Actually it would go into a lower orbit. It’s a bit counter-intuitive I know but slower objects go into a higher orbit or, the lower you are to the earth the faster you need to move to stay in orbit.
Just out of curiosity if I threw a ball up at 25,039 mph (ignoring air resistance and such) how fast would the ball be moving when it left the solar system? Would it be slowed down to nearly zero speed? Also, how far away does an object need to be to be considered free of earth’s gravity?
Uh, no.
The faster object goes into a higher orbit. The orbital period is longer, but the velocity is higher. If it worked the way you say, home runs would come back to the stadium, and a gentle toss would reach the space shuttle.
Tris
You’re thinking of the paradox where you catch up with an object in your orbit by slowing down … When you speed up you do go into a higher orbit, and orbit at a faster speed. However, your angular speed is lower. A circular orbit at twice the distance of another has an orbital velocity of about 1.4 times that of the lower orbit. (See Kepler’s Second Law.) However, since it has to go twice as far to get through one orbit, it completes only two orbits while the slower-moving satellite in the lower orbit finished three & has started on its fourth. This creates an illusion that that satellite is actually moving faster than the first.
When you slow down to catch up with another satellite, you do, in fact, slow down. But your slower speed is less than the decrease in your orbital distance, and you ‘catch up’ along the inner track, as it were. But that’s real hard to see when you’re on the merry-go-round, you know. All you see is that you’re passing your target, it’s not as obvious how much shorter of a track you’re on. An outside observer, at reast with respect to the Earth, can see that you’re just ‘cutting off a corner’, though.
I think you are wrong.
Consider our ball thrown at 29,000 mph (just below escape velocity). If I throw the ball with the proper trajectory instead of falling back to earth it will orbit the planet but its speed will be (at a guess) only a few miles per hour. That same ball orbiting at a much lower altitude will still be hauling ass at several thousand miles per hour.
Basically, when you are WAY out you don’t need much energy to keep from falling to the planet but the closer you are the harder you will have to work to stay aloft.
Mind you, I am NOT talking about a geosynchronous orbit or anything like that…just the minimum necessary to stay in orbit.
I didn’t see SCSimmons’ post before submitting mine. Thanks for the clarification.
You know what, I think did that math wrong … the speed of the higher orbit does seem to be slower, I had a reciprocal reversed in the equation. I think. Let me try that again …
Yep, Whack-a-Mole was right the first time. Two morals to this story- 1) don’t extend parabolic intuitions to elliptic or hyperbolic trajectories, and 2) double-check your math before posting.
Kepler’s third law works better for this, BTW. 
Ahhh…vindication!
I remember reading about this awhile back but not being able to provide a cite and not being a rocket scientist myself I was ill-equipped to defend my assertion. Fortunately others on this board can usually be relied upon to put the smack-down on the truth. Thanks SCS for your work!
Actually, it wouldn’t escape, it would go into orbit around the Sun.
The escape velocity of an object is the square root of two (1.41) times the orbital velocity. The orbital velocity of the Earth around the Sun is about 30 km/sec, so escape velocity of the solar system starting at the distance of the Earth = 1.4 x 30 = 42 km/sec. So you’d need to add over 30 km/sec to your rocket to get it to escape the solar system. That’s why we use gravitational assists from planets. That’s a lot of extra speed.
Not quite, but I can see the source of the confusion. Let’s look at three orbits: One is a circular orbit close to the Earth (like the Space Shuttle’s), and one is a circular orbit farther from the Earth (geosynchronous, perhaps). The third orbit is highly elliptical, with perigee (closest point to the Earth) being at the Shuttle’s height, and apogee (furthest point) being at geosynch. The Shuttle is, in every sense, going faster than the geosynchronous satellite. However, the elliptical orbit is faster than the Shuttle at the bottom, and slower than geosynch at the top.
In other words: We’ve got the Shuttle going in its usual orbit. Then, it fires its engines to increase its speed, and goes into that elliptical orbit (yes, I know that the Shuttle isn’t capable of that much [symbol]D[/symbol]V). This new orbit is never lower than the original one, and except for a single point, higher. It’s fair to call it a higher orbit. Now, as the Shuttle is following this orbit, it’s climbing partway out of Earth’s gravitational well. It’s gaining potential energy, so it must be losing kinetic energy: In other words, it’s getting slower. By the time it’s up to the top of the orbit, it’s going too slow to be in a circular orbit at that height, so it’ll start back down again. Alternately, it could fire its engines again at the top, to increase speed again, and go into geosynchronous orbit.
The net result of this is that the Shuttle has used its engines to increase speed twice, but in between it’s lost speed to gravity, so it’s now going slower.
And I was all set to mention that pet peeve of mine, but here, I’ve gone and beat me to it. Thanks, Shagnasty.
Yes, if you shoot something with tangential escape velocity it will escape but note that, as the object get farther away, the velocity vector becomes more vertical. I guess what I mean is that escaping means getting away which only happens in the vertical direction. The tangential component does not help escape except that the part of the tangential component which is greater than the circular orbit speed at that distance is moving the object away from the earth and the vector becomes more vertical. (Well, I know what I mean even if it sounds confusing).
