Orbit = Mind blown

[brentee]

Holy crap!!
I asked what I thought was a simple question with a simple answer.
I was thinking along the lines of Henning's answer, but noooooo.
Now I have some reading to do
Physics is fun, though

 

[wanttaja]

Or if you are in a Fly Baby, the plane is motionless with respect to a car on a city street in a school zone.

Ron Wanttaja

 

[Dan Thomas]

Ron Wanttaja

Kinda cold for the open cockpit, ain't it?

Dan

 

[wanttaja]

Kinda cold for the open cockpit, ain't it?

Dan

If necessary, I just crank up the heater a bit more.

Besides, that's an OWT. Space is vacuum, so it can't have a temperature.

Rom Wanttaja

 

[Matthew]

Class E space. Did you get flight following?

 

[MAKG1]

If necessary, I just crank up the heater a bit more.

Besides, that's an OWT. Space is vacuum, so it can't have a temperature.

Rom Wanttaja

Nope. Not a vacuum. Just really rarefied.

The temperatures can get very high, but there is very little energy involved due to the miniscule density.

 

[FormerHangie]

If you want to experience this "weightlessness", these folks can set you up: http://www.gozerog.com/ .

 

[Captain]

If you want to experience this "weightlessness", these folks can set you up: http://www.gozerog.com/ .

Hey! I used to fly those planes. B727's operated by AmeriJet. Pretty crap operator IMO, I'd hold out for Virgin Galactic.

 

[JeffDG]

OK, want your mind further blown.

The vast majority of the energy that spacecraft use to get to orbit is not in the "up" direction, it's in the "Go east" direction.

 

[EdFred]

For some more orbit fun, suppose you threw a killer fastball off the international space station, for the sake of argument parallel to Earth's surface. Where would it go?

That would depend on which direction the ball was thrown. Since parallel in relation to an oblate spheroid could mean a few things.

 

[Bill]

For some more orbit fun, suppose you threw a killer fastball off the international space station, for the sake of argument parallel to Earth's surface. Where would it go?

Which direction am I throwing it?

In the direction of orbit I'd expect it to climb to a higher orbit. Away from the direction of the orbit, I'd expect it to fall to a lower orbit. Sideways, I'd expect it to remain at the same orbit height as the station.

Of course I could be wrong!

 

[Bill]

Which direction am I throwing it?

In the direction of orbit I'd expect it to climb to a higher orbit. Away from the direction of the orbit, I'd expect it to fall to a lower orbit. Sideways, I'd expect it to remain at the same orbit height as the station.

Of course I could be wrong!

Although now that I think about it more:

Guess 1a: It would climb to a higher altitude no matter the direction because you've added more energy to the system (baseball) and the only result can be a higher orbit.

Of course I could be wrong!

 

[EdFred]

Which direction am I throw it?

In the direction of orbit I'd expect it to climb to a higher orbit. Away from the direction of the orbit, I'd expect it to fall to a lower orbit. Sideways, I'd expect it to remain at the same orbit height as the station.

Of course I could be wrong!

If you threw it prograde (forwards), the orbit would no longer be circular, and where you threw it from (or very close to it) would be the perigee of the orbit, and the apogee of the orbit would be 180º from where you threw it. If retrograde, the opposite would happen. If you threw it "sideways" the inclination would change and unless you countered it by also throwing slightly "inward" or "outward" the perigee/apogee would also change slightly.

 

[SixPapaCharlie]

It would fall back down to earth unless you threw it how ever many 1000s of miles per hours needed to keep it in orbit right?

 

[Henning]

Which direction am I throwing it?

In the direction of orbit I'd expect it to climb to a higher orbit. Away from the direction of the orbit, I'd expect it to fall to a lower orbit. Sideways, I'd expect it to remain at the same orbit height as the station.

Of course I could be wrong!

I think 'sideways' would still make it climb since you are still adding velocity while changing the vector a bit.

 

[JeffDG]

For some more orbit fun, suppose you threw a killer fastball off the international space station, for the sake of argument parallel to Earth's surface. Where would it go?

