Escape Velocity
Published on Nov 29, 2006 at 6:26 pm.
8 Comments.
Filed under physics.
In From the Earth to the Moon, by Jules Verne, space travelers are propelled to the Moon by a giant cannon in Florida. Well, that would have to be one heck of a cannon!  Written in 1865, this was one of the first science fiction novels written, and it is still a classic that everyone interested in science fiction should read. Using figures given by Verne, I figure that the occupants of his projectile would experience a force of over 61,000 g’s (a force over 61,000 times their weight). That would leave them as little wet spots on the back of the projectile. At least, it would leave them there for a fraction of a second. The projectile would leave the cannon moving at Mach 45 (45 times the speed of sound). Given that the description of the projectile was that it was constructed of a thin aluminum shell … well, it would be crushed by air pressure immediately and then burn up. But, the problem is that the projectile couldn’t have been launched much more gently or slower. Basic physics provides certain constraints. And, by basic physics, I mean first semester physics.
 The problem is gravity. Gravity is an attractive force between bodies having mass. So, if you toss something upwards, the gravitational attraction between that body and the Earth will slow it down. Eventually, the object will stop, and then fall back to the ground. Hence, the famous saying, “What goes up must come down.” If you simply throw it up faster to start with, it takes longer for gravity to stop it, so it goes higher, but eventually falls back. If you throw it up really fast, then you can imagine it going so high that the Earth’s gravity is weaker, so it doesn’t pull back quite so hard. That makes it take even longer to slow down, but eventually it does, and then it stops, and then it falls back down. But, you can imagine it moving so fast to start with that it gets so far from the Earth before it stops that Earth’s gravity is simply too weak to stop it. In this case, the object has escaped Earth’s gravitational influence, and so it goes on forever. The minimum speed needed to accomplish this is called the escape velocity. For Earth, the escape velocity is about 11.2 km/s (for my readers here in the US, that is about 25,000 mph).
Note that this is really a thought experiment. In real life, you can’t just hurl something upwards at 25,000 mph and expect it to leave Earth. First of all, it would likely burn up in the atmosphere (and be torn asunder by atmospheric forces). If by some miracle it survived, the atmospheric forces would slow it down so that it would be moving too slowly to escape Earth. So, the 11.2 km/s figure given in textbooks all over the Earth is really a joke. It would only be true if Earth had no atmosphere! But, the Moon and Mercury have no appreciable atmospheres, nor do asteroids or many other bodies in the Solar System. So, the calculations for their escape velocities are fine. But, you don’t have to start on the ground! If you are already in Earth orbit, then you can calculate how fast you’d have to be going to get away from Earth’s gravitational clutches, and that calculation would be pretty good (Again, there are complications, since there are other things in the universe besides the two bodies in question, so it would depend on where the Moon was, the Sun, what direction you were going relative to these bodies, etc.).  But, the interesting thing is that the calculations for escape velocity can be done in first semester physics, and they are not at all difficult.
Basically, the escape velocity, ignoring external influences, can be determined simply by application of conservation of energy. When an object is moving, it has energy of motion. We call this kinetic energy, and in introductory physics we use the symbol K to represent the kinetic energy of a body. When you throw an object upwards, it slows down. This means that its kinetic energy becomes smaller. When an object falls, gravity pulls on it making it speed up, thus increasing its kinetic energy. But, energy is conserved. That is a basic principle in physics. The total energy must remain the same.  So, we come to the conclusion that there must be some other energy than kinetic energy.  Since gravity is the key here, we call this gravitational potential energy, and the symbol that we normally use in the introductory physics classes is the letter U. The farther apart two bodies are, the greater their gravitational potential energy. So, when you throw a ball upwards, it slows down the higher that it goes. Its kinetic energy is getting smaller, but its gravitational potential energy is getting bigger. The sum of the two remains constant. This is conservation of energy.
 Kinetic energy for an object of mass m moving at a velocity of v is given by the equation:
and the gravitational potential energy between two bodies is given by the equation
where m is the mass of the smaller body, M is the mass of the larger body, and r is the separation between them. For a spherical object, r is measured from its center. The term G is the Universal Gravitational Constant, and it is given by 6.67×10-11 N m2 /kg2 , in SI units. The gravitational potential energy is defined as negative by convention (so that the potential at infinite distance between objects, where they can’t interact at all, will be zero). The total energy is given by simply adding the kinetic energy and the potential energy together.
