# What is the escape velocity for the star VY Canis Majoris?

Asked by poisonedantidote (21604) December 13th, 2011

If you accelerate to about 25000 miles per hour you will go fast enough to escape Earth’s gravitational pull. If you accelerate all the way up to 1381600 miles per hour you will go fast enough to escape the gravitational pull of our sun.

VY Canis Majoris is the largest sun/star known to man, at a massive 1.9 billion miles in diameter you could fit our entire planet inside of it about 11 quadrillion times over.

How fast would you need to go in order to escape it’s gravity?

Bonus Question: Is it possible for a star to be so big that you can’t escape it even at the speed of light? (without it turning in to a black hole)

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V = sqrt(2GM/r)

No the internal pressure would be insufficient to resist collapse due to the stars own gravity, and because of the size the mass would be compressed within the schwarzschild radius – basically the definition of a black hole.

Also check out the chandrasekhar limit if you’re interested for alternatives in smaller stars.

wonderingwhy (10770)

I’ll do some calculations for your first question when I have time but right noow I’ll just answer your second question: no, if you couldn’t escape from a body even at the speed of light, the body would necessarily be a black hole; that’s the definition of a black hole.

Mariah (22836)

@poisonedantidote And after reading your link about VYCM (thank you!) here’s a link to the most massive star! R136a1 at 265 solar masses and estimated to be around 320 solar masses at birth O.o! (vs. 30–40 for VYCM)

wonderingwhy (10770)

Alright, so @wonderingwhy provided the formula for escape velocity and from there it’s just a matter of plugging a couple of values. G is the gravitational constant and it’s equal to about 6.67 * 10^-11. M is the mass of the star, which according to your link is not known exactly but estimated to be about 30 to 40 solar masses. A solar mass is about 2 * 10^30 kg, so if this star is, say, 35 solar masses, that’d put it at 70 * 10^30 kg. Finally, r is radius of the star, which according to the wikipedia article is about 1.5 * 10^9 km. We need that in meters to do this calculation, so add three more zeroes: 1.5 * 10^12 m.

So v = sqrt(2GM/r). Plug in the above values:

v = sqrt(2 * 6.67 * 10^-11 * 70 * 10^30 / 1.5 * 10^12)
v = 78,900 meters per second, or 176,500 miles per hour, or about 7 times more than earth’s escape velocity.

So why is it less than the sun’s? Because you chose the largest star by volume, not by mass. This star is only about 35 times more massive than the sun, but its radius is about 1000 times larger. Since escape velocity increases with the square root of mass/radius, having such a large radius means a smaller escape velocity even though the mass is also larger.

Interestingly, this must mean that this star has very low density. If it were the same density as the sun and had a radius 1000 times larger, its mass would be one billion times more than the sun’s (mass increases proportionate to the cube of radius). However, if it were this dense, I’m pretty sure it would be a black hole.

Mariah (22836)
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augustlan (47222)

Need one necessarily reach escape velocity, though? I seem to recall reading once that it would be possible, with enough fuel, to simply crawl out of the gravity well instead of leaping all at once.

Nullo (21899)

@Nullo You’re right, wikipedia sums up pretty much the same thing you said. Reaching escape velocity is only necessary for a a ballistic, meaning an object that, unlike rockets, does not exert a force while it’s traveling.

Mariah (22836)

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