How Fast do I have to Go to Become a Black Hole?

Becoming a Black Hole

For science fiction fans, a disappointing result of general relativity is that the speed of light is the ultimate velocity.  In fact, even to propel anything with mass just to the speed of light would require infinite energy because, as the object approaches light speed, its mass increases without bound, according to the theory.

A more exciting (or at least unexpected) prediction of general relativity is the possibility of black holes.  Black holes can arise when a sufficient mass is compressed into a sufficiently dense core such that the escape velocity on the surface of the mass (or some distance from the surface even) reaches or exceeds the speed of light.

Escape velocity does not depend solely on mass.  It also depends on distance.  If the earth had a smaller radius (without changing its mass), the escape velocity from the surface would be greater.  Technically, the mass of the earth could be a black hole, if it were compressed into a sphere of a few centimeters diameter.

The proposition that an object traveling close to the speed of light increases in mass without bound the closer it approaches the speed of light gave me an idea.  If I could add mass to myself simply by increasing my velocity, then, in theory, at some velocity less than light speed, I should turn into a black hole because my mass would be sufficiently large for my radius.

Warning: Do not try this at home.

Just kidding.  All you need is a pencil and some imagination.  We’ll begin by introducing the equations we will use to determine the necessary velocity to add enough mass to an object to turn it into a black hole.

The Equations

In this corner, we have the relativistic equation relating mass and velocity:

{m = \frac {m_0} {\sqrt{1 - \left(\frac {v}{c} \right)^2}}}

In the other corner, we have the Schwarzschild radius equation for a black hole of given mass:

{{R_s} = \frac {2G_n m} {c^2}}

Solving for mass m yields:

{m = \frac {R_s c^2} {2G_n}}

We now have two equations for mass m.  We will set the two equivalent so that our final mass is equal to the critical black hole mass.

{\frac {R_s c^2} {2G_n}  = \frac {m_0} {\sqrt{1 - \left(\frac {v}{c} \right)^2}}}

Now we solve the equation for the ratio of velocity v to the speed of light c and express our result as a fraction of the speed of light.

  1. {{R_s c^2} {\sqrt {1 - \left(\frac {v}{c} \right)^2}} = {2G_n m_0}}
  2. {\sqrt {1 - \left(\frac {v}{c} \right)^2} = {\frac {2G_n m_0} {R_s c^2}}}
  3. {{1 - \left(\frac {v}{c} \right)^2} = {\left( \frac {2G_n m_0} {R_s c^2} \right)^2}}
  4. {\left(\frac {v}{c} \right)^2 = 1 - {\left( \frac {2G_n m_0} {R_s c^2} \right)^2}}
  5. {\frac {v}{c} = \pm \sqrt {1 - {\left( \frac {2G_n m_0} {R_s c^2} \right)^2}}}

Since we only care about speed (the magnitude of velocity) we drop the plus or minus when taking the square root.

{\frac {v}{c} = \sqrt {1 - {\left( \frac {2G_n m_0} {R_s c^2} \right)^2}}}


What do each of these terms represent?  Some of the terms are constants, like the gravitational constant; others are parameters, such as initial mass.


{{G_n} = {6.674 \times 10^{-11}} \frac {m^3}{kg \ s^2}}

Newton’s gravitational constant (meters cubed per kilogram seconds squared)

{c = 2.99792458 \times 10^8 \ \frac {m}{s}}

The speed of light (meters per second)


{m_0 = 68.0389 \ kg}

Initial mass.  Let’s say 150 lb, or a bit over 68 kg.

{R_s = 0.2286 \ m}

Usually, the Schwarzschild radius is calculated for a given mass.  In this case, I’m going to estimate that my “radius” is about 9 inches (0.2286 meters) if I were to curl into a sphere.


With the parameters in place, we should be able to plug all the numbers into our equation and come up with a velocity.

