What is the minimal velocity to throw an object (material point) to the Sun from Earth, with no specific restrictions?

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    $\begingroup$ The TL;DR for those who don't read all the way thru the answers: far less speed (aside from the vector part of velocity) required if you just want to hit the sun sometime in the future, as opposed to falling "straight down." $\endgroup$ Mar 1, 2018 at 18:56
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    $\begingroup$ “Accordingly” to what? $\endgroup$
    – DaG
    Mar 1, 2018 at 21:53
  • $\begingroup$ Is one permitted to "throw" a solar sail-like object? $\endgroup$
    – ceilingcat
    Mar 2, 2018 at 3:32
  • $\begingroup$ It takeprobly more Delta v I bet than to leave solar system. $\endgroup$ Jul 5, 2018 at 15:30

3 Answers 3


The limitation to hit the sun is that the object has to have very little angular momentum. The reason for this is that as the distance to the sun gets smaller, the velocity in a direction perpendicular to the sun gets larger, thanks to conservation of angular momentum:

$$ L = mv_\perp r\rightarrow v_\perp={L\over mv}$$

A good first-order approximation can be found just by assuming you throw the object so that it has zero angular momentum. To do this, you have to throw the object as fast as the earth is traveling around the sun, just in the opposite direction. So, roughly $30~\rm{km\over s}$. There are two effects that change this a little and one that changes it a lot:

  • Earth's gravity will slow the ball down, so you have to throw it a bit faster at the start. This requires you to throw the ball about $7\%$ faster, as when the object leaves the earth's gravity well it loses $\frac12mv_{\rm escape}^2$ of its kinetic energy. This means the initial kinetic energy must be $\frac12mv^2_{\rm required\ speed\ after\ escape}+\frac12mv^2_{\rm escape}$, so $v_{\rm throw}^2=v_{\rm ignoring\ escape}^2+v_{\rm escape}^2$
  • The sun has a finite extent, so the ball can have a small angular velocity and still hit the sun. This lets you throw the ball a little slower (but not much: the sun is a small target as far as orbits are concerned)
  • Air resistance is enormous at $30\ \rm{km\over s}$, so you're going to have to throw it a lot faster if you're not ignoring air resistance (So throw it from orbit, not from the ground)
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    $\begingroup$ It's important to note that, at least in terms of $\Delta v$, this isn't the cheapest way to reach the Sun. You can do better by transferring to a close-to-parabolic orbit with its perihelion at the Earth's radius and its aphelion as far away as you can, coasting to the aphelion (which can admittedly take a long time) and then doing a small correction at aphelion to lower the perihelion down to zero. The correction is cheap (because your velocity is low) and the burn to $\lesssim$ escape velocity is cheaper than your procedure, so the whole thing is cheaper. $\endgroup$ Mar 1, 2018 at 10:22
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    $\begingroup$ I take the meaning of "throw" as it's forbidden to do course corrections: apply Δv once, and crash into the Sun in finite time. $\endgroup$
    – user27542
    Mar 1, 2018 at 15:56
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    $\begingroup$ @EmilioPisanty: and then doing a small correction at aphelion to lower the perihelion down to zero That would no longer make it a throw, as per the question. Throwing implicitly entails that the object is not actively steering itself after it has been launched. $\endgroup$
    – Flater
    Mar 1, 2018 at 15:57
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    $\begingroup$ @Flater Indeed it doesn't, which is why I added it as a comment instead of a separate answer. It's not an answer to the question as posed but it's still relevant. $\endgroup$ Mar 1, 2018 at 15:59
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    $\begingroup$ @user27542 You don't have to add the escape velocity, just the kinetic energy required to escape. So the required speed is about $\sqrt{30^2+11^2}~\rm m/s$. $\endgroup$
    – Chris
    Mar 1, 2018 at 16:07

The moment you throw stuff into space, orbital mechanics apply. Specifically, an object will go into an elliptical trajectory around the sun. The best way to reach the sun is to throw it retrograde from Earth's orbit. The other answers take this to an extreme level, throwing it retrogade at Earth's orbital velocity. The object will fall straight down. Luckily, we don't have to go that far.

