Odd questions

[ManifestedISO]

Well then, backstroke your way to that salon again and get a frontstroke. Do it for the audience, if not for yourself. 

[Woldan]

That made me smile. Almost. 

[Woldan]

Can anyone with some physics and aviation knowledge explain me how parable flights work? Today I saw a video of a fighter pilot in his jet flying a parable which made his chocolate bar float around in the c0ckpit. 

Pretty straight forward of course. But something puzzles me, I know that an object pulled by gravity continues to accelerate till air resistance and gravity pull even each other out, it reaches terminal velocity. But since the chocolate bar in the c0ckpit does not experience any air resistance how does it not immediately fall forward and onto the front of the c0ckpit glass of the diving plane? If the c0ckpit was incredibly long, shouldn't the bar then start to fall inside the plane till it reaches terminal velocity itself? Does the plane perfectly out-accelerate the chocolate bar for some time?  And if it does, why doesn't the chocolate bar smack onto the rear c0ckpit glass if the plane accelerates just a little too fast? 

 

Which leads me to another question. According to science, in perfect vacuum a feather and a led ball will hit the ground at the same time. Makes sense, but not really. Objects with more mass create more gravity, which means a lead ball should make the earth move towards it. A really tiny amount but still right? 

Lets say If we have two really long vacuum tubes on earth, in one a feather, in the other the moon. Wouldn't the moons gravitational forces created by its much larger mass pull the earth towards itself hitting it before the feather does? So heavier objects will hit the ground first after all? 

 

Just some questions that just popped into my head. 

Edited March 9, 2015 by Woldan

[Amentep]

I'm not a physicist or an airplane buff, so take my understanding with a grain of salt.

Remember that anything in the plane is going at the "speed" of the plane; when it hits enough thrust to counter (temporarily) the gravitational pull objects in the ****pit would have no force that would act on it to cause it to fall until it loses that countering force.  Since all of the air in the ****pit is also moving at the "speed" of the plane, the air itself won't exert a force on the candy bar.  In other words, it can't reach terminal velocity until the plane alters the forces (by reaching a point in the parabola that they have to alter thrust) in such a way that the force acting on the candy bar no longer is sufficient to negate gravity.

 

The example of the feather and the ball is actually talking about weight not mass (in a frictionless environment, a feather and a metal ball will drop at the same rate). When you look at mass, while the ball may have more mass that the feather, the mass of both the feather and the ball are negligible in relation to the mass of the earth and therefore the slight more pull the ball exerts on the earth doesn't change the elapsed time significantly enough to be measurable against that of the feather.  Obviously this isn't the case between a feather and a celestial body.

 

Again not a physicist so someone will surely correct me if I'm wrong.

[213374U]

Let's do it in inverse order, I think the explanation is easier to follow that way.
 
As per Newton's universal gravitation law, the gravitational force between two bodies is given by: F = G m₁m₂ /r² (1)
 
G is the gravitational constant, m₁ and m₂ are the masses of the bodies involved and r is the distance between them. In this example, m₁ is the mass of a random object and m₂ is the mass of the Earth.
 
We also know that, as per Newton's second law, F = ma (2)
 
The force is the same in both equations, as it's the gravitational force an object is subjected to by the Earth. "a" is the acceleration the body experiences as a result of force. So substituting the value of F in the first equation to solve for "a" you get:
 
ma = G m₁m₂ /r²
 
Now, divide both sides by "m" (in this case, m = m₁), and you get the following expression: a = G m₂ /r²

 

For objects near the surface of the Earth (where r → 6,400 km), that amount is essentially a constant, and it's approximately 9.8 m/s². This means that even though the force applied on an object by Earth's gravity depends on its mass, the acceleration it experiences in a free fall doesn't and is the same for all. This is a simplification useful for most calculations, but it's not strictly correct because it ignores, as you suspected, the effect of the gravitational force caused by the other body in the system. But consider that even asteroids have a mass that, compared to Earth's, is insignificant. If, for Earth's mass (5.9x10²⁴ kg) we get a value for a of 9.8 m/s², imagine the ridiculously low values of acceleration Earth experiences for human-scale objects.

 

You wanted to consider the Moon. Yes, in the experiment you suggested, the Earth would in turn accelerate towards by the Moon at ~1/6 the value of g, so they would accelerate towards each other at ~11.4 m/s². Again, this is only correct for values of separation equal or very similar to Earth's radius; the acceleration would be much lower at distances relevant to objects of planetary mass.
 
