# Escape energy required for an orbiting body?

1. Dec 29, 2013

### Ascendant78

My current level of knowledge is based on Physics I and my own readings, so bear with me if this is something relatively simple that I'm overlooking...

Anyway, my question is regarding the energy required for an orbiting body to escape orbit? I understand the formula of escape velocity where the kinetic energy exceeds the gravitational energy. However, I'm wondering how sensitive that value is and if there is a basic method of calculating it?

For example, if a very small meteor impacted the Earth, it is not going to provide Earth with enough kinetic energy to escape orbit. So, I am wondering exactly how the needed value is determined when you break it down to infinitesimal increments?

2. Dec 29, 2013

### SteamKing

Staff Emeritus
If you want a body to escape orbit, then the body must achieve and exceed the escape velocity, first and foremost.

http://en.wikipedia.org/wiki/Escape_velocity

3. Dec 29, 2013

### Ascendant78

Thanks. So, from what I was able to gather, it seems as if velocities below the escape velocity would change the orbit of the body, but it would still remain in orbit. So, relating this back to my meteor example, a shift in orbit would occur, but it would be so slight as to not make any noticeable difference, right? I guess where I was not getting this was that the professor kept emphasizing how delicate the balance was between the kinetic and gravitational, but now I see that it isn't so much "delicate" as it is "proportionate."

4. Dec 29, 2013

### SteamKing

Staff Emeritus
From the formula for calculating escape velocity, the mass of the body is an important factor. Even the largest meteors known to have struck the earth (even the one at Chicxulub which wiped out the dinosaurs) have masses which are much, much less than that of the earth.

http://en.wikipedia.org/wiki/Chicxulub_crater

It's hard to say what effects this impact had on the orbital characteristics of the earth. The environmental catastrophe which resulted was enormous, long-lasting, and world-wide.

5. Dec 30, 2013

### iScience

:

just realized my earlier post didn't answer your question at all lol oops

the total energy of a circular orbiting system (ignoring spin for simplification):

$$\sum E = -\frac{GMm}{r^2} + \frac{1}{2}mv^2 = Constant$$

well, with this, assuming you don't already know the potential energy (the first term..) of the system, you can i suppose find it with this.

and let's say that a planet in orbit gets hit with a rock and its kinetic energy increases. the two terms in the equation swap off in magnitude trends, ie since the total energy is conserved, if the kinetic energy gets bigger, the pot. energy gets smaller and vise versa. so let's say that the planet gets struck with a rock such that it increases the velocity of the planet+rock system, when this happens to a circular orbit (which does not have any vertical velocity components), the planet gains a vertical velocity component, and the potential energy of the system will get larger and larger because the vertical velocity component means that the planet will start getting farther and farther away from the star, and there are two cases here.

1.) if the vertical velocity component reaches zero eventually, then, since there still exists a potential energy between the planet and star, ie since there is still a force exerted between the star and the planet, the planet will start to head towards the star again; all the meanwhile, the planet still has its horizontal velocity component so it won't be going into the star. So this case describes an elliptical orbit. the more distance the planet has to travel to get to that $$v_y=0$$ point, the more eccentric the orbit is said to be. This is common for the case of comets. planets i think usually have a very low eccentricity.

2.) so, the vertical velocity component will decrease no matter what, because gravity is an attractive force and so the gravitational force will make $$v_y$$ decrease. but if it asymptotically decreases, never to reach zero, then we have here the condition for a planet escaping its orbit, and this type of "orbit" is said to be a hyperbolic orbit.

Last edited: Dec 30, 2013
6. Dec 30, 2013

### Ascendant78

Well thanks iScience for such a thorough response. It definitely helped to solidify what I was thinking about the matter.

7. Dec 30, 2013

### jbriggs444

For circular orbits, this is obviously true, because both terms are constants. For non-circular orbits it remains true because of energy conservation. More interesting is the fact that the kinetic energy of an object in circular orbit is equal to half of its potential energy deficit. It follows that (for a circular orbit):

$$\sum E = -\frac{1}{2}mv^2$$

Double the kinetic energy of an object in circular orbit and it'll be at escape velocity.

If the planet is hit by a rock coming from outside the system, energy within the system is not conserved. Energy conservation applies in closed systems. The above equality need not and usually will not hold.

8. Dec 30, 2013

### iScience

yes, my mistake; the $$\sum E$$ value changes, it increases. but having taken this into account, this relationship should definitely hold. i was not using the equation to find specific quantities of how much energy is needed to escape any orbit, hence i did not bother to mention the virial theorem. instead i wanted to enhance the op's qualitative understanding.