8
$\begingroup$

Confusing title, I know. Imagine a perfect, homogenous sphere with an exact radius of $1,000 \text{km}$ and an exact mass of $8 \times 10^{15} \text{kg} $.

If you use the formula for escape velocity $v_e = \sqrt{\frac{2GM}{r}}$ and plug in the numbers, you get $v_e = 1.033386 \text{m/s}$, which is just a tiny bit over $1 \text{m/s}$.

When a human jumps on Earth, they reach a certain maximum height before coming back down. So while someone would theoretically feel less gravitational attraction to the Earth when they're at the maximum height of the jump, the difference is so comically low that it's not worth taking into account.

The same thing applies to smaller bodies like Mars, Pluto, and Ceres. In each of these worlds the maximum height of a jump would increase, but even then it would still be too small for it to be taken into account.

But when it comes to very small bodies, like the one I described, someone jumping at an exact velocity of $1 \text{m/s}$ (that's only marginally lower than the escape velocity) would reach gigantic heights before theoretically coming back down. And the gravitational acceleration felt at the maximum height would be drastically lower than the one felt at the surface of the body.

So, my question is: Does the formula for escape velocity $v_e = \sqrt{\frac{2GM}{r}}$ take into account how an object's distance increases from its primary when it's traveling at speeds very near escape velocity? Or, in other words, would someone jumping at a velocity just $1 \text{cm/s}$ less than $v_e$ eventually come back down, all other forces outside this 2-body system ignored?

$\endgroup$
7
  • 8
    $\begingroup$ Did you have a look at the derivation of the escape velocity? That will answer your question. $\endgroup$ Nov 10, 2023 at 0:59
  • 1
    $\begingroup$ Note (to everyone) that it is the escape speed, since you are taking the square root of an equation for $v_{esc}^2$. The direction in which the object is launched does not matter (in Newtonian physics, and so long as it is directed away from the surface), so it is not an "escape evelocity". $\endgroup$
    – ProfRob
    Nov 10, 2023 at 12:16
  • 7
    $\begingroup$ @AtmosphericPrisonEscape - Not all AstronomySE users (nor all AstronomySE readers) have mastered integral calculus to the point where the answer of this question is obvious or the derivation of escape velocity is easy to follow and understand. I wouldn't close the question because it's useful for a subset of readers. $\endgroup$
    – Pere
    Nov 10, 2023 at 14:50
  • 3
    $\begingroup$ Something that always felt funny to me was that clearly the direction matters in practical situations. You won't escape if you plow into the planet. But the formulation of escape velocity is based around a point-source gravitational object (or, more completely, that the radius of the object is much smaller than the dimensions of the orbit). Remembering that the surface of the planet doesn't exist was helpful for me in accepting that direction doens't matter. $\endgroup$
    – Cort Ammon
    Nov 10, 2023 at 16:03
  • 1
    $\begingroup$ @Pere This was in relation to the rudeness of OP's comments, which were abusive of anyone prospectively helping them. They, by now, put their tail between their legs and deleted those comments. $\endgroup$ Nov 10, 2023 at 19:40

5 Answers 5

16
$\begingroup$

The escape speed is defined in Newtonian physics simply by demanding that the sum of kinetic energy at launch (ballistically, with no power applied thereafter) and gravitational potential energy at launch (which is negative) is greater than zero.

If that is the case, then the body will escape. There is no need for any calculus and no need for any consideration of the decrease in gravitational field strength with distance, since total energy is a conserved quantity that does not change as the object moves away. i.e. For escape $$ \frac{mv^2}{2} - \frac{GMm}{r} > 0\ , $$ where $r$ is the radius from which it is launched.

It is an escape speed rather than an escape velocity, because the $v^2$ is a scalar quantity. So long as the launch is away from the surface it does not matter in what direction the launch is (again, in Newtonian physics).

$\endgroup$
1
  • 2
    $\begingroup$ “No need for any calculus”, well, the way that I learned this in (my country’s equivalent of) high school, the $\frac{GMm}r$ was obtained by integrating $\frac{GMm}{r^2}$, so there was some calculus in it. $\endgroup$
    – Carsten S
    Nov 12, 2023 at 14:53
12
$\begingroup$

Yes, escape speed is an instantaneous calculation at the distance 'r' from the center of the object, as that changes you have to recalculate your escape speed.

For example, these are the escape velocities at 4 points from the sun (calculator):

  • On the sun: 617.5 km/s
  • 1 AU from the sun: 42.13 km/s
  • 200 AU from the sun: 2.98 km/s
  • 1774 AU from the sun: 1 km/s

So say you are at 200 AU from the sun. If your speed is 2.98 km/s you will eventually escape from the sun's gravity (barring any collisions or other force acting on you such as a Jupiter fly-by or plunging into the sun :)). It doesn't matter what direction you are travelling.

Let's say your trajectory will take you close to the sun. As you pass 1 AU from the sun, your speed relative to the sun will have increased due to it's gravitational influence and you should be going about 42.13 km/s. As you get nearer the sun your speed will keep increasing. As your trajectory passes the closest point to the sun you will be going the fastest relative to it. As you pass back by the 1 au mark you will again be going about 42.13 km/s relative to the sun but in an outward direction. As you get to 200 AU away from the sun your relative speed will be only about 2.98 km/s relative to it. When you get to 1774 AU from the sun you will be going only 1 km/s relative to it.

