# Escape velocity and gravitation

1. Aug 28, 2004

### Cyrus

Hi, me again. I have another question concerning the gravitation equation. My physics book shows through a proof that the escape velocity of an earth satelite is just $$sqrt(2)$$ greated than the orbital velocity. But I am having trouble seeing how this is so. My instinct tells me that if I increase the velocity by $$sqrt(2)$$, I will start to move out farther in space, but as a consequence I will also increase my radius, until which point I have achived the right speed/distance to become a satelite again, but this time at some further distance from the earth. I dont see how increasing my speed by the square root of two will allow me to drift off into space and no longer stay in orbit.

Part 2.

Lets say I have a delta V rocket that can go the escape velocity of 25kmi/hr. Now, lets also say I have an estees model rocket that can go 50mi/hr. Clearly the delta V has the required velocity to go off into space. What happens to the model rocket at 50mi/hr. If fuel is not an issue, it will always be going at 50mi/hr. Wouldent this small model rocket go off into infinity also. Is there a conflict with the equation to find the escape veloctiy then?, since I could send a delta V at 25kmi/hr or a meager model rocket at 50mi/hr off into infinity?
Cyrus Abdollahi.

Last edited: Aug 28, 2004
2. Aug 28, 2004

### Janus

Staff Emeritus
The greater the radius of the Orbit the smaller the orbital velocity needed So, if you increase velocity you start to pull away from the Earth. The further you are form the Earth, the slower you have to be moving to maintain orbit. Now as you pull away from the Earth you will also lose some velocity, but if you are moving fast enough you never lose all of it.

For instance, you are orbiting at a velocity of V, you increase your velocity by x and start increasing your distance from the Earth. When you reach a radius of Y, you would need to have a velocity of less than V by a certain amount to maintain orbit. You have lost some velocity climbing to that new radius, but the your new velocity is still more than you need to maintain orbit, so you climb further to a higher orbit with even a lower orbital speed, but you still haven't lost enough velocity and have some left over and climb to a even higher orbit. If your value of x was large enough, the required orbital velocity will continue to drop faster than you lose speed, and you will continue to keep climbing to an "higher orbit" forever.
The 50mi/hr velocity for the Estees rocket is the velocity it will reach after burning up all its fuel. The rocket continues to accelerate until it uses up all the propellant. So its top speed is dependant on its fuel.

But the question I think that you are asking is what happens if you were to climb at a constant 50 mph, and how do you escape Earth's gravity without exceeding the escape velocity?

The further you are from the Earth, the smaller the escape velocity. For instance, at the distance of the moon it is down to 1.4 km/sec. If you keep climbing at 50 mph, you eventually reach a height where the escape velocity is less than 50 mph. At which point you could shut your engine off and continue away from the Earth forever without ever having the reach the 11 km/s escape velocity needed from the surface of the Earth. Of course, you will burn a lot more fuel reaching that distance at 50 mph than you would if you just accelerated up to 11 km/sec right at the start.

3. Aug 29, 2004

### Cyrus

WOW! that is the best and clearest response i have gotten from a person yet! thanks soooooooooooooooooo much ;-) ! :-). Yippie im happy now. hehehe. If you could, would you please explain how come it would require alot more fuel to go the steady 50mph than the 11km/sec from the start. Why is it alot easier and use less fuel if you do it all at the start than over some period of time? I think my first problem was that I was forgetting that the velocity will decerease with distance. I was assuming you would have to go faster to orbit further away, which is opposite what is true. Your example made things very clear. THANKS! So I guess things like comets which 'seem' to go off forever, but really dont since they return back to the earth, are just in very very large orbits and have very small speeds, (elliptical orbits) i take it, and so they seem to come and go, but in actuality they go off into an ellipse which is too big to see them when there gone, and then they return again. Would the thing with the escape velocity leave in the earth in a straght line, or in a parobolic arch or what? If it were launched straight up and just continued on straight from there, it would seem to go straight forever. But if it were in orbit, it would seem to me based on your analogy, it would orbit bigger and bigger forever, so it would go off into infinity in some arched space curve. Or perhaps, would it move off into space in some striaght line tangent to the orbit, at that instant when the rocket was fired? Now my brain hurts, Ill leave it to the experts like you :-).

