# Isn't escape velocity against classical physics?

1. Jan 5, 2006

### eosphorus

"In physics, for a given gravitational field and a given position, the escape velocity is the minimum speed an object without propulsion, at that position, needs to have to move away indefinitely from the source of the field, as opposed to falling back or staying in an orbit within a bounded distance from the source. The object is assumed to be influenced by no forces except the gravitational field"

"On the surface of the Earth the escape velocity is about 11.2 kilometres per second. However, at 9000 km altitude in "space", it is slightly less than 7.1 km/s."

this is taken from the wikipedia: http://en.wikipedia.org/wiki/Escape_velocity

so if i have 1 kg mass at sea level at a speed of 11200 m/s according to physics the more altitude it can reach is 11200 m, conservation of energy,

so why this says that once reached the scape velocity it can go away indefinitly?

2. Jan 5, 2006

### Staff: Mentor

Not quite. Note that the kinetic energy is given by $$1/2 m v^2$$.

3. Jan 5, 2006

### dicerandom

Also keep in mind that when calculating escape velocities you have to account for the fact that $g$ is not constant, it changes with the height of the planet. This means that you have to use the full form of the gravitational potential energy equation rather than the $mgh$ approximation we usually make for situations near the surface of the Earth.

4. Jan 5, 2006

### eosphorus

right i realized of it just after sending it so it would be 1/2*1*11200*11200=1*10*h so h would be 6272 km and taking into account the gravitational field weakens with distance it would still go higher but still limited

theres no way that at 11200 km/s you get infinite potential energy as this web says

in fact according to newton is imposible to break away from a gravity field you can just cross by other bigger

so either objects orbit as clasical physics say because this escape velocity suggest the objects accelerate oposite to falling(the only way to beat a falling constant atraction) or theres no such thing as scape velocity

so can anyone explain me whats this scape velocity staff?

5. Jan 5, 2006

### LeonhardEuler

As Doc Al pointed out the kinetic energy is given by
$$KE=\frac{1}{2}mv^2$$
For objects close to the surface of the earth the potential energy is
$$PE=mgh$$
So, if an object is close to the surface of the earth, the maximum height change it can attain would be
$$mg\Delta h=\frac{1}{2}mv^2 \rightarrow \Delta h=\frac{v^2}{2g}$$
Which would seem to indicate that if we wanted h to approach infinity, then v would have to approach infinity as well, however this is not the case. The force of the earth's gravity is given by:
$$\vec{F}=\frac{GM_em}{r^2}\hat{r}$$
The PE, then is given by:
$$PE=\int_{h_o}^{h}F\cdot d\vec{r}=\int_{h_0}^h\frac{GM_em}{r^2}\hat{r}\cdot d\vec{r}$$
$$=\int_{h_0}^h\frac{GM_em}{H^2}dH$$(because the position vector points straight up, so the projection of $d\vec{r}$ onto $\hat{r}$ is just the change in height, and the hieght of an object above the center of the earth is the same as the distance from it)
$$=-\frac{GM_em}{h}+\frac{GM_em}{h_0}$$
From this we see that as h approaches infinity the potential energy does not approach infinity as it does in the expression PE=mgh. Now consider a change in hoeght:
$$\Delta PE=-\frac{GM_em}{h_2}-(-\frac{GM_em}{h_1})$$
$$=-\frac{GM_em}{h_2}+\frac{GM_em}{h_1}$$
$$=GM_em(\frac{-h_1+h_2}{h_1h_2})$$
$$=GM_em(\frac{\Delta h}{h_1(h_1+\Delta h)})$$
$$=GM_em(\frac{\Delta h}{h_1^2+h_1\Delta h)})$$
Now, if h_1 is much greater than delta h, as is the case for objects close to the surface of the earth, then:
$$\Delta PE=GM_em(\frac{\Delta h}{h_1^2+h_1\Delta h)})\approx m\frac{GM_e}{h_1^2}\Delta h$$
Putting in appropriate values for G, M_e and h_1, you will find that the constants in the middle come out to about 9.8m/s^2, or g.

