Imagine a space rock that starts at rest at the Lagrange point between the sun and our nearest stellar neighbor, Alpha Centauri. That would put it at about 2 light years away from the sun.

Now, it gets nudged slightly toward the sun. How fast will it be going when it nears the sun?

I used the formula for instantaneous velocity of a falling object that has traveled distance over a large fall distance, found on this Wikipedia page:


Using the following values:

G = 6.674 × 10^−11 N·m^2/kg^2

M = 1.989 × 10^30 kg

r = 695,700,000 m

d = 2 light years = 9.4607 × 10^15 m

I calculate that by the time the space rock reaches the sun, it is traveling at 617,752 m/s. I would think that this would be the minimum speed of any interstellar object that reaches us because, presumably, it would already have some nonzero speed when it enters the sun's gravitational influence.

However, news reports say that ʻOumuamua is only traveling at 26,330 m/s. Why am I off by a factor of 23?


It is travelling at that speed now after being decelerated by the Sun's gravitational field as it climbs out of the Sun's gravitational potential.

The maximum speed at perihelion was about 87 km/s.

The discrepancy between this and your 617 km/s figure is just that perihelion (closest approach to the Sun) was around 0.25 au. If its trajectory had taken it much closer to the Sun, its speed would have approached the figure you calculated.

Roughly: for a body with zero kinetic energy at infinity, it will travel with a speed $\sqrt{2GM_{\odot}/R}$ when at a distance $R$ from the Sun (just conservation of energy).

For $R=0.25$ au, we get a speed of 84.3 km/s. Because Oumuamua started with a speed of about 26 km/s at infinity, its maximum speed was a touch greater.

The main misunderstanding here is just applying conservation of energy and assuming the asteroid can get arbitrarily close to the Sun. This isn't the case because angular momentum must also be conserved and this limited the closest approach to 0.25 au.

  • $\begingroup$ Okay, got it. To a physicist, there is an obvious difference between free-falling straight into the sun (my example) and zooming very near and around the sun in a hyperbolic trajectory (ʻOumuamua). But to a non-physicist, it all just looks like accelerating toward the sun over a very big distance, and therefore building up tremendous speed. Thanks for the answer! $\endgroup$ – SlowMagic Nov 10 '18 at 1:10
  • 1
    $\begingroup$ @SlowMagic Both energy and angular momentum must be conserved. $\endgroup$ – Rob Jeffries Nov 10 '18 at 8:29

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