Escape Velocity

Escape velocity is the velocity that something must be going to be able to pull free of the gravitational field of another body. For example, if you want to go to the moon, you need to achieve escape velocity in order to do it.

We can calculate escape velocity by starting with Newton's law of gravity:

F_g = GM₁M₂ / r²

Here F_g is the force due to gravity between the 2 bodies, M₁ and M₂
G is the gravitational constant = 6.6726^-11 /(kg sec²)

dE = GM₁M₂dx / r²

dx is the tiny distance moved (so that the force of gravity is essentially a constant) and dE is the tiny amount of energy it took to move the bodies apart by dx.

If we could add all the tiny dxs up as the bodies move apart (to essentially an infinite distance apart) we would know the energy we'd have to impart to the bodies (this is known as their Gravitational Potential Energy) in the first place in the form of kinetic energy to allow them to achieve escape velocity.

How do you add an infinite number of dxs? What seems impossible here is accomplished by a mathematical wonder called integration. Isaac Newton was the first to do this (Gottfried Leibnitz did it a bit later, independently of Newton) and it is one of the reasons for his great fame. For a more detailed discussion of how integration works, click here

When the above equation is integrated, we get:

E = GM₁M₂ / r

evaluated between r and infinity.

Now that we have the energy required we can calculate the escape velocity. The energy of a moving body is given by its kinetic energy:

E_k = MV² / 2

E_k is kinetic energy. V is velocity

To achieve escape velocity, the kinetic energy must equal the gravitational potential energy, so we set our equations equal to each other and solve for V. We'll also assume that M₁ is very small, and M₂ is very large. By doing this, we can ignore the effect that M₁ has on M₂, which simplifies the math considerably.

M₁V² / 2 = GM₁M₂ / r

The M₁s cancel, and if we multiply through by 2, we get:
V² = 2GM₂ / r

Just for fun, how fast does a rocket have to go to escape earth's gravitational field? When we plug in the numbers, we get:

V = sqrt(2 x 6.6726 x 10^-11 x 5.97 x 10²⁴ / 6367000) = 11,184m/sec.
(That's about 25,000mph, for the metrically impaired)

The Schwartzchild Radius

We just rearrange the last equation so we are solving for r, not V:

r = 2GM₂ / V²

r = 2GM₂ / C²

Note that our derivation did not mention the word relativity once. Yet using the full blown theory of General Relativity, with its attendant tensor complexity, Karl Schwartzchild was able to derive the above. The fact that classical methods and the theory of General Relativity arrive at the same conclusion is, to say the least, remarkable.