$\endgroup$
2
  • $\begingroup$ This is a specific application of @ProfRob's answer. $\endgroup$
    – EvilSnack
    Nov 11, 2023 at 16:56
  • $\begingroup$ I actually found that answer confusing because it talks about launch, and the question seem to be asking about speed after launch and how that change affected the escape speed. There doesn't even have to be any known 'launch' as in the case of what may be a long period comet or a body from outside the solar system. If you know the distance at any point in time you can calculate the escape speed which changes with distance from the center. If you know the actual speed at that time you can then tell if it will escape the system. $\endgroup$ Nov 12, 2023 at 21:40
3
$\begingroup$

The escape velocity is based on how much energy it takes to go "infinitely" far away from the object. That is, the escape velocity is a velocity such that for any $r$, the amount of work against the gravitational field that it takes to move an object to a distance $r$ away is less than the energy kinetic energy of the escape velocity.

If we didn't adjust for the gravitational force getting weaker, then the gravitational potential energy would be $mgh$. Since $h$ is "infinity", the energy would be infinite as well.

The reason the escape velocity exists in the first place is because the gravitational force become negligible once you get far away. The energy that takes to move something from 10 light years to 100 light years is less than the energy that it take to go from 1 light year to 2. Because the force decreases so quickly, the total energy ends up being finite, even over an "infinite" distance.

$\endgroup$
2
$\begingroup$

Yes, it does.

A good way of looking at this is to take not the difference in the object's velocity and the escape velocity at a given height, but at the difference between the object's velocity squared and the escape velocity squared. This is a constant.

In particular:

  • If the object's initial velocity squared was less than the escape velocity squared at its initial height, the object will come to rest at a height where the escape velocity squared is still positive: that is, at a real height. So the object's trajectory has a highest point.
  • If the object's initial velocity squared was more than the escape velocity squared at the initial height, the object will come to rest at a height where the escape velocity squared is negative. Which is impossible! So the object will never come to rest but carry on going.

If you prefer, you can tell the same story differently. At an infinite height the escape velocity squared is zero. If the object's initial velocity squared was greater than the escape velocity squared at the initial height, then the object's velocity squared at infinity will be positive: the object will still be going. But if if was less, then the object's velocity squared at infinity will be negative: which is impossible - in other words, the object will never get to infinity.

"Velocity squared" isn't a rabbit out of a hat. The velocity squared of the object is related to its kinetic energy, and the escape velocity squared at a given height is related to the potential energy.

$\endgroup$
0
$\begingroup$

When a human jumps on Earth, they reach a certain maximum height before coming back down. So while someone would theoretically feel less gravitational attraction to the Earth when they're at the maximum height of the jump, the difference is so comically low that it's not worth taking into account.

...

Or, in other words, would someone jumping at a velocity just $1 \text{cm/s}$ less than $v_e$ eventually come back down, all other forces outside this 2-body system ignored?

I think this part of the question is best at address first, because it helps frame the rest of the question. What does it mean to "come back down?" Being "down" means you are no longer in free fall, and have the normal force of the ground pushing up on you. It's a statement that must be made with respect to the body of a planet, not just its gravity. Were all of the mass of the Earth to be collapsed down to a black hole of that mass (not actually possible, but bear with me), you and I would find that we were in an orbit. It'd be an elliptical orbit with an apoapsis (furthest point from the gravitational body) at the altitude you were at when the Earth collapsed. Our perigee (nearest point to the gravitational body) could be calculated using the rotation rate of the Earth and our latitude. The actual calculations could take a bit of work, but regardless, we would be in some elliptical orbit.*

The only thing that makes jumping and landing special is the surface of the earth. Otherwise, you are merely on an elliptic orbit. You may have been told that when you jump, you follow a parabolic path. This is actually false. You follow an elliptical path. You would follow a parabolic path if the force of gravity was a constant and always pointed in the same direction. Because gravity varies by altitude and points to a central point, the path you take is elliptical. As an exercise to the reader, you can calculate how much the ellipse and parabola differ at sea level for a reasonable jumper. They're astonishingly similar until you get to the massive power of a modern rocket.

This is true regardless of the size of the planet. It can be a 1000km sphere. However, what you will indeed find is different as you change the size of the planet is how easy it is to impart enough velocity as to be in an orbit that doesn't intersect the surface of said planet.

So you ask about something traveling just under escape velocity at their particular altitude. They will "come back down," as gravity brings them back. However, assuming they don't hit the surface of the planet on the way down, they'll reach a perigee (point closest to the gravitational body), and then slingshot right back out to the same point. I point this out because the concept of "coming back down" has a slightly different feel when the planet isn't in your way. It's orbital. While they'll come back down, they'll also come back up, 1/2 of their orbital period later.

If they jumped just a little bit stronger, they would have enough energy that the planet could never quite bring them back. They'd be going up forever.

All of this is true regardless of the mass of the gravitational body... at least up until it gets too close to the mass of the jumper itself. The normal "orbital" behaviors are simplifications -- we assume that the orbiting body does not have enough mass to change the position of the gravitational body. When this is not true, we have to use more complicated equations, and the orbits look a little different. You no longer orbit "around a body," but rather both bodies orbit around their barycenter (the center of mass of both bodies). As a very rough rule of thumb, it's typical to assume these simplified rules apply as long as the gravitational body is at least 1000x more massive than the orbiting body. In your case, with a 8000000000000kg gravitational body, a "jumper" who might weigh 100kg clearly fits in this category.

*. Just for completeness, there would be two points that don't result in an elliptical orbit: the geographic north and south pole. At this points, you would get a degenerate "line" orbit that falls directly into the black hole. This would be quite unpleasant, so forgive me if I do not address those degenerate cases any further.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .