However, for any body projected with a velocity less then the escape velocity, the final maximum height will vary with variation in angle of projection. But why is this not the case in case of escape velocity?

  • 1
    $\begingroup$ A better name would be escape speed. $\endgroup$
    – Qmechanic
    Dec 20, 2018 at 7:24
  • 4
    $\begingroup$ It does depend on the direction to an extent, you shouldn't launch toward the ground. $\endgroup$
    – Aetol
    Dec 20, 2018 at 9:04
  • $\begingroup$ Aetol , so funny . $\endgroup$ Dec 24, 2018 at 12:38

4 Answers 4


There are a couple of incorrect answers here - you are right that the escape velocity is independent of the angle of projection.

The energy of the launched projectile can be written $$E = T + U = \frac{1}{2}mv^2 - G\frac{Mm}{r}$$

There are three types of orbits depending on the sign of $E$:

  1. $E<0 \implies $ Elliptical orbit (red)
  2. $E=0 \implies $ Parabolic orbit (green)
  3. $E>0 \implies $ Hyperbolic orbit (blue)

enter image description here

Of the three, only elliptical orbits are closed; hyperbolic and parabolic orbits correspond to the projectile escaping off to infinity. The parabolic orbit is the "boundary case" - the first value of the energy for which the projectile escapes to infinity. We therefore define the escape velocity as the velocity needed to make the total energy zero:

$$\frac{1}{2} m v_{esc}^2 - G\frac{Mm}{r} = 0 \implies v_{esc} = \sqrt{\frac{2GM}{r}}$$

If the particle is projected radially away from the planet with the escape velocity, then it will follow a straight line trajectory while gradually slowing down, asymptotically approaching zero speed at infinity.

On the other hand, if the particle is projected horizontally (tangent to the Earth), the orbit will look something like this: enter image description here

In both cases, however, the projectile escapes to infinity, a fact which depends only on the sign of the total energy.

  • $\begingroup$ I would like to add something that might help intuition. If you shoot towards the other mass (assuming point masses and that they can pass through each other) it will also be a straight line escaping to infinity, because it will first accelerate towards the center and reach the other side at the same distance with the same speed $\endgroup$
    – user65081
    Dec 20, 2018 at 14:29
  • $\begingroup$ If the projected body is projected tangentially , then according to my intution . At the highest point (infinity) , it will have a horizontal velocity and thus its kinetic energy will not be zero . Can you make me clear regarding this statement of mine ? $\endgroup$ Dec 24, 2018 at 12:47
  • $\begingroup$ There is no highest point. When we say that the particle has zero velocity at infinity, we mean that $\lim_{t\rightarrow \infty} |\vec r'(t)| = 0$. For any finite time you are correct, but in the limit that speed goes to zero. $\endgroup$
    – J. Murray
    Dec 24, 2018 at 15:45

I have tried to answer the question in a different way.

Suppose the object, mass $m$, is projected at the escape speed $v_{\rm e}$ at a tangent to the surface of the Earth radius $R$ and mass $M$.

In terms of energy $\frac 12 mv^2_{\rm e} -\frac{GMm}{R}=0 $ and $\frac 12 mv^2 -\frac{GMm}{r}=0 $ at any distance $r$ from the centre of the Earth when travelling at a speed $v$. This tells us that at every distance $r$ from the centre of the Earth the object is travelling at the escape speed for that distance.

Combining these two equations gives $v^2=v^2_{\rm e} \frac Rr$

The gravitational force is central and so angular momentum is conserved
$$mRv_{\rm e} = mrv_{\theta}\Rightarrow v_{\theta} = \frac{v_{\rm e} R}{r}$$ where $v_{\theta}$ is the speed at right angles to the radius vector, the tangential speed.

If $v_{\rm r}$ is the speed along the radius vector then $$v_{\rm r}^2 = v^2 - v^2_{\theta}= v^2_{\rm e} \frac Rr-\left(\frac{v_{\rm e} R}{r}\right)^2=v^2_{\rm e}R\left(\frac 1 r - \frac {R}{r^2} \right)$$

To see the implication of the equations for the tangential and the radial speeds lets look at some graphs and for simplicity make $v_{\rm e} =1$ and $R=1$ so we have $$v_{\rm r} = \sqrt{\left (\frac 1 r - \frac {1}{r^2} \right)}=\frac 1 r \sqrt{r-1}\text{ and }v_{\theta}= \frac 1 r$$

enter image description here

Note that for $r>2$ the radial speed is greater than the tangential speed with the radial speed becoming more and more dominant as the distance from the Earth increases.
Thus the trajectory of the object gets closer and closer to being along a radius vector.
The initial tangential speed has not been "wasted" rather it has contributed to the object being able to escape from the Earth.

