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Messages - Janus

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41
Physics, Astronomy & Cosmology / Re: How much does the moon weigh?
« on: 08/06/2019 22:33:01 »
Quote from: chris on 08/06/2019 11:06:09
I was asked on the radio this week "how much does the Earth weigh?"

More accurately, we were of course discussing the mass of the Earth, and the answer, which I happened to know, is about 6 x 1024kg.

Later, driving to work, I began to think - as you do - what I would have said if the questioner had enquired instead about the mass of the Moon.

Naively, I reasoned that gravity is about 20% of that felt on the Earth's surface, so the Moon must have a mass 20% that of Earth.

Intrigued, I later looked it up. The stated mass of the Moon is about 7 x 1022kg. That's only about 1% of the mass of the Earth.

Is the acceleration due to gravity on the lunar surface as large as it is (almost 20% of the Earth's despite the lunar mass being only 1% of Earth mass) because the Moon is correspondingly smaller, so the inverse square law being what it is (GmM/r2) it works out that way?

Lazily, I've not done the maths to check...
Assuming a constant density, Mass goes up by the cube of the radius, and since acceleration due to gravity decreases by the square the distance, surface gravity would increase at the same rate as the radius.
The Moon's radius is 1738 km, ad the Earth's is 6378.   6378/1738 = ~3.7,  so you might expect surface gravity on the Earth to be 3.7 times that of the Moon. However, the Moon is not as dense as the Earth, ( the Earth is roughly 1 2/3 times more dense.   3.7 * 5/3 = 6    The Moons surface gravity is roughly 1/6 that of the Earth's.
The following users thanked this post: chris

42
Physics, Astronomy & Cosmology / Re: Proxima Centauri b and the red dwarf?
« on: 01/06/2019 16:25:50 »
Quote from: Europan Ocean on 31/05/2019 19:24:29
If Proxima b is bigger than the earth, it may have a bigger magnetosphere, could this trap water?

We can compare Mercury and Venus, the latter is close to the Sun and has a very thick atmosphere. Smaller than the Earth, yet inside maybe more Iron?
The size of a planet has nothing to do with its magnetic field, which is generated by its spinning core.  One of the issues with orbiting so close to the star is that the planet is going to be tidally locked so that it rotation and orbit have the same period. Since Proxima b takes about 11 days to orbit, it would also take 11 days to rotate. The slower rotation doesn't bode well for generating a strong field. 
It is estimated that  Proxima loses mass to solar wind at a rate about 1/5 of that of our Sun, but Proxima b is only about 1/20 of the distance from its star than the Earth is from the Sun. This ends up with the Solar wind being more than 80 times stronger for Proxima b than it is for the Earth. 

While this does not completely rule out Proxima b from having a viable biosphere, it does but it at very long odds.
The following users thanked this post: Europan Ocean

43
Physics, Astronomy & Cosmology / Re: The Northern Stars and Exoplanets
« on: 31/05/2019 18:02:37 »
The present Northern star Polaris is actually a triple star system. the Primary star is a yellow giant some 2000 times brighter than our Sun, which also is a Cepheid variable, meaning its output varies over a set period, varying by about 8% over 4 days.  Its overall luminosity is also slowly increasing, having increased by ~2.5 times in the last 2000 yrs or so. 
The other two stars are also brighter than our sun, but only by about 3-4 times.
The main star is not a good candidate for planets due to its unstable nature, and the smaller of the other two is too close to the main star.   This leaves one star that could even possibly have some type of planet system.   However,  larger more massive stars spent less time in the main sequence. This star would only be expected to do so for about 1.5% as long as our sun.  Likely too short a time for any complex lifeforms to form.
Kochab, the "old" northern star is also more massive and brighter than our Sun.  It has also reached that part of its life cycle where it is beginning to expand into a giant.  Any planets it may have had are cooked by now.
The following users thanked this post: Europan Ocean

44
Physics, Astronomy & Cosmology / Re: How do galaxies keep a stable size?
« on: 23/05/2019 16:40:07 »
Individual galaxies do not expand due to the expansion of space because they are compact enough that the mutual gravitational attraction holds them together. This effect actually extends all the way out to include groups of galaxies.  The individual galaxies of our local group are not receding from each other but orbit each other as a gravitationally bound system.  It the larger groups of galaxies that are moving apart from each other.

