Greetings, fellow earthlings.
My apolgies for what is about to be a long post, but I have been curious about
a number of things for a while, and have not gotten any answers at any of the
astronomy sites I have been to:
I know that an object in geosynchronous orbit effectively "hovers" over a specific
point on earth, due to it actually orbiting the planet at the same relative speed as the
earth is revolving.
Yet the object does not actually "fall". So how high up does an object need to be
before it can do this without "falling"?
Also, what would be the minimum speed an object can be moving before it falls,
probably dependant on its altitude.
Also, I have heard that the force of the earth's gravity does not actually diminish very much
even at an altitude of several hundred miles. So, theoretically, if you can "hold" a spacecraft motionless
over the earth with no orbiting movement at all, "holding" it so it does not fall down, if you were standing
inside it, would you feel normal earth gravity? Would it just be a little less?
I know the feeling of no gravity aboard a spacecraft is actually just an illusion caused by centrifugal force
putting you into freefall around the planet.
Hopefully someone with more physics knowledge than myself will stumble upon this and answer.
Thanks! Joe Cap
Question about gravitational field

 Ensign
 Posts: 99
 Joined: Mon Jul 26, 2004 8:55 pm
 Location: Michigan Tech
Well, a geosynchronous object is "falling" just as any other object in orbit is "falling." Thus, an obverver inside would feel microgravity just as they do on the spaceshuttle (microgravity being essentially zero gravity).
As for the altitude, it depends on several things.
The main consideration to take: The object needs to be high enough up that friction from the atmosphere is negligible. Thus, it is quite possible that one certain other planets geosychronous orbit would be impossible if the atmosphere extended outward far enough. (Yes, the atmosphere is rotating with the planet, but there will still be frictional effects that will eventually slow the object down enough that it will lose orbit)
Once you're out of any significant atmosphere, you simply need to equate the tangent of the rotational speed of the planet with the minimum transversal speed you need to travel so to stay in a circular orbit according to the planet's gravitational attraction.
As for the altitude, it depends on several things.
The main consideration to take: The object needs to be high enough up that friction from the atmosphere is negligible. Thus, it is quite possible that one certain other planets geosychronous orbit would be impossible if the atmosphere extended outward far enough. (Yes, the atmosphere is rotating with the planet, but there will still be frictional effects that will eventually slow the object down enough that it will lose orbit)
Once you're out of any significant atmosphere, you simply need to equate the tangent of the rotational speed of the planet with the minimum transversal speed you need to travel so to stay in a circular orbit according to the planet's gravitational attraction.
Dan Cordell, Giant Space Cow

 Ensign
 Posts: 37
 Joined: Wed Aug 11, 2004 4:51 pm
 Location: Longmont, CO
Re: Question about gravitational field
Joe Cap:
"Greetings, fellow earthlings."
Howdy, Joe. Very nice to hear from you.
"My apolgies for what is about to be a long post, but I have been curious about a number of things for a while, and have not gotten any answers at any of the astronomy sites I have been to:"
Personally, I never think you need to apologize for long posts. The great thing about web forum as a communication method is that your "audience" is always capable of skipping over posts they find to be not to their liking. Let *them* deal with the length of your post as they will. *I* tend to like long posts.
"I know that an object in geosynchronous orbit effectively "hovers" over a specific point on earth, due to it actually orbiting the planet at the same relative speed as the earth is revolving.
Yet the object does not actually "fall". So how high up does an object need to be before it can do this without 'falling'?"
Dr. Brian Suits PH204 lectures of Spring '91, don't fail me now!
The intuitive answer is (and you've mentioned it in your post) that it depends on how fast the object is moving (tangentially to the direction of the force of earth's gravity). The faster this tangential velocity is, the lower the object must be for gravity to keep it locked in on the earth. However, if it's to be geosynchronous, it's velocity is also perscribed to be the *same speed* as the rotation of the earth below it (well, same angular velocity, at least). So when you say "geosynchronous" (but, you know, actually spell it correctly), you're actually fixing both the velocity *and* height of the object. Think of it, for a moment as a different frame of reference Lagrangian point (and then abolosh that thought from your brain before the trained Physicists come hunt you down!).
So how do we determine this velocity and height? Sir Isaac, this is your cue!
F=ma (I'm a mechanical engineer, so I'm forced to start all of my analyses this way).
Ft (tangential force, the component of our force causing acceleration required for geosynchronous orbit) = R*alpha (the radius from the point mass center of earth's gravity multiplied with the angular acceleration {the rate at which the object speeds up or slows down})
If you followed all that, then you should guess that for our geosynchronous orbit the "alpha" must be zero (the object doesn't really change speed as it "hovers" over a point on earth), which means the tangential component of the force is zero! Woo!
Fc (centripedal force, the component of the force that causes the object to constantly change direction, i.e., "fall" towards earth at just the correct rate that it keeps falling downward at just the right speed that as it hurtles forward it never really hits the earth) = m*v^2/R (the object's mass times it's forward velocity squared divided by the radius at which it's orbiting).
But what *causes* this m*v^2/R force? Sir Isaac again comes to our rescue with ye olde Law of Universal Gravitation, stating
F=G*m1*m2/R^2 (or in intuitive terms, the force due to gravity is proportional to the masses of the objects in question and inversely proportional the the square of the distance between their centers of gravity).
So we'll call m2 Earth, and replace it with Me (well, not *me*  I've been losing weight you see...), and we already refered to the mass of the object as m, so m1=m.
So the duelling Newtonian determinations of the force on the object can be set on separate sides of the equation:
m*v^2/R = G*m*Me/R^2
And calling in for algebraic backup, we can simplify things to:
v = sqrt(G*Me/R)
G is a constant (thank you Mr. Cavendish for figuring out this value!), and Me is something that we can look up in a Pysics book. So what we've got is a generic equation relation between v and R. What we haven't yet used in our equation is the fact that we're specifically after geosynchronous motion.
Or the v=R*omega (velocity at some radius is equal to the radius times the rotational speed).
omega is one we can figure out ourselves, as it turns out the earth rotates once every... (wait for it)... 23 hours and 56 minutes! Okay, just kidding. Once every 24 hours. In terms friendly to calculations that means one revolution (2*pi radians) per (24 hours/day * 60 minutes/hour * 60 seconds/minute) 86400 seconds.
So now we've got two equations with two unknowns which we can solve a couple of different ways. I'm not going to go through the bother of describing the actual mechanics (the process is to set one equation such that one unknown is isolated and then plug that equation into the other), but what you end up with is
v = sqrt(G*Me*omega/v)
v^2 = G*Me*omega/v
v^3 = G*Me*omega
v = [(6.67 x 10^11 N*m^2/kg^2) * (5.98 x 10^24 kg) * (2*pi/86400 s)]^1/3
v = 3070 m/s = 6870 mph
we can then put that into either of our equations to determine R:
R = v/omega = 42.3 x 10^6 m (the distance from the center of the earth) or, subtracting off the radius of the earth (say at the equator),
d = R  Re = 35.8 x 10^6 m = 35800 km or 22200 miles above the equator.
"Also, what would be the minimum speed an object can be moving before it falls, probably dependant on its altitude."
Indeed! In our calculation above, we showed that for an object to stay in orbit (Kepler, eat our dust!) v = sqrt(G*Me/R). The bigger the altitude (R), the faster she's got to fly around the earth.
"Also, I have heard that the force of the earth's gravity does not actually diminish very much even at an altitude of several hundred miles."
Well... Kinda.
"So, theoretically, if you can 'hold' a spacecraft motionless over the earth with no orbiting movement at all, 'holding' it so it does not fall down, if you were standing inside it, would you feel normal earth gravity? Would it just be a little less?"
You're absolutely right when you say that something orbiting the earth is firmly feeling Earth's gravity (otherwise it'd sail off into the depths of the galaxy at large). But the cool thing about something orbiting is that it's falling (getting pulled down by gravity) just the same amount that it's hurtling forward each second. Someone along for the ride on that satellite or object is going to be doing the same thing  hurtling and falling. And they're very likely to lose their lunch in the process. In our geosynchronous example, they're falling 3 km each second (but fortunately for them, they're also hurtling forward 3 km each second, otherwise they'd hit the earth and *really* have a bad day). For them, gravity may as well not exist. Wee!
"I know the feeling of no gravity aboard a spacecraft is actually just an illusion caused by centrifugal force putting you into freefall around the planet."
Physics guys aren't going to like seeing "centrifugal force" too much (and we engineers are really going to tug at our hairs). For whatever reason, technical minded folks like to think of the inward force that causes an orbiting object to change direction as "centripedal force." Depending on which object you're talking about forces acting on, "centrifugal force" can be an appropriate way to describe things. But that's really neither here nor there at the moment, as what you're talking about is the illusion of being weightless despite the fact that gravity is tugging away on you. And you're absolutely correct!
"Hopefully someone with more physics knowledge than myself will stumble upon this and answer."
Include "not wanting to start working after a 3day weekend and feeling like stretching out the old greymatter muscles" to your list of qualifications, and you've nailed me to a T!
"Thanks! Joe Cap"
My pleasure,
~The Meal
"Greetings, fellow earthlings."
Howdy, Joe. Very nice to hear from you.
"My apolgies for what is about to be a long post, but I have been curious about a number of things for a while, and have not gotten any answers at any of the astronomy sites I have been to:"
Personally, I never think you need to apologize for long posts. The great thing about web forum as a communication method is that your "audience" is always capable of skipping over posts they find to be not to their liking. Let *them* deal with the length of your post as they will. *I* tend to like long posts.
"I know that an object in geosynchronous orbit effectively "hovers" over a specific point on earth, due to it actually orbiting the planet at the same relative speed as the earth is revolving.
Yet the object does not actually "fall". So how high up does an object need to be before it can do this without 'falling'?"
Dr. Brian Suits PH204 lectures of Spring '91, don't fail me now!
The intuitive answer is (and you've mentioned it in your post) that it depends on how fast the object is moving (tangentially to the direction of the force of earth's gravity). The faster this tangential velocity is, the lower the object must be for gravity to keep it locked in on the earth. However, if it's to be geosynchronous, it's velocity is also perscribed to be the *same speed* as the rotation of the earth below it (well, same angular velocity, at least). So when you say "geosynchronous" (but, you know, actually spell it correctly), you're actually fixing both the velocity *and* height of the object. Think of it, for a moment as a different frame of reference Lagrangian point (and then abolosh that thought from your brain before the trained Physicists come hunt you down!).
So how do we determine this velocity and height? Sir Isaac, this is your cue!
F=ma (I'm a mechanical engineer, so I'm forced to start all of my analyses this way).
Ft (tangential force, the component of our force causing acceleration required for geosynchronous orbit) = R*alpha (the radius from the point mass center of earth's gravity multiplied with the angular acceleration {the rate at which the object speeds up or slows down})
If you followed all that, then you should guess that for our geosynchronous orbit the "alpha" must be zero (the object doesn't really change speed as it "hovers" over a point on earth), which means the tangential component of the force is zero! Woo!
Fc (centripedal force, the component of the force that causes the object to constantly change direction, i.e., "fall" towards earth at just the correct rate that it keeps falling downward at just the right speed that as it hurtles forward it never really hits the earth) = m*v^2/R (the object's mass times it's forward velocity squared divided by the radius at which it's orbiting).
But what *causes* this m*v^2/R force? Sir Isaac again comes to our rescue with ye olde Law of Universal Gravitation, stating
F=G*m1*m2/R^2 (or in intuitive terms, the force due to gravity is proportional to the masses of the objects in question and inversely proportional the the square of the distance between their centers of gravity).