{\frac  {v}{c} = {\sqrt {1 - \left( \frac {2 \left( 6.674 \times 10^{-11}\ {\frac {m^3}{kg \ s^2}} \right) \left(6.80389 \times 10 \ kg \right)}{\left(2.286 \times 10^{-1} \ m \right) \left(2.99792458 \times 10^8 \ {\frac {m}{s}}\right)^2}\right)^2}}}

Since this is a ratio of velocity to the speed of light, we would expect all of the units to cancel such that we have a simple scalar representing a fraction of the speed of light.  Indeed, the units in the numerator, when multiplied, are cubic meters per square second, as are the units of the denominator. (Making sure we end up with the right units can help verify the computation is at least sane).

Already, there is something to notice.  The exponents in the denominator are positive and the exponents in the numerator are negative.  This would indicate we’re going to end up with the square root of a number very close to 1.

Continuing on…

{\frac {v}{c} = \sqrt {1 - \left( \frac {9.081832372 \times 10^{-9} \ {\frac {m^3}{s^2}}}{2.05455433859263 \times 10^{16} \ {\frac {m^3}{s^2}}} \right)^2}}

{\frac {v}{c} = \sqrt {1 - \left( 4.42034177505848 \times 10^{-25} \right)^2}}

{\frac {v}{c} = \sqrt {1 - 1.9539421408295 \times 10^{-49}}}

Assuming I did the math right, this is something of a disappointing result.  Plug this into a calculator, and you will find it deviates so little from 1 that the calculator rounds it to one.  With a paper and pencil, we can write out all the zeroes after the decimal (some 48 of them) and perform the subtraction, which gives:

\frac {v}{c} =\sqrt {0.99999999999999999999999999999999999999999999999980160578591705}

This is a square root I don’t really feel like trying to compute.  Taking the square root would produce a number slightly larger than the one above and slightly less than 1.  A calculator would call it 1, but that’s not actually quite correct.  My mass at the speed of light would technically be infinite, whereas I would become a black hole at a speed just slightly beneath that.

What does it mean?

This result is telling me I would become a black hole at somewhere around 99.9999999999999999999999999999999999999999999999% the speed of light, give or take a few decimal places.  I have to admit, I was hoping for a more interesting result.  If it was 95% or even 99.5% the speed of light, it would be a little more quantified than what I computed, which is “virtually the speed of light”.

Where could I have gone wrong?

One assumption that is probably invalid is that I can maintain my curled up radius of 9 inches, even as my mass increases to extraordinary proportions.  Even still, it would take extraordinary contraction of my radius to make any significant difference in the calculation.  There are about 25 magnitudes to overcome between the gravitational constant and the inverse square of the speed of light.  If I contracted down to the size of an atom, my radius would be on the order of picometers or 10^{-12} meters.  That would still leave 13 orders of magnitude.

What if I just wanted to double my mass?

The outcome for becoming a black hole suggests it takes a lot of velocity for a significant mass increase and that the mass increase curve is very steep.  What if I just wanted to double my mass?  How fast would I have to travel?  In this case, we only need the first equation, and we’ll set our final mass to twice our initial mass.  Let’s start by solving for velocity v again as a ratio to the speed of light c.

{m = \frac {m_0} {\sqrt{1 - \left(\frac {v}{c} \right)^2}}}

  1. {\sqrt {1 - \left(\frac {v}{c} \right)^2} = \frac {m_0}{m}}
  2. {1 - \left(\frac {v}{c}\right)^2 = \left(\frac {m_0}{m}\right)^2}
  3. {\frac {v}{c} = \sqrt {1 - \left(\frac {m_0}{m} \right)^2}}

Note if we wanted the black hole mass, we could substitute our earlier Schwarzschild radius value for m, i.e.  {m = \frac {R_s c^2} {2G_n}}.  We would have gotten the same equation we produced earlier.  Since we want to double our mass, however, we set m = 2{m_0} which gives:

{\frac {v}{c} = \sqrt {1 - \left(\frac {m_0}{2m_0} \right)^2}}

{\frac {v}{c} = \sqrt {1 - \left(\frac {1}{2} \right)^2}}

{\frac {v}{c} = \sqrt {\frac {3}{4}}}

Thus, to double my mass just by increasing my velocity, I have to accelerate from rest to a bit more than 86.6% of the speed of light.

About the Featured Image

The featured image is an artist’s rendering of the black hole Cygnus X-1 produced by NASA and is in the public domain.


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