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I was reading a popular thread about the delta-v required to escape the solar system compared to the delta-v required to crash into the sun. I get it: the earth itself already has a high speed (29.7km/s) so you just need to keep pushing forward to escape. But you need to lose all the "earth speed" (from 29.7 to 0 km/s) to crash into the sun. If you compare both, it's "cheaper" to escape the solar system.

The part I don't get is: why does one need a velocity of 0 km/s to crash into the sun? Wouldn't you inevitably spiral down to the Sun's surface even if you were going faster than 0 km/s?

You don't really need to "drop in straight line" (which would require, indeed, 0 km/s), or do you?

Glorfindel
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ker2x
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    Your earth velocity is perpendicular to the sun. If you don't get rid of all of it you always miss the sun and end up in an elliptical orbit. – user3528438 May 05 '20 at 21:06
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    It's also possible to be put into a hyperbolic orbit where the periapsis encounters the Sun. This would be done by using gravity from other planets. – Star Man May 05 '20 at 21:12
  • Possibly some important info. The Earth and all other planets are actually spiraling away from the Sun. So you wouldn't spiral towards the Sun. At least not for billions of years. – Star Man May 05 '20 at 21:16
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    @user3528438 but that would only be true if the "sun" was an infinitely small point, doesn't it ? the sun is big and i don't understand why you need 0m/s to reach it's surface. – ker2x May 05 '20 at 21:17
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    @user3528438 The sun isn't a point object. If you put your periapsis in the photosphere you'll burn in (you can't really crash into the sun, even not counting the radiant energy, entry heating will destroy you before you come to anything solid enough to stop you.), not come back up. – Loren Pechtel May 06 '20 at 03:48
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    @LorenPechtel yep, in my question reaching the photosphere count as "crashing" :) – ker2x May 06 '20 at 04:40
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    Please note that there is no such thing as "spiraling into something". Neither stars, planets, nor black holes are cosmic vacuum cleaners actively sucking in things. Unless some external force acts upon a body (or it encounters significant drag), orbits stay as they were. – vsz May 06 '20 at 06:48
  • Technicaly, how can one "crash" into the Sun? There is no hard surface for a descent "crash". There is even no dense enough gas to slow you down before the heat evaporates you out. – fraxinus May 06 '20 at 07:40
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    I think what really should be mentioned crashing (or plummeting if you prefer) doesn't have to be that expensive if you're not in a hurry. It's called bi-elliptic transfer. You accelerate forward, into almost escape trajectory (~12.7km/s, 8 out of which are needed to get out of Earth, into low orbit), and at the apoapsis you'll be really slow - getting down to that "zero" will be really cheap. Then you'll plunge into the Sun from somewhere in the Oort cloud, But it will take a couple decades at least. – SF. May 06 '20 at 08:21
  • The Parker Solar Probe is an example of the work needed to slow something down to get it close to the Sun. –  May 06 '20 at 10:17
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    Note that orbits are entirely reversible - any orbital trajectory will play out exactly the same in reverse if you run the clock backwards (landing a rocket is just takeoff in reverse). If you could really "spiral into the sun" without doing anything, that would mean you could spiral out of the sun "for free", which is intuitively impossible. But it requires the exact same energy transfer as spiraling in, just backwards. – Nuclear Hoagie May 06 '20 at 13:11
  • There's a reason the classic crash-into-the-sun trajectory is a Jupiter gravity assist. – Joshua May 06 '20 at 14:46
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    @Nuclear Wang: That "no spiraling" would be true in an ideal universe. In this one, the Sun, like most of the planets, has an atmosphere. Get close enough, and you experience friction, which means you will gradually lose energy, and eventually crash into the body. That's why satellites in LEO eventually de-orbit themselves, if they are not re-boosted like the ISS. – jamesqf May 06 '20 at 15:42
  • @jamesqf Yes, the reversibility of orbits only works in situations without non-conservative forces like atmospheric friction. It's a good approximation of reality for most of space, but it does break down when you're close enough to an atmosphere to experience drag. The sun's atmospheric thickness is <0.1% of its radius, so if you're experiencing drag, you're practically already in the sun, from the scale of the solar system. – Nuclear Hoagie May 06 '20 at 15:48
  • @Nuclear Wang: That of course depends on what you consider as atmosphere. By one definition, it might be said to extend well beyond the orbit of Pluto. I admit that you're going to have to wait a while for friction to have an effect at that distance :-) – jamesqf May 07 '20 at 03:02
  • @ker2x: The sun is big, but in comparison to your distance from the sun (i.e. Earth's orbital height), the sun is a tiny dot. You don't need exactly 0 km/s, but it doesn't take a lot of speed for you to miss the sun. Essentially, if you move sideways > the radius of the sun over the time it takes you to "fall into" the sun, then you miss it. It's a long journey so even a low velocity is liable to make you miss. So you need to be close to 0 km/s to prevent moving sideways too much. – Flater May 07 '20 at 12:36
  • @ker2x please note that most fo the answers below are not actually correct. The speed you go at has nothing to do with whether you crash into the sun or not, as long as your speed is non-vero. The comments are refering to reducing your "velocity", but that's nonsense too. Velocity is a vector, it has magnitude (your speed) and a direction (a tangent to the orbit). You need to apply energy to rotate that vectors direction, so its not a tangent, but a line intersecting the sun. This can be done regardless of the vectors size. A speed of 0km/s means you are stationary, and wont collide at all. – Innovine Jun 08 '20 at 13:26
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    @Innovine - It's your comment that is incorrect. The answers are correct, assuming a single burn at Earth orbit and no gravity assists from other planets. Re "A speed of 0km/s means you are stationary, and wont collide at all." This is incorrect. You seem to be forgetting gravity. An instantaneous velocity of zero means the object will collide with the Sun as the object is on a radial trajectory. – David Hammen Jul 29 '20 at 10:50