It would enter an elliptical orbit. Depending on the direction of the throw, it would have either a lower perigee, or a higher apogee, Assuming it started from a circular orbit, however, it would return to the altitude thrown from on the next orbit.

 

[JeffDG]

Guess 1a: It would climb to a higher altitude no matter the direction because you've added more energy to the system (baseball) and the only result can be a higher orbit.

If your thinking was correct, then spacecraft would not do retro-burns to get out of orbit, because those burns would "add energy to the system" and result in higher orbits.

 

[JeffDG]

Folks need to spend more time with the KSP:
https://kerbalspaceprogram.com/

 

[EdFred]

I must be on ignore.

 

[RotorDude]

For some more orbit fun, suppose you threw a killer fastball off the international space station, for the sake of argument parallel to Earth's surface. Where would it go?

If "it" is the ISS itself, it could actually de-orbit and fall back to earth, if your fastball were fast enough pro-grade.

 

[Bill]

I must be on ignore.

5x5 here

 

[Captain]

It would fall back down to earth unless you threw it how ever many 1000s of miles per hours needed to keep it in orbit right?

Sorta right, but not really.

The orbit would change slightly but it would stay in orbit. Remember the premise is you started at the ISS so you're already 'in orbit'.

However, technically nothing stays in orbit FOREVER so eventually it would fall back to earth whether or not you throw it. Even the moon is slipping out of orbit, albeit in the other direction...it's getting further and further away. Eventually it'll be gone.

 

[Capt. Geoffrey Thorpe]

East takes you out, out takes you west, west takes you in, in takes you east.

http://en.wikipedia.org/wiki/The_Integral_Trees

 

[Seanaldinho]

It would fall back down to earth unless you threw it how ever many 1000s of miles per hours needed to keep it in orbit right?

The ball is already moving at those thousands of mph. As is the ISS, as is the person throwing the ball. However, the balls relative speed to the ISS and pitcher is zero until it is thrown. After it was thrown it would be moving around 90-100 mph faster than the ISS and thus have a slightly more elliptical orbit.

I think.

 

[MAKG1]

I must be on ignore.

Nope, not on ignore.

I just had other things to do for several days.

The answer was close. It's actually even simpler. Regardless of the direction thrown, the ball will enter a slightly elliptical orbit that intersects the ISS at the thrown point. After one orbit period (about 90 minutes on the ISS), it will hit the thrower in the back of the head at the same velocity it was thrown with. If you remember to duck 90 minutes later, you had better do it again after 180 minutes….

The perturbation solution does not really require any particular direction. You can throw straight down if you want, and then get hit from above 90 minutes later. That was a red herring. The solution is most obvious if you throw straight up, but it works in any direction. Note you don't get hit as it passes 45 minutes later because you've thrown it into a higher orbit and it will pass behind you. The lag goes away on the second half of the orbit.

Orbital mechanics can be fun. Lots of counterintuitive ideas.

Like, if you want to catch up to something ahead of you in the same orbit you slow yourself down (i.e., drop to a lower, faster orbit).

 

[EdFred]

Nope, not on ignore.

I just had other things to do for several days.

The answer was close. It's actually even simpler. Regardless of the direction thrown, the ball will enter a slightly elliptical orbit that intersects the ISS at the thrown point. After one orbit period (about 90 minutes on the ISS), it will hit the thrower in the back of the head at the same velocity it was thrown with. If you remember to duck 90 minutes later, you had better do it again after 180 minutes….

The perturbation solution does not really require any particular direction. You can throw straight down if you want. That was a red herring.

I was replying to the rest of the people who said almost the same thing I did.

Are you sure the periods will be exactly the same? I would expect the period to be slightly longer if throwing the ball prograde, and the only point that the orbits will have the same altitude is at the ball's perigee. So the next time around the ISS will arrive to that point prior to the ball since the period of the ball is now longer. Each time around the ISS will get slightly further ahead of the ball's orbit, and eventually there will be a time that they will both be at the exact same point. But I wouldn't expect them to share the same period.

