41

General Science / Re: Do we pee more than we drink?

« on: 19/11/2018 18:06:22 »

When your body burns food for energy, one of the by-products is water.   So even if you actual fluid intake is low, you can end up with a fair amount of water in your body that you need to get rid off. 
If it wasn't for the fact that we get rid most of this through exhaled moisture in your breath, you'd be be peeing a lot more than you are now.

42

Physics, Astronomy & Cosmology / Re: Would particles orbit a falling body?

« on: 17/11/2018 19:47:12 »

No, at that close to the Earth, even a steel marble would be within the Roche limit.   Tidal forces generated by the Earth would exceed the gravitational attraction of the marble. For a steel marble you would have to be at least 708 km above the surface for this to not to be the case. 

For objects to orbit the marble, they would have to be within the marble's Hill sphere. A 0.5 cm radius steel marble would have to be ~1689 km above the surface of the Earth in order for its Hill sphere radius to be larger than it own radius, so this is the closest it could be to the Earth and hold anything in orbit around itself.

43

Physics, Astronomy & Cosmology / Re: Can gravity be said to act in two opposite directions at once ?

« on: 21/10/2018 16:44:31 »

Here's a map (not to scale), marking the lines of gravitational potential for the Sun-Earth system.
https://map.gsfc.nasa.gov/media/990529/990529.jpg

And here's another showing a cross section of the gravity wells of the planets of the solar system within the gravity well of the Sun.
https://imgs.xkcd.com/comics/gravity_wells.png

44

Physics, Astronomy & Cosmology / Re: Is acceleration downwards due to gravity less when the Moon is overhead?

« on: 14/10/2018 23:17:46 »

We can apply F=GMm/r² to do a back-of-the-envelope calculation.
To get the relative strength of Earth & Moon's gravity on you, we can eliminate half of these parameters, since G & m (your mass) is common.

The mass of the Moon is about 1/80 mass of the Earth.
And the radius of the Earth at 6000km is around 1/60 the distance of the Moon.
So the relative gravitational force of the Moon on you is about (1/80)*(1/60)², or 1 part in 300,000.
So if you mass 70kg, the position of the Moon makes a difference of about 0.2grams in the weight you feel.
This is much less than the forces on your body every time you take a step or have a sip of water.
And because the change occurs over 12 hours, the effect is imperceptible.

The net weight difference is less than that.  At the the center to center distance between Earth and Moon, the acceleration due to gravity caused by the Moon is 3.3e-5 m/s^2  at the near side of the Earth to the Moon it is 3.44e-5 m/s^2.  Since the center of the Earth is in free fall with respect to the Moon, it is the difference between these values, or  ~1.4e-6 m/s, that would result in a net lessening of weight.  This is ~1.4e-7g.    For a 70 kg person you get a net weight difference of ~ 0.01 gram.

45

Physics, Astronomy & Cosmology / Re: How does dark matter affect galaxy rotation?

« on: 26/09/2018 06:18:48 »

First off, the "halo" is not a ring. It is a spherical volume in which the galaxy is embedded. A ring shape is not what the word "halo" means in astronomy.
  Also, within such a spherical volume, at any given point The only gravitational effect would be caused by that material closer to the center than that point (Newton's shell theorem).
The visible matter of the galaxy is mainly confined to the bulge and disk.  Once you get away from the bulge and into the disk region, moving further out does not add much to the contribution that the visible matter contributes to to your orbital speed. The dark matter, however extends well above and below the visible matter disk and the thus the amount of DM closer to the center of the galaxy grows quite fast.
 For example, the density of DM in the vicinity of the Solar system is so low that it would only adds up to the equivalent of a small asteroid within the entire volume of the Solar system. At that same density, the spherical volume of the region closer to the center of the galaxy than the solar system would hold a significant fraction of the entire visible galaxy's mass. 
Thus extra mass, plus the mass of the visible matter closer to the center of the galaxy is what determines the orbital speed of the solar system ( and makes it higher than it would be is there just was the visible mass.)

46

Physics, Astronomy & Cosmology / Re: What spun-down the inner planets ?

