## Escape Velocity – Part 2

In the previous post, I derived the formula for finding the escape velocity from a spherical mass (e.g., the Earth):

$v_{escape} = \sqrt{\frac{2GM}{R_0}} \tag{1}$

$$G$$ is the universal gravitational constant, $$M$$ is the mass of the attracting body, and $$R_0$$ is the starting distance from the sphere’s center. Using the mass and radius of Earth, the formula gives  11.7 km/sec for the escape velocity.

In this post, I will go over some implications of this important formula.

#### Escaping the Sun

To leave the Solar System, we must escape the Sun’s gravity, as well as the Earth’s. Let’s compute the Sun escape velocity, ignoring for the moment the Earth’s effect on our spacecraft. In formula (1), we let $$R_0$$ be the Earth’s distance from the Sun, and $$M$$ be the mass of the Sun. The result is:

$v_{S} = 42.3 \text{ km/ sec}$

which is considerably larger the Earth’s escape velocity.

#### Escaping the Solar System

How do we combine the Sun escape velocity and the Earth escape velocity to get the  net velocity, call it $$v_{total}$$, needed to get out of the Solar System? We can compute this number as follows:

• Suppose that $$E_E$$ is the escape energy for Earth, and $$E_S$$ is the energy to escape the Sun (starting from the Earth’s orbit distance).
• The total energy required is thus $$E_E + E_S = 1/2\, v_E^2 + 1/2 \, v_S^2 = 1/2 \, v_{total}^2$$
• Cancelling the common factors, we have $$v_{total} = \sqrt{v_E^2 + v_S^2}$$

Therefore,

$v_{total} = \sqrt{11.17^2 + 42.3^2} = 43.8 \text{ km/sec}$

This is an alarmingly high value, and the velocity required to reach the outer planets is only a little less than this. In the early days of space exploration, it was thought that this huge velocity would prohibit us from sending spacecraft to the outer Solar System, at least for a long time. Then, in 1962, an engineer conceived the “gravity slingshot” technique, which made such missions far easier. I will describe this beautiful idea in a future post.

#### Black Holes in 1783

The English scientist John Mitchell (1724 – 1793) believed, as did Isaac Newton, that light consisted of particles. Mitchell speculated that these particles, as they left a star, might be acted on by the star’s gravity, thus slowing them down. If so, a large star with an escape velocity equal to the speed of light would prevent the light particles from being able to escape. In that case, the star would be dark, and we would have what we now call a black hole. His premise was wrong, and, of course, he didn’t know about General Relativity, but it was a remarkable conception for his time.

Mitchell described his idea in a 1783 letter. In the letter, he says that he had calculated that if our Sun had a radius 500 times larger, with density held constant, the escape velocity would be the speed of light, and no light could escape.

He didn’t outline how he computed this number, but we can readily check his arithmetic. Using (1), and the known mass $$M_s$$ and radius $$R_s$$ of the sun, we get $$6.18 \times 10^5$$ m/sec as the velocity to escape, starting from the Sun’s surface. Now imagine the Sun with some larger radius $$R$$ and mass $$M$$ (density held constant), such that its escape velocity  is the speed of light, ($$3 \times 10^8$$ m/sec). That is,

$3 \times 10^8 = \sqrt{\frac{2GM}{R}}$

Writing the escape velocities as a ratio:

$\frac{3 \times 10^8}{6.18 \times 10^5} = 485 = \frac{\sqrt{\frac{2GM}{R}}}{\sqrt{\frac{2GM_s}{R_s}}} = \left( \frac{M}{M_s} \right)^{1/2} \left(\frac{R}{R_s} \right) ^{-1/2} \tag{2}$

Now, mass increases as the cube of the radius ( $$V = 4 \pi R^3 / 3$$), so that the ratio of masses  in (2) is $$(R / R_s )^3$$. Equation (2) is then

$485 = \left( \left( \frac{R}{R_s} \right)^3 \right)^{1/2} \left(\frac{R}{R_s} \right) ^{-1/2} = \frac{R}{R_s} \tag{3}$

Therefore, with constant density, escape velocity increases in direct proportion to the radius. So, our 485 value means that the Sun would need to be 485 times larger to have an escape velocity equal to the speed of light. That’s close enough to Mitchell’s value of 500.

#### Escape Velocity and Atmosphere

The molecules of a gas have some average kinetic energy that is proportional only to the absolute temperature. Thus, lighter molecules have a higher average velocity that massive molecules.

However, the temperature only determines the average energy; the energies of the molecules are distributed according to a statistical law called the Maxwell Distribution. At any moment in time, a few molecules will have a very high velocity, and this velocity may exceed the escape velocity of the planet. If such a molecule is in the outer atmosphere, it may not collide with another molecule (which would slow it down), allowing it to escape into space.

This phenomenon explains why less massive planets, like Mars, have trouble keeping an atmosphere. It is also one reason why hydrogen and helium, which have higher average velocities, are not present in our atmosphere.

#### Small Masses

Consider a typical asteroid, such as the one that killed off the dinosaurs. With a radius of about 5 kilometers, and a mass of $$1.4 \times 10^{12}$$ kg, the escape velocity works out to 0.2 meters/sec. An astronaut on such an asteroid would need to be careful; an energetic jump would be enough to reach escape velocity.

As a more extreme example, imagine a lead ball one meter in diameter (mass = 5942 kg). Escape velocity from its surface would be about 1 mm / sec

#### Relation to Orbit Velocity

It can be shown that a satellite’s orbit velocity must be:

$v_{orbit} = \sqrt{\frac{GM}{R}}$

Comparing this to equation (1), we see that the velocity to escape a planet is $$\sqrt{2}$$ times the velocity required to orbit it.

### One Response to “Escape Velocity – Part 2”

1. Heather Says:

That was so brilliant, deriving formula, solution and answers could be very exciting specially when you know your answer is correct. I got so impressed from your work.