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I read that the escape velocity of a body is independent of the direction of projection.

For example, I could throw a ball at $11.2$ km/s velocity horizontally, and it would still leave earth.

I am unable to visualise or understand this. Could someone explain?

Edit: I think I should put my exact problem down more clearly.

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At point A: I release the ball. It has horizontal velocity of 11.2 km/s.

From A to B: Gravity acts on the ball. One component slows down the horizontal velocity and one component constantly attracts the ball to earth.

At B: the ball is trying to leave earth with less than escape velocity.

What I realised: B might not be exactly on earth, the ball might have already covered some height.

So right now, I would just like someone to confirm if what I realised is right. (Or correct me.)

Thank you.

Community
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Mahathi Vempati
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3 Answers3

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Because it is a energy balance between a potential $V(r)$ and the kinetic energy of the object $ T = 1/2mv^2$, so the work to do provide in order to escape: $T > V$ does not depend on the path followed by the object (it only depends on the module of it’s velocity).

Serge Hulne
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  • I am not sure your answer fully clarifies the issue. Why is the condition that the kinetic energy be larger than the potential energy enough to ensure that the projectile ‘escapes’? If this had been clear to the OP, he may not have posed the question in the first place. – Guillermo BCN Aug 08 '21 at 05:11
  • Think of it as jumping a fence. In order to jump a fence, one merely has to jump in the air high enough (the threshold being the height of the fence). The result does not depend on the angle at which one runs towards the fence. – Serge Hulne Aug 08 '21 at 10:26
  • With a fence, it depends on the vertical component of the velocity. It won’t be of any use running very fast toward the fence if your jump speed is low; you will just hit the fence very fast. So it does not seem to be the modulus of the velocity that is crucial. – Guillermo BCN Aug 08 '21 at 11:00
  • It is just a metaphor of a case where a single parameter matters (height) and another don’t (angle). I was trying to give a graphical metaphor for a potential barrier. – Serge Hulne Aug 08 '21 at 11:05
  • Ok, I get that. And I am sure the analogy is there. However, I think the source of his confusion (as is mine) is the fact that some escape trajectories actually intersect the planet, which seems to contradict the idea of ‘escape’. – Guillermo BCN Aug 08 '21 at 11:14
  • If you think about Gauss’s theorem, it appears that the gravitational potential of the Earth is the same as the one you would get from a body with the same mass as the Earth, located at the center of the Earth (basically a small black hole). The potential is exactly the same in the both cases. In the case of a black hole, there would be basically no forbidden trajectories (except a for a head-on collision with its center or rather its event horizon). – Serge Hulne Aug 08 '21 at 12:25
  • Here is another explanation by Neil de Grasse Tyson : https://youtu.be/7vXoaiu4zFI (the mountain thought experiment by Newton) – Serge Hulne Aug 10 '21 at 18:04
  • Thank you. I think his qualitative explanation is basic. He speaks of ‘a velocity’ at which escape happens. He does not say anything about its magnitude being independent of the angle though. In any case, Thank you for your efforts. – Guillermo BCN Aug 10 '21 at 19:45
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Ignoring atmospheric friction and buoyancy, if you threw the ball at a and it arrived at b, the velocity you threw the ball at would have to be about 8000 m\s, and the ball's velocity at b would not have changed.

If you threw the ball at escape velocity it would follow a parabolic path, with the vertex of the parabola at the throwing position. The distance of the ball from the center of the Earth would be constantly increasing, and the velocity would be constantly decreasing, without ever reaching zero.

Nick
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Of course, the escape velocity is all about energy, and a very special 'thing' about $1/r^2$ forces is that energy of an orbit does not depend on the eccentricity. Hence, if you are at a $r_0$ with energy $V(r_0)<0$, you just need:

$$ T(\vec v) = T(|v|) =\frac 1 2 mv^2 \ge -V(r_0) $$

to escape.

In principle, the independence of $E$ from $\epsilon$ can be related to the conservation of the Laplace-Runge-Lenz vector:

$${\bf A} = {\bf p}\times {\bf L} -mk{\bf {\hat r}}$$

but at this moment, I am not having a "Eureka" moment as to how that applies here.

JEB
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