Possibly with appropriately scaling of the graphs you can show the effect of changing the angle at which the object is fired.
For example when the angle is $45^\circ$ the initial radial and tangential speeds are equal and as the distance from the Earth increases the radial speed will dominate more and more.


Escape velocity is defined so that any object, no matter which direction it is traveling, will be able to reach an infinite distance from another body (Earth, for example) despite that body's gravitational pull. The launch angle does not matter because there is no maximum height. The object will just keep going forever to farther distances.

If you launch the object at escape velocity vertically, then it will always be moving away from Earth for the rest of time. It will be gradually slowing down, but it will never stop.

If you launch the object at escape velocity parallel to the ground, then it will take a parabolic path out to infinity. It will always be slowing down and slowly turning towards 90° from its original launch angle, but it will never stop increasing its distance and never complete the turn.

  • $\begingroup$ Any direction other than straight up at the escape speed will result in a parabolic path. At greater speeds the trajectory will be a hyperbola. $\endgroup$
    – Farcher
    Dec 20, 2018 at 7:45

Because the angle for escape velocity is implicitly ninety degrees in the formulas you have seen. You can express escape velocity utilizing the initial angle of launch, but the equations you've been exposed to assume that a projectile is shot straight up for escape velocity.

EDIT: J. Murray's answer is not quite complete and criticizes mine as being wrong. I don't have enough rep to respond directly since I'm new, so I will here. He writes the basic energy equation:


But this equation effectively treats v and r as scalars, and not vectors. In other words, the equation is being treated as one dimensional. From Wikipedia:

"escape velocity is the minimum speed needed for a free object to escape from the gravitational influence of a massive body."

Thus, to have a projectile escape Earth's gravity parallel to the plane of the Earth, it will need a greater initial velocity, but by definition the escape velocity remains the same. On another semantic note, the definition says its a "speed" but the name is a velocity, the misnomer is probably a great source of confusion.

Here is the "calculus derivation" from Wikipedia.

The total work needed to move the body from the surface $r_0$ of the gravitating >body to infinity is then

$W = \int_{r_0}^{\infty} -G\frac{Mm}{r^2}\,dr = -G\frac{Mm}{r_0} = -mgr_0$

This is the minimal required kinetic energy to be able to reach infinity, so the escape velocity $v_0$ satisfies:

$W+K=0\Rightarrow {\frac {1}{2}}mv_{0}^{2}=G{\frac {Mm}{r_{0}}}$

This equation more explicitly states that r and v are scalars indicated by $r_0$ and $v_0$ Furthermore, we go from work (W) to a scalar kinetic energy term (K), when vector kinetic energy terms are possible. Here is the Wikipedia definition of Kinetic Energy:

"It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity."

Thus, the definition that Murray and Wikipedia are using for Kinetic energy is only a convenient case where only 1D kinetic energy (utilizing speed, not velocity) is considered. Furthermore, you cannot simply substitute work for kinetic energy. Gravity is a conservative force, meaning it does no work in a closed path. However, Kinetic energy does do work in a closed path, so to equate the two in all instances is yet another fallacy.

  • $\begingroup$ It will need less velocity at best. $\endgroup$
    – Alchimista
    Dec 20, 2018 at 11:54
  • $\begingroup$ for a projectile to escape planar to the Earth? I don't think so. Recall that escape velocity is the "minimum speed needed" so to have a slower speed which escapes seems contradictory at face value. Furthermore, a projectile thrown tangentially with less velocity will expend some energy curving toward the Earth (perhaps spiraling into it) which doesn't get used in rectilinear (straight line) motion away from the Earth. $\endgroup$
    – n30n6r33n
    Dec 20, 2018 at 12:01
  • $\begingroup$ Right. Indeed what you wrote is non sense. Physically and semantically. $\endgroup$
    – Alchimista
    Dec 20, 2018 at 12:13
  • $\begingroup$ Please do tell, we're all here to learn. $\endgroup$
    – n30n6r33n
    Dec 20, 2018 at 12:15
  • $\begingroup$ Straight up means it is curving down. Using your way of reasoning bring exactly into such things ;) $\endgroup$
    – Alchimista
    Dec 20, 2018 at 12:16

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