Cheers,

Cyrus

Last edited: Aug 29, 2004
4. Aug 29, 2004

### enigma

Staff Emeritus
If you're in a bounded orbit (less than escape velocity), then you're in an elliptical orbit. e<1

If you've got escape velocity, you're in a parabolic orbit. e=1

If you're going faster than escape velocity, you're in a hyperbolic orbit. e>1

5. Aug 29, 2004

### Janus

Staff Emeritus
If you burn all your fuel early on you don't have to lift that fuel. If you burn your fuel slowly, you have to lift at the early part of the trip the fuel you are going to burn later, which means you need more fuel to lift that fuel, and even more fuel to lift that fuel, etc.
An object traveling at exactly escape velocity would follow a parabola. If it was traveling at greater than escape velocity, it would travel in a hyperbola.

6. Aug 29, 2004

### Cyrus

Another reply, sorry :-). You said at first that you will loose some velocity to go to a higher orbit. I was wondering how is this so. Since there is nothing in space to retard your motion, if you go further from the earth and increase your radius, how would your velocity around the earth decease slightly? Would you not keep that velocity you gained when you fired your engine and just keep gaining distance forever with that same constant velocity?

7. Aug 29, 2004

### Janus

Staff Emeritus
Earth's gravity slows you down, for the same reason that when you throw a ball up into the air gravity slows it to a stop and pulls it back down. Just because it is moving too fast for gravity to ever completely bring it to a stop doesn't mean thart gravity still can't slow it down as it climbs away. The object still exchanges kinetic energy for potential energy.

8. Aug 29, 2004

### Cyrus

Ok I see what your saying. So if I left the earth in my rocket at escape velocity from the getgo, does that mean by the time I reach the point at infinity, I will have used up all my velocity and will just sit there at infinity with zero velocity and wont ever be sucked back into the earth. Also, lets say I had a rocket that failed in the middle of the burn, and I only was able to go half the speed of escape velocity. I would go half way to infinity. But then, would I orbit around the earth at a distance of half of infinity, or would I be sucked in and crash and burn in the atmosphere some time later? Because I know you need a certain velocity in order to orbit dependent on your radius, but it would seem that this velocity has to be directed tangential to the earth, while If i were traveling off into space, i would have a perpendicular velocity, so this might mean i would come back to earth, since my velocity does not allow me to fall around the earth, but fall straight towards the earth, like a stone thrown up falls straight down, instead of a stone thrown side ways falls sideways. Also, I know I asked this before but I want to make sure. If I launch my rocket from the ground and only go straight up with the proper escape velocity. Will I travel into space forever straight up from where I started, or will I still have to follow a parabola or hyperbola even though I launched straight up at 90 degrees with respect to the ground. I could see how if I was in orbit, and I used my booster, I would go off into space, but since I had some component of velocity that is radial, It would cause me to swing into a really wide arc. Ive kinda been asking the same question over and over again, but I have been thinking of different situations in which it would apply. Hope you dont think im wasting your time by asking the same thing so often, just trying to get a handel on it by beating the problem to death.

Last edited: Aug 30, 2004
9. Aug 30, 2004

### BobG

If you're going exactly the escape velocity, the spaceship traces out a parabola. In other words, if you took the distance the object was at launch, doubled it, and drew a line perpendicular to the (doubled) launch radius, the object would always be the same distance from the line and the center of the Earth. (If the object eventually came to rest, it would be pulled back into the strongest source of gravity).

If you launched straight up relative to the ground, you would have a lateral velocity, as well, since the launch site is moving with the Earth's rotation. The only exception would be if you launched from the North or South pole. If you launched in a direction that took you directly away from the Earth, it would not be a parabolic orbit. It would be a line. It is somewhat similar to a parabolic orbit. Parabolas are planes that intersect a cone parallel to the slant height. A line would be a plane that both intersects the cone parallel to the slant height and passes through the vertex of the cone. The object would continue on a straight line forever (excepting the influence from the Moon, Sun, planets, etc.)

If your booster failed and only got you halfway to escape velocity, your rocket would crash into the Earth. Technically, it would put you into a closed orbit (elliptical or circular). But, at only half the speed, the specific energy would be -46.87. That means the semi-major axis of your orbit would only be 4252 km, which is less than the radius of the Earth.

A velocity just barely below escape velocity would get you a really large orbit, though.

