Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Petrochemicals on 20/10/2018 20:01:42

As in the gravity on the surface of the earth, you are attracted to the earth, but can you be attracted to the sun simultaniously OR is it correct to say that the gravitational attraction toward the earth is altered by the sun ?
This has a bearing in my mind on the action of gravity bodies. That is to say is the atmosphere all attracted to the sun or is the atmosphere less compressed by the gravity of earth.
Or is a third way correct, is none of the above true and attraction is wrong either way and the correct way of understanding it is that space is less curved in someway ?

The force of gravity acts as a vector, which means that it has a strength and a direction.
To find the total force of gravity acting on you, you have to add up all the forces acting on you, and there are mathematical rules for doing this.
The biggest force is the force of Earth's gravity.
 The next biggest is the force of the Moon. But we are on the Earth, and the Earth is orbiting the Moon, ie the center of the Earth is in "Free Fall" around the Moon. So we have to subtract the force of the Moon on us from the force of the Moon on the center of the Earth.
 The next biggest force is from the Sun. Similarly, the Earth is in free fall around the center of the Solar System.
 Much smaller forces come from the mountain chain near your city, the person in the next room, Mars, Jupiter, Saturn, etc. And these forces have different strengths, and are all pointing in different directions
Mathematics provides the method to add up these 10 (or more) different forces to produce a total force.
But the gravitational force of the Earth is by far the largest  it takes very sensitive equipment to detect any deviation from the force of Earth's gravity.
If you were an astronaut in the International Space Station, you are in free fall around the Earth, and you are near the center of gravity of the ISS, so the gravity of the Earth, Moon and the Sun exert no measurable force on you.
See: https://en.wikipedia.org/wiki/Euclidean_vector

That is to say is the atmosphere all attracted to the sun or is the atmosphere less compressed by the gravity of earth.
If the sun put out a uniform gravitational field, both the atmosphere and the Earth would be equally attracted to the sun and there would be no compression effect on the atmosphere. But the field is not entirely uniform, and the pull is greater than the pull on Earth on the atmosphere on the sun side, and it is pulled less than Earth on the dark side, so both sides are less compressed than elsewhere on Earth. This is the tidal effect, the same thing that drags the water up and down.
The moon, while exerting a considerably weaker gravitational pull on Earth compared to the sun, has a considerably larger gradient from one side of Earth to the other, so the tidal effect of the moon is stronger than the effect from the sun.
The force of gravity acts as a vector, which means that it has a strength and a direction.
To find the total force of gravity acting on you, you have to add up all the forces acting on you, and there are mathematical rules for doing this.
Vector addition, yes.
The biggest force is the force of Earth's gravity.
 The next biggest is the force of the Moon.
The sun exerts a considerably higher force than does the moon, as is brought up repeatedly in the twotidesaday thread.
If you were an astronaut in the International Space Station, you are in free fall around the Earth, and you are near the center of gravity of the ISS, so the gravity of the Earth, Moon and the Sun exert no measurable force on you.
See: https://en.wikipedia.org/wiki/Euclidean_vector
You accelerate due to the sum of those forces, so the force is measurable, just not locally. You have to look out of the window to measure the effect of the force.

Can gravity be said to act in two opposite directions at once ?
My australian friends and I hope so.

The gravitational field around an object is three dimensional. As Bored Chemist has explained quite succinctly. If we define spheres around the earth of increasing radius we would find that the force of gravity diminishes with increasing distance from the earth. The sun is so much further away from the earth than the moon that it's overall influence is diminished. This is determined mathematically as others have demonstrated here and elsewhere in other topics. It is worthwhile to pick up some books on the subject. Expand your mind!

Another thing to consider. The force on an object at any point in a gravitational field does not vary with time. That is, if the source object is stationary, in other words moving with inertial motion. This means that you do not require time to describe a geodesic and line elements suffice. This reduces to geometry. If we introduce time then it cannot be absolute and does not fit with nonrelativistic quantum mechanics.