7 Answers7

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Wouldn't i inevitably spiral to sun surface even if i was faster than 0km/s ?

No. On reasonable timescales, an orbit will have a fixed distance of closest approach, called "periapsis." (These timescales shorten if you're close enough to what you're orbiting that an atmosphere can drag you down).

You don't really need to "drop in straight line" (which would require, indeed, 0km/s), or do you ?

True. 0 km/s would be necessary to hit the center of the sun. We can solve for the necessary velocity to lower your periapsis below the sun's radius. Per Wikipedia, the first burn for a Hohmann transfer takes a delta-V of $$ \Delta v = \sqrt{\frac{\mu}{r_1}} \left( \sqrt{\frac{2r_2}{r_1+r_2}} -1 \right) $$

For the transfer we're considering

Plugging all that into Python, I find we need a delta-V of -26.9 km/s to graze the sun's surface. Assuming your figure of 29.7 km/s was correct, we've shed 90% of our sun-centric velocity to do this.

Erin Anne
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  • Thank you, i understand now. As it was pointed by @user123, it doesn't make much difference in the grand scheme of thing. One quick question : you give a "negative delta-v". Should i understand it mean that, to crash the surface of the sun, you need to decelerate by 26.9km/s and therefore need a 1km/s (or less) velocity. correct ? – ker2x May 05 '20 at 21:53
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    @ker2x you're correct that the negative indicates the delta-V is a deceleration (likewise that, for r_2 > r_1, an acceleration is required), but 29.7 - 26.9 is 2.8. So not 1km/s or less. – Erin Anne May 05 '20 at 21:55
  • i really need to sleep ... thank you again ! (good night!) – ker2x May 05 '20 at 21:57
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    Ey, that's not bad, actually! With the rocket equation being what it is, 3 km/s is nothing to sneeze at. – John Dvorak May 06 '20 at 06:08
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    @ErinAnne, the other option is to accelerate towards the Sun so you're going fast enough that the lateral velocity doesn't matter. Ignoring gravity, a delta-V of 6387 km/s inwards towards the Sun should do it, for a flight time of just over 6.5 hours. – Mark May 06 '20 at 19:48
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    @Mark I tend to think of the ascend-first options as "the other option," but a 2% c delta-V directly sunward is certainly somewhere on the list, lol. Though while you're at it, you may as well cancel out that lateral velocity, just to be sure you're pointed correctly. – Erin Anne May 06 '20 at 23:18
  • This is entirely incorrect. You can crash into the sun at any velocity other than 0, as lung as yeur velocity is in the appropriate directieo. If you have a velocity of zero, you are motionless, therefore not about to crash into anything – Innovine Jun 06 '20 at 13:50
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    @Innovine we're talking here about starting in an solar orbit at the Earth. You have to dramatically reduce your initial velocity to be able to direct your velocity in "the appropriate direction." Reducing that velocity to zero means you fall directly into the sun under its gravitational influence. Having a tangential velocity greater than stated above (2.8 km/s) at 1 AU from the sun means that solar gravity alone is not sufficient for you to graze the sun (here, coming 695 700 km from its center). – Erin Anne Jun 06 '20 at 20:36