10. Aug 30, 2004

### Cyrus

Yes, But what if i launched from the north pole as you said, and my engine failed. Now I would be going off in a straight line into space, but not have enough speed. Would I still return in an elliptical orbit, because would I not just fall straight back down like a stone thrown up will fall straight back down? (at the north pole there would be no rotational effects so the stone would just come right back down and not be turned sideways a small amount due to the rotation, same as a rocket launched at the north pole.) I would still like an anwser to my other question as well please. If I have the right escape velocity to leave earth, does that mean that when I reach the point infinity, will my velocity now be zero, and I can stay at infinity without being pulled back in. Likewise, If I only have half the escape velocity, and I get halfway to infinity, will I eventually get to a velocity of zero, and then start drifting back down towards earth slowly going faster and faster? Just like a stone thrown upwards.

-Cyrus Abdollahi

Last edited: Aug 30, 2004
11. Aug 30, 2004

### BobG

Yes, if your path were directly away from the Earth, and you failed reach escape velocity, the return path would be a straight line back toward the Earth. In a theoretical world, where the Earth wouldn't stop the object, the object would oscillate back and forth in a straight line reaching the same distance away from Earth on opposite sides. In the real world.....

The only problem with the idea of reaching a zero velocity at 'infinity' is that you can never reach 'infinity'. Infinity is just away of saying an infinite never ending distance - and it takes an infinite never-ending amount of time to reach it. In a parabolic orbit or hyperbolic orbit, an object would continue getting further and further away from the Earth forever (ignoring the effects of other objects inside and outside the solar system).

But, infinity is the start point for measuring the potential energy of the Earth's gravitational pull and, even though non-existant, it is the point where a zero velocity would result in an object remain stationary instead of being pulled towards the Earth.

12. Aug 30, 2004

### Cyrus

Thanks BobG and Janus I am, you dont know how greatful I am to the both of you for anwsering all my questions. These questions were troubling me for a bit of time, as I could not find a direct anwser to them, but thanks to the both of you for helping me! Hopefully, however doubtfully, I can repay the favor. ( You guys are just too smart for me :-) ).

13. Aug 30, 2004

### Janus

Staff Emeritus

If you only attain 1/2 escape velocity you would reach a maxmum altitude of 2126 km above the surface of the Earth before you start to fall back. And ignoring air resistance, when you hit you would be going at 1/2 escape velocity.

14. Aug 30, 2004

### Cyrus

thats amazing, you would think that with half of the escape velocity you would be able to go much further into space. I also have a side question. In doing my physics I did the proof for keplers law and how sector area is constant with time, which proves that the magnitue and direction of the angular momentum is conserved, and the book says this helps explain why the planets orbit along a common plane. But I somehow dont buy into this explination, becuase from the start, we assumed that the planets were traveling along a plane when we calculated sector area. Suppose the planets moved around some funny zizaged circle, then it would not be a sector area, but some line integral (or is it surface integral) since as it goes around in a circle it might also go up or down in the same processs, the area of the shape it traces would no longer be planar, which is what we assumed was true when we did keplers law.

15. Aug 31, 2004

### Tide

But the planets don't just zig zag along. Their orbits are essentially elliptical.

To see how they go coplanar think about this example. Suppose two planets orbit a star in nearly circular orbits and further suppose that these orbits are perpendicular to each other. What influence do these planets have on each other's orbits? They will, in fact, tend to pull each other toward a common plane!

16. Aug 31, 2004

### Rogerio

As you can see at thread https://www.physicsforums.com/showthread.php?t=39745 , even 99% escape velocity is not enough to reach 384000km (aboute the Earth-Moon distance) .

17. Aug 31, 2004

### BobG

Good point. The tendency for geosynchronous satellites to want to re-orient their orbital plane with the Moon/Sun is one of the major perturbations satellite operators have to compensate for.

Except, this is an external force acting to change angular momentum (at least when considering individual planets) instead of a demonstration of conservation of angular momentum.

Unless, you want to look at the whole system and how a change in angular momentum of one object causes a change to another object in order to keep the system's overall angular momentum constant. The end result is the same regardless.

As a demonstration of conservation of momentum of the individual planets, they are relying on the idea of the Sun and planets all being created from the same cloud of rotating dust. Even after coalescing into big balls of matter, the overall angular momentum of the dust particles remain constant, hence the planets all starting out in the same plane from the get-go.

Obviously not a perfect demonstration, since there are very small differences in each planet's and each moon's orbital plane. And Pluto's orbit and the orientation of Uranus's rotational axis record some very significant events (external interference) in the solar system's history.