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.

  • 1
    $\begingroup$ Have you taken into consideration that the ground will eventually curve away from the horizontal trajectory, and at such a speed(if constant) the particle will escape. $\endgroup$
    – Abhinav
    Commented Dec 26, 2015 at 6:04
  • 5
    $\begingroup$ It's simply a matter of energy conservation. If the kinetic energy exceeds the (negative) potential energy at the launch point, the body will escape. $\endgroup$
    – CuriousOne
    Commented Dec 26, 2015 at 7:34
  • $\begingroup$ Please see the edit, and thank you both for getting me thinking. $\endgroup$ Commented Dec 26, 2015 at 9:03
  • 2
    $\begingroup$ related: physics.stackexchange.com/q/193497 and physics.stackexchange.com/q/60515 $\endgroup$ Commented Dec 26, 2015 at 11:12

3 Answers 3


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).

  • $\begingroup$ 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. $\endgroup$ Commented Aug 8, 2021 at 5:11
  • $\begingroup$ 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. $\endgroup$ Commented Aug 8, 2021 at 10:26
  • $\begingroup$ 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. $\endgroup$ Commented Aug 8, 2021 at 11:00
  • $\begingroup$ 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. $\endgroup$ Commented Aug 8, 2021 at 11:05
  • $\begingroup$ 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’. $\endgroup$ Commented Aug 8, 2021 at 11:14

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.


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.


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