Trouble understanding the idea of escape velocity

[subzero0137]

I'm having trouble understanding the idea of escape velocity. How can an object escape the gravity of a massive object like the Earth? No matter what the velocity the object is, doesn't Newton's law of gravity imply that eventually, the force of gravity will cause the object to decelerate, and then fall back to earth?

 

[dextercioby]

Not trully so, it may happen that for some initial conditions, the conic solution of the Kepler problem is a parabola or hyperbola.

 

[subzero0137]

Not trully so, it may happen that for some initial conditions, the conic solution of the Kepler problem is a parabola or hyperbola.

Hmm...I haven't come across the "conic solution of the Kepler problem" before, so I'll have to look that up or wait for my lecturer to get to that part. But could you explain why I can't simply use the Newton's law of gravity, which states that F=\frac{GMm}{r^2}, and argue that no matter how fast the object travels away from Earth, it will always experience a deceleration and thus eventually fall back to Earth?

 

[dextercioby]

The law you wrote is the scalar version, but forces are vectors. The acceleration vector (the force vector if you prefer) is not collinear to the velocity one for motion along a curve. The 2-body problem involves motion along a curved trajectory (a conic section).

 

[subzero0137]

The law you wrote is the scalar version, but forces are vectors. The acceleration vector (the force vector if you prefer) is not collinear to the velocity one for motion along a curve. The 2-body problem involves motion along a curved trajectory (a conic section).

I see. But what if the trajectory of the object is not a curve? What if a rocket simply lifted off with the escape velocity radially away from the Earth?

 

[dextercioby]

The Newtonian potential is conservative. Even for the motion you propose, you're free to increase the KE of the rocket as much as you like, which from a velocity value upwords would be higher than the Potential energy which the rocket has at the surface of the earth. The rocket would then evade the gravitational field of the Earth and would not orbit it.

 

[subzero0137]

The Newtonian potential is conservative. Even for the motion you propose, you're free to increase the KE of the rocket as much as you like, which from a velocity value upwords would be higher than the Potential energy which the rocket has at the surface of the earth. The rocket would then evade the gravitational field of the earth and would not orbit it.

Why would that happen? Sorry if I'm being difficult, but I just don't understand it!

 

[russ_watters]

No matter which direction you start from, the further away from Earth you get, the closer to directly away from it your trajectory points.

 

[SteamKing]

Why would that happen? Sorry if I'm being difficult, but I just don't understand it!

Well, start your understanding by reading this article:

http://en.wikipedia.org/wiki/Escape_velocity

 

[subzero0137]

Well, start your understanding by reading this article:

http://en.wikipedia.org/wiki/Escape_velocity

So basically, we can't use Newton's law of gravitation on its own. We have to use it alongside the conservation of energy law, so when the total mechanical energy is 0, the object pretty much travels at a constant speed, despite being in a force field?

 

[SteamKing]

Who said anything about the object traveling at constant speed? When the velocity of the object makes its kinetic energy equal to the gravitational potential energy, then the object is said to have reached escape velocity. Objects can't achieve any velocity, let alone escape velocity, without accelerating from rest.

 

[HallsofIvy]

If an object leaves the earth, or any other massive body, the "escape velocity" is the initial velocity ("escape velocity" is defined assuming no additional force so the velocity is steadily decreasing) needed that the velocity will stay non-zero to "infinity".

 

[subzero0137]

If an object leaves the earth, or any other massive body, the "escape velocity" is the initial velocity ("escape velocity" is defined assuming no additional force so the velocity is steadily decreasing) needed that the velocity will stay non-zero to "infinity".

I see. So an object that has 'escaped' the gravity of a massive object will still decelerate. I understand that bit. But in the wiki article, it says that the velocity will drop to 0 at infinity, and not stay non-zero. Am I missing something?

 

[jbriggs444]

I see. So an object that has 'escaped' the gravity of a massive object will still decelerate. I understand that bit. But in the wiki article, it says that the velocity will drop to 0 at infinity, and not stay non-zero. Am I missing something?

For practical purposes, this "at infinity" can be taken as shorthand for "far enough away that it doesn't mater" and this 0 can be taken as "close enough to zero that it doesn't matter". Realistically, there is, no such thing as actually being at infinity.

