Escape Velocity and Earth's gravational pull

1. Dec 30, 2008

stanwal

I am a historian[ PhD] who specializes in the History of Science and consequently reads a great deal of physics. Since my childhood I have been fascinated by space travel and have always read that the escape velocity from the Earth's gravational field is 25000 mph. I recently read a book in which a very prominent physicist declared that this figure was an idealization and because of air resistance the actual velocity was much higher. I must admit that this has me totally puzzled. I can understand that air resistance would make the achieving of escape velocity more difficult but considering the extreme thinness of the upper atmosphere and that the Earth's gravational pull is a constant quantity I would think that the escape velocity is the escape velocity and not a higher figure , let alone a much higher figure. Am I missing something someone with rigorous training in Physics would spot at once ? Is the escape velocity from Earth really higher than 25000 mph and if so how much? Thanks, SW

Last edited: Dec 30, 2008
2. Dec 30, 2008

tiny-tim

Welcome to PF!

Hi stanwal! Welcome to PF!

(which book? which physicist?)

Escape velocity is the speed you need to achieve after you turn the rocket off …

if you turn the rocket off before you've left the atmosphere, then certainly you need a higher speed …

if you turn the rocket off after you've left the atmosphere, then the escape velocity is the figure given.

3. Dec 30, 2008

stanwal

The book is THE BLACK HOLE WAR by Leonard Susskind. Footnote on page 26 "The escape velocity is an idealization that ignores effects such as air resistance , which would require the object to have a far greater velocity". As I said, I do not understand this. My naive interpretation of Newton is that you can treat the mass of the Earth as being concentrated at a point at it's center and then apply the inverse square law to get the velocity necessary to climb out of its gravity well. I have never seen any other factors mentioned and he mentions none except air resistance.

4. Dec 30, 2008

A.T.

Why only "the upper atmosphere"? I guess, this escape velocity is for earth's surface. If you want to build a gun and shoot at Mars, you need more than that as your muzzle velocity, because the projectile will be slowed down by air drag.
The escape velocity depends on where you start your climb.

Last edited: Dec 30, 2008
5. Dec 30, 2008

stanwal

"The escape velocity depends on where you start your climb" So if you were to start at 10000 Ft. you would have to achieve a lesser velocity than if you started at the surface? If you are in Earth orbit you do not have to achieve 25000 mph?

6. Dec 30, 2008

Staff: Mentor

Think in terms of energy. The escape velocity is derived from the amount of kinetic energy a projectile at the earth's surface needs in order to climb out of the earth's gravitational well. (See: http://hyperphysics.phy-astr.gsu.edu/Hbase/vesc.html#ves") Of course, that ignores the additional energy you'd need to overcome air resistance. Thus, at the earth's surface, you'd have to give a projectile an even greater speed to overcome both gravity and air resistance.

Last edited by a moderator: Apr 24, 2017
7. Dec 30, 2008

tiny-tim

hit and forget?

Hi stanwal!

Was Susskind writing about something like a cannon, that just "hits and forgets", or something like a rocket that has a continuous "burn"?

If it was the former, then air resistance is very important.

You're completely right to start with Newton, and that we can treat the mass of the Earth as being concentrated at a point at its centre …

that gives the correct figure for the gravitational force on the missile …

but Newton's second law … Fnet = ma … also requires us to take into account all forces, so we have to add in the force of air resistance.
Yup … have a look at the PF Library article on escape velocity …

the formula is vexcape = √(2GM/r).

8. Dec 30, 2008

stanwal

I had assumed that the escape velocity and the energy needed to achieve it were separate questions. That of course it would take more energy to lift off from the surface of the Earth than higher in the atmosphere but that the terminal velocity would be the same in both cases. This is apparently a faulty assumption. Susskind's remark is appended to his discussion of light being unable to escape from a black hole.

9. Dec 30, 2008

mgb_phys

They are linked escape velocity assumes you use all the rocket's energy near the surface to generate kinetic energy equal to the energy you will need to overcome gravity. It's purely an engineering practicality - you could fly to space at walking speed if you wanted but it would be a lot less efficient.

It takes less energy to launch from higher in the atmosphere because you are already starting with some potential energy. Of course if you include the energy to lift the payload to that higher altitude then it comes out the same. There are practical reasons to launch from a higher altitude (see the Pegasus XL project) to do with the efficiency of jet engines vs rockets and other factors.

The classical picture that you can't escape a blackhole because the escape velocity is faster than light is misleading. Escape velocity only applies to cannons, or rockets that use up all their fuel quickly. You can climb out of a planets gravity well on a ladder without approaching escape velocity.

10. Dec 30, 2008

Staff: Mentor

Well, if you did fly out of the atmosphere (and kept going) at walking speed, the total potential energy gain would still be the same as if you blasted out at escape velocity and you'd still (eventually) reach escape velocity (or, rather, it would reach you). It's just answering a slightly different question.

11. Dec 30, 2008

mgb_phys

Yes the energy gain would obviously be the same. I suppose 'escape velocity' goes to zero at a far enough distance. I though the OP was replying because he was confused about escape velocity and energy.

The idea that you can't escape a black hole because escape velocity > c, gets a lot of people.

12. Dec 30, 2008

Naty1

The 25,000 mph figure is the necessary speed at the surface of the earth, and refers to a projectile, and is clearly referred to as such in the book:

Hence you have, say 100,000 ft or so of air resistance to overcome.

Alternatively you could start at only 1 mph and with enough continuous power to maintain that speed, eventually escape to another world...

Correct.
In fact, were you close close enough to, say the moon or the sun, their gravity will drag you away from that of the earth....no velocity is required "out there".

13. Dec 30, 2008

stanwal

I have grasp the concept that escape velocity can be less than 25000 mph. My original problem was with the footnotes assertion that it could be MUCH GREATER than 25000 mph. If that is the surface escape velocity and it is at the surface where the highest escape velocity is required where are the places that require higher velocity? What was the actual velocity attained by the Saturn 5 moon launches?

14. Dec 30, 2008

D H

Staff Emeritus
The 25,000 mph figure ignores air resistance. This is one reason why it is an idealization. Another reason:

We can't change the velocity of some object from 1000 mph (our speed from Earth rotation, at the equator) to 25,000 mph. It takes time to do that. The only way we know how to do this is to carry the fuel needed to accelerate the vehicle with the vehicle. Some (almost all!) of the work performed by a rocket is simply accelerating the fuel and the parts of the rocket dropped off when they run out of fuel. If you want the gory details, google "Ideal Rocket Equation". Note well: Even that is an idealization.

Regarding the Saturn 5 vehicles. Saturn 5 was a three stage engine. The first stage lifted the vehicle off the Earth and got the vehicle going. The second stage lifted the vehicle into orbit. The tiny third stage sent the vehicle on the way to the Moon. At no time was the Saturn 5's (or Apollo's) velocity equal to escape velocity. There was no need to do so. The Apollo vehicles never did truly "escape" Earth's gravity well. Think of it this way: The Moon is in orbit around the Earth.

15. Dec 31, 2008

tiny-tim

I haven't seen that footnote, but I assume Susskind is starting with the Earth as an example, and working his way from there, through smaller and smaller (but more massive) objects, to a black hole …

in that case, I think the footnote is simply saying that if the Earth got smaller (but had the same mass), then the escape velocity would be far higher, and ultimately would approach the speed of light.

btw, reaching escape velocity (or higher) means being in a parabolic (or hyperbolic) orbit

but all our rockets have far less energy than that, and remain at all times in elliptical orbits