The individual stars in a galaxy are in orbit around the their mutual center of gravity.  Orbits are not as fragile as some people tend to think they are.  If you slow down or speed up an object in orbit, it will not necessarily "fall out" or "fly off" of orbit.  Unless you make a large change in its speed, it just settles into a new orbit.   So for example,  if you wanted to make the Moon "fall" into the Earth,  you would have to reduce its orbital speed to less than 1/5 of its present orbital speed, and to make it fly away into space, you would have to increase it by almost 41%.   
Galaxies are made of stars that fall somewhere between.  It's a bit more complicated with galaxies, but the general idea is the same. Stars in the galaxies don't have to maintain a delicate balance to remain in orbit and have a fair amount of "wriggle room". *
This is not to say that galaxies don't evolve over time.  Early in their lifetimes, things are a lot less stable.   You still had material falling in towards the center feeding the black hole their, etc.   And even later, and occasional close encounter will fling a star clear of its parent galaxy.  But the vast majority of stars in a mature galaxy are in that "comfort zone" as far as their orbit goes, and would take quite a bit to knock them completely out of it.

* As an analogy, think of a rocking chair.  If you give it a push, it will rock back and forth, but won't fall over. It takes quite a large shove to make it completely topple over.
The following users thanked this post: RobC

45
Physics, Astronomy & Cosmology / Re: Is spacetime always understood in relation to the speed of light (em radiation)?
« on: 19/05/2019 16:26:18 »
Quote from: jeffreyH on 19/05/2019 02:28:53
Einstein argued that if you were inside a box you would not be able to tell if you were in a freely falling frame in a gravitational field or in an inertial frame. In an accelerating frame an object would be emitting electromagnetic radiation. So does the free falling object in the box emit electromagnetic waves? Does the radiation increase with increasing speed? If it emits no radiation due to acceleration then it can be considered truly inertial. In which case Einstein is correct. Which is it?
https://en.wikipedia.org/wiki/Paradox_of_radiation_of_charged_particles_in_a_gravitational_field
The following users thanked this post: jeffreyH

46
Physics, Astronomy & Cosmology / Re: Does a photon self propagate due to the interactions of the electric and magnetic fields?
« on: 15/05/2019 16:26:35 »
To add to what Colin2B has already said.  A photon, being a quantum of electromagnetic radiation, is the smallest divisible unit of radiation energy you can have for EMR of a given wavelength. 

And, as pointed out, if a single photon strikes photographic paper, it produces a single localized spot. ( like a particle would).
However, if you send a series of single photons, one at a time, through a double slit interferometer, While each individual photon will produce a localized dot on the paper,   The pattern formed after many photons have passed through will be an interference pattern like you would expect from a wave.
A classical particle would not create the interference pattern,  while with a classical wave, you'd get the interference pattern, but it wouldn't be formed from individual dots.( the whole pattern would form at once and just get more distinct as more light passed through the slits. 
While you will hear it said that exhibits both the properties of particles and waves, it is probably more accurate to say that the classical distinction between wave and particle just doesn't hold, and that this becomes more apparent at the quantum level.
This characteristic is not limited to light alone.  Electrons( original considered classical particles) show the same behavior.  You can pass them through a diffraction grating and produce an interference pattern composed of individual dots also.
It is something built into the universe at that scale and not something exclusive to this or that "particle". 
The following users thanked this post: Bogie_smiles

47
Physics, Astronomy & Cosmology / Re: How do black holes evaporate?
« on: 17/04/2019 16:36:06 »
Quote from: 5ASE on 17/04/2019 14:36:54
As far as I understand the popular science explanation of black hole evaporation, quantum fluctuations of vacuum at the event horizon create a pair of matter-antimatter particles. One escapes thus reducing the mass of the black hole. So far so good. However, just outside the horizon, the same effect must be continually feeding new particles into the black hole. The outside sphere is somewhat bigger, there must be more mass entering than leaving.

I suspect he popular explanation must be leaving out some important details?
One thing to keep in mind is that the formation of virtual particles just outside the event horizon comes from "borrowed" energy. Under normal circumstances, these particles only exist briefly, before rejoining and returning this borrowed energy. It is a quirk of QM that allow this to happen without violating energy conservation. 
If, however, one of the particles crosses inside the event horizon before the rejoining can occur ( it does not matter whether it is the particle or anti-particle),  You are left with a lone particle.  The energy books must be balanced in order for this particle to continue to exist, and this come from the mass of the BH.  One of the virtual particle pair has become a real particle at the expense of the BH.  This particle may escape, or more likely interact with another particle which produces light radiation that does escape the region near the event horizon. You end up with some "leakage" of radiation.
That being said, only black holes below a certain size would end up experiencing net shrinkage through Hawking radiation.  One of the features of Hawking radiation is that it increases as the mass of the Black hole decreases. Large black holes lose mass through Hawking radiation at a much lesser rate than they gain it from other sources.  Any stellar sized black hole, even if far away from any matter that it could absorb would still be gaining mass from just the cosmic background radiation faster than it lost it through Hawking radiation.   
For a black hole to lose mass faster than it gains it from the CMBR, it could not be any more massive than our Moon.
This is much smaller than any black holes that form today, but there is some speculation that small or "quantum" black holes could have formed during the early universe.  It is possible that some of them survived to our time.