Any dV given to the ball should change its orbital period slightly. Now if I threw the ball without giving any prograde, retrograde, or radial dV then yes, every orbit we would intersect twice. Once at the ascending node, and once at the descending node.

 

[MAKG1]

OK, want your mind further blown.

The vast majority of the energy that spacecraft use to get to orbit is not in the "up" direction, it's in the "Go east" direction.

Excellent point.

And now these guys can understand why the European Space Agency's launch complex is not in Europe. It's in French Guiana.

AND why Kennedy Space Center must be in Florida if it is to be in CONUS. Puerto Rico may be a better site from an orbital dynamics perspective, but it has some significant logistics problems.

 

[MAKG1]

I was replying to the rest of the people who said almost the same thing I did.

Are you sure the periods will be exactly the same? I would expect the period to be slightly longer if throwing the ball prograde, and the only point that the orbits will have the same altitude is at the ball's perigee. So the next time around the ISS will arrive to that point prior to the ball since the period of the ball is now longer. Each time around the ISS will get slightly further ahead of the ball's orbit, and eventually there will be a time that they will both be at the exact same point. But I wouldn't expect them to share the same period.

Any dV given to the ball should change its orbital period slightly. Now if I threw the ball without giving any prograde, retrograde, or radial dV then yes, every orbit we would intersect twice. Once at the ascending node, and once at the descending node.

Good thinking, but whatever gain you get from throwing prograde is lost due to the higher orbit. In the perturbation solution, it comes back to exactly the same point. None of us can throw fast enough that the perturbation solution doesn't work at that orbit, at least for the few-hours periods we're talking about.

The orbit doesn't intersect twice. You can see this with a prograde throw -- the baseball is tangent to the ISS at the point thrown. From an inertial point of view, it forms an ellipse tangent to the ISS at the point thrown, and is at a higher orbit throughout until it comes back to the ISS after one full orbit. From the ISS point of view, the baseball always follows a 2:1 ellipse.

 

[EdFred]

Good thinking, but whatever gain you get from throwing prograde is lost due to the higher orbit. In the perturbation solution, it comes back to exactly the same point. None of us can throw fast enough that the perturbation solution doesn't work at that orbit, at least for the few-hours periods we're talking about.

The orbit doesn't intersect twice. You can see this with a prograde throw -- the baseball is tangent to the ISS at the point thrown. From an inertial point of view, it forms an ellipse tangent to the ISS at the point thrown, and is at a higher orbit throughout until it comes back to the ISS after one full orbit. From the ISS point of view, the baseball always follows a 2:1 ellipse.

It does intersect twice if I throw 100% normal or anti-normal as I've changed the inclination of the orbit. But if I do throw the ball prograde, (or retrograde) I have changed the semi-major axis slightly, meaning I have changed the period slightly. I shouldn't have to duck. It should allllllmost hit me in the back of the head (prograde), but not quite. If the calculation works for a 90mph throw, then it should also work the same for a 900mph throw, or 9000mph throw. Though at 9000 that might reach escape velocity.

 

[Henning]

Excellent point.

And now these guys can understand why the European Space Agency's launch complex is not in Europe. It's in French Guiana.

AND why Kennedy Space Center must be in Florida if it is to be in CONUS. Puerto Rico may be a better site from an orbital dynamics perspective, but it has some significant logistics problems.

PR logistics are simple, everything needed infrastructure and equipment wise on the logistics already exists.

 

[MAKG1]

PR logistics are simple, everything needed infrastructure and equipment wise on the logistics already exists.

That may be, but it has no powerful congress critters.

Mustn't forget why Johnson Space Center is in Texas….there is no good technical reason to put it there, and a number of logistical and financial reasons to co-locate it with the launch center.

 

[Henning]

That may be, but it has no powerful congress critters.

Mustn't forget why Johnson Space Center is in Texas….there is no good technical reason to put it there, and a number of logistical and financial reasons to co-locate it with the launch center.

LOL, yeah, but that's a political problem, not a logistical one.

