Perlan 2: The glider that will slip the surly bonds of Earth – and touch the edge of space Euro airliner firm Airbus is sponsoring a glider capable of soaring to greater altitudes than the famous SR-71 Blackbird spy aircraft. The Perlan 2 project aims to get a conventional sailplane up to altitudes of 90,000 feet, surpassing the SR-71's declared service ceiling of 85,000ft. Sponsored by Airbus, the Perlan 2 glider …
if it could be used to reduce the cost of sending 'stuff' into orbit.
I suppose the limiting factor is that for a greater payload you need bigger and bigger wings and presumably at some point there just isn't enough air mass at those altitudes moving fast enough to generate the necessary lift. But then you could, perhaps, use some hybrid large-wing, low-power engine for the 'last mile'. But then maybe the extra complexity negates the initial advantages.
But then it also seasonal so perhaps overall just not worth the effort. Anyone 'know'?
Anything that you carry up there is likely to be quite heavy - it will need an engine/fuel to allow it to continue its journey. Given that the glider itself is so light, I would expect that you would be increasing the overall package's mass by an order of magnitude by adding a launchable thing to it.
The payload of this glider is two people and life support for these in addition to instruments for sampling air. So my guess is 300-400 kg. That could be enough to carry a small rocket that could reach space, but probably not enough to get anything into orbit. Using a balloon to carry a rocket to the edge of the atmosphere seems more practical.
A long cable with clever hooks at each end and a central mass could be used to pick up such a glider, and by carefully choosing when to let go route it onward. Engines still required, but less mass.
For a fictional treatment see Neal Stephenson's Seveneves.
>What if you go straight up, as slowly as you like, to geostationary orbit?
Because when you get there, the sideways velocity you inherited from the surface of the earth would be 1600kmh if you had started on the equator. The sideways velocity required to stay there is 11000kmh which - a delta-v of ~10000kmh requires a huge amount of energy to reach.
Using your proposal, when you turn your engines off when you get there, the result will be that you will fall back down again and crash somewhere else on earth as it has rotated underneath you.
@Ben 44
You're not very good at thought experiments... This should have been intuitively obvious.
Two bowling balls. Put the 1st into geostationary orbit in the usual manner (involving high speed). Carry the 2nd straight up to the exact same geostationary orbital slot by means of a really tall ladder; reach out and place the 2nd bowling ball right beside the 1st. Give the 1st bowling ball a pat (just to make sure it's real) while you're at the top of the ladder. Leave them there. Two bowling balls, both in precisely the exact same state.
Ben wrote "...when you turn your engines off when you get there, the result will be that you will fall back down..."
Einstein would ask, how does God know which bowling ball is which? How does He determine which one to bring crashing down?
Are bowling balls like holistic water? Do they have memory?
Think about it. Take as much time as you need.
Hopefully, by means of this trivial thought experiment, you'll realize that your post was completely and utterly wrong.
Getting into orbit isn't just about getting high, erm, I mean reaching a high altitude. You also need to be travelling sideways fast enough that you fall towards the Earth at the same speed it's falling away underneath you.
Orbital speed at about 200km (below about 150km there's too much atmosphere) is about 7000m/s, so to get into orbit from Perlan 2 you'd have to raise your altitude from ~30km up to ~170km and your speed from ~180m/s to ~6500m/s.
For a 1kg satellite you're going to need something like an extra 20 mega joules of energy to get into a very low orbit. Looking at some energy densities, you might be able to get that out of a rocket weighing 5-10kg, which I guess isn't out of the realm of possibility (but I'd check with an actual rocket scientist rather than going with my back-of-the-envelope guesses).
Not entirely true - can't remember the figure, but standing still on the earth you already have a certain rotational speed, and if you go high enough, that _will_ be enough to stay in orbit. Much higher than normal orbits, but then we all know that standard orbits are getting very crowded these days.
I believe you are referring to geostationary orbit. The altitude for this is VERY high.
Geostationary orbit is at 255614285.7143 Linguine, or 35786 KM for our less enlightened readers.
