This is just a short discussion about black holes which does not use any of Einstein’s complicated field equations and relies (mainly) on school level physics with a few digressions on the way.

Escape Velocity

Firstly, I am going to discuss the ‘escape velocity’ – the speed that we would need to throw an object so as it escapes the Earth’s gravitational pull. To do so, the kinetic of the object must be greater than the energy needed to go from the Earth’s surface to outer space (infinity).

Kinetic energy is given by

where m₁ is the mass of the object and v is its velocity.

Newtons tells us that the force of gravity is given by

where G is Newtons gravitational constant and m₁ and m₂ are two masses separated by a distance r. The energy needed (again from Newton) is the force times distance. However, since the force on our mass changes with distance, we cannot just multiply by the distance.

Inverse Square Laws – digression

Just a quick diversion into the reasons for inverse square laws. It is not that the ‘force gets weaker’ – it is just spread over a larger area as you move away from the Earth. The area it is spread over is the area of the sphere distance r from the Earth i.e 4Ï€r². This is explicit in the usual formula for the electrostatic force between two charges q₁ and q₂:

In Newtons gravitational equation, the 4? is incorporated into Newton’s constant G.

Back to Escape Velocities

If we do a bit of calculus we find that the actual energy need to move a mass m₁ from a distance r₀ from a mass m₂ is given by¹:

To escape the Earth’s gravitational pull this must equal the kinetic energy therefore:

cancelling the mass of our object – m₁ – rearranging we can find the escape velocity:

We now can bring in some real numbers

Mass of Earth	5.97×1024	Kg
Gravitational Constant	6.67×10-11	N·(m/kg)2
Speed of light	2.999×108	 m/s
Radius of Earth	6.371×106	m

Putting these figures into the formula above gives the escape velocity for the Earth as 11,185 m/s (11.2Km/s).

The Sun’s Gravity – digression

This figure is much, much smaller than would actually be needed since the object would have to escape the gravitational pull of the Sun which is much larger. If it is so large, then why do we not feel it on Earth? The reason is that we do and we are actually in free fall towards the Sun. However, since the Earth is also moving it continually misses – i.e. the Earth is in orbit.

Black Holes

Now if we keep the mass of the Earth the same but decrease its radius do we get to a point where the escape velocity is greater than the speed of light? That would mean that not even light could escape the gravitational pull – i.e. we would have a black hole.

If we go back a few steps and replace our escape velocity v by the speed of light c then we have:

We can now rearrange to find the radius r where this will happen

If we plug in some numbers again then we find that if the mass of the Earth was concentrated in a sphere of radius 0.0088m (8.8mm) then it would be so dense that light could not escape – i.e. it would be a black hole.

If we consider a body the mass of the Sun, 1.989 x 10³⁰ Kg, then the radius is 2.95Km.

Although this simple classical analysis does not capture all the intricacies of black holes, the radius calculated agrees well with those calculated by Schwarzschild using Einstein’s field equations.

If you want to play about with the numbers then my spreadsheet is here.

---

If you don’t mind a bit of calculus then energy is given by

putting in Newtons gravitational force expression we get

evaluating this integral gives

While we are here, we can also work out the kinetic energy. If we consider motion in the x direction and using F = ma then the energy becomes

we can now change the integration variable since we know that

this gives

Notice that we have an integration constant. We now know from Einstein that this is mc². What most people know about Einstein’s special theory of relativity (E=mc²) is in fact not relativistic – it has always been hidden there, however Einstein also gives more terms in the expression of kinetic energy alongside E = mc2 + 1/2mv²