The velocity at any point in orbit is determined by the equation

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

with $r$ the distance from the sun's center, and $a$ the semi-major axis, which is half the 'cross-section' of the elliptic orbit. In our case, we have an orbit which has the aphelion (furthest point) intersecting with Earth's orbit, and the perihelion (nearest point) intersecting exactly the surface of the sun. We take the sun's radius to be 695700 km, and the Earth's orbital radius $r$ 150 million kilometers. This results in $a=150\times 10^6-695700\approx 149.3\times 10^6$ kilometers.

The resulting velocity requirement is 2,8 km/s at aphelion. The Earth travels around he sun at about 29.8km/s. So, we need to throw the object with at least 27km/s.

Of course, we also need to escape Earth's gravity; the moment we throw something retrograde, the Earth starts pulling it prograde, so the object gains kinetic energy again. To counteract this, we must throw the object with the extra kinetic energy required in excess of the gravitational potential energy. The escape velocity of Earth is 11.2km/s. Since kinetic energy grows with the square of the throwing velocity, we find the new throwing velocity $v_\mathrm{throw}=\sqrt{11.2^2+27^2}=29.2\mathrm{km/s}$.

(side note: the Earth's trajectory is not exactly circular, so try and throw it at aphelion if you're planning on doing this for real).

  • $\begingroup$ Would it be possible to use e.g. Jupiter for gravity assist to drop your orbital speed to near zero? $\endgroup$
    – user27542
    Mar 1, 2018 at 15:59
  • $\begingroup$ @user27542 I think so. However, the question 'what is the ideal mission profile to crash into the sun' is a bit different than 'how hard should I throw'. The precision required to get the throw right for a gravity assist without any correction during the mission would be astronomically high. $\endgroup$
    – Sanchises
    Mar 1, 2018 at 16:43
  • $\begingroup$ @Sanchises It's astronomically high anyway ;) $\endgroup$
    – Chris
    Mar 1, 2018 at 18:46
  • $\begingroup$ @Sanchises I think in the theoretical world full of spherical cows and massless pulleys, a perfectly precise throw in the direction of Jupiter won't be a problem. $\endgroup$ Mar 1, 2018 at 18:57
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    $\begingroup$ A gravity assist off Jupiter would results in a huge savings: If we go with NASA level shenanigans, where Cassini did 4 gravity assists on the way out, it's very likely you could do it with half that dV budget or less. $\endgroup$ Mar 1, 2018 at 21:56

It's very hard to hit the sun. With the current technologies it's actually more cheaper to escape the solar system than to reach our own sun!

Going to the scientific perspective, the earth orbits the sun at roughly 30 km/s (relative to the sun). So in order to crash into the sun we would have to accelerate our object until it has 30 km/s velocity (relative to us) in the opposite direction so it would have zero velocity relative to the sun. The sun's gravitational pull just then pulls the object towards itself, but
CAUTION: Even if our object has even a little velocity it would probably miss the sun and start a complicated elliptical orbit around it.
So we need the velocity to be exactly 30 km/s.

On the contrary the escape velocity from earth is 11 km/s which is much lesser than 30 km/s. so it's a lot harder to crash into the sun.
As a little evidence to back up by statement NASA in 2005 proposed a solar probe in which the satellite would be first sent out to Jupiter (where the 30 km/s on earth is just 13 km/s on Jupiter) where it much easier to get that velocity and then towards the sun.

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    $\begingroup$ "So we need the velocity to be exactly 30 km/s." No, we need the velocity to be exactly (the opposite of) the earth's, which is approximately 30km/s. Also, since the sun is a large object, not a point, there is a range of velocities that will result in impact somewhere on the sun. $\endgroup$ Mar 1, 2018 at 19:57
  • $\begingroup$ @DavidRicherby good point, I would like to add here; That since earth's orbit isn't exactly a circle it would have different velocities at different positions in orbit, hence it would vary maybe just a little bit above or below 30 km/s. $\endgroup$ Mar 2, 2018 at 12:47

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