22235.8867 Miles for our bass ackwards American friends.
All thanks and credit/blame belong with The Reg online standards converter.
" if you go high enough, that _will_ be enough to stay in orbit"
Well, much higher is one way to put it, but basically yes. According to the first online orbital calculator I could find, approximately two million kilometers up the orbital speed happens to be almost exactly the 1600 km/h we have standing on the equator. That of course means pretending the Sun or the Moon have nothing to say about the issue (I'm sure they both do), but, basically, two million kilometers straight up you can apparently indeed stay up indefinitely...
The original post was about launching with a rocket, not a space elevator/ladder. With a rocket your initial orbital velocity is the surface velocity of the earth at the launch point. You then have to increase that orbital velocity to reach stable LEO or GEO. Launching straight up without increasing orbital velocity to GEO does NOT result in a stable orbit. The ladder in your thought experiment does this increasing by robbing angular momentum from the earth and transferring that through the ladder to increase the velocity of the bowling ball. Launching straight up without increasing velocity along the orbit only results in a stable orbit some 2million km from earth where the stable orbit velocity matches initial launch surface velocity.
Agreed wrote "Launching straight up without increasing velocity along the orbit only results in a stable orbit some 2million km from earth where the stable orbit velocity matches initial launch surface velocity."
Wrong.
You're very confused about the altitude of geostationary orbit.
They're not at "2million km".
You also missed the key point about 'bowling balls' not having memory.
This whole thing was a counter example rebuttal to the partially true claim that "orbit is sideways speed". Typically true, but not necessarily true.
Perhaps you could read up on space elevator theory to convince yourself that they're difficult enough without replacing the geostationary height number with 'two million'.
Wrong.
He's not.
You're very confused about the altitude of geostationary orbit.
You're very confused about orbital velocity.
You also missed the key point about 'bowling balls' not having memory.
You missed the key point about the second ball being accelerated sideways during its passage up the ladder. You know the angular velocity of a geostationary object - that's the bit that stays the same for the ball you're carrying up the ladder. But the velocity of any object of constant orbital velocity will necessarily increase as the radius of the orbit increases. This is basic geometry...
This whole thing was a counter example rebuttal
It really wasn't...
Perhaps you could read up on space elevator theory
Perhaps you should consider what happens when the velocity of the top of the elevator is the same as the velocity of the bottom, despite having a much larger path to traverse each rotation.
Vic.
It's as if none of you have ever heard of the Space Elevator concept, a.k.a. a tall ladder.
Relative to the Earth (why do I even need to type that?), the bowling ball on the ground isn't moving. The one in geostationary orbit isn't either. Both are zero mph relative to the Earth's surface,and the Earth itself.
The bowling ball halfway up the ladder, same thing. Zero mph relative to the Earth. Move up the ladder as slowly as you like. Orbit will be achieved. And high speed never happened.
This entire subthread was in response to: "Space is up, Orbit is sideways." Yes. But...
The counterexample being the Space Elevator concept.
Orbit can be achieved, in theory, without any high speed ever happening. Fact. Obviously.
Downvotes are just an admission of not comprehending the above. Since it's quite trivial, it indicates a gap in basic understanding on your part, or a monumental communications failure between us.
Relative to the Earth (why do I even need to type that?)
Because it's the irrelevance that makes you think you have a point. you don't. Try thinking relative to something else - e.g. the sun. You will see that *all* these objects are moving.
Both are zero mph relative to the Earth's surface,and the Earth itself.
Had you thought about taking your theory to any of the major space agencies? Because they all expend an awful lot of energy making spacecraft go sideways, and according to you, that's unnecessary. You have just revolutionised orbital mechanics. Or maybe you're just wrong. Hmmm.... let me think about that one.
NASA's opinion on the subject reckons that geostationary orbit requires a speed of 11,100 KM.hour. Perhaps you'd like to tell us all why they're wrong.
Since it's quite trivial, it indicates a gap in basic understanding on your part, or a monumental communications failure between us.
