eli5: I’ve heard orbit described as continuously falling past or missing the Earth, how then do objects in geosynchronous orbit above a single point not fall out of the sky?
In: 2204
I just want to add that the geosynchronous orbit that stays in one point is the geostationary orbit, a specific kind that is circular and matches the equatorial plane.
But there are other kinds of geosynchronous orbits that don’t stay in one point, instead they do a figure eight with a loop on every hemisphere and the crossing point on the equator. The figure stays in place though that’s why it’s still geosynchronous.
This is because orbits can’t stay at a fixed latitude (other than the equator), they have to cross the equatorial plane periodically.
I just want to add that the geosynchronous orbit that stays in one point is the geostationary orbit, a specific kind that is circular and matches the equatorial plane.
But there are other kinds of geosynchronous orbits that don’t stay in one point, instead they do a figure eight with a loop on every hemisphere and the crossing point on the equator. The figure stays in place though that’s why it’s still geosynchronous.
This is because orbits can’t stay at a fixed latitude (other than the equator), they have to cross the equatorial plane periodically.
Think of a giant frozen lake. There’s a post in the middle with a rope tied to it.
You’re on the edge of the lake, on ice skates, holding the rope.
Try to haul yourself to the post, it’s easy: just pull on the rope.
But now imagine you’re skating hell-for-leather at right-angles to the direction of the rope.
Try to haul yourself in now, and it won’t work, all you’ll do is swing around. No matter how hard you haul on the thing, you just can’t reach the post – *your turning circle is too big to let you*.
You physically cannot by any means reach the centre of the lake without slowing down. Especially as you’re not really on ice-skates, but big blocks of wet ice that give you no traction whatsoever.
That’s orbit. That’s all it is: a big turning circle, and no way to dump your speed. You can’t just suddenly pull a 90-degree turn, so you can’t ever hit the thing you’re orbiting.
Now for geostationary orbit:
Imagine that on top of the post, there’s a carousel. One of those stately fairground ones with the horsies, which takes like an hour to go round.
Is it possible to swing around the post, but stay lined up with one particular horse?
Sure it is – if you’ve got a really huge lake and a really long rope.
If you’re skating far enough out that it takes you an hour to do a full circuit – exactly the same time as it takes one of the horsies to go round – then you’ll stay lined up. From the horsie’s perspective, you’re not moving at all.
And that’s geostationary. You have a big enough orbit, it takes 24 hours to go round the earth, the same as the continents on the surface. From their perspective, you’re not moving at all, just hovering in the air (even though you’re hurtling through space at terrifying speeds).
Think of a giant frozen lake. There’s a post in the middle with a rope tied to it.
You’re on the edge of the lake, on ice skates, holding the rope.
Try to haul yourself to the post, it’s easy: just pull on the rope.
But now imagine you’re skating hell-for-leather at right-angles to the direction of the rope.
Try to haul yourself in now, and it won’t work, all you’ll do is swing around. No matter how hard you haul on the thing, you just can’t reach the post – *your turning circle is too big to let you*.
You physically cannot by any means reach the centre of the lake without slowing down. Especially as you’re not really on ice-skates, but big blocks of wet ice that give you no traction whatsoever.
That’s orbit. That’s all it is: a big turning circle, and no way to dump your speed. You can’t just suddenly pull a 90-degree turn, so you can’t ever hit the thing you’re orbiting.
Now for geostationary orbit:
Imagine that on top of the post, there’s a carousel. One of those stately fairground ones with the horsies, which takes like an hour to go round.
Is it possible to swing around the post, but stay lined up with one particular horse?
Sure it is – if you’ve got a really huge lake and a really long rope.
If you’re skating far enough out that it takes you an hour to do a full circuit – exactly the same time as it takes one of the horsies to go round – then you’ll stay lined up. From the horsie’s perspective, you’re not moving at all.
And that’s geostationary. You have a big enough orbit, it takes 24 hours to go round the earth, the same as the continents on the surface. From their perspective, you’re not moving at all, just hovering in the air (even though you’re hurtling through space at terrifying speeds).
I just want to add that the geosynchronous orbit that stays in one point is the geostationary orbit, a specific kind that is circular and matches the equatorial plane.
But there are other kinds of geosynchronous orbits that don’t stay in one point, instead they do a figure eight with a loop on every hemisphere and the crossing point on the equator. The figure stays in place though that’s why it’s still geosynchronous.
This is because orbits can’t stay at a fixed latitude (other than the equator), they have to cross the equatorial plane periodically.
A sustainable orbit (one that doesn’t collide with the planet) around a planet is almost completely independent of the concept of geosynchronous (specifically geostationary) orbiting.
What I mean by that is, a random planet could have very different rates of spin. Some planets could have no spin. Others can be rotating around at thousands of miles per hour. Outside of some tiny influences from things like General Relativity (that we can ignore), this spinning of the planet has no influence on the sustainable orbits around the planet.
