# Escape velocity help!

1. Dec 24, 2005

### Mixolydian

I keep reading that escape velocity from the earth is about 7 miles per second or 25,000 miles an hour... I know escape velocity is determined by the size of the object with the gravitational pull (earth) and how far away you are from the center. What I can't seem to understand is how 2 objects trying to leave earth would have to have the same escape velocity even if there sizes varied.
If I have a machine that exerts the same amout of thrust on two objects one is a golf ball one is a bowling ball they both have the same thrust but the golf ball will go higher obviously. So why is the escape velocity the same for a golf ball as it is for a rocket ship? Shouldnt the escape velocity be lower because the golf ball has less mass??
I know thrust on a golf ball couldn't reach the speed needed anyway because it would have to carry its own fuel which is added weight but theoretically if i could get a golf ball to exert a constant force of atleast 200 MPH wouldnt that be enough to leave earth?
If not why would a smaller object still need to travel at 25,000 MPH to leave earth?

2. Dec 24, 2005

### Staff: Mentor

Welcome to PF...

1. Escape velocity isn't escape thrust.
2. Escape velocity is ballistic: there is no thrust.
3. You could "escape" by moving at a constant 1mph and under constant force, but that isn't what escape velocity is all about.

3. Dec 24, 2005

### skywolf

to get an object to move at a certain velocity you have to give it energy proportional to its mass, yes a larger object would require more energy.

about the force question, it is a little weird since as far as i know force is measured in newtons not mph

if you get stuck on a physics problem, use F=ma it ussually works for me

4. Dec 24, 2005

### Mixolydian

Then why do Space shuttles leaving earth have to travel at 25,000 MPH to leave earth?

Everytime a read about that it says they have to travel that fast to escape earths gravity because that is the escape velocity....

i dont understand

And I was wondering
Do you have to travel faster and fast the higher you get from the earth?
I did an experiment with a kind of hovering object a made its was thrust off the ground came from a fan I had fixed in the center, it was very light and hovered about 1 foot off the ground. So it had enough thrust to pick it self off the ground. I made the fan spin faster which resulted in its hieght off the ground to increase by about 4 inches. So
it already had enough thrust to hold it self up how come by adding more thrust it didnt just keep going up? it just stayed a uniform height just a little higher off the ground... So does this mean a fighter jet traveling vertical at 500 MPH eventually would get to high and fall down? would he need to go faster and faster to leave earth? (off course assuming this jet has its own air supply for the fuel and is not piloted by a man that needs oxygen) Just wondering about that jet also because I have seen jets fly vertical and i was just wondering if they could keep going.

Last edited: Dec 24, 2005
5. Dec 24, 2005

### Tide

Space shuttles do not achieve 25,000 mph. They enter orbit at speeds of about 17,000 mph. They are not equipped or capable of interplanetary flights.

You may be thinking of spacecraft that do travel out to the other planets (Mars, Jupiter, etc.).

Russ' point was that with a constant application of force just sufficient to balance the gravitational force one could leave the Earth at any designated speed. There is, in fact, no such spacecraft as the fuel requirements to maintain that kind of thrust for great periods of time are technically and economically beyond our means.

6. Dec 24, 2005

### Tide

Mixo,

Regarding your experimental craft, the answer depends on the circumstances (you provided no detail on the construction or conditions of the experiments). In principle, if a given thrust causes the craft to remain at a fixed level then increasing the thrust will cause the craft to accelerate upward. However, you're dealing with fluid flows here so you must account for at least two factors: (a) ground effects (how the air flow of the craft near the ground affects the pressure/density gradient of air near the craft) and (b) pressure variation as a function of height.

In (b), for example, increasing the rotation rate of the fan will cause the craft to rise but as it does so the ambient pressure decreases thereby reducing the thrust. The craft will level off when the thrust equals its weight. (a) is more complicated and I have no information about your particular set up.

7. Dec 24, 2005

### Mixolydian

Thank you tide that was very helpful giving the little information I gave you thought about it in the right way. You answered my question thnx.

8. Dec 24, 2005

### Mk

Another satsified customer! GO TEAM PF!!

9. Dec 25, 2005

### Staff: Mentor

Just to make sure we're clear: escape velocity is the ballistic(unpowered) way to get away from earth, but there are other non-ballistic ways to get away from earth.

10. Dec 25, 2005

### KingNothing

In other words, escape velocity is how fast you'd need to throw a baseball to get it to leave earth, correct?