It's very simple to grasp, and I cannot help but be astounded at your failure so to do. But your evaluation of the possible causes omits one very important one: that you're utterly and completely wrong. And that being the correct assessment, you're not going to get any further on this until you go and read some factual material, rather than just making it up as you go along. You assertion that a geostationary object is not travelling sideways at great velocity is as wrong as asserting that a car doing 70mph is stationary, because there's another car alongside also doing 70mph; you appear to have forgotten that the Earth is spinning.
Vic.
So how does God tell the difference between the bowling ball that reached geostationary orbit in the traditional high speed manner, and the other bowling ball that was slowly carried straight up and gently placed into geostationary orbit, just beside the first? Two bowling balls in precisely the same state. One got there by "Orbit is (sideways) speed". The other got there by means of a really tall ladder (a.k.a space elevator). Orbit achieved without moving quickly at any point. Energy installed by slow lifting, not speed.
Any thinking person can understand the above.
The point we are all trying to make is that the ladder is NOT just increasing altitude it is increasing the velocity of the second bowling ball to match that of the first. The end result of your thought experiment is indeed 2 bowling balls in the same energy state (i.e orbit). You are confused as to the process to GET there. Lets take a closer look at the space elevator example you give. Indeed both ends are at the same angular velocity relative to the earths Center of Gravity. They are however not at the same velocity along the orbital path. To understand why lets look at another object, the rotor blades on a helicopter. The blades are perfectly straight and (apart from a lag hinge on some helicopters for the pedants out there) remain perfectly straight when spinning around the shaft. The blades attach to the hub with a short stub some distance R from the main shaft before the airfoil profile starts. Lets call this the root of the blade. The blade has a length L to the tip of the blade. As the rotor blade spins the root describes a circle with radius R. And a circumference of 2*Pi*R. The tip makes a circle with a radius of R+L and a length of 2*Pi*(R+L). Because the tip covers the same full revolution as the root to keep the blade at the same angular velocity and the blade straight that means each revolution the tip of the blade covers 2*Pi*L more distance relative to the root. Because the tip makes the revolution in the same time as the root is means it covers more distance in the same time and thus HAS to move faster. Now lets imagine R is equal to the earths radius and L is equal to GEO altitude. Now we are looking at the idealalised version of a space elevator. The same basic equations still apply and so the space end of the space elevator has to have a higher orbital velocity relative to the earths CoG than the earth attached end.
The same basic equations still apply and so the space end of the space elevator has to have a higher orbital velocity relative to the earths CoG than the earth attached end.
Watch yourself - arguing against our very own Orbital Mechanics Genius will attract downvotes for you like it did for me...
Vic.
Agreed wrote "...the ladder... ...is increasing the velocity of the second bowling ball..."
This is the root of the argument.Please review the meaning of the word "geostationary" and get back to me. If we use the Earth's surface as our reference, which is not at all uncommon in human affairs, then the bowling ball on the ladder never goes fast (relative to the Earth). But it gets to orbit.
Which was my point.
To reinforce the point, either bowling ball can come back down to the Earth's surface without any heat shield. Just come back down on the ladder, slowly as you wish.
FWIW, I do understand your points. It's just that my point is still perfectly valid.
Please review the meaning of the word "geostationary" and get back to me.
Everyone except you knows the meaning of the word "geostationary". For your edification, I shall give you a simple definition: it means the orbiting object has the same angular velocity as the Earth. This only means it is moving at the same speed if either the distance above the Earth is zero (i.e. it is on the ground) or of the angular velocity is zero (i.e. the Earth is not spinning). Neither of these situations relates in any way to this situation.
So here - I wasn't going to do this, but you appear to need a worked example. Here are my initial data:-
Your super-ladder sits with its base on the equator, and (initially) points directly upwards).
The circumference of the Earth is given by 2*pi*r = 40,000 km, near enough. It makes one full rotation per day, meaning its speed is 40,000 km/day or nearly 1700km/h. This is the speed at the bottom of your ladder.