--
 
Now, for parabolic flights. We know that all bodies fall towards the center of the Earth at a rate of 9.8 m/s² regardless of their mass and, in principle, regardless of their speed as well. That is the rate at which the mars bar inside the cоckpit is accelerating downwards, if you take the center of the Earth as your frame of reference (and not the cоckpit). Now, if the cоckpit is also accelerating downwards at the same rate of 9.8 m/s² and the initial speed of both items is the same, they won't move relative to each other. This is the same as when you drive next to another car. So long as you are both accelerating at the same rate and moving at the same speed, you will appear not to be moving relative to each other. It's the same for astronauts in orbit, they aren't in an environment without gravity — they are in perpetual free fall.

 

If you give the mars bar in free fall a slight upwards nudge, you momentarily decrease its downwards acceleration, and the cоckpit will, for a slight while, gain downwards speed faster than the bar. Once the bar regains its maximum 9.8 m/s² acceleration value, it will no longer keep losing speed relative to the cоckpit, but the speed difference will remain, so the bar will, slowly but steadily fall towards the back of the cоckpit. Drag for the bar isn't a factor because it isn't moving relative to the air inside the cоckpit.

 

 

disclaimer:

 

Edited March 9, 2015 by 213374U

[Hurlshort]

I have no idea what you guys are talking about, but I like it.

[Gromnir]

woldan is, predictably, wrong about speed of elementary backstroke compared to other strokes.  however, in case folks is genuine interested, the fastest swim style requires the swimmer to stay submerged and use a dolphin kick.  unfortunately, every time some joker finds a loophole in the rules to take advantage o' such, the governing bodies quickly close such gaps.  why? because staring at a pool with folks swimming underwater is even more boring than watching traditional competitive swim events... and there were some genuine fears about folks accidentally asphyxiating themselves.  o' sure, those handful o' us old enough might recall the following with mirth

 

 

however, it turns out that patrick duffy were onto something.

 

HA! Good Fun!

[ManifestedISO]

Where is Walsingham, anyway. 

[Woldan]

So the plane indeed has to be accelerating the same speed objects accelerate in vacuum, since the objects are not subject to drag because the air in the plane is also accelerating (falling) the same speed / alongside the objects. 

 

But once the plane has reached terminal velocity due to air resistance and cannot accelerate the objects should slowly start to hover to the front of the c0ckpit, though slower because the air is now creating friction on them, also giving the objects a terminal velocity. 

 

Thank you for taking your time for writing such an elaborate helpful answer, by the way. 

Edited March 9, 2015 by Woldan

[ShadySands]

Where is Walsingham, anyway. 

 

Haven't seen anything from him or Nepenthe since like January

 

/stalker mode

[Gfted1]

I think Nep was moving or something. Don't quote me on that.

[Hurlshort]

Wals tends to get busy with work as a government assassin.  He should be back soon, now that Boris Nemtsov is dead.

[ManifestedISO]

Probably popped over to Monte's flat for tea and strumpets and scotch-soaked cigars. Say that three times fast.

 

 

My odd question is, is it possible when someone thinks of someone, could it also make them think of them

[Guard Dog]

Wals tends to get busy with work as a government assassin.  He should be back soon, now that Boris Nemtsov is dead.

Or M is explaining to him that Putin was the target he was supposed to go for. That would be... awkward.

[Valsuelm]

Been wondering this for awhile.

 

What does it mean when a person's name is in italics in the 'reading this forum' / 'reading this topic' section below?

[Gfted1]

^Iirc, that they are currently typing a post.

[Malcador]

And who is this Google(1) person

[Hurlshort]

And who is this Google(1) person

The ghosts in the machine.

[Valsuelm]

^Iirc, that they are currently typing a post.

 

Huh. If that's correct, then it's a bugged feature. As people appearing in italics is uncommon, whereas people posting is common, and I've never seen myself in italics. 

[213374U]

 

^Iirc, that they are currently typing a post.

 

Huh. If that's correct, then it's a bugged feature. As people appearing in italics is uncommon, whereas people posting is common, and I've never seen myself in italics. 

 

No, it works fine, it's mostly just outdated.

 

The thing is, it's a feature from before the quick reply option was implemented. Back then, people had to go into full reply mode to type up a post. Nowadays I guess everyone just uses the quick reply window unless they are involved in a full-on quote war, or need to embed a ton of images or whatever. The quick reply mode isn't recognized by the forum as an "action" like, say, starting a new thread or browsing someone's profile. If you hover your mouse over a user name, you get a popup window with the latest info the forum software has on that user's activity. I used to avoid appearing in italics when in the aforementioned quote wars by opening a new tab and browsing the forum main page.

