Escape velocity is the velocity required to project an object from an object's (here, the Earth's) surface so that it "escapes" the gravity of that object (or in our case, the Earth's gravity). Now, the Moon is under constant free fall, which means that Earth's gravity acts even until the Moon. So, does that mean that the escape velocity makes the object go even further than Moon's orbit?

  • $\begingroup$ I've removed some comments that were answering the question, and their responses. $\endgroup$ – David Z Sep 20 '17 at 21:19

You are misunderstanding the escape velocity and gravity concepts entirely.

Now, the Moon is under constant free fall, which means that Earth's gravity acts even until the Moon.

Earth's gravitational field extends to Infinity and it slows down any object going against it.

Visualize this: You throw a stone upward with an initial speed. Gravity will slow it down to 0m/s and the stone will fall down again. You throw the stone with even higher speed, it will go little bit higher, but gravity will slow it down to 0m/s.

What happens when you throw a stone at escape velocity (11.16Km/s at sea level) or higher: Gravity will start to slow the stone even this time, but the stone will never return. The stone will approach 0m/s, but it'll never actually reach 0m/s. This is an asymptomatic relationship. The stone will hit 0m/s at Infinity.

So, does that mean that the escape velocity makes the object go even further than Moon's orbit?

Yes. The object would go to the Infinity.

Note: I haven't taken friction, drag, other's gravity or other resistance forces into account.


Technically (and classically) the gravity is extended up to infinity, because the value

$G\frac{M}{r^2}$ is only 0 when $r\rightarrow+\infty$. However, since it's almost void for just large $r$'s, and there are so many more objects, many of them much more massive, it can be compensated.

So, gravity rigorously reaches the infinity, and that's why the escape velocity is defined as "the speed with which the particle reaches the infinity with 0 velocity". That's why it's the minimum one. It can escape $g$ with even more velocity, but we want the minimum one, which is 0. That's why we impose 0 kinetic energy at the ending, and that's how you solve it.

So yes, it's much further than the moon's orbit (theoretically, because there are more objects in space). You don't need to cross the orbit because you can go in other direction, but it's far away from the Moon's distance.


Earth's Gravity acts even until the Moon

Actually, Earths gravity extends forever! It just becomes weaker and weaker with distance. The gravitational force is:

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

The force is always there, just very tiny (and possibly negligible compared to other forces) at large distances $r$.

So, does that mean that the Escape Velocity makes the object go even further than Moon's orbit?

I think you misunderstood the concept of escape velocity. An object (a satellite, a moon...) may orbit our planet with a specific speed. Increase this speed and the orbit becomes larger. But the object still returns to finish it's orbit and repeat it over and over, even if the revolution takes a long time.

But at some specific velocity the object does not come back. It escapes completely. While the object before was "caught" by Earth's gravity and never broke free of it's (maybe large) orbit, then now with a speed higher than the escape velocity it does break free and does not come back.

"Not coming back" essentially means that it can travel to infinity if nothing else stops it.

So, does an object that has the escape velocity or even higher speed go further than the Moon? Well, it definitely can. Because it has escaped.

  • $\begingroup$ So, the Escape Velocity just needs to take you until the shortest possible orbit. Right? $\endgroup$ – Aravind k Sep 20 '17 at 15:05
  • 2
    $\begingroup$ @Aravindk Shortest possible orbit? No, it has to take you out of orbit. $\endgroup$ – Steeven Sep 20 '17 at 15:17

Let's approach this from a conservation of energy viewpoint in a frame of reference tied to the surface of the Earth. Any object with mass $m$ moving with some speed $V$ with reference to this frame has a (classical) kinetic energy of $$K(V)=\frac{1}{2}mV^2.$$

We can also define a potential energy of the system of the Earth and the object due to gravity:$$U(r)=\frac{-GmM_E}{r},$$ where $M_E$ is the mass of Earth, and $r$ is the distance from the center of Earth to the object.

Now, we propel the object vertically at some speed $V$, and then let gravity take over. Gravity will neither add nor remove mechanical energy from this system, but will change $K$ of the object into $U$ of the system, and vice versa. That means that $$K(V_1)+U(r_1)=K(V_2)+U(r_2).$$

As the object moves higher, away from Earth, $U(r)$ become less negative which means that $K$ must get less positive, and the smallest that $K$ can be is zero. If this happens before $U(r)$ reaches zero, the object will return to Earth. This would happen if $$K(V_1)+U(r_1)<0.$$

If, on the other hand $$K(V_1)+U(r_1)>0,$$ the object will still have some kinetic energy when the object reachs "infinity and beyond," and will never turn around.

That transition point, when $K(V_1)+U(r_1)=0$ is the condition at which $V_1=v_{\mathrm{escape}}$. For Earth that works out to be about $11200$ m/s. To reach the moon, you only need to be going $11100$ m/s, if straight-line motion was the only consideration, and ignoring the moon's gravity.


protected by Qmechanic Sep 20 '17 at 21:21

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