Hypothetically, suppose a person had the power to teleport. This “teleportation” works by shrinking the intervening distance between their current location and their destination to 0, and they just step there.
Momentum is conserved, kinetic energy is conserved, etc.
They choose to teleport to the altitude of a geosynchronous orbit next to a communications satellite located there(22, 236 miles). They make sure to don a space suit first.
Will the person remain at rest relative to the satellite, or will they start rapidly falling?
Alright, that makes sense. What was throwing me off was that if you draw a line between your location and the ground after teleporting, the ground would be completely stationary relative to this line if you were in geosynchronous orbit.
Hence, just “appearing” in geosync should work.
This is a real situation in a video game that realistically simulates orbits, where you use an in game hack to actually teleport.
So he doesn’t get shot there, like an instaneous transportation system could impart a velocity… that would contribute toward achieving orbit.
The only speed he gains in this movement is from his regular single foot step.
The geosynchronous satellite has a much larger radius than earths surface, so the circumference is much further… but it traverses that larger distance in the same time (24 hours) so it must be faster than earths surface.
In fact, if the person is taken from the north pole, then the person has no orbital velocity at all … The geosynchronous satellite is doing 3 km/sec
He would still have the velocity from the rotation of the earth (that’s why you launch rockets from the equator), but that’s not enough to allow you to remain in orbit. At the equator, your velocity is about 460m/s, and this decreases as you go up in latitude, until you hit the equator, where it’s zero.
A geostationary satelite has to be above the equator, and at an altitude of 35,786 km, and have a velocity of 3.07 km/s. That’s a lot more than the 460m/s you’re going at the surface of the earth, and is not sufficient to keep you in orbit at that altitude. Or at any altitude, really.
Not enough for geostationary orbit, but will he hit the Earth, or will he be in an elliptical orbit above Earth’s atmosphere? What would his perigee be?
Good question. Velocity from the rotation of the earth isn’t enough for us to be in ‘orbit’ at sea level, so my immediate thought is ‘It definitely won’t be enough for an orbit above the atmosphere.’
I suspect the answer might be more complicated than that. Will the effective velocity from that small drift become larger by the time our guy has fallen to LEO?
If I looked up the right numbers from wikipedia, the earth’s surface velocity at the equator is 465.1 m/s. In order to maintain a geosynchronous orbit, you need to have an altitude of 35,786 km and a velocity of 3.07 km/s. You’d only have about 1/6th the velocity you need to maintain orbit, so you’d drop back to earth.
This of course assumes that you are standing on the equator and you teleport yourself to a spot directly overhead. Doing it in other ways could be rather unpleasant. For example, if you chose to teleport yourself to a spot 35,786 km above the earth, but 90 degrees off of your present angle with respect to the center of the earth, you will end up racing into the atmosphere at about 465 m/s, which I suspect would get rather unpleasant as the atmosphere grows more dense during your descent. If you did the same thing at 180 degrees you would be in a retrograde orbit of sorts, but you’d still fall back to earth roughly the same as you would if you teleported yourself directly overhead. At the 270 degree mark you’d fling yourself up into space, and then you would fall back to earth. While your path would be straight up and down, it would look more like a parabolic arc to those on the surface since the surface will be rotating away under you.
Just to pick a semantic nit, you probably mean geostationary, which is a subset of geosynchronous orbits.
The former are the ones where the object stays in the same spot in the sky all the time. They’re always over the equator, and at a specific altitude.
The latter are the ones where the object is in the same spot in the sky at (at least) one time, every day. There are many variations on that theme, even including some polar orbits (orbits that pass over the poles).
If instead of geostationary orbit you teleported to a Lagrangian point would you stay in position relative to the bodies in question, or do you need the correct orbital velocity first?
Although the OP stated that momentum is conserved, any teleportation device would need to have a way to compensate for the relative motions of the two locations. It does us little good to transport to Europa if we are just going to smack into the surface or get left floating behind once we get there. Even moving form one spot on earth to another is going to require a change in momentum.
It would then seem that this device should be capable of placing a person in orbit, with the velocity necessary to stay there.
Again, you would drop towards Earth, but it’s not clear if you would impact or not. I tried looking for an online calculator like the one you linked to in that other thread, but which handles elliptical orbits, but couldn’t find any.
Ultimately, any teleportation device is already outside the bounds of known physics, and so any question like this is going to depend on the precise nature of the new physics needed to explain it.
It isn’t a device. The person is able to just hack the universe’s variables and set a position update. This example comes from a video game where you can in fact do exactly this.
In other words, “any teleportation device would need to have a way to compensate for the relative motions of the two locations,” just isn’t true. This proposed mechanism for teleportation does not compensate for the relative motions. So while “It does us little good to transport to Europa if we are just going to smack into the surface or get left floating behind once we get there. Even moving form one spot on earth to another is going to require a change in momentum,” is true, the proposed mechanism for teleportation fails to account for it.
Unless your proposed teleportation coordinates allow you to enter a trajectory as well as a position.
I calculate a minimum distance from the center of the Earth of about 480 km (if the surface of the Earth were not inconveniently in the way), and a maximum velocity of 40 km/s.
Are you talking about Kerbal Space Program? In that case what does actually happen in KSP when you teleport a Kerbin to GeoSynch height? Does he plummet or stay in orbit?