  • @ErinAnne "You have to dramatically reduce your initial velocity to be able to direct your velocity" is totally incorrect. You are confusing velocity with tangential velocity. Velocity is a vector. 2.8kms is a speed (a scalar). You can take that initial velocity, and by applying an acceleration along the nadir direction, retain the exact same magnitude of velocity, while changing the direction into a collision course with the sun. At all times, the initial 2.8km/s can be maintained. The amount of Kerbals posting incorrect answers is astounding. – Innovine Jun 08 '20 at 13:19
  • Why is this still an upvoted and accepted answer? " 0 km/s would be necessary to hit the center of the sun" is physically impossible. A speed of 0 means you are not moving, and cannot collide with anything. – Innovine Jun 08 '20 at 13:23
  • @Innovine please feel free to expand on your arguments in a proper answer. I won't pretend my answer is the only correct one (though I think it addresses the question), but what you've laid out so far isn't terribly compelling to me. You can absolutely add radial velocity instead of working with your tangential velocity, but I think you'll find that those delta-Vs are much higher than the ones laid out in the other answers. Keep in mind that we're describing initial conditions that then evolve under the influence of the sun's gravity, and please stop calling people Kerbals. – Erin Anne Jun 08 '20 at 21:41
  • i think we're all answering the question by assuming that "velocity" mean "tangential velocity", relative to the sun. we could argue all day about it, because the sun doesn't have a null velocity either so ... i accepted this answer because the first one and i understood what was written :) – ker2x Jun 09 '20 at 06:30
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You need below 2866 m/s of orbital velocity at 1 AU to crash into the Sun.

You technically don't need to slow down exactly to 0 m/s relative to the Sun in order to crash into it. Let's calculate the approximate velocity required to graze the "surface" of the Sun. This is an excellent answer on how to calculate apoapsis and periapsis of an orbit.

So first, the Earth is about 150,000,000 km from the centre of the Sun. We want to obtain a perihelion of 700,000 km from the centre of the Sun (radius of the Sun is about 697,000 km, so that's about 3,000 km above the "surface").

So let's work backwards. To calculate eccentricity, use: $$e=\frac{r_a-r_p}{r_a+r_p}$$ which is $$e=\frac{1.5 \times 10^{11}-7 \times10^8}{1.5 \times 10^{11}+7 \times10^8}$$ therefore, $e = 0.99071$. Now let's find what velocity we need at apoapsis (starting point) to have a periapsis of 700,000 km. Let's work backwards. $$a = \frac{r_p}{1-|e|}$$ which is $$a = \frac{7 \times 10^8}{1-0.99701}$$ and therfore, $$a=7.535 \times 10^{10}\space m$$ Calculate orbital specific energy (we need to use the Sun's GM which is $1.327\times 10^{20}$): $$E=\frac{-GM}{2a}$$ so, $$E=\frac{-1.327 \times 10^{20}}{2 \times (7.535 \times 10^{10})}$$ and therefore, $E = -880557398.8$. Now we just calculate velocity at 150 million km. $$V=\sqrt{2(E+\frac{GM}{r})}$$ substitute values (remember, $r$ is 150 million km). $$V=\sqrt{2\bigg(-880557398.8+\frac{1.327 \times 10^{20}}{1.5 \times 10^{11}}\bigg)}$$ and $V = 2866.8$ $m/s$.

