If the Space Shuttle were to stop all its forward motion while in low earth orbit (that is kind of orbit the shuttle is in correct?) would it fall from the sky like a rock and crash into the ground? OK I doubt it could make it intact through the atmosphere. My question is I’m assuming that the gravitational force at the height of the Shuttle would be more than enough to make this a very unpleasant trip for those on board. Is this correct?
The only thing keeping it in orbit against gravity is its motion. If it were to make a huge thrust opposite to its orbital path (I mean huge–it’s going, what, 20,000 MPH?), then it would indeed fall like a rock.
Thanks, that’s what I thought.
Actually, you could say that it is always falling like a rock. It is just that due to the forward motion, the earth is falling away from the shuttle at the same rate that the shuttle is falling.
In other words, the shuttle is in perpetual freefall.
rsa, this has always confused me. I don’t really understand how an orbit works. The Shuttle for instance is falling toward the Earth, but Earth is falling away from the Shuttle, correct? Since the Shuttle circles the Earth, why doesn’t the moon cross into path the Earth is falling? How does this work for the Earth and the Sun?
Does the Shuttle has to a maintain a certain speed? What if it goes faster?
…which is what causes the lack of gravity on board the shuttle. Everything inside it is falling at the same rate. Like a really fast elevator going down. I only mention this because a lot of people seem to think that the space shuttle orbts far enough away to be free of the Earth’s gravity, causing weightlessness.
You’ve got several separate orbits going on, here. It helps to think of only one at a time. First, you’ve got the Earth orbiting the Sun. It’s in continual free-fall, as described above. Then, there are things orbiting the Earth. They have almost the same motion as the Earth, on the grand scale, so they can be considered to be orbiting the Sun, but they’re also orbiting the Earth, which is much closer and therefore (usually) is the more obvious effect.
As for orbital speed: At any given height, there is one and only one speed which will result in a circular orbit, but orbits don’t need to be circular. If the Shuttle were to fire its thrusters in orbit so as to increase its speed, it would go into an elliptical orbit, with the perigee (closest point) at the height it started at. This is a perfectly stable orbit, and sometimes has advantages, but it costs more fuel, and most things you can do in orbit, you can do in circular orbit, so that’s all they bother with usually.
Meanwhile, if you’re in a circular orbit and decrease your speed, you’ll also end up in an elliptical orbit, but this time with the apogee (furthest point) at your original height. Starting at the Shuttle’s typical orbit, though, there’s not much room to go further down, so you only do this when you want to land.
BZ00000,
Don’t think of the entire Earth as falling away from the shuttle. The shuttle is falling, but it follows a curved path. Just like if you stand on the earth and throw an object forward, it will curve downward toward the ground. Throw it forward faster, and it will go further before hitting the ground. Throw it REALLY fast, and as it curves toward the ground, the surface of the earth (because it is a giant sphere, sort of) curves away from it. So the object continues “falling” in a curved path, but never reaches the earth because the earth’s surface is curved, too. That’s how things orbit the earth. Many physics textbooks include a sketch of this example drawn by Isaac Newton.
So let’s say there is a stationary sphere somewhere, that is not being effected by any other object. We could take another sphere put it into orbit around the 1st sphere so long a the speed where enough to keep it from crashing into the 1st object and the 2nd object would be considered to be “falling?” And the closer the orbit is the faster the 2nd object would have to move?
Since the Moon’s orbit is speeding up, could it break away from the Earth at some point in the very distant future if nothing destroys the Earth and/or Moon before then?
So what speed would I have to attain to achieve orbit one foot above sea level? Would that be considered orbit?if I first rid us of the moon so tidal variances could be negated, and dig a large trench through those pesky continents that get in the way
Yes, BZ00000, that’s about it.
Last I heard, the moon’s orbit was slowing down due to the effect of the tides (cite) - and therefore the orbit was getting bigger for the reason Chronos stated. As to whether it could eventually break away… well, it’ll take a long time 
The gravity in low-earth orbit is only lesser by about a hundredth, therefore, the speed required to maintain a foot-above-sea-level orbit would be greater by only about a tenth. The figure itself would be about 20,900 mph, assuming an orbital velocity of 19,000 mph. But you’d trigger devestating sonic booms that would be better described as a continuous blast wave and burn your ship to a cinder in the friction of the atmosphere, so only try this after using Mega-Maid to suck the air from the planet :).
And yes, it would be considered an orbit if there were no atmosphere. If there were, you’d need a power source to maintain your circle around the Earth, and that isn’t free-fall; it’s called flying really fast and waking the neighbors.
Okay, let’s ignore air resistance, ignore perturbations by all third bodies, ignore the oblateness of the Earth, and ignore topography (mountains &c), and just calculate a simple Newtonian circular orbit.
The acceleration due to gravity of an object effectively at sea level on Earth is 9.8 metres per second per second.
