Understanding escape velocity [closed]

Question

Suppose we have an object thrown up from earth, then, it's energy is given by,

$$ E = \frac{1}{2} mv^2 - V(r)$$

Where V(r) is a potential dependent on distance from centre of earth. By conservation of energy, and some nice assumptions, E is a constant.

Now, suppose we could write velocity as a function of 'r', then, if we took the limit as r goes to infinity, then

$$ \lim_{r \to \infty } E = \lim_{ r \to \infty} [ \frac{1}{2} mv^2 - V(r)]$$

So, at the final state if we assume that only 'enough' energy is given that the body escapes, then,

$$ \lim_{r \to \infty} E = 0 $$

But since, $E$ is a constant,

$$ E=0$$

So, this means that the path followed by a particle which has just enough velocity to escape is given by,

$$ \frac{1}{2} mv^2 = V(r)$$

or,

$$ v = \sqrt{ 2 \frac{V(r)}{m} }$$

Now, this would mean that suppose took a certain distance from centre of earth, if the particle was on trajectory to escape, then the velocity at each point from the earth is directly related to potential.

So by this logic, we can take the velocity of a give particle at any point in it's trajectory and predict whether it could escape or not, as in , if it's velocity obeyed the inequality as shown below,

$$ v_{particle} (r) \geq \sqrt{ 2 \frac{V(r)}{m} }$$

Now, how do I intuitively interpret the last inequality, as in it seems very strange to me that you can just take velocity at a point compared it to potential and see if the particle would finally escape or not? What is "intuition" behind why this would work other than the mathematical one?

Answer 1

Any object moving in the gravitational field of a spherical mass traces out an orbit that is a conic section i.e. either an ellipse, a parabola or a hyperbola. It may not look as though a stone is travelling in an ellipse when you throw it, but it is. If the ground weren't in the way (i.e. if all the Earth's mass was concentrated in a point at its centre) the stone would move in an ellipse, and you are seeing only the short section of the ellipse in between the stone leaving your hand and hitting the ground.

Orbits can have a range of eccentricities and the axes can be at any angle, so for example there are an infinite number of ellipses in which an object could orbit. However it is always true that:

• an elliptical orbit has a negative total energy

• a parabolic orbit has a zero total energy

• a hyperbolic orbit has a positive total energy

where the total energy is the sum of the negative potential energy and the positive kinetic energy. And since energy is conserved the total energy of the particle is the same everywhere along its orbit.

And this is the reason for the relationship that you have discovered. Since the energy is constant we can take any point anywhere along the orbit, calculate the total energy at that point, and the result gives us the energy of the orbit and hence tells us what shape orbit it is. It then just remains to point out that a particle in an elliptical orbit is bound, a particle in a hyperbolic orbit is unbound and a particle in a parabolic orbit is exactly on the borderline between bound and unbound.

Answer 2

Technical point: Gravitational potential energy is mV, and V (well above the earth) is generally taken as (- GM/r) (with V = 0 at infinite r). When you throw something up, its potential energy increases [in this case from a (– mV) to a smaller (-mV)] and the velocity decreases. If you start with the escape velocity (at any point above the atmosphere), then the escape velocity depends on the height from which you start.