So for a planet with no spin, there is no geostationary orbit that doesn’t cause the satellite (or whatever object) to fall into the planet. The planet doesn’t rotate and so the object can’t move in its orbit if it wants to stay geosynchronous. The only way to do this is to fall straight down into the planet and that’s not a sustainable orbit.
If the planet is spinning, then there is a chance at a geostationary orbit existing. That possibility will depend on the gravitational force of the planet and the speed of its surface. If the surface is too slow it won’t meet the escape velocity needed to overcome the gravitational pull of the planet. And so the orbit will end up intersecting the planet.
If the speed is high enough, then there will be some altitude away from the surface that a satellite can sit that will be a sustainable orbit and allow the surface and the satellite to have the same orbital period.
The [formula](https://en.wikipedia.org/wiki/Geosynchronous_orbit?wprov=sfla1) that relates all these is:
r = [ (G * M * T^2) / (4 * π^2) ]^(1/3)
Where T is the period (length of the day), r is the orbits radius, and G is the gravitational constant and M is the mass of the planet. Taking a step back from this complication, you can see that, at a high level r is positively correlated with T. So if T is 0, then r must also be 0. Which is the behavior we expected.
Separate the two movements and it makes it easier to visualize.
If you put a yardstick on a basketball that would represent the forward motion of a satellite. You can see how eventual that forward motion would mean it would shoot off somewhere into space never to be seen again. So there has to be a downward motion around to wrap around the basketball. For example two units forwards and one unit down.
In a satellite it’s the downward is what we call falling. In a geosynchronous orbit the motion forward and the motion down is the same. So on earth it appears to be in the same spot because the earth is also rotating
A sustainable orbit (one that doesn’t collide with the planet) around a planet is almost completely independent of the concept of geosynchronous (specifically geostationary) orbiting.
What I mean by that is, a random planet could have very different rates of spin. Some planets could have no spin. Others can be rotating around at thousands of miles per hour. Outside of some tiny influences from things like General Relativity (that we can ignore), this spinning of the planet has no influence on the sustainable orbits around the planet.
So for a planet with no spin, there is no geostationary orbit that doesn’t cause the satellite (or whatever object) to fall into the planet. The planet doesn’t rotate and so the object can’t move in its orbit if it wants to stay geosynchronous. The only way to do this is to fall straight down into the planet and that’s not a sustainable orbit.
If the planet is spinning, then there is a chance at a geostationary orbit existing. That possibility will depend on the gravitational force of the planet and the speed of its surface. If the surface is too slow it won’t meet the escape velocity needed to overcome the gravitational pull of the planet. And so the orbit will end up intersecting the planet.
If the speed is high enough, then there will be some altitude away from the surface that a satellite can sit that will be a sustainable orbit and allow the surface and the satellite to have the same orbital period.
The [formula](https://en.wikipedia.org/wiki/Geosynchronous_orbit?wprov=sfla1) that relates all these is:
r = [ (G * M * T^2) / (4 * π^2) ]^(1/3)
Where T is the period (length of the day), r is the orbits radius, and G is the gravitational constant and M is the mass of the planet. Taking a step back from this complication, you can see that, at a high level r is positively correlated with T. So if T is 0, then r must also be 0. Which is the behavior we expected.
A sustainable orbit (one that doesn’t collide with the planet) around a planet is almost completely independent of the concept of geosynchronous (specifically geostationary) orbiting.
What I mean by that is, a random planet could have very different rates of spin. Some planets could have no spin. Others can be rotating around at thousands of miles per hour. Outside of some tiny influences from things like General Relativity (that we can ignore), this spinning of the planet has no influence on the sustainable orbits around the planet.
So for a planet with no spin, there is no geostationary orbit that doesn’t cause the satellite (or whatever object) to fall into the planet. The planet doesn’t rotate and so the object can’t move in its orbit if it wants to stay geosynchronous. The only way to do this is to fall straight down into the planet and that’s not a sustainable orbit.
If the planet is spinning, then there is a chance at a geostationary orbit existing. That possibility will depend on the gravitational force of the planet and the speed of its surface. If the surface is too slow it won’t meet the escape velocity needed to overcome the gravitational pull of the planet. And so the orbit will end up intersecting the planet.
If the speed is high enough, then there will be some altitude away from the surface that a satellite can sit that will be a sustainable orbit and allow the surface and the satellite to have the same orbital period.
The [formula](https://en.wikipedia.org/wiki/Geosynchronous_orbit?wprov=sfla1) that relates all these is:
r = [ (G * M * T^2) / (4 * π^2) ]^(1/3)
Where T is the period (length of the day), r is the orbits radius, and G is the gravitational constant and M is the mass of the planet. Taking a step back from this complication, you can see that, at a high level r is positively correlated with T. So if T is 0, then r must also be 0. Which is the behavior we expected.
They fall at the right distance away around the earth that their velocity matches that of the spot on the earth as it rotates. Or another way of thinking about it is that the orbital period is 24 hours around the axis the earth spins. You cannot have geosynchronous orbit by orbiting the poles north south, the farther you deviate from the rotation of the earth the less synchronous it becomes.
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