11. Dec 25, 2005

### DH

ok, to understand the concept of escape velocity, you must know about the concept of gravitional potential.
The gravitional potential causes of the gravity force
F = M1*M2*G/r^2
gravitional potential is the energy that require to bring a object from infinitive to the distance r from the the earth or...
because of the gravity force, when you bring the object toward the earth, you must do the work due to gravity force.
so that the gravitional potential will have the same with the work but opposive sign. At the distance r from the earth, object will have a gravitional potential = -W = - (M1*M2*G/r^2)*r = -M1*M2*G/r
now, in the mechanism view, we have to know that every object will have potential and kinetic energy
so any object that have the distance r from the earth and travel with the speech v to escape from the earth will have the energy
Total energy = -M1*M2*G/r + 1/2 M2*v^2
M1: mass of earth
M2: mass of object
assuming that when the object is at the infinitive possition, it means that object escapes from the earth
-> gravitional potential = 0 since distance is infinitive
kinetic energy > or = 0 however, we are trying to find the escapes velocity, it means that the minimum speech need for the object to escape --> we only set for kinetic energy = 0 at infinitive possition
--> Total energy = -M1*M2*G/r + 1/2 M2*v^2 = 0
--> v = (2*M1*G/r)^(1/2)
--> the escape speech is not dependent on the mass of object
now you got it?

Last edited: Dec 25, 2005
12. Dec 25, 2005

### Dmstifik8ion

Escape velocity refers to the velocity of a object moving parallel to the surface of a planetary body or more precisely perpendicular to its center of gravity. At a certain velocity (the escape velocity) this perpendicular circular orbital motion will exceed the rate at which the centripetal force of gravity is pulling the object back toward the center, thus, the object will proceed more in a straight line and move away from the center of gravity.

I welcome anyone who can expound on or clarify this ‘explanation’.

Last edited: Dec 25, 2005
13. Dec 25, 2005

### ZapperZ

Staff Emeritus
Where in the world did you get this definition of an escape velocity?

Zz.

14. Dec 26, 2005

### Dmstifik8ion

My guess is that you were not impressed with my ‘definition’. I apologize for what was a poor attempt to satisfy the need for a precursory understanding of this that was apparent in the question originally posed in this thread.

Since you have not yet supplied a satisfactory explanation for this yourself, here is my second attempt at this. Escape velocity refers to a velocity greater than the maximum orbital velocity required at any given radius from the center of gravity of a spherical planetary body; At a certain velocity (the escape velocity), the inertial component (the tendency of inertia to keep an object moving in a straight line) exceeds the rate at which the centripetal force of gravity is pulling the object down toward the center of gravity and the orbiting object begins to drift further away.

Escape velocity from a planetary body is aproximately the sqrt(2) times the velocity for an orbit equal to the radius of that planetary body and decreases as an object moves further from the surface. Since this velocity is not constant a better definition might be the initial speed required in a direction directly away from a center of gravity from a given distance to completely escape the gravitational pull or attraction of a gravity source disregarding all other forces or friction, (such as atmospheric drag).

Again, I encourage anyone to improve on this explanation even if your amusement at this would bring you to prefer that I make a third attempt.

Last edited: Dec 26, 2005
15. Dec 26, 2005

### ZapperZ

Staff Emeritus
Wait... let me get this right. You MADE UP your own definition of an escape velocity, rather than looking it up in a standard physics text? And I've just given you a hint on where you could find an improvement to your explanation.

Zz.

16. Dec 26, 2005

### Dmstifik8ion

Would you care to read from your ‘standard physics text’ your definition for escape velocity and then explain what it means? When we take another’s word (yours, mine or Einstein’s), for anything without understanding it we are no better off than when we raise the question.

Here's my third attempt: ‘Escape velocity’ is the minimum initial difference in velocity, regardless of vector, of one object, at a specific initial distance, relative to the other, that by virtue of inertia alone, (ignoring obstacles and all other forces), will eventually separate them from the influence of each others gravitational field.

Note that some vectors will initially bring these two objects closer to each other prior to their eventual separation and that their speed relative to each other will vary throughout this process. Of course collisions with objects, including each other, should be avoided.

Because of the difficulty of obtaining escape velocity near the Earth's surface, due to the hindering effects of the atmosphere, the escape velocity is generally obtained from a low earth orbit where the influence of gravity, and therefore the escape velocity, is a little less. Using the rotational velocity of the Earth can contribute significantly to obtaining this orbit and the required escape velocity as well. Also the gravitation pull of the moon is useful for obtaining, and its motion can help to accelerate objects beyond, the overall required escape velocity.