Well the spheres is the area i am thinking about. If we are trapped within the gravitational field of earth (no escape velocity), is the sun putting a very thin long range point attraction, or is the gravitational sphere of earth altered in a more circular way.
If it is a sphere singular to the earth, it is weird to think that even though the mass of earth does not change, the gravity field of earth is somehow less when the sun is over head. There must be a point where the gravitational attraction of earth is over come and neither the gravitational sphere of the sun nor earth is gravitationaly a certain attraction( the escape of the solarsystem not withstanding)

At any point in the universe, if you put a tiny test mass there it will move off (perhaps very slowly) in some direction or another.
If you happen to be near Earth then it falls pretty much "down".
If you are at a lagrange point then WRT the Earth and Sun it moves in orbit with the Earth.
In deep space you might have to wait a long time to be able to detect the movement at all.
The direction in which it moves is the (vector) sum of all the gravitational forces acting on it.
If you do this lots of times and draw a small plane perpendicular to the direction the test mass falls, and then you join all the planes together you get a surface which is an equipotential surface (roughly speaking).
Near the Earth that's roughly spherical. But there is only 1 gravitational field, and it fills the universe.

Here's a map (not to scale), marking the lines of gravitational potential for the SunEarth 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

As in the gravity on the surface of the earth, you are attracted to the earth, but can you be attracted to the sun simultaneously OR is it correct to say that the gravitational attraction toward the earth is altered by the sun ?
I think the comments are that gravitational attraction of sun earth moon all act at same time & are additive & subtractive.
I like to consider that earth's escape v is 11.2 kmps, the sun's is 617.5 kmps but at earth's orbit it is 42 kmps, the moon's escape v is 2.38 kmps but at earth it is 0.16 kmps. I reckon that the gravitational forcesaccelerationsg's on the surface of earth are due to the sun (42) then earth (11.2) then moon (0.16) in those proportions (or the square root of)(or the square of)(i am too lazy to think about this)(but anyhow u get the drift).
However Allais i think suspected that it aint that simple  he thort that praps at eclipse the alignment of the sunmoonearth somehow briefly did something peculiar to gravity & nett g on earth. However i think that the peculiar Allais effect on pendulums might be simply due to the moon's shadow moving across the earth & cooling the atmosphere, hencely changing the disposition of the mass of the atmosphere (i dont thing that gravity is changed in some peculiar way at alignment).This has a bearing in my mind on the action of gravity bodies. That is to say is the atmosphere all attracted to the sun or is the atmosphere less compressed by the gravity of earth.
I suspect that the three gravitational effects on the atmosphere might be fairly similar to the three effects on water etc that gives us the water tides that we see every day.Or is a third way correct, is none of the above true and attraction is wrong either way and the correct way of understanding it is that space is less curved in someway ?
Understanding gravity is an impossibility for most. Saying that gravity is due to curvature of spacetime due to the presence of mass doesnt explain much even if true (which it aint). This supposed curvature is supposedly due to the effect of mass on time (time)  plus the supposed effect of mass on dimension (space), due to radial length contraction near mass. I think here that Alby means that objects always go straight, but due to TD & LC we seethink that they go in a curve (i might have this wrong). So there u have it, now u understand gravity. No need for a micro theory as to how the smallest massive particle can possibly affect time & distance, no need for a micro theory as to what time & distance are. Einsteinians help u to swallow that krapp by throwing in a sweetener about the fabric of spacetime. Or  hey everybody look over there at that blackhole  what blackhole, i dont see anything  yes, exactly, proof, anyhow aint it wonderful about LIGO finding gravitational waves  now if u look over here ladies & gentlemen blah blah blah....
Now all u need do is to teach that spacetime krapp to skoolkids to perpetuate the cycle of ignorance.

