Do you need 0 km/s velocity to crash into the sun?

Question

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?

Answer 1

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

• $\mu$ is the gravitational parameter of the sun, $1.32712440018(9)×10^{11} \text{km}^3 \text{s}^{−2}$
• $r_1$ is the orbital radius we're starting from, i.e. the semi-major axis of the Earth's orbit, 149,598,023 km
• $r_2$ is the orbital radius at periapsis (perihelion here, closest-to-the-sun), where we'll use the radius of the sun, 695,700 km

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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

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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

• Is it possible to reach the Sun without expending any fuel/reaction mass?
• What is the functional form for r(t) for a solar-sail deorbit into the Sun?
• What is the optimal angle for a solar-sail deorbit towards the Sun when radial thrust is included?
• Maximum velocity achieved by solar sail

Poynting–Robertson drag

• What is the origin of the dust near the sun?

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.

Answer 6

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

Answer 7

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.