The condition for someting escaping from a larger body will be given by
  Â
which makes sense if you think about it. The farther apart that the two objects get, the smaller the kinetic energy, because some of that energy is going into potential energy.  At infinite distance, the gravitational energy is zero. So, if the sum of the two terms is positive, that means that the object slows down, but once infinitely far away U is zero, so K must be some non-zero value, meaning that the object is still moving away.  The object has escaped the gravity of the larger body.  But, if the sum of the two terms is negative, then K gets smaller until it becomes zero. There is still a potential energy, so the smaller body begins to fall back. The critical case is when the sum of the two terms is equal to zero. Then, the object slows down, but at an infinite distance it reaches a speed of zero. But, the two bodies are infinitely far apart, and thus not able to interact. The object has barely managed to escape. The speed at which this ballance occurs we call the escape velocity.
So, all we need to do is to plug our equations for K and U into this energy equation and do a bit of algebra. Move the U term to the right hand side of the equation and we get
.
Here, we note that the mass of the smaller moving body is on both sides of the equation and can thus be canceled. Also, I’ve substituted R, the radius of the larger body, for the separation r. If you are starting from the surface of a planet or moon, then the separation distance from the body trying to escape and the center of the planet is simply the radius of the planet. If you are starting from orbit, then R would be the radius of the orbit. Now, we just do a little bit of algebra. In this equation, if we isolate v2 on the left side of the equation and take the square root, we get
as an expression for the escape velocity from a body. If we plug values for Earth into this equation (the mass of the Earth is 5.974×1024 kg and the average radius of the Earth is 6.371×106 m), we come up with a value of 1.12×104 m/s, or 11.2 km/s, as the escape velocity of Earth.
But, as I said, this isn’t really a very realistic number, because of the effects of Earth’s atmosphere. Now, you could always try to leave Earth from low Earth orbit, say at an altitude of 350km. In that case, R=6.721×106 m, so when you plug this into the equation then you get a value of vescape=10.9 km/s. That is a little less than if you were to take off from the surface of the Earth.
Since the Moon, asteroids, etc, have no appreciable atmosphere to get in the way, then the escape velocity formula works just fine for shooting off from their surfaces, with a few qualifications. The smaller asteroids are non-spherical, so that needs to be accounted for. Also, a body on the surface of a fast rotating object will already be moving if it launches from the equator, so it takes less additional speed to get away from the equator than from the poles.Â
But, even ignoring the Earth’s atmosphere, firing a cannon to shoot something out away from Earth simply is not realistic. The forces needed to provide an acceleration to escape velocity over the distance of the cannon’s barrel is simply too great. That is why we use rockets. A rocket can sustain a lower acceleration for long enough to get something moving at escape velocity. Even better, the rocket’s acceleration is low enough that the object is moving at a slow enough speed in the atmosphere not to be torn apart of to burn up. But, the rocket is picking up speed as it gets higher, where the air is thinner and the forces are less.Â
So, now you know.
-Astroprof
Â
A Ler…-- Rastos de Luz on December 1, 2006 at 1:42 pm: 1
[…] “Escape Velocity“, no Astroprof’s Page. […]
jerry on March 5, 2007 at 7:10 am: 2
what i dont understand is that if a rocket manages to lift up slowly traveling below the escape velocity-an so far overcoming gravity- y would it need to increase speed to get in orbit? a rocket has a continuous propulsion-unlike a cannon- so y not proceed slowly?
Astroprof on March 5, 2007 at 7:33 am: 3
Yes, that is exactly why rockets can lift off slowly — they have continuous propulsion, so they are always gaining speed as they go. The acceleration needed for something like a gun to shoot something into orbit (or beyond) would crush the payload (see the first paragraph of the post).
John Dumas on April 8, 2007 at 3:40 am: 4
Can you tell us what is the smallest rocket that can achieve escape velocity with current technology. I imagine the answer would be the rocket’s mass, which might be mostly the mass of it’s fuel.
Astroprof on April 8, 2007 at 9:51 am: 5
Unfortunately, that depends upon the type of fuel, engine efficiency, etc. So, it isn’t just a simple figure. It also depends upon the object that you are trying to escape from. But, it generally needs to be quite large, if you are talking about escaping Earth. One problem is that you have to get through Earth’s atmopshere, and that complicates the equations (drag, etc).
John Dumas on April 8, 2007 at 11:55 am: 6
Thanks, I realize it is a complex and multi-pronged question. Excuse my naivete, I have just begun to get interested in this subject.
Dora Leibu on April 4, 2008 at 8:41 am: 7
Hi,
I was wondering, why is it that the escape velocity of Earth is greatest at the poles but smallest at the equator? I realize that this is the reason why rockets are launched into space at places such Florida’s Cape Canavrel, and French New Guinea since they are close to the equator.
Astroprof on April 4, 2008 at 9:33 am: 8
The escape velocity is not significantly different at the poles. However, the rotation of the Earth means that the closer to the equator something is, the faster it is already moving. So, that means getting to orbit requires less rocket power.