« on: 26/09/2018 00:39:03 »

The time it takes for a body orbiting another body to tidally lock is found by the equation:
T = wa^6 mQ/(3GM^2 K R^3)
w is the starting angular velocity of the body's rotation
a is the semi-major axis of its orbit
Q is the dissipation factor
m the body's mass
G the  universal gravitational constant
M the mass of tht body it is orbiting
K the "Love" number.
R the radius of the body.
Q and K are not well known except in the case of the Earth and Moon system.  So for the sake of argument, we will assume they are the same for all three planets and the Sun.
Since we are comparing planets orbiting the Sun G and M will be the same.
If we assume that w starts out as the came for all three then the difference in tidal locking times can be simplified to the relationship of
a^6 m/R^3   
It also turns out that the ratio of m/R^3 comes out to be fairly close to each other for all three planets.  So in the end, the major deciding factor is a^6
 Earth is 2.58 times further from the Sun than Mercury, so it would take nearly 300 times longer to tidal lock to the Sun.
The Earth is 1.38 times further than Venus and would take 7 time longer to lock.  Assuming all other things are equal.

One thing to keep in mind is that tidal forces fall of by the cube of the distance.  So even though the Moon's tidal effect of the Earth is ~ twice that of the Sun's, At the distance of Mercury, the Sun's tidal force has increased by a factor of 17, and be 8.5 times stronger than the Moon's effect on the Earth and at the distance of Venus it would be 1.3 times stronger than the Moon on the Earth. 

But we don't know that all things started out equal or stayed that way. For example, the Earth is believed to have been struck by a Mars sized body in its past which initially formed the Moon, and likely spun the Earth up quite a bit.  This would have given it a lot more rotational energy to shed.
Venus actually rotates slower than it orbits.  This also may have been due to a collision; one that robbed it of spin rather than giving it more spin. 

So without knowing the full history of each planet it is hard to say what all the influences were that contributed to their present rotations.

47

Physics, Astronomy & Cosmology / Re: How fast does the atmosphere rotate?

« on: 25/09/2018 16:17:28 »

I am not quoting that long long post Janus !

If you have two tanks, linked by a 10m tube,  each tank 10000km  wide and 3knm high to shift 10m of water from one tank to the other by lowering the one tank over 6 hours the water has to flow through the pipe at 1,666 km per hour. Now 10 m is only .3333 percent of 3000m so the turbulance and the mass of the water in the intake will subdue the energy, equivilant to 5.3 km per hour of the entire tank moving, but it does flow at 1600km inward

That is in no way analogous to how ocean waters redistribute themselves to form the tidal bulges.

The top image show water with no bulge.  The next image the same water with a bulge.  The ends have moved in slightly ( this image has the bulge and movement exaggerated to make them more noticeable.) 

In forming the bulge, water just below it rises, the water below that takes its place, the water next to that water does the same, etc.  By the time it gets to the end of the take, the water there just moves in a little to fill the space left by water that vacated that space.   The water missing at the ends does not flow all the way to make up the water in the bulge.
That's like saying that when the row of railway cars shown in the next image moves one car length to the left, it is because the rightmost car passed through all the other cars to become the rightmost car. 


bulge.gif (2.77 kB . 500x300 - viewed 3376 times)

48

Physics, Astronomy & Cosmology / Re: What is Important About Instantaneous Age Changes in the Twin "paradox"?

« on: 05/09/2018 16:06:23 »

My reasoning may be wide of the mark, here, so I would value comments.

Acceleration involves a change in speed, direction or both, thus, the “turn-around” constitutes acceleration.

The “twins paradox” comes about as a result of time dilation and length contraction. 

The Lorentz equations, which gave mathematical veracity to time dilation and length contraction, feature only time, length, velocity and the speed of light.  There is no mention of acceleration.

Einstein incorporated time dilation and length contraction into SR, and SR is concerned only with uniform motion.  There seems to be no need to venture into general relativity.  So why do we have to use acceleration to solve this particular paradox? 

You are forgetting the relativity of simultaneity.  While time dilation and length contraction are only dependent on the speed difference between two frames,  the relativity of simultaneity is dependent on the velocity difference, which involves direction.
For example,  if you are traveling between two clocks A and C while passing a third clock B, and assuming that A, B, and C are synchronized in their own rest frame, then:
If you are traveling from A to C, then clock C will be, according to you, ahead of clock A in respect to what time it reads and if you are traveling from clock C to A then according to you, clock A will be ahead of clock B.

A numerical example:
You're traveling at 0.6c relative to the three clocks, for which the proper distance between the clocks is 1 light hour.
You pass clock B while it reads 12:00, while traveling towards A,  At that moment, according to you, Clock A reads 12:36 and Clock B reads 11:24
On the other hand, if you pass B at the same speed while going from A to C, then Clock A reads 11:24 and Clock C reads 12:36 at that moment.*
If you were to suddenly reverse direction as you pass B, while initially traveling from A to B, then clock A will "jump" from reading 11:24 to 12:36*

Acceleration comes into play with time dilation and length contraction when it involves a change speed, as changing speed changes the magnitude of these two effects.   Acceleration additionally effects the relativity of simultaneity when it involves a change of direction.