With respect to orbits: For every distance there is a speed which gives a circular orbit. An object moving with that speed tangentially will remain in that circular orbit. An object with a tangential speed higher than that will start to climb and lose speed until it reaches an apogee and will begin to fall back and gain speed until it reaches perigee. It is swinging up and down. If you want it to stay in the higher orbit when it reaches apogee you need to give it enough energy to stay there because it is not moving fast enough and it will fall back down again otherwise.
Obviously, if some part of the orbit crosses the atmosphere or the Earth, that’s the end of the story.
Note (to answer at least one earlier query): you have effectively escaped the gravitational field when you reach escape velocity. There’s no such thing as someone being able to travel slower than the escape velocity with respect to the source and be able to escape from it.
Nope. Whack-a-Mole was wrong the first time, when they said “Actually it would go into a lower orbit.” Remember, there is no such lower orbit. Lithospheric friction, you know.
Third moral: 3) You guys need to go on the road together.
Now you’ve gone and confused me again so let me restate what I think you are saying and then let me know if I’m right or wrong.
In order to move to a higher orbit you need to expend energy (ala fire your rocket engines). However, when you get to your higher orbit you will be moving slower than you were at the lower orbit. In effect, you weren’t speeding up so much as you were using energy to climb farther out of a gravity well. So, it is still proper to say a higher orbit is a slower orbit but you need to use more energy to get there (odd thinking you expend energy to achieve a lower speed).
Do I have? if not go ahead and let me have it 
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Huh?
Why isn’t there a lower stable orbit and why would you suppose our satellite or whatever would necessarily crash into the ground? Satellites and the shuttle and whatnot don’t skim the surface of the earth’s atmosphere but are generally well above it allowing a fair amount of room to move into higher or lower orbits.
If the satellite is moving 10 km/sec (that was the speed in Achernar’s original post, no?), and it is in a lower orbit than the shuttle (what you said), it is auger-in.
For circular orbits, the relationship between speed s and altitude a (from the center of the Earth) is approx. s = sqrt(G M / a) where G is the gravitational constant and M is the mass of the Earth. At the surface of the Earth, s is about 8 km/sec, close to the shuttle speed. The space shuttle orbit is close to the surface of the Earth, after all. A lower orbit would have higher speed, true, but that case would not be very interesting–it would be boring.
It can go to a higher non-circular orbit, like Chronos described, but it’ll be going much slower when it gets to the top of the orbit, and it’ll return to its lower altitude at perigee.
Sorry for the earlier ambiguity. When I said that a 10 km/sec object would go into a higher orbit than the shuttle, I meant that an object that was launched at 10 km/sec would go into a higher orbit, not that it would orbit at that speed. ATMC, if it had a circular orbit, then the following relation holds:
v[sub]ORBIT[/sub][sup]2[/sup] = v[sub]ESC[/sub][sup]2[/sup] - v[sub]LAUNCH[/sub][sup]2[/sup]
For Earth, v[sub]ESC[/sub] = 11.182 km/s. So for v[sub]LAUNCH[/sub] = 10 km/s, you’d get an orbital speed of v[sub]ORBIT[/sub] = 5.004 km/s. As has been mentioned, there is no such thing as a circular orbit around the Earth with an orbital speed of 10 km/s.
Just wanted to add some clarification.
In the first case, if you throw a ball very fast from the surface of the earth, and ignoring air resistance, the orbit the ball is in will pass through the point you threw it from - at the same velocity.
So, your orbit is ellipitcal and at apogee may be going very slowly, but at perigee it’s going 29,000 mph.
Next, there’s no such thing as having to “work to stay aloft” in orbit. There’s just orbits, and it’s free to stay in one. What costs is changing orbits. To go from a circular orbit to a higher elliptical orbit, you must speed up. Your apogee speed will be slower, but your perigee speed, at the same distance as your old circular orbit will be faster.
Chronos did a good job explaining that bit, but I wanted to point out the first bit about perigee speed.
Well direction does matter a little. I mean if you launch a rocket at escape velocity straight down it probably won’t even orbit
If you have two objects travelling, one behind the other at the same distance from the earth, and in a trajectory that will not intersect with the earth, and one is travelling faster than the other, the faster object will be in an orbit that reaches a greater distance from the earth.
The faster object will be in a higher orbit. Orbital speed is a variable, yes, but the specific velocity of an object at a given position with respect to the earth is implied when one makes the statement that “it goes into a lower orbit” that is just wrong.
xash asked what would happen to an object moving at one speed, compared to the same object moving at another speed, Achernar answered “it goes into a higher orbit” Whacksaid, “no, it goes into a lower orbit.”
It goes into a higher orbit. There may well be points along that orbit which have lower speeds with respect to Earth than the one it is in now, but the orbit will be higher if the speed now is higher for the same object at the same place.
“If God had intended for man to go to Mars, He would have given us more money.” ~ an unnamed NASA official ~