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.

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.

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

• 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.

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.

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

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.

Many good answers here, though some strategies require very good timing of the delta V's to arrive at precisely the right time at an outer planet.

A launch to a distant aphelion and a tiny delta V to shed angular momentum is not much different than launch to solar escape ... and both missions will take many years.

If you want QUICK results, retrograde delta V to a high eccentricity Sun-skimming orbit is the delta-V-expensive-but-simple way to do it.

Most of our orbit calculations ignore gravitational effects of the Moon and Jupiter, and the non-zero eccentricity of the Earth's orbit (almost 2%!).

Hence, four decimal place calculation is "measure with a micrometer, mark with chalk, cut with an axe", and some methods require equations and parameters that are difficult to remember.

So, let's start with three numbers: the average radius of the Earth's orbit (149.6 million kilometers = 1.496e11 meters), the 1557600 second length of the year, and angular diameter of the Sun as viewed from Earth 0.0093048 radians.

Now make that two different numbers ... "circular" Earth orbit velocity is Ve = 2 π A.U. / year = 2 * π * 149,597,870,700 meters / 31,556,926 seconds = 29785.89 meters per second.

The aphelion velocity of a Sun skimming orbit is Vsa = 29785.89 m/s * sqrt( 0.0093048 ) = 2873 m/s

That is a 0.2% match to the other calculations ... and that is not a coincidence, it is algebra, and Kepler's second law, with more simplifications. "It is better to vaguely right than exactly wrong" (first written by Carveth Read in "Logic", not John Maynard Keynes).

I'm not skilled with stack exchange mathjax, or I would show the algebraic derivation. The intermediate equations involve perihelion Sun radius (angular size times distance), perihelion velocity (very slightly less than solar system escape, so I assumed equal to escape), and the Solar standard gravitational parameter (derived from Earth orbit velocity and distance).

Fortunately, many terms drop out on the way to Vsa (imagine Karen Carpenter singing that ...).