Now let's look at the top of your ladder. The distance from the centre of rotation (the core of the Earth) is 35,786 + 6378 = 42184km. The distance the top travels in one complete rotation is given by 2*pi*r = 265,000km approx.
Now if your assertion about the speed being the same at the top as at the bottom, then the distance travelled in a day would be the same at the top as at the bottom. Thus the top of your ladder would travel a mere 40,000km a day - which is rather less than a sixth of the distance it needs to travel to make a complete rotation. It falls over very quickly; it is *not* geostationary.
Conversely, if it were geostationary, it would travel the entire 265,000 km in a day. This would leave the top of the ladder above the bottom, as required - but the speed at the top is now our 11,000 km/h as declared in the NASA figures, and is very much not the same as the speed at the bottom, as you have asserted.
So tell us - which of these do you prefer? That your "geostationary" ladder is no such thing, or that your assertion is plain wrong? Either makes you look decidedly foolish.
Your position is untenable. You have chosen to ignore the fact that NASA's factsheet declares you to be completely wrong, I notice. Now I've had to hand-hold you through basic arithmetic to prove that your theory cannot hold water. Please stop bullshitting and read up on the geometry of circles - it's really very easy and would mean you don't make such monumental cock-ups in future.
FWIW, I do understand your points. It's just that my point is still perfectly valid.
No, it isn't. Your point is wrong. Your continued assertion that it has any value demonstrated that you have no understanding whatsoever of the arguments presented here. Your continued avoidance of the discrepancy between your claim and the data offered by an agency that has put satellites into orbit speaks volumes.
Vic.
So how does God tell the difference between the bowling ball that reached geostationary orbit in the traditional high speed manner, and the other bowling ball that was slowly carried straight up and gently placed into geostationary orbit, just beside the first?
There is no need to tell any difference between the two, since they have both been accelerated to the same velocity.
Two bowling balls in precisely the same state
Yes - precisely the same state. Both have been accelerated to orbital velocity. They are both going sideways at quite a lick.
Orbit achieved without moving quickly at any point. Energy installed by slow lifting, not speed.
No. Completely, horrifically, abjectly incorrect.
Can you really not cite the formula for the circumference of a circle? That's the length of the orbital path. If the ball were doing the same speed as the ground beneath it, it would take far more time to complete that orbit than would the ground beneath - meaning it would not be geosynchronous, and the top of your ladder would be falling behind. This means the ladder falls over, along with your theory.
If, however, both the top and bottom of your ladder have the same angular velocity - which is a requirement for it to be upright, then they cannot have the same speed for any length of ladder >0. The top of your ladder is going faster than the bottom, or else it falls over. Similarly, a ball lifted up the ladder will be accelerated by that ladder too match the speed of each bit of the ladder during the ascent.
This is basic geometry. It amazes me that you continue to argue something which, as I have pointed out, NASA tells you is wrong. Now it's always possible that you are right and both NASA and I are wrong - so let's see some evidence. How many satellites have you put into stable otbit, about his planet or any other?
Any thinking person can understand the above.
Yes. It troubles me that you do not.
Vic.
Vic "Try thinking relative to something else - e.g. the Sun."
Reductio ad absurdum follows: Why the Sun? Why not the Moon? The Earth's Moon is a lot closer, and packs a greater gravitation punch on Earth (tides). Let's choose the Moon instead. Using that as your randomly chosen reference gives everything a 3700 kmh speed boost. Yay!
Thought experiment follows: Somebody stole the Sun. The Earth was left isolated. Besides being cold and dark, what about orbital mechanics changes?
Thinking - you're doing it wrong.
Why the Sun? Why not the Moon?
Because the Sun can be considered a fairly stable reference point for the system - a datum line between Earth and Sun moves about 1 per day. You can do it with the Moon if you prefer, but there's a whole lot more mathematics in that, and you seem to be having difficulty with basic multiplication at the moment.
Thinking - you're doing it wrong.
One of us is - and you're yet to explain why your theory disagrees with NASA's. You might want to start there first.
Vic.
Vic offered that geostationary orbit requires "11,100 kmh".