Last edited: Dec 26, 2005
17. Dec 26, 2005

### ZapperZ

Staff Emeritus
I think you should speak for yourself when you accuse someone of not understanding the accepted definition of "escape velocity".

If you wish to make something up, please name it as something else. What you are describing is not "escape velocity". None of us, and certainly not you, have the freedom to take a physics terminology and define it any way we like. Even YOU can see how confusing this can become.

If you wish to introduce something new and go beyond the accepted physics and terminology, please do so in the IR forum per our Guidelines. If not, please do not respond to things like this using stuff that you've made up.

Zz.

18. Dec 26, 2005

### D H

Staff Emeritus
Dmstifik8ion, one of your problems apparently results from the not-so-good terminology in naming the escape velocity "escape velocity". It would be better to call this concept "escape speed". Escape velocity depends only upon position. Escape velocity from near the surface of the Earth is about 25,000 MPH, period. It does not depend on current velocity.

BTW, I am not the DH who doesn't use LaTeX and can't spell.

19. Dec 26, 2005

### rbj

yes, and never return to the earth. since no one else is (oooops, the "DH" who doesn't use latex did do this), i'm gonna bring on a little of the math (with inferences to calculus, but without the specific notation). the classical universal gravitational force equation (Newton's) is:

$$F = G \frac{M m}{r^2}$$

assuming that the force is roughly constant as you move a mass $m$ upward for each little distance $d$ (say, a meter), then the work done (or energy invested) for that little movement is

$$F \cdot d = d \cdot G \frac{M m}{r^2}$$

but now the force of gravity is a little less because you moved away from the center of the planet a little. now the force is:

$$F = G \frac{M m}{(r+d)^2}$$

so if you move the mass $m$ up another little distance $d$, the work is a teeny bit less:

$$F \cdot d = d \cdot G \frac{M m}{(r+d)^2}$$.

so, to bring mass from the planet surface (at distance $r$) to infinity meters away from the planet, we gotta add up all those little amounts of work done

$$d \cdot G \frac{M m}{r^2} + d \cdot G \frac{M m}{(r+d)^2} + d \cdot G \frac{M m}{(r+2d)^2} + d \cdot G \frac{M m}{(r+3d)^2} + ...$$

which is

$$G M m d ( \frac{1}{r^2} + \frac{1}{(r+d)^2} + \frac{1}{(r+2d)^2} + \frac{1}{(r+3d)^2} + ...)$$

so (here comes the calculus), if you let that little distance $d$ shrink to virtually zero (you're still adding an infinite number of terms representing a small energy contribution needed to lift the mass $m$), and that sum does not sum to infinity, but sums to this finite amount of (potential) energy:

$$PE = G \frac{M m}{r}$$

now (ignoring air/wind resistance), if i throw a ball (of mass $m$) up at velocity $v$, i need to put in this amount of (kinetic) energy:

$$KE = \frac{1}{2} m v^2$$

if that kinetic energy is equal to or greater than the energy needed to bring that ball out past our galaxy ("to infinity and beyond"), then the velocity it has is sufficient to escape without any additional rocket or anything pushing it.

$$KE \ge PE$$

$$\frac{1}{2} m v^2 \ge G \frac{M m}{r}$$

notice the little mass $m$ gets kill off on both sides of the equation, and we get:

$$\frac{1}{2} v^2 \ge G \frac{M}{r}$$

$$v^2 \ge 2 G \frac{M}{r}$$

$$v \ge \sqrt{2 G \frac{M}{r}}$$

that is the classical escape velocity. if the velocity is at least as much as $$\sqrt{2 G \frac{M}{r}}$$ that ball ain't never coming back. if the velocity is equal to $$\sqrt{2 G \frac{M}{r}}$$, then when the ball finally gets to its place infinity meters away from Earth, it will have no KE or velocity left. if it is strictly greater than $$\sqrt{2 G \frac{M}{r}}$$, it will still have some speed left when it gets to infinity. if the initial velocity is less than $$\sqrt{2 G \frac{M}{r}}$$, then the initial kinetic energy is less than what it takes to get to infinity and at some finite point, it's upward velocity will be zero and then gravity will start pulling it back to the planet.

Last edited: Dec 27, 2005