How does a mass ejected at escape velocity keep moving till infinity?

Question

I've studied that if a mass is ejected from the surface of the Earth at 11.2 km/s (or at any point above the surface at a lower Ve), it 'escapes' to infinity.

Does that not imply a violation in the law of conservation of energy? In such a case I'm giving an object only definite energy to move it out of the earth's Gravitational field, but it apparently goes on until infinity?

I also read that at some point, the velocity of the ejected object turns zero. And that springs up the following doubt:

At $v=0$ at infinity, the object can once again be influenced by the Earth's gravitational intensity, as it technically extends upto infinity (Considering an empty universe, with only Earth and that object.).

Help in clarifying my concepts would be appreciable.

Answer 1

Motion through space does not require energy. As Newton pointed out, an object continues in its state of motion forever in a vacuum far from any gravitational sources. In a gravitational field, kinetic energy is traded for gravitational potential energy and hence we require a minimum kinetic energy to escape to infinity. The difference in gravitational potential energy between e.g. the surface of the Earth (with radius R) and infinity is equal to $$\int_{\infty}^{R} \frac{GMm}{r^2} \, dr = \frac{GMm}{r} \Bigg|_{\infty}^{R} = \left(\frac{-GMm}{R} + \frac{GMm}{\infty}\right) = \frac{GMm}{R}$$, which is not equal to infinite energy. Hence, we do not need infinite energy to get to infinity.

At v=0 at infinity, the object can once again be influenced by the Earth's gravitational intensity, as it technically extends up to infinity (Considering an empty universe, with only Earth and that object).

Upon arrival at 'infinity' :-) the force due to gravity is: $$\frac{GMm}{r^2} = \frac{GMm}{\infty^2} = 0$$ so there will no longer be any force pulling the particle back.

Answer 2

You are used to gravity being uniform. Suppose your mass is $102$ kg. If you climb a hill $1$ m high, the energy it takes is $E= mgh$, so it works out to $10000$ Joules. If you climb another $1$ m, it takes another $10000$ joules. If you keep going to infinity, obviously it takes an infinite amount of energy.

But gravity gets weaker as you get farther from the earth. Every meter takes less energy than the one before. It works out that it takes a finite amount of energy to climb to infinity.

To get the idea, we will use some fake numbers. Suppose it takes $1$ joule to climb the first meter, $1/2$ Joule to climb the next, $1/4$ ...

The total energy would be $E = 1 + 1/2 + 1/4 + ...$. There is a famous proof that this adds to $2$. It goes like this.

$2E = 2 + 1 + 1/2 + ...$.

$E = 2E - E = (2 + 1 + 1/2 + ...) - (1 + 1/2 + 1/4 + ...)$

If you look at all the terms, the $1$'s cancel, the $1/2$'s cancel, the $1/4$'s cancel, the ... All the terms cancel except the leading $2$.

So $E = 2$.

Not all infinite sums add up to a finite total. But the energy to get from the surface of the earth to infinity does, as DKP has shown.

Answer 3

In the abstraction, it approach $v=0$ asymptotically, and just creeps along forever.

In real life there is no infinity, and it never gets there, and something else happens, such as the tiny radiation pressure drag from the CMB (and it falls back to earth), or the Hubble Expansion puts it past the cosmic horizon, or we transition to a true vacuum.