 

stalker level = John Hinckley Jr.

[Blarghagh]

Mr. Too Elite For Me hit the nail on the head there. Essentially a leftover.

[Woldan]

An odd question just popped into my head.
Ok, so we know that the planets orbit the sun in fixed paths because like with satellites, the gravitational pull from the sun, the velocity and the side force equal each other out. Makes sense.
 
But what about drag? For example the moon Europa is so affected by Jupiter's gravity it gets massively deformed which, combined with its rotation creates heat that may make a liquid ocean under its thick ice surface possible. 
 
The sun must have the same effect on the planets that orbit it, obviously on a much much smaller scale but still. Shouldn't all the planets eventually slow down (and get pulled closer and closer to the sun? And the rotational speed of the sun should change too, right?
Otherwise we would have a perpetual motion machine, which is impossible.

Oh, its a really hot day and my brain is getting baked, I'm sure I just missed a small detail that explains it all.

Edited August 14, 2015 by Woldan

[kgambit]

An odd question just popped into my head.

Ok, so we know that the planets orbit the sun in fixed paths because like with satellites, the gravitational pull from the sun, the velocity and the side force equal each other out. Makes sense.

 

But what about drag? For example the moon Europa is so affected by Jupiter's gravity it gets massively deformed which, combined with its rotation creates heat that may make a liquid ocean under its thick ice surface possible. 

 

The sun must have the same effect on the planets that orbit it, obviously on a much much smaller scale but still. Shouldn't all the planets eventually slow down (and get pulled closer and closer to the sun? And the rotational speed of the sun should change too, right?

Otherwise we would have a perpetual motion machine, which is impossible.

 

Oh, its a really hot day and my brain is getting baked, I'm sure I just missed a small detail that explains it all.

 

 

 

You're not talking about drag but tidal deceleration.      https://en.wikipedia.org/wiki/Tidal_acceleration

Edited August 14, 2015 by kgambit

[213374U]

An odd question just popped into my head.

Ok, so we know that the planets orbit the sun in fixed paths because like with satellites, the gravitational pull from the sun, the velocity and the side force equal each other out. Makes sense.

 

But what about drag? For example the moon Europa is so affected by Jupiter's gravity it gets massively deformed which, combined with its rotation creates heat that may make a liquid ocean under its thick ice surface possible. 

 

The sun must have the same effect on the planets that orbit it, obviously on a much much smaller scale but still. Shouldn't all the planets eventually slow down (and get pulled closer and closer to the sun? And the rotational speed of the sun should change too, right?

Otherwise we would have a perpetual motion machine, which is impossible.

 

Oh, its a really hot day and my brain is getting baked, I'm sure I just missed a small detail that explains it all.

 

Nope, you are actually on to something. Gravity and, more specifically, tidal forces, are a mechanism through which systems of celestial bodies lose energy over time. For instance, due to the tides caused on Earth by the Moon, the length of Earth days is increasing, and the orbit of the Moon is being boosted. Friction between the surface of the Earth and the mass of water being constantly displaced by the gravity of the Moon is causing Earth's angular momentum to decrease, the difference in energy being dispersed as heat. These effects operate over astronomical periods, so even if you're right in that it cannot be an example of true perpetual motion, in human scales, it may as well be.

 

Another, more exotic example is the slowdown of pulsars. Pulsars are the super dense cores of stars that have exploded in a supernova. Owing to the conservation of angular momentum, they spin much more rapidly than their progenitor stars did (the ice skater that spins in place with arms wide open and then pulls its limbs towards the body gaining rotational speed is an apt analogy). Neutron stars have a powerful magnetic field that, due to the rapid rotation (as fast as 1/4 of the speed of light on its surface), generates an electrical field that excites protons and electrons present on the pulsar. These, in turn, release electromagnetic radiation along the poles of the magnetic field, resulting in the emmissions we receive. This process eventually slows down the rotation of pulsars until they no longer "broadcast".

Edited August 14, 2015 by 213374U

[Woldan]

Thats extremely interesting, thanks for the reply 213374U! 

 

I wonder if its possible to measure or calculate the theoretical electromagnetic power output of a young pulsar, I bet the numbers would be absolutely mind boggling.