuver in practice would only cost about whatever it takes to get to Jupiter. $\endgroup$ Commented Jul 29, 2020 at 19:33

Escaping the solar system requires adding orbital velocity to the spacecraft. Similarly, getting closer in the solar system requires removing orbital velocity. It turns out Earth is more out of the Sun's gravity well than it's in it.

In other words, the simple answer is that Mercury is "farther away" in terms of the change of velocity that's required to reach it.

There are a few ways to visualize this. One is this subway-style map made by ucarion on Redit:

enter image description here

You'll notice in either case first the craft has to get into orbit, and then escape Earth's gravity, which means getting to the point labeled "Earth Intercept". From there, it's 8650 ms/s to get to a Mercury intercept, but only 5390 m/s to a Neptune intercept.

Unfortunately the chart does not have a point for solar system escape, but it's not too much farther from intercepting Neptune.

Neither does it have anything for reaching the Sun. For that we'd need to define what that means exactly: do you want to orbit the Sun, or "land" on it, if such a thing were possible? Let's say you just want to fly right to the middle of the Sun because that's easy to calculate: to do that you must cancel all the orbital velocity you started with because you took off from Earth. Earth's orbital velocity is about 29700 m/s, which is a lot more than even getting to Mercury. (And that's in addition to escaping Earth first.)

As a side-note, adding more delta-v to a spacecraft is expensive due to the tyranny of the rocket equation. So while the difference between 5390 m/s 8650 ms/s might seem like "not even 50% harder", it's actually significantly harder, because to gain that additional 3260 m/s of delta-v your craft must carry more fuel, but that makes the craft heavier so then you need more fuel to accelerate your extra fuel.

Another way to visualize the "distance" to space travel was published on XKCD:

enter image description here

It's pretty intuitive that to go up you must "jump", but what's less intuitive is that to go down you must "anti-jump", that is lose orbital velocity, which takes fuel just like gaining orbital velocity. Imagine a ball on a trajectory: if you want it to land shorter you have to slow it down. You can't just change the initial trajectory because by virtue of starting from Earth, the initial trajectory is fixed.

If you get out a ruler and measure the vertical distance from Earth to Mercury, you'll see it's significantly more than the vertical distance from Earth to the highest "hill" at the edge of the page, which isn't too far from escaping the solar system.

  • $\begingroup$ Great answer! I wish that "metro map" had the lengths of lines to match the expense in delta-v. Right now it seems like travelling to Venus vs Mercury is about the same but when you look at the numbers, things are pretty different. $\endgroup$ Commented May 14, 2022 at 17:55
  • $\begingroup$ Voting down for using "xkcd" as a serious reference. Let's hope we won't progress to sourcing 9gag and alike in the answers. $\endgroup$ Commented May 19, 2022 at 21:44
  • $\begingroup$ Voting up for using XKCD as a visualization (not a reference.) And sure, cite a joke if it gets the point across. $\endgroup$ Commented Feb 19 at 22:36

Based on the calculations presented by @uhoh I generated a plot showing the necessary delta-V for

  • a fly-by mission, i.e. entering into a Hohmann transfer with a far point intersecting the orbit of a planet
  • to get into a circular orbit at the same radius as a planet

dv required to reach different orbits

Note that this does not include any methods to save fuel (aero-braking, swing-by) and ignores complicating details like eccentricity and inclination of orbits as well as the gravitational wells of the planets themselves.

I cut the y-axis because entering a circular Low Sun Orbit (LSO?) is just ridiculously expensive at about $\Delta v = $ 200 km/s.

And for your curiosity: If your intuition tells you that it should be much easier to reach Sun once you are at Mercury - your intuition is wrong: enter image description here If you want to travel to the Sun cheaply, you have to pay the price and live on Pluto. But that offer doesn't include circular orbits, they are even more expensive from out there.