The centripetal acceleration of an object following a circular path is v^2/r, where v is the velocity of the object, and r is the radius of the path. Now, in a orbit gravity provides the centripetal acceleration, so we know that
v^2/r = 9.8 ms^-2
therefore
v^2 = 9.8 ms^-2 * r
v = SQRT (9.8 ms^-2 r)
A reference book tells us that the radius of the Earth is approximately 6 * 10^6 m. Substituting, we get
v = SQRT (9.8 * 6 * 10^6 m^2s^-2)
and the calculator reports that this gives
v = 7.7 * 10^3 ms^-1
ie. 7.7 kilometres per second (4.8 miles per second). It is no coincidence that this answer is equal to escape velocity at sea level divided by the square root of two.
Regards,
Agback
The moon’s distance from the earth is increasing. Friction between the earth surface and the tidal bulge of the ocean tends to make the bulge rotate with the earth. However, once the bulge advances slightly ahead of a line from earth to moon, the gravitationial attraction between the bulge and the moon stops the rotation of the bulge which then stands still relative to the moon. This gravitational force pulls the moon faster in its orbit so that for its current distance it is now going too fast so the moon recedes slightly from the earth and slows down. In other words the pull of the earth’s tidal bulge increases the kinetic energy of the moon slightly, while decreasing that of the earth. But this increased kinetic energy is converted to potential energy by moving the moon slightly further away against the pull of the earth’s gravity. And this change from kinetic to potential energy occurs because there is a fixed ratio between the orbital velocity and the distance of a satellite from its primary.
Here’s an applet someone wrote that allows you to put satellites into the Earth-Moon system (sun is not involved).
http://www.colorado.edu/physics/2000/applets/satellites.html
To add one, just click in the black area and move the mouse a bit before releasing the mouse button. There doesn’t seem to be a limit on how many satellites you can have at once. Either that, or I get tired of adding them before I reach the limit.
Anyway, see if you can get one to orbit in a figure 8. It won’t be stable, but if you’re lucky it will go around a few times.
Actually, I doubt that the atmosphere would be that much of a hinderance for the Shuttle. I mean, think about it: It is able to smack into the atmosphere at 17,000 mph and slow down due to friction, so I think starting from 0 mph and simply falling through it, it would reach terminal velocity before boiling away to nothing.
Some back-of-the-envelope calculations: I get an acceleration due to gravity to be about 8.8 m/s^2 at an altitude of 350 km. Falling from that height, it would hit the ground at 2480 m/s (Mach 7.5) if there were no atmosphere. Considering that the Shuttle routinely slams into the atmosphere at Mach 24, and also considering that there is a substantial amount of atmosphere to slow it, I don’t think the Shuttle dropping from free fall in LEO would burn up.
Thoughts, anyone?
-b
Well, since you didn’t specify anything like a stable orbit, it’s easy. Just jump. You will go into orbit for a very short time.
Anyone remember the Nike Ads with the professor describing the physics of Michael Jordan as attaining a low earth orbit? Hilarious, and true.
BZ00000
OK, imagine that you’ve throw a baseball in a nice curving arc. Since you didn’t throw it straight up, it goes both up and out. As it travels horizontally, it gains altitude, levels off, then starts to descend, finally thumping to the ground over yonder in center field somewhere.
Now let’s try it from a tiny little perfectly spherical asteroid, one that’s, say, 5000 yards in diameter. Ignoring the fact that in this tiny gravity you could fling the baseball completely OUT of orbit, you very carefully throw the baseball with just enough force to achieve the exact same arc as the one you threw on earth. As the baseball travels horizontally, the surface of the tiny little asteroid below it isn’t sitting below it like a flat plane, like the earth did–instead, because it is so small, as the baseball continues on its trajectory towards “center field”, the curvature of the asteroid means that the surface of the asteroid is curving AWAY from the path of the baseball. Center field on the asteroid is as far around the globe of the asteroid as some distant continent is to you on earth. Anyway, the baseball stops its vertical climb and starts descending towards the surface, but as it continues its horizontal trajectory the curve of the asteroid below it keeps curving away. By the time the baseball thumps down in the asteroid’s version of center field, it has gone an appreciable way around the asteroid, and the curve of its arc, instead of looking like a parenthesis-- ( --turned on its side, looks more like the letter C with the asteroid between its “tongs”.
So…what’s going to happen if you fling it again, a bit harder?
Well, if you fling it too hard, it’s going to fly out into space. But somewhere between that and your “letter C” trajectory, you end up with a letter O trajectory, and the baseball never lands.
That’s orbit.
I’m with you bryanmcc. I don’t think that this hypothetical fall from “orbit” would be much of a problem for the shuttle. Your calculation seems like a pretty good upper bound on speed, and it’s clearly far less than the Shuttle handles routinely. I’d even go on to speculate that the Shuttle would become maneuverable and able to make a fairly normal landing (with advance planning, that is).
All that re-entry heat trouble comes from all that speed, and that’s what we’ve wished away here.
It does every time it comes home and lands safely, don’t it?