Even present day large black holes are subject to eventual evaporation.  As the universe continues to expand, it cools which decreases the intensity of the background radiation.  At some point it will cool enough for stellar mass black holes to begin to evaporate as they lose mass faster than they gain it even through the CMBR.   But we are talking about really, really, really long time scales.
The following users thanked this post: 5ASE

48
Physiology & Medicine / Re: Why do optometrists dilate our eyes then do the exam?
« on: 05/04/2019 16:47:40 »
The wider your pupils are, the shallower your depth of field.  Depth of field is the distance range over which objects in your vision will appear relatively in focus. With your pupils contracted and focused on an object 10 ft away, objects 7 to 13 ft away might look in focus, but with them dilated, it might drop to objects between 9.5 and 10.5 ft 
 By dilating your eyes and reducing your depth of field, they are making your eyes more sensitive in small changes in the prescription.  In other words, when they ask you if  "1" or "2" looks better, You'll see a larger difference with your eyes dilated than with them not, which allows them to fine tune the prescription more accurately.
The following users thanked this post: benm

49
Physics, Astronomy & Cosmology / Re: What would air and water do on a rotating space craft?
« on: 03/04/2019 20:43:49 »
Quote from: jimvideo on 03/04/2019 18:47:10
I think your example is missing a step, overcoming inertia. A spinning space craft must constantly overcome inertia, which is especialy important with liquid.

When a drop of water hits a water surface another drop of water is released from the spash. On Earth gravity will bring the drop down, but in the space craft lake the drop will continue upward for several minutes. The wayword drop will continue until it hits something or dissipates.


* www.maxpixel.net-Nature-Water-Drops-Of-Water-Liquid-578897.jpg (49.9 kB . 640x426 - viewed 2183 times)

But that drop of water will have the same tangential momentum it had when it it was part of the body of water.  Thus it will travel in a straight line along a vector determined by its initial inwards velocity and its tangential velocity.  Since the surface of the space ship is curved inwards as is the surface of the water, this path will intersect the water again.   For anyone moving along with the rotating craft, will also have a tangential speed that stays pretty much equal to that of the drop of water. So from their perspective, the water rises from the water and then falls back down.  There will be a small apparent drift due to Coriolis effect, the larger the "inward" velocity of the drop, the more apparent this will be.   But unless you completely remove all of the tangential motion of the drop ( effective throwing "backward" against the rotation at exactly the rotation speed), it will not fail to return to the water surface. 
The following users thanked this post: jimvideo

50
Physics, Astronomy & Cosmology / Re: What would air and water do on a rotating space craft?
« on: 30/03/2019 22:44:12 »
Quote from: jimvideo on 30/03/2019 19:32:36
Quote from: Halc on 29/03/2019 11:53:48
Water would be 'down' just like water on Earth which is attracted to Earth by gravity.  So a spinning object would have 'down' feel like the outside of the craft and 'up' be towards the center of rotation.  The closer to the edge you get, the stronger the artificial gravity.  Thus there would be no 'gravity' at the center.

Thank you for answering. This is a good description of it works in movies, but I don't believe this is correct. While gravity has an area of effect, a spinning space craft can only effect what it touches. Without gravity a thrown baseball will continue it's trajectory until it is acted on by an outside force, even if that force is air. Throw that ball in the exact opposite of the rotation and it might stay in one spot. I don't know what the answer to this question is, which is why I'm asking.
If you threw the ball at exactly the right speed, then yes, you could make it standstill relative to the station axis and not "fall to the ground.  However the station would be still turning, Thus for you the person throwing the ball, it would appear that the ball was traveling in a circle at a constant distance from the floor.( ignoring any air resistance).
If you could ignore air resistance and any obstacles in the way, if you could throw a ball at ~7.9 km/sec in the horizontal while standing on the Earth, it would travel in circle around the Earth maintaining a constant distance from the ground. It would be in orbit around the Earth.   Throwing the ball at just the right velocity on the station is the equivalent putting it in orbit from the rotating station's frame of reference.*

But you can't ignore air resistance.  If add air to the station, that air will expand to fill the station, when it does so, it comes in contact with the rotating walls air molecules striking the wall will be be given some of the wall's these molecules will go on to transfer it to other molecules, etc.  Eventually, you will end up with the air in the station having the same angular velocity as the walls, and rotate with it.   If you throw your ball now, it will have to deal with the drag caused by this air which worls to drag the  ball around it with the station.  The ball will not be able to maintain is fixed position and will drift, until it hits the "floor" picking up more motion from it.   

The same happens for water, it would be eventually dragged to have the same velocity as the floor.  So while there might be a bit of "settling down" to do at first, eventually you'll end up with water hugging the floor and air pressing against it.

