# Escape velocity of an artificial satellite

1. Sep 11, 2014

### Crutchlow13

Hey, I'm doing a research on artificial satellites, and I'm really confused about the escape velocity required for an artificial satellite to escape from the gravitational attraction of the Earth.

I know that the equation for it is v^2 = 2GM/r , and with that, the rocket should launch at that speed, but it could go much slower spending much more fuel to escape from gravity right? arghh Wouldn't it be easier to calculate it with energy?

Pol.
PS: Sorry for my english but I'm spanish u.u.

I don't know where to post this thread, move it if it's necessary please.

2. Sep 11, 2014

### rcgldr

Assuming that the amount of fuel is limited, eventually the rocket runs out of fuel. If it hasn't reached or exceeded escape velocity by then, it ends up in an orbit of some type. Once beyond the atmosphere, it's wasteful to spend fuel opposing gravity, so it's best if the thrust is perpendicular to the pull of gravity, so all of the spent fuel is used to increase velocity. Although the thrust is perpendicular to the pull of gravity, the path of the rocket would be similar to an outwards spiral with ever increasing radius, due to the increasing velocity. Another way to explain this is that what is an acceleration perpendicular to the pull of gravity at one moment in time, will result an increased outwards component of velocity at a later moment in time.

3. Sep 11, 2014

### AbhiNature

Not excatly.

V^2 = 2GM/R gives the initial speed required for the mass or rocket in your case to move out of gravity field. If an object has this speed in an opposite direction to that of the force of gravity then no further energy is required to propel that object out of that gravity field.

Rockets are never given all of that energy once. Rockets therefore never gain escape speed.

For example: If a stone is thrown such that it has initial speed of that of escape velocity then that stone will move out of gravity field.

4. Sep 11, 2014

### D H

Staff Emeritus
You don't need to launch with the escape velocity to escape. One could escape by firing a tiny bit more than one g upwards, and then back off on the thrust so the velocity remains some constant at some small upward velocity. Eventually, you'll reach an altitude where that small velocity is the escape velocity. The amount of energy needed to escape via this approach is very, very large.

Prior to GPS, launches put the upper stage and payload into low Earth orbit. Any follow-on use of thrusters had to await a determination of the orbit into which launch had inserted the vehicle. With the advent of GPS, vehicles can now follow a more direct route, but many still do use the launch insertion orbit concept.

So why do we use escape velocity? The answer is simple. It's a very useful concept.

5. Sep 11, 2014

### Crutchlow13

Thanks to all of you. I guess it's not worth talking about it as it's never taken into account when building an artificial satellite, isn't it?

Last edited: Sep 11, 2014
6. Sep 11, 2014

### Murtuza Tipu

You've made the common mistake of thinking that the velocity needed to launch a satellite is the (initial) velocity needed to raise it to its orbital radius.

If you raise a satellite to e.g. 300km then let go the satellite will immediately fall straight back to Earth. You need to do two things:

raise the satellite to 300km

increase its tangential velocity to
square root of GM/r

7. Sep 11, 2014

### Staff: Mentor

A satellite, by definition, moves slower than the escape velocity. If you want to go to other planets, however, you have to get a speed above the escape velocity. Usually this is done from low earth orbit, as using thrusters at a low altitude is more effective than using them elsewhere (this is called Oberth effect).

8. Sep 11, 2014

### Crutchlow13

Orbital radius? Aren't most of the orbits elliptical? Therefore it should be the semi major axis right?

9. Sep 11, 2014

### D H

Staff Emeritus
You're missing Murtuza Tipu's point. A launch vehicle doesn't go straight up. A launch goes straight up for a short time and then starts turning so that the vehicle is moving more or less horizontally at the end of the launch process. That means a high velocity. The vast majority of the energy from the launch goes into increasing velocity. The energy that goes into increasing altitude is but a tiny fraction of the total.

10. Sep 11, 2014

### Crutchlow13

Yeah I understood that but I was just pointing that he said orbital radius, and I've been told that when artificial satellites move around earth their orbit is elliptical, so how would it be considering that it's not a perfect circumference?

11. Sep 11, 2014

### D H

Staff Emeritus
Most satellites are in orbits that are nearly circular. Notable exceptions include the Molniya and Tundra orbits used by Russia and Sirius Satellite Radio.

12. Sep 11, 2014

### BobG

(bolding mine)

Yes. Minimum escape velocity is the velocity required to raise your specific energy* to 0 (closed orbits have a negative specific energy).

In other words,

$$0=\frac{v^2}{2} - \frac{GM}{r}$$

Rearrange that equation to solve for v, and you get the equation you stated.

And, as others noted, you don't launch into an escape trajectory. Even if it were possible to instantaneously accelerate to that speed, going that fast through the atmosphere would be problematic. You'd generate an awful lot of heat due to atmospheric friction - in fact, most objects entering the atmosphere, such as meteors and dead satellites, at similar, or even slower speeds (in the case of re-entering satellites) burn up from the heat.

You launch into a parking orbit and then fire a second booster to accelerate you to escape velocity from your parking orbit. (Meaning r is going to be some distance into outer space; not the radius of the Earth.)

*specific energy per unit of mass. To determine how much energy (and how much fuel you need), you multiply your specific energy per unit of mass by your spacecraft mass.

13. Sep 13, 2014

### Murtuza Tipu

Any launch profile will suffice (as long as you do not try to go through the Earth of course) as long as the velocity at the end meets the following criteria,

∥v⃗ ∥≥2GM/∥r⃗ ∥−−−−−√

where r⃗  is the radius (position relative to the center of mass of the Earth) at that moment.

For this I also assume that its trajectory will not go through the Earth as well and is sufficiently out of its atmosphere.

You could few escape velocity from an energy point of view, since when escape velocity is reached the specific orbital energy becomes zero:

ϵe=v^2/2 - GM/r

because the gravitational potential is defined such that it goes to zero when r approaches infinity. So at escape velocity, if all kinetic energy would be converted into potential energy, then you would have to go infinitely far away.