*Keep in mind that this is the time it "is" at these clocks and not the time you would "visually see" on these clocks.  In all these examples, when you are next to clock B when it reads 12:00, both you and an observer sitting by clock B would be visually seeing images of Both Clocks A and C reading 11:00.

49

New Theories / Re: Discussion on Reactionless drive (extracted)

« on: 26/08/2018 18:01:43 »

To:Kryptid
Must have logic , live by it !
Speaking of planetary systems , how do you like that Earth-Moon system ?  Much of Earth's initial angular momentum has been transferred to Luna , BUT much of it was converted to heat in the process .  Were it a perfect transfer , the E.-M. system would be in a different location , with a different momentum .  If you now spun the Earth in the opposite direction , the opposite effect would occur . Bottom line , destroying energy of momentum works as well as removing mass.  Sounds like a propellantless rocket to me ! 
 Hokay , next victim !      P.M.

 P.S.-" l'm freakin' Bobby Fisher ! "

As already noted the Earth Moon barycenter doesn't shift due to the tidal acceleration of the Moon.
If we consider the angular momentum of the Earth -Moon system:
The Moon's orbital radius grows by some 4 cm per year, and if we use 384,000 km as its average orbital radius, We find that moving out by 4 meters ( the expected distance in one century) puts it into an orbit with an orbital velocity of 1018.9608624m/sec , vs its starting velocity of 1018.9608677 m/s with the Moon's mass, that works out to a gain of angular momentum of 1.348e26 kgm^2/s  ( the Moon will also gain orbital energy by the relationship of E_0 = mv^2/2-u/r = -u/2a where u is the gravitational parameter for the Earth, GM= 3.987e14 m^3/s^2, and a is the semi-major axis of the orbit.)
The Earth, in that same century, increases its period of rotation by 1.7 milliseconds.  Using the Earth mass, and assuming it is a sphere of uniform density in order to calculate its moment of inertia, we get a loss of 1.39e26 kgm^2/s
Pretty damn close considering we took a number of shortcuts along the way. (treating the Moon as a point mass, and assuming uniform density for the Earth, etc)
If we were to accurately account for all the factors we would get agreement in the numbers to whatever level of accuracy we had for those factors.

Now the Earth does heat up in this exchange, so some part of its remaining angular momentum will be due to the "effective" mass increase due to that thermal energy, And the Earth will radiate that energy.   But this does not mean that there is any net change in total angular momentum.  Because, as I already mentioned earlier Electromagnetic radiation exhibits the property of momentum and this "lost" angular momentum is being carried off by this radiation.   Thus, if you take the Moon, Earth and the radiation emitted by the system all into account, there is no net gain or loss of momentum, angular or otherwise.
There is nothing "reaction-less" about this at all.

50

Physics, Astronomy & Cosmology / Re: Would an asteroid hitting the moon have any impact here on Earth?

« on: 22/08/2018 22:26:46 »

I estimate that there is an equal balance point between Earth and Moon (Lagrangian Point L1) that is about 43,000km from the Moon, so something flung off the Moon in this direction would only need to start with about 1.9km/s velocity, which is less than Lunar escape velocity. But that is also a very specific trajectory.

Comments welcome!

L1, some 326054 km from the Earth, is a point where, with the help of the Moon's gravity, an object could be placed and it would stay (if not further disturbed) even though it is traveling at a speed slower than what would normally be needed to maintain a circular orbit.  Reaching L1 does not mean you now have a free trip to Earth.
Moon orbital velocity is ~1019 m/s.   In order to maintain relative position with the Moon, an L1 object would be moving at ~864 m/s, which is -241m/sec slower than the  1106 m/s it would have to be moving in order to maintain the same path around the Earth in a free orbit without the aide of the Moon. (V_o = sqrt(u/r), where u is the gravitational parameter for the Earth of 3.987e14 M^3/s^2)
If we were to remove the Moon, an object at L1 distance from the Earth and moving at L1 point velocity would be at the apogee of an elliptical orbit with a perigee of ~143250 km from the Earth.
(orbital energy/kg = v^2/2-u/r = u/2a,   where a is the semi-major axis of the orbit.
solving solve for and then use a2-r to find the perigee distance)

 Even without the Moon's aide, an object placed at the L1 point would not "fall" all the way to the Earth.