I was thinking about inclination and space elevators. Would they need to be at the equator? I realize the extra atmosphere issues of the incline, but what would be the orbital issues?

 

[MAKG1]

It does intersect twice if I throw 100% normal or anti-normal as I've changed the inclination of the orbit. But if I do throw the ball prograde, (or retrograde) I have changed the semi-major axis slightly, meaning I have changed the period slightly. I shouldn't have to duck. It should allllllmost hit me in the back of the head (prograde), but not quite. If the calculation works for a 90mph throw, then it should also work the same for a 900mph throw, or 9000mph throw. Though at 9000 that might reach escape velocity.

Your head isn't that small. If you move the intersection point a few mm, it's still a hit.

No, perturbation solutions do not work for arbitrary parameters. They work as long as the resulting orbit is close to the original. Now, if you can really throw at 9000 MPH, we might have to revisit that assumption.

If you throw "100% normal," you've changed the inclination, but that's not all. The orbit is no longer circular.

 

[MAKG1]

LOL, yeah, but that's a political problem, not a logistical one.

I was thinking about inclination and space elevators. Would they need to be at the equator? I realize the extra atmosphere issues of the incline, but what would be the orbital issues?

Well, if there were an actual material you could make space elevators out of, the orbital connection would need to be over the equator, but the other end could be anywhere in sight of it.

Space elevators have the unfortunate problem that the material properties needed are not even close to feasible.

Aside from the materials problems, any cable would be the equivalent of one of those 2000 foot TV towers located at an airport. It would be vulnerable to collisions from lower orbits, and would be one hell of an obstacle.

 

[EdFred]

Your head isn't that small. If you move the intersection point a few mm, it's still a hit.

No, perturbation solutions do not work for arbitrary parameters. They work as long as the resulting orbit is close to the original. Now, if you can really throw at 9000 MPH, we might have to revisit that assumption.

If you throw "100% normal," you've changed the inclination, but that's not all. The orbit is no longer circular.

You're so ****ing pedantic. If I change the inclination and adjust my -grade and radial dV so the ball has the same period (as to appear that the only change was 100% normal/anti normal) as the previous circular orbit it will intersect twice. Once at the AN and once at the DN.

Is that such hard concept to grasp?

 

[Henning]

Well, if there were an actual material you could make space elevators out of, the orbital connection would need to be over the equator, but the other end could be anywhere in sight of it.

Space elevators have the unfortunate problem that the material properties needed are not even close to feasible.

Aside from the materials problems, any cable would be the equivalent of one of those 2000 foot TV towers located at an airport. It would be vulnerable to collisions from lower orbits, and would be one hell of an obstacle.

Yeah, I was just thinking about the dynamics involved.

 

[MAKG1]

You're so ****ing pedantic. If I change the inclination and adjust my -grade and radial dV so the ball has the same period (as to appear that the only change was 100% normal/anti normal) as the previous circular orbit it will intersect twice. Once at the AN and once at the DN.

Is that such hard concept to grasp?

It's not pedantry.

The only solution that doesn't make a noncircular orbit is dV=0. It's not possible to change one circular orbit into another with one impulse.

This is fundamental, and it's not easy to grasp, so please leave the insults at home.

You're trying to apply point kinematics to orbits. It doesn't work (well, at least not in the same way). This is no longer plane geometry, and you're assuming a symmetry that isn't there.

 

[EdFred]

It's not pedantry.

The only solution that doesn't make a noncircular orbit is dV=0. It's not possible to change one circular orbit into another with one impulse.

This is fundamental, and it's not easy to grasp, so please leave the insults at home.

I'm pretty sure it is. You add normal/anti-normal, while making a tweak to radial and -grade vectors at the same time and you have changed your inclination without de-circularizing your orbit. So, you still have the same semi major axis. Wait a second. The same semi-major axis? That means the periods are the same. That means the orbits other than the inclination are the same size! What witchcraft is this?!?!

Oh yeah, and the orbits intersect twice, at the ascending and descending node.

Granted, it is easier to do it with 2-3 adjustments, but it can be done in one.