More formally, an object at escape velocity has enough velocity so that no matter how far you want it to go, if you wait long enough, it will get that far. And for any finite distance, you will not have to wait infinitely long.

If you go far enough in mathematics you'll find things like the extended reals or non-standard analysis where infinities [or infinitesimals] can be put on a firm enough footing so that it can be formally meaningful to talk about something being zero "at infinity". Until then, it would be wise to avoid naive reasoning about infinity as if it were just another place or another number.

 

[ASG141]

unless there is another field, you cannot escape the earth.

if you go too far off, you enter the Mars or venus sphere of influence or may be zones where the sun's field is peaking.

mapping the space G field wil give you an idea of where an object will be directed at a given (x,y,z,t).

 

[D H]

I see. So an object that has 'escaped' the gravity of a massive object will still decelerate. I understand that bit. But in the wiki article, it says that the velocity will drop to 0 at infinity, and not stay non-zero. Am I missing something?

You never reach infinite distance, so your velocity is always non-zero. It is however tending toward zero, and becomes zero in the limit r→∞.

That's the case for a parabolic trajectory. For a hyperbolic trajectory, the velocity tends toward some non-zero quantity, typically called v_∞.

 

[ASG141]

the parabolic as per galileo is a general form for linear field L >>> O object with horizontal velocity x and vertical velocity y. the Earth's field is considered to be a line of infinite length and constant G.

at greater distance, the Earth's field becomes a point source with decreasing G and spherical density.

 

[mpresic]

I think I see the confusion. Newton's law of gravity tells us the force of gravity diminishes as the inverse square of the distance. That is doubling the distance from the source of gravity (center of the Earth) allows the gravity force to decrease to one-fourth it's former strength. The gravity is weakening. Now if the "rocket"
is given a sufficient initial velocity, the rate at which it covers distance (velocity) is greater than the rate at which the gravity field is weakening (as a result of the increased distance). Consequently, the rocket is allowed to escape

 

[snorkack]

Newton´s original laws do not express energy conservation, actually. Vis viva does come out, but it was almost two centuries after Newton that energy conservation came out.

But converging infinite series were known to Newton. So above escape speed, an object is indeed always in gravity field, and always slowing slightly, but never will stop nor slow beyond a certain nonzero speed.

 

[dauto]

The OP's mistake seems to be the belief that if something is constantly decelerating then its speed must eventually reach zero. That's just not true. (This is kind of a reversed Zeno paradox. In the classic paradox the distance between Achilles and the tortoise is mistakenly believed never to reach zero because it consists of an infinite number of steps. The OP thinks the speed must reach zero because of the infinite number of steps taken reducing the speed by the continuous action of gravity. They are both wrong)

 

[A.T.]

But in the wiki article, it says that the velocity will drop to 0 at infinity, and not stay non-zero.

If the velocity was initially exactly the escape velocity, it approaches zero at r→∞.
If the velocity was initially greater than the escape velocity, it approaches a non-zero value at r→∞.

 

[A.T.]

So basically, we can't use Newton's law of gravitation on its own.

You can. The energy computations are based on potentials derived directly from Newton's law of gravitation. But you can just as well integrate the gravitational acceleration along the path, which lead to the same result.

 

[mpresic]

A.T. is correct. (Do not need to invoke energy conservation (directly) or worse Kepler's problem in a plane, or space)
Easiest way is to write down (F=ma=) m dv/dt (LHS) = -GMm / r squared (RHS). Because v = dr/dt (here is the trick); multiply LHS by v, and RHS by dr/dt (note the equality is still valid since v = dr/dt). Now cancel the dt's (I hope the mathematicians will forgive me), and you have a differential equation (DE) for v as a function of r. Integrate the DE and apply vf = 0 and v0 = initial (i.e. escape) velocity for the upper and lower limits respectively for v and r = infinity and r = R (Surface of Earth) for the upper and lower limits for r and you should be OK. Last solve this equation for v0.

I suspect this is not in introductory textbooks since authors generally include energy before Newtons Universal law of gravitation, and therefore feel free to present escape velocity through energy.

By the way in my last post, I said allow to escape. Snorkack is more precise. The final velocity will approach a nonzero speed in the limit and it is always slowing slightly.