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    $\begingroup$ Why does the delta required to reach orbits beyond 11 AU get smaller? $\endgroup$ Commented Jul 29, 2020 at 23:36
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    $\begingroup$ @Speedphoenix Because the circularization burn gets smaller faster than the transfer burn gets larger once you're that far out. $\endgroup$ Commented Jul 29, 2020 at 23:41

Many quantitative questions about orbits can be answered using the vis-viva equation

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

where $a$ is the semi-major axis, $r$ is the current distance to the central body and $v$ is the velocity at $r$, and the vis-viva equation comes straight from the principle of conservation of total energy which is the sum of kinetci and potential energy:

$$E = T + P$$

and the following two equations, one for each. These are written for reduced energy which is just energy per unit mass $m$ since it just divides out:

$$T = \frac{1}{2}v^2$$ $$P = -\frac{GM}{r}$$

All of the equations for delta-v required for Hohmann transfers and what not can be gotten from the vis-viva equation.

The tool we have to change orbits is impulse, a change in momentum. Whether we need to increase or decrease our velocity, it still requires momentum and therefore delta-v. The only question is "how much?"

We can work the problem without units if we just think of 1 AU as a distance of 1, the Sun's gravitational parameter $GM$ as 1 and the units of velocity as 2 $\pi$ AU/year, so let's do that.

Starting in Earth's orbit with a velocity for a circular orbit, our velocity is

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

So in 1 year we will travel 2 $\pi$ AU.

If we want an elliptical orbit that goes from 1 AU to 0.4 AU (close to Mercury's orbit) then our new semi-major axis is $(1 + 0.4)/2$ or 0.7. At aphelion our velocity is now

$$v = \sqrt{\left( \frac{2}{1} - \frac{1}{0.7} \right)} = 0.76$$

so our delta-v is 0.24.

Instead if we want to leave he solar system we need a semimajor axis of $\infty$ so let's put that in:

$$v = \sqrt{\left( \frac{2}{1} - \frac{1}{\infty} \right)} = 1.41$$

which is a delta-v of 0.41, which is more than what we needed to just touch Mercury's orbit at perihelion.

But what if "get to Mercury" means establishing a circular orbit with $a=0.4$? That means we need a second impulse.

Our velocity at perhihelion in our elliptical transfer orbit is

$$v = \sqrt{\left( \frac{2}{0.4} - \frac{1}{0.7} \right)} = 1.89$$

and if we want to circularize it would have to be

$$v = \sqrt{\left( \frac{2}{0.4} - \frac{1}{0.4} \right)} = 1.58$$

so we need a second delta-v of 0.31 for a mission total of 0.24+0.31=0.55

So the reason we need 0.55 to reach a circular orbit at the distance of Mercury and only 0.41 to escape the solar system is...

Energy is always conserved.

Homework left for the reader:

  1. Once we reach infinity on our parabolic trajectory, how much delta-v is necessary to circularize? :-)
  2. How do these check out compared to delta-v tables? My velocity units are 2$\pi$AU/year, but those can be converted to km/s and compared directly to tables.
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    $\begingroup$ Answer to #1: Assuming you reached a distance of $\infty$ with a velocity of zero, you're already in a circular orbit of infinite radius and thus an orbital period of $\infty^2$ . Yes that's all purely humorous. $\endgroup$ Commented Jul 29, 2020 at 12:17
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    $\begingroup$ While I'm at it, $\frac{2 \pi AU}{year}$ is equal to 179 094 946 furlongs/fortnight . $\endgroup$ Commented Jul 29, 2020 at 12:21
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    $\begingroup$ Addendum to homework #1, note that at ∞ it also takes 0 m/s to null your orbital velocity of 0 m/s to drop something directly into the Sun. So at the limit, if you do a bi-elliptic transfer, it takes 'only' the escape velocity of the Sun to drop something into the Sun if you're willing to wait a really long time. In practice you wouldn't go all the way to infinity, just raise your perihelion very high. Without gravity assists, this would be the cheapest way to drop something into the Sun, cheaper than trying cancel Earth's orbital velocity. $\endgroup$
    – tylisirn
    Commented Jul 29, 2020 at 16:45
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    $\begingroup$ Sorry, meant *aphelion in the previous rather than perihelion. $\endgroup$
    – tylisirn
    Commented Jul 29, 2020 at 16:56
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    $\begingroup$ @CarlWitthoft: Shouldn't the period corresponding to $r = \infty$ be $T = \infty^{2/3}$ by Kepler's Third Law? $\endgroup$ Commented Jul 29, 2020 at 18:37

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