Last edited: Jan 5, 2006
6. Jan 5, 2006

### Staff: Mentor

As dicerandom mentioned, you need to use an expression for potential energy that is accurate beyond the surface of the earth. Here it is:
$$\mbox{PE} = - \frac{G M_e m}{r}$$

This gives the gravitational PE for a mass (m) at a distance r from the center of the earth (as long as r > earth radius). Note that the PE is negative at the surface of earth since this formula sets the zero point of PE at r = infinity. To escape the earth, you need a KE at the surface that is equal or greater than:
$$1/2 m v^2 \ge \frac{G M_e m}{R_e}$$

where R_e is the radius of the earth.

7. Jan 5, 2006

### Staff: Mentor

It will keep going and going, just like the energizer bunny.

You certainly don't gain infinite potential energy as you go higher and higher. Even wiki wouldn't make that mistake!

"Escaping" the earth's gravitational field just means that you have enough energy so that you don't fall back. If your takeoff speed at the surface just equals the "escape velocity" you will make it to infinity, but slowly, as your speed will be zero by the time you get there. (We're talking about an idealized case here.) If your speed is greater than the escape velocity, then you will have some non-zero speed as you go to infinity.

You don't really need to go to infinity to escape the earth's gravity for all practical purposes. After all, it drops off as an inverse square with distance.

Objects under gravity still accelerate towards the earth, as always. As they get higher, the acceleration decreases, but it's still there.

8. Jan 5, 2006

### Staff: Mentor

Wrong - escape velocity is mathematically derived from Newton's law of gravity.
You aren't under the impression that it stops decelerating, are you?

Btw, we're not grammar Nazis here, but too many errors and it gets distracting. The word is "escape".

9. Jan 5, 2006

### Staff: Mentor

Hmm - according to this site, you don't even need Newton's law of gravity (though it is kinda in there, with "G"), just his energy equations:
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Gravity/EscapeVelocity.html [Broken]

Very simple.

Last edited by a moderator: May 2, 2017
10. Jan 5, 2006

### Staff: Mentor

But his expression for the gravitational PE is derived from Newton's law of gravity!

11. Jan 7, 2006

### LURCH

I have a related question.

The answers contained in this thread suggest a direct correlation between escape velocity and the velocity of an escaping object. I'm curious as to whether or not this correlation can be expressed as a simple ratio.

For example: if I am coasting away from Earth at some speed > escape velocity, both my distance and my velocity will constantly be changing. If I calculate escape velocity for a particular altitude (EV), then measure my own velocity when a reach that altitude, I might find that my velocity is escape velocity for that altitude plus 1% (1.01 EV). If I continue to repeat the same calculation and new altitudes, will I always find my velocity to be 1.01 EV (where EV equals the escape velocity for that new altitude)? Or is the relationship more complex than that?

12. Jan 7, 2006

### Staff: Mentor

The relationship can be described more easily in terms of kinetic energy rather than speed. For example, if you started with a speed exactly equal to the escape velocity then your kinetic energy would exactly equal the escape KE (see my post #6 above). Your total mechanical energy (KE + PE) would equal zero. This would, of course, remain true as you rose up to a greater altitude. And if you had any extra KE (above the minimum), that would remain true as you rose up also.

13. Jan 10, 2006

### rcgldr

The short and easy answer is that there is a maximum gravitational potential energy. Even if an object was an infinite distance away from the earth, it would have a finite speed as it impacted into the surface of the Earth (assuming no atmoshpere). Starting at the surface of the earth, if the object had this same amount of finite speed (and again assuming no atsmosphere), it's path would be parabola. If it's speed was higher, it's path would be a hyperbola, and if it's speed was less, it's path would be an ellipse. The exception being if the object was headed directly away from the center of the earth, then it's path would be a straight line.