To get such an object to reach Earth surface  you would have to reduce its L1 point velocity to ~217 m/s or by an additional 647 m/s.
V_r = sqrt(u(2/r-1/a)   where a=  (r+r_e)/2 and r__e is the radius of the Earth. 

 Again, this velocity change would need to be applied in the opposite direction of the orbital velocity.
From the actual L1 point, its going to take a bit more. That same gravitational "help" the Moon gives an L1 point object in allowing it to hold position, is now going to be a hindrance in trying to reach Earth.   
Think of it this way, without the Moon, the object would drop to 143250 km from the Earth, with the Moon, it maintains a constant 326054 km distance.   That means with the Moon, you would have to give the L1 object a push in order to force it into that elliptical orbit that it would naturally assume without the Moon.  The same is true for putting it on an Earth intersecting trajectory, its going to take more effort to put it on that path with the Moon than it would have without the Moon for the Same object starting at the same point and speed.

This is orbital mechanics that we are dealing with. Intuition often leads you astray.

51

New Theories / Re: What is the applied hour in the theory of tides?

« on: 15/08/2018 16:46:55 »

To add to what Bored chemist has said. The OP seems to think tidal bulges move like this:

tides1.gif (292.01 kB . 512x384 - viewed 3034 times)
With the blue inner circle the Earth and striped oval the oceans, and the Moon dragging the ocean around with it.

However, what it really happening is this:

tides2.gif (131.91 kB . 512x384 - viewed 2948 times)

The gravity of the Moon distorts the shape of the ocean, but the body of the ocean does not follow the Moon, as shown by the fact that the lines representing the ocean don't move relative to the body of the Earth, just the bulge,

52

Physics, Astronomy & Cosmology / Re: How fast would the earth spin if the moon was closer?

« on: 10/08/2018 23:21:47 »

I do ot see why moving the moon closer would cause it to spin any faster, other than as a byproduct of the increaced orbital velocity around the mutual centre of gravity between the moon and earth . If the rotation of earth is merely a balanced motion that is dependant on the observer, ie measured from the sun. If you move the earth and moon closer however they would biorbit each other at an increased formulaic velocity. Could you explain why the spin would have to have a defacto increace?

Initially, putting a Moon-massed object into a close orbit around the Earth would not cause an increase in the Earth's rotation rate.  But as it stays there, tidal interaction between the object and the Earth will transfer angular momentum from the object to the Earth. (This is the opposite effect from what is now happening with the Moon, where angular momentum is being transferred from Earth to Moon.  The difference in how the tidal interaction affects the Earth is due to whether the rotational period of the Earth is greater than or less than the orbital period of the object. If the object was placed at geosynchronous orbit distance, there would be no speed up or slow down.)

53

Chemistry / Re: Are there molecules inside the atom?

« on: 22/07/2018 15:57:32 »

Are molecules inside the atom?Does that mean that the electrons and protons are outside the atom?I don't know a lot about chemistry but I know that the protons are in the nucleus.Can someone explain to me how a proton is inside a molecule?

Molecules are made from atoms when atoms bond together.   One oxygen and two hydrogen atoms bound together makes 1 water molecule.  Add another hydrogen atom and you get a hydrogen peroxide molecule. 
Atoms consist of a nucleus which is made up of neutrons and protons, surrounded by a cloud of electrons.   The number of protons determines what type of atom you have. The neutrons help stabilize the nucleus so it doesn't fly apart.  The number of protons also determines how many electrons there will be in the surrounding cloud.  It is the interaction between the electron clouds that forms the bonds that hold a molecule together.
 How atoms bind to others and form certain molecules has to do with how the electrons arrange themselves around the nucleus.  They arrange themselves in "shells" with each shell capable of containing a given number of electrons.  Shells "prefer" to be full, and if the number of electrons for a isn't enough to fill the outer most shell, two or more atoms can share electrons between them in an attempt to fulfill this need. This sharing of electrons forms a bond between the atoms and creates a molecule.

54

General Science / Re: Is zero an even number, odd number, or neither?

« on: 20/07/2018 18:42:05 »

Also, by the rules of multiplication:
Odd x Odd = Odd
Odd x Even = Even
Even x Even = Even

Since  N x 0 = 0  whether N is odd or even, then 0 must be even.

55

Physics, Astronomy & Cosmology / Re: What will happen when these two stars collide in 2022?