*This only works if you throw the ball at precisely the right speed. Any slower or faster and it will drift in a straight line until is hits the curved floor of the station.
The following users thanked this post: jimvideo

51
Physics, Astronomy & Cosmology / Re: Do you think Tau Ceti e is inhabited?
« on: 10/02/2019 18:58:47 »
Quote from: chiralSPO on 10/02/2019 17:32:01
Quote from: Janus on 10/02/2019 15:47:37
Call it 99%. 99% of 150° is  148.5°.

Sorry to be the pedant, but are you scaling a temperature? In ° F?!?

Quote from: Janus on 09/02/2019 16:55:29
150° F

At the very least, please use an absolute temperature scale (Kelvin or Rankine) before multiplying temperatures by any factor, otherwise we end up with such silly results, like: "50% hotter than 32 °F (0 °C) is 48 °F (8.9 °C), but 50% hotter than 0 °C is, still 0 °C?" or even worse: "50 % hotter than 20 °F (–6 °C) is 30 °F (–1.1 °C)..."

Even with absolute temperatures, scaling is somewhat nonsensical, but at least the maths work out.

Yeah your right.  Silly mistake on my part.


  Using the same numbers gives ~ 144° F.  For difference of 6°F.  But then again, I was being generous with the 99% estimate.   At the highest resolution of the graph in the paper, you might get 99% of circular orbit equilibrium temp at an eccentricity of 0.4, better than twice the eccentricity of Tau Ceti e, and its on a curve that flattens out as you approach 0 eccentricity. 

So despite my embarrassing lapse, I still don't think that this is enough to significantly increase the chances of Tau Ceti e being habitable
The following users thanked this post: chiralSPO

52
Physics, Astronomy & Cosmology / Re: What is the origin of Earth's magnetic field?
« on: 09/02/2019 23:06:05 »
Quote from: evan_au on 09/02/2019 21:39:47
It is a mystery why Earth has a magnetic field, and Mars doesn't (any more).
Mars is slightly smaller than Earth, so it would have cooled slightly quicker - but not much quicker.

Perhaps the collision that produced the Moon might have injected a pulse of energy into the Earth's interior, keeping it hotter for longer? The iron core of the impactor would have sunk towards the center of the earth, merging with Earth's core.
See: https://en.wikipedia.org/wiki/Giant-impact_hypothesis
Earth's radius is ~1.87 times that of Mars,  This gives it ~3.5 times the surface area, but  ~6.5 times the volume of Mars.  Earth also has a density of 5.52 g/cc to Mars 3.93g/cc,  which makes the Earth over 9 times more massive.   All in all, Earth's mass to surface area ratio is ~2.66 times that of Mars.   Surface area determines how fast the heat can radiate away and mass determines the total heat contained for objects of the same temperature and like make up.  Thus Mars should be expected to lose its heat fairly quickly compared to the Earth
The following users thanked this post: evan_au

53
Physics, Astronomy & Cosmology / Re: How long could a 400kg space probe last in an antimatter plasma?
« on: 08/12/2018 22:09:15 »
Quote from: acsinuk on 08/12/2018 19:29:25
Thanks Evan, Is there any sign of a signal yet as I think 4 December was mentioned as date of expected reconnection??
Perihelion took place on on Nov 5, and the probe passed it first post perihelion health test on Nov 16.
It is presently heading away from the Sun before looping back for the next perihelion pass in April

Here is a link showing where the probe is at this time:

http://parkersolarprobe.jhuapl.edu/The-Mission/index.php#Where-Is-PSP
The following users thanked this post: evan_au

54
Plant Sciences, Zoology & Evolution / Re: Why so many flowers have 5-fold simmetry?
« on: 06/12/2018 16:09:52 »
The number of petals on many flowers follow the Fibonacci series: 1,2,3,5,8,13,21,34,58...
This has to do with the most efficient way for the flower to grow out from the center.  In many cases the center of the flower consists of seed that grow outward from the center.  The way to make the best use of the space is to grow in a pattern called a Fibonacci spiral, like this.
https://www.mathsisfun.com/numbers/images/phi-flower.jpg
This results in the out-most layer consisting of a number on the Fibonacci series, and this is were the petals originate. It is a geometry issue, but one sealing with how to maximize use of space, and not molecular bonds.
The following users thanked this post: evan_au

55
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.
The following users thanked this post: Petrochemicals

56
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.
The following users thanked this post: Bill S, jeffreyH, Colin2B

57
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
The following users thanked this post: Petrochemicals

58
Physics, Astronomy & Cosmology / Re: Is acceleration downwards due to gravity less when the Moon is overhead?
« on: 14/10/2018 23:17:46 »
Quote from: evan_au on 14/10/2018 21:59:48
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.
The following users thanked this post: evan_au, Petrochemicals

59
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.)
The following users thanked this post: David Cooper

60
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.
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