We can conclude that we need about 2867 m/s of velocity at the distance of 150 million km to obtain a periapsis of 700,000 km which is just above the surface of the Sun. Meaning you need a $\Delta V$ of $-26.914$ $km/s$ because Earth's velocity is about 29 km/s. Since 26 km/s of delta v is A LOT, what most spacecraft do is go to one of the outer planets (like Jupiter) and use a gravity assist to decelerate. Orbital velocity decreases with distance.

And Earth would lose its orbital energy and spiral and crash into the Sun but that would take billions of years. Satellites take many years to de-orbit Earth because of the atmosphere and the Sun's activity. But before Earth even loses its orbital energy, the Sun would expand into a Red Giant and possibly swallow Earth.

Star Man
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  • You note in the comments on the question that Earth is spiraling out, so why contradict that in the last paragraph of this answer? – Erin Anne May 06 '20 at 05:38
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    There are few effects that make Earth spiral in and out. The dominant now is probably the tidal effect that pushes Earth out. At some point (a great deal of bilions of years from now) the gravitational energy emission will predominate and Earth will spiral down. – fraxinus May 06 '20 at 07:32
  • I think I'm misunderstanding the "that would take billions of years" in the last paragraph here. It's the same statement and I'm just parsing it wrong (parsing this one as "it is currently spiraling, but hitting the sun will take billions of years.") My mistake. (Though I thought the sun was due to expand to red giant before a spiral down would happen) – Erin Anne May 06 '20 at 09:48
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    @fraxinus Gravitational wave emission is completely negligible and will never become dominant. Jupiter emits on the order of 200W, the Earth probably only a handful of W, and decreasing with orbit increase. Running into space dust and rocks head on at 30km/s will likely exceed the effects of GW forever (until the Earth is vaporized by the Sun). – Jens May 07 '20 at 09:10
  • Please, could you change this "2,866 m/s" to something less ambiguous (to a non-native-English reader)? I only found that it's $2.866,\mathrm{km/s}$ and not $2.866,\mathrm{m/s}$ after I had read to the end of your answer, and until that I was very surprised that one needs such a small velocity (not knowing exactly what order of magnitude it should actually be)! – Ruslan May 07 '20 at 13:53
  • @Ruslan Okay. I've removed the comma. In the U.K, U.S, Canada, we use commas to separate digits and a period to denote a decimal. I know that other countries do the opposite. – Star Man May 07 '20 at 14:43
  • BTW, your use of perigee and apogee is wrong: they are the apsides of an orbit about the Earth. For an orbit about the Sun the proper terms are perihelion and aphelion. See the table in the description of the first figure in this page. – Ruslan May 07 '20 at 16:23
  • This answer is incorrect. You can change orbit from the initial circular orbit into a collision course with the sun without reducing your speed (the magnitude of the velocity vector). You just need to change the direction of the vector (apply a force in the nadir direction). – Innovine Jun 08 '20 at 13:22
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And note that if you want to hit the sun the cheaper (but slow!) way to do it is to head out. 12.32km/sec will take you to infinity, at infinity a burn of 0m/sec will kill your orbital velocity and you'll come straight in. Of course this will take infinite time, but even going only as far as Jupiter's orbit means you use less energy to drop your periapsis than if you had done it directly.

The cheapest way is to head for Jupiter and use it to slow you down.