NASA offered that geostationary orbit requires "11,100 kmh". You say they're wrong. Can you not see why this is a poor starting position?
Or use a 'really tall ladder' (space elevator), which is not *yet* practical. But orbital mechanics theory says is possible.
No it doesn't! You're mistaking angular velocity for orbital velocity. There is a radius multiplication you've missed out, which is why you're getting the wrong answer time after time after time.
If you were to read up on this, rather than simply reposting your ignorance, you might actually learn something.
Vic.
Vic offered "...geostationary orbit requires 11,100 kmh."
Clue 1 = Geo
Clue 2 = stationary
Geostationary requires 35,786 km. How you get there is up to you.
There's a GPS stuck to bowling ball. It shows zero kmh on the ground, not 1040 kmh as your view would indicate.
As it slowly goes up the ladder, the GPS Speed is stuck on zero kmh. But the Altitude is increasing.
Eventually it reaches geostationary orbit and it never went above zero kmh.
FWIW, I understand your points, but my point remains perfectly valid.
How fast is the base of my ladder moving? It's set in concrete. 1040 kmh? Or zero kmh?
Your position and Vic's both depend on the base of my ladder, set in a concrete foundation, "moving" at 1040 kmh.
The seeming discrepancy between our statements just comes down to that. It's not the slightest bit complicated.
Which is perhaps why Vic has gone silent, and you actually have no rebuttal.
THERE IS NO POINT IN POSTING A REBUTTAL! I have plenty but I refuse to waste my time on someone who refuses to even THINK for a second! You REALLY can't understand why we say the foundation is moving (relative to a fixed reference frame located around the earths center) at 1040 km/h? You REALLY can't understand why the top of your hypothetical ladder HAS to be moving relative to that same reference frame FASTER than the foundation? You REALLY insist on rotating the entire reference frame along with the ladder around the reference frame? Because we can STILL make the same calculations, showing the same damn thing, it's just going to involve a lot of very complicated mathematics that is frankly entirely pointless because the same sums in a fixed reference frame barely cover an A5 sheet of paper.
Again, I'm done. Good luck trying to troll Vic. I'm pretty sure he is done.
Fuck, I just wrote another post in reply....
Yep. If the top of the geostationary ladder is "moving" at 11,100 kmh, then the base (which is set in concrete) is therefore "moving" at 1040 kmh. That's exactly what you're saying.
Quite a few Earthlings use WGS-84 as their "reference frame". Which what I was using.
Your points are trivial. Don't think for one second that I don't understand them.
My point is also perfectly valid. An item can achieve orbit without ever moving quickly, relative to the WGS-84 (essentially the Earth's surface) reference frame.
Anyone left here should by now see my point.
As someone much cleverer than me once said: the problem with getting into orbit is not getting high enough it's going sideways fast enough. In low earth orbit your talking about going at around 7km/s (16,000mph) that takes a lot of fuel which, necessarily, weighs a lot. This is the same reason we don't try to launch things from balloons which have already taken people over 41,000m (135k feet).
This post has been deleted by its author
Yes: I know that aviators traditionally measure altitude in feet, but that means nothing to me, especially when we are always told that the edge of space starts at 100 km (the Kármán line). The other thing that is of interest is how much air pressure is left at that height.
So:
90,000 feet is 27.4 km or 17 miles, where the air pressure is 2 kPa which is 2% of what it is at sea level.
By way of comparison: in 2012 Felix Baumgartner jumped from his balloon at 39 km (24 miles, 128,000 ft).
> Yes: I know that aviators traditionally measure altitude in feet...
"Interesting" factoid: While general (and commercial) aviation (well, the engine* carrying ones) in Germany measures height in feet, distance in nautical miles and speed in knots, glider pilots use meters, kilometers and kilometers per hour. They also use meters / second as climb rates in contrast to the feet / minute used elsewhere... you get to be quite quick in converting between the two systems.
(icon 'cause it is sort of unnecessary and confusing and...)
* we call them FNCs, fuel to noise converters
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