« on: 20/07/2018 16:55:48 »

We're talking about a pair of stars some 1800 ly distant, which presently are too dim to see by the naked eye (~ magnitude 12)
The explosion should make it naked eye visible to the point that it will rival some of the brighter stars.   In other words, people will be able to see a new star (nova) in the constellation of Cygnus for a short while

56

Physics, Astronomy & Cosmology / Re: The Graviton?

« on: 05/07/2018 21:46:55 »

However, real gravitons propagate away to infinity, as oscillations on the gravitational field. This is what LIGO detected. (And real photons propagate away to infinity, as oscillations on the electromagnetic field. This is what telescopes detect.)

LIGO detected gravitational waves as predicted by EFE not gravitons. Gravitons are theorized to exist and have such low energy levels that we can not detect them with todays technology.

If gravitons exist (Gravity can be explained with a quantum theory), then gravitational waves would consist of gravitons.

Bosons can not escape a BH, a graviton is a 2 spin boson how can it get out of a BH. A gamma ray cannot escape the event horizon of a BH and it has significantly more energy.

Gravitons can't,  but since gravity would be mediated by virtual gravitons, this is no different from the fact that photons can't escape a BH, yet a BH can have a charge and an electromagnetic field mediated by virtual photons.

Gravitational waves definitely exist gravitons might not exist, and might not have enough energy to escape a BH.

Gravitational waves can't escape a BH either (By this I mean that gravitational waves cannot get out from inside the event horizon anymore than electromagnetic waves can.)

57

Physics, Astronomy & Cosmology / Re: How much is the air drag energy at 60mph?

« on: 01/07/2018 17:44:04 »

Can anyone do some sample calculations to give me an idea?

I am struggling with my discalculia.

From the link, for a 1997-1999 Acura CL
CD = 0.34
frontal area (A) = 21.6 ft^2 = 2 m^2
V = 26.8 m/s
p = 1.225 kg/m^3

Thus drag force is 0.5 x 1.225 x 26.6^2 x 0.34 x 2 = ~ 295 N
Energy is force x distance, and in one sec the car travels 26.8 m so
295 x 26.8 = 7906 joules. 
 watts are joule/sec so this works out to 7906 watts = 10.6 hp

So a 1999 Acura CL requires 10.6 hp to overcome air drag at 60 mph.

58

Physics, Astronomy & Cosmology / Re: How much is the air drag energy at 60mph?

« on: 24/06/2018 22:46:15 »

You can calculate it from the drag equation:

F = 0.5 * p * u^2 *  Cd * A

Where p is the density of the air (depends on altitude, 1.225 kg/m^3 at sea level), u is the speed in metres per second (26.8 m/s at 60 mph), Cd is coefficient of drag, and A is the cross-sectional area of the car, viewed from the front.

And the power is F * u

So you just need the frontal cross-section and the Cd factor, they are often quoted for a car.

Sounds good, any sample calculation for a usual car?

http://ecomodder.com/wiki/index.php/Vehicle_Coefficient_of_Drag_List

59

Physics, Astronomy & Cosmology / Re: Which force is stronger...gravitation or dark energy?

« on: 12/06/2018 21:52:56 »

Which is stronger depends on the circumstances.  Dark energy has a uniform density, while gravity depends on the amount of mass and the distance from it.   The more you concentrate mass, the stronger the local gravity.   When it comes to things like stars, galaxies and even clusters of galaxies, the mass density is still enough for gravity to dominate.  At scales larger than this, the overall density is low enough that dark energy dominates.

60

Physics, Astronomy & Cosmology / Re: Does general relativity encompass quantum mechanics?

« on: 09/06/2018 16:57:34 »

Well if GR and QM are BOTH inconsistent near to a black hole then that is something they have in common. So it is not necessarily a problem.

It's not that simple.  What it really comes down to is that each theory does make predictions when pushed beyond certain limits, but these predictions don't match reality.  When you try to apply GR to the realm where Quantum effects become noticeable, it doesn't match what we actually see happening.  On the other hand,  QM fails to make accurate predictions (such as gravitational effect) when you get to larger scales.  It not enough to just say "Well, this theory only applies to this point and no further".  The only time this is a valid argument is if this limitation is a built in part of the Theory itself. An example would be FTL speeds and SR.  c as a limiting speed is built into SR.   There is nothing in GR  itself that says it shouldn't give accurate predictions below a certain scale.  What is being looked for is the reason that these two theories don't make the correct predictions under certain conditions, and yet give very accurate results otherwise.