So we'll call m2 Earth, and replace it with Me (well, not *me*  I've been losing weight you see...), and we already refered to the mass of the object as m, so m1=m.
So the duelling Newtonian determinations of the force on the object can be set on separate sides of the equation:
m*v^2/R = G*m*Me/R^2
And calling in for algebraic backup, we can simplify things to:
v = sqrt(G*Me/R)
G is a constant (thank you Mr. Cavendish for figuring out this value!), and Me is something that we can look up in a Pysics book. So what we've got is a generic equation relation between v and R. What we haven't yet used in our equation is the fact that we're specifically after geosynchronous motion.
Or the v=R*omega (velocity at some radius is equal to the radius times the rotational speed).
omega is one we can figure out ourselves, as it turns out the earth rotates once every... (wait for it)... 23 hours and 56 minutes! Okay, just kidding. Once every 24 hours. In terms friendly to calculations that means one revolution (2*pi radians) per (24 hours/day * 60 minutes/hour * 60 seconds/minute) 86400 seconds.
So now we've got two equations with two unknowns which we can solve a couple of different ways. I'm not going to go through the bother of describing the actual mechanics (the process is to set one equation such that one unknown is isolated and then plug that equation into the other), but what you end up with is
v = sqrt(G*Me*omega/v)
v^2 = G*Me*omega/v
v^3 = G*Me*omega
v = [(6.67 x 10^11 N*m^2/kg^2) * (5.98 x 10^24 kg) * (2*pi/86400 s)]^1/3
v = 3070 m/s = 6870 mph
we can then put that into either of our equations to determine R:
R = v/omega = 42.3 x 10^6 m (the distance from the center of the earth) or, subtracting off the radius of the earth (say at the equator),
d = R  Re = 35.8 x 10^6 m = 35800 km or 22200 miles above the equator.
"Also, what would be the minimum speed an object can be moving before it falls, probably dependant on its altitude."
Indeed! In our calculation above, we showed that for an object to stay in orbit (Kepler, eat our dust!) v = sqrt(G*Me/R). The bigger the altitude (R), the faster she's got to fly around the earth.
"Also, I have heard that the force of the earth's gravity does not actually diminish very much even at an altitude of several hundred miles."
Well... Kinda.
"So, theoretically, if you can 'hold' a spacecraft motionless over the earth with no orbiting movement at all, 'holding' it so it does not fall down, if you were standing inside it, would you feel normal earth gravity? Would it just be a little less?"
You're absolutely right when you say that something orbiting the earth is firmly feeling Earth's gravity (otherwise it'd sail off into the depths of the galaxy at large). But the cool thing about something orbiting is that it's falling (getting pulled down by gravity) just the same amount that it's hurtling forward each second. Someone along for the ride on that satellite or object is going to be doing the same thing  hurtling and falling. And they're very likely to lose their lunch in the process. In our geosynchronous example, they're falling 3 km each second (but fortunately for them, they're also hurtling forward 3 km each second, otherwise they'd hit the earth and *really* have a bad day). For them, gravity may as well not exist. Wee!
"I know the feeling of no gravity aboard a spacecraft is actually just an illusion caused by centrifugal force putting you into freefall around the planet."
Physics guys aren't going to like seeing "centrifugal force" too much (and we engineers are really going to tug at our hairs). For whatever reason, technical minded folks like to think of the inward force that causes an orbiting object to change direction as "centripedal force." Depending on which object you're talking about forces acting on, "centrifugal force" can be an appropriate way to describe things. But that's really neither here nor there at the moment, as what you're talking about is the illusion of being weightless despite the fact that gravity is tugging away on you. And you're absolutely correct!
"Hopefully someone with more physics knowledge than myself will stumble upon this and answer."
Include "not wanting to start working after a 3day weekend and feeling like stretching out the old greymatter muscles" to your list of qualifications, and you've nailed me to a T!
"Thanks! Joe Cap"
My pleasure,
~The Meal
BSME, Michigan Tech 1995
MSME, Michigan Tech 2000
MSME, Michigan Tech 2000