Loren Pechtel
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  • Isn't Mars good enough? – fraxinus May 06 '20 at 07:33
  • @fraxinus that could be an interesting new question – uhoh May 06 '20 at 11:06
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    For a real life example (not crashing, but pretty damn close) we use Venus https://en.wikipedia.org/wiki/Parker_Solar_Probe#Trajectory – Viktor Mellgren May 06 '20 at 11:38
  • It will take infinite time.... unless you bring a spare aerosol can with you and give yourself a tiny push towards the sun to get started. – J... May 06 '20 at 13:13
  • @J... Yeah--note my 0m/s burn (which obviously isn't possible) at infinity, vs less than the direct path if you kill your velocity at Jupiter. The farther out you go the smaller the needed burn. – Loren Pechtel May 06 '20 at 18:47
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    @ViktorMellgren, note that the Parker Solar Probe is using seven Venus flybys, while the original mission plan called for a single Jupiter flyby to get the same periapsis drop, plus an inclination change into a polar orbit. – Mark May 06 '20 at 19:54
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    This is known as a bi-elliptic transfer. (https://en.wikipedia.org/wiki/Bi-elliptic_transfer). This isn't the same as the Parker trajectory (which flew in on a direct trajectory and is using gravity assists to further reduce perigee). – lwr May 06 '20 at 22:59
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1. Math

Another version of @StarMan's answer using only the prolific† vis-viva equation to find the minimum velocity at 1 AU that will graze the Sun:

$$v_{1 AU}^2 = GM_{Sun}\left(\frac{2}{1 AU} - \frac{2}{r_{peri} + r_{apo}} \right)$$

where $GM_{Sun}$ is $1.327 \times 10^{20} \ \text{m}^3 / \text{s}^2$, $a = (r_{peri} + r_{apo})/2$ and $r_{peri}$ is the radius of the Sun.

It's no coincidence that this looks exactly like @ErinAnne's answer as well; there's only so many ways to enforce conservation laws.

The minimum of $v^2$ will be where $r_{apo}$ is also 1 AU ($1.496 \times 10^{11} \ \text{m}$).

With $r_{Sun}=6.957 \times 10^8 \text{m}$ that gives 2865 m/s confirming the other answers.

†https://space.stackexchange.com/search?q=%22vis-viva%22

2. Physics

Wouldn't it inevitably spiral down to sun surface even if it was faster than 0 km/s?

That could happen passively if the object had certain peculiar characteristics either by design or by coincidence.

Solar sail

Poynting–Robertson drag

A object orbiting near the Sun could, under some special circumstances slowly spiral into the Sun, but it would take a very long time even for a speck of dust, much longer than for a solar sail.

uhoh
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    yes, i realized after the fact that it need to slow down over time (eg : drag) in order to spiral down to the sun. i was very tired when i posted this -_- ... thank you for all the link, that's some i interesting information :) – ker2x May 06 '20 at 04:38
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Already a lot of very good answers, but one simple explanation might be worth adding:

If you want to hit the sun, you have to be heading quite straight to the sun, otherwise you'll miss it.

And in space missing the sun on the first attempt means that you'll never hit it. You either have enough speed to leave the solar system on a parapolic course, or you'll end up in an elliptical orbit that either touches the sun or misses it, on every turn. Without active thrust, in space there is no such thing as a spiral trajectory.

That said, the Earth orbit gives you a lateral speed of 29 km/s, so if you want to head straight into the sun, you have to compensate that speed.

Ralf Kleberhoff
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You don't need to slow down all the way but the difference between lowering your periapsis to the core of the sun compared to it's surface is not that much in the grand scheme of things

User123
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    Additionally, even getting close to the sun will have enough radiation to cause problems. The "surface" isn't really a defined change, its much more gradual. – Criggie May 07 '20 at 01:16
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The sun is TINY compared to 1 AU, the distance from Earth to the Sun. If you really want to reach the core, 0 km/s is the way to go. If you just want to hit the sun (for example, if you want to dump nuclear waste there for whatever reason), you just need to slow down... a lot. But not precisely to 0 km/s. Of course, this assumes you're using pure rockets. You could slow down, albeit very slowly, with some form of solar sail. There also might be some other form that may be known or not that is more efficient for sun-smacking endeavors.

EDIT 1

An easier way to hit the sun than ~0km/s is to go to the outer region of the solar system, as this makes it easier to slow down… and take the final dive.

Someone
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