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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?

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    $\begingroup$ 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. $\endgroup$ – user3528438 May 5 at 21:06
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    $\begingroup$ 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. $\endgroup$ – Star Man May 5 at 21:12
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    $\begingroup$ @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. $\endgroup$ – ker2x May 5 at 21:17
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    $\begingroup$ @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. $\endgroup$ – Loren Pechtel May 6 at 3:48
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    $\begingroup$ 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. $\endgroup$ – vsz May 6 at 6:48
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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.

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    $\begingroup$ @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. $\endgroup$ – Erin Anne May 5 at 21:55
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    $\begingroup$ Ey, that's not bad, actually! With the rocket equation being what it is, 3 km/s is nothing to sneeze at. $\endgroup$ – John Dvorak May 6 at 6:08
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    $\begingroup$ @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. $\endgroup$ – Mark May 6 at 19:48
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    $\begingroup$ @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. $\endgroup$ – Erin Anne May 6 at 23:18
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    $\begingroup$ @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). $\endgroup$ – Erin Anne Jun 6 at 20:36
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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.

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  • $\begingroup$ You note in the comments on the question that Earth is spiraling out, so why contradict that in the last paragraph of this answer? $\endgroup$ – Erin Anne May 6 at 5:38
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    $\begingroup$ 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. $\endgroup$ – fraxinus May 6 at 7:32
  • $\begingroup$ 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) $\endgroup$ – Erin Anne May 6 at 9:48
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    $\begingroup$ @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). $\endgroup$ – Jens May 7 at 9:10
  • $\begingroup$ 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)! $\endgroup$ – Ruslan May 7 at 13:53
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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.

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  • $\begingroup$ Isn't Mars good enough? $\endgroup$ – fraxinus May 6 at 7:33
  • $\begingroup$ @fraxinus that could be an interesting new question $\endgroup$ – uhoh May 6 at 11:06
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    $\begingroup$ For a real life example (not crashing, but pretty damn close) we use Venus en.wikipedia.org/wiki/Parker_Solar_Probe#Trajectory $\endgroup$ – Viktor Mellgren May 6 at 11:38
  • $\begingroup$ 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. $\endgroup$ – J... May 6 at 13:13
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    $\begingroup$ @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. $\endgroup$ – Mark May 6 at 19:54
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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.

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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

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  • $\begingroup$ ho... ok then... thanks :) $\endgroup$ – ker2x May 5 at 21:30
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    $\begingroup$ 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. $\endgroup$ – Criggie May 7 at 1:16
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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.

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  • $\begingroup$ I tried to put a beginner's answer since I am not very good at the physics of this situation. $\endgroup$ – Someone May 6 at 13:56
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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.

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    $\begingroup$ 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 :) $\endgroup$ – ker2x May 6 at 4:38

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