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The biggest force is the force of Earth's gravity.
 The next biggest is the force of the Moon.
The sun exerts a considerably higher force than does the moon, as is brought up repeatedly in the twotidesaday thread.
I have to say i agree with evan despite the maths. Im sure the black hole in the galaxy exerts more force than the sun! But it just doesnt seem right

The Sun is about 2* 10^30 Kg, and is about 150,000,000 km away
The moon is about 7 *10^22 Kg and is 384,400 km away.
So the Sun is about 10^8 times heavier, but 400 times further
Even allowing for squaring the distance, the Sun's attraction is about 600 fold bigger.
(The moon's effect on tides is bigger because the gravity gradient is bigger)
The black hole at the centre of the galaxy is (apparently) 4.3 million times the mass of the Sun. And it's about 1,000,000,000,000,000,000 km away
So the effect is
(1,000,000,000,000,000,000 /150,000,000) ^2 i.e.
44,444,444,444,444,444,444.
times less due to the distance and only
4,300,000 times bigger due to the mass.
So the distance wins out the Sun's effect is much bigger (about ten million million fold, if I got the arithmetic right) than the black hole's

The sun exerts a considerably higher force than does the moon
the Sun's attraction is about 600 fold bigger (than the Moon)
We need to clarify something that is tripping up some of the discussion and calculations above (and tripped me up, too, a few days ago).
Newton's gravity is described as an "Inverse Square Law", which means the force is proportional to 1/r^{2}. By this measure, the Sun's gravitational influence on Earth is much bigger than the Moon's gravitational influence on the Earth.
But you can only directly measure Newton's gravitational force with giant bathroom scales if the two objects are at rest (stationary) relative to each other.
In reality, any planets, asteroids, comets or moons that we see today were not at rest relative to each other in the past, or they would have crashed together and be no more. All these remaining Solar System objects are in elliptical orbits, in "free fall". So the effects of Newton's gravity are cancelled out, when measured at the center of gravity of the object (eg at the center of the ISS).
Now we don't live at the center of gravity of the Earth (it's rather hot and pressured for my tastes), so we get a small gravitational deviation from free fall, called a tidal effect. As the difference between two points on Newton's gravity, it is an "Inverse Cube Law", proportional to 1/r^{3}. By this measure, the Sun's tidal influence on Earth is slightly smaller than the Moon's tidal on the Earth, due to the Sun's greater distance.
So some of the contradictory conclusions above were both right  but some were talking about gravitational attraction (1/r^{2}) which we don't directly feel on the surface of the Earth because most of it is cancelled by Earth's orbit.
While others were talking about tidal effects (1/r^{3}) which we do feel on the surface of the Earth, even though the effect is slight.
In future posts, please clarify whether you are talking about gravitational forces or tidal forces, and why you think they are more important in this context.
Can gravity be said to act in two opposite directions at once ?
Both gravitational forces and tidal forces can be handled as a vector addition at a single point.
And tidal forces can be calculated as a vector subtraction of gravitational forces between the center and surface of the Earth (two different points).

Ok, ill try and explain it better.
if the moon affects the gravity acceleration of the earth, the earths gravitational field must either extend less, at which point you have to wonder where it has dissapeared to, likewise the moon which should be extending less far, so inbetween the moon and the earth there would be a place where the gravitational attraction of both bodies would be smaller and perhaps a place of denser Space that was gravitationally neutral regards both planets. This would mean gravitational acceleration
Or it is less powerful for a greater distance around the rear side of the moon acting in unison with the moons gravitational field and acceleration working in two opposing directions at the same place and time at the near sides

Ok, ill try and explain it better.
if the moon affects the gravity acceleration of the earth, the earths gravitational field must either extend less, at which point you have to wonder where it has dissapeared to, likewise the moon which should be extending less far, so inbetween the moon and the earth there would be a place where the gravitational attraction of both bodies would be smaller and perhaps a place of denser Space that was gravitationally neutral regards both planets. This would mean gravitational acceleration
Or it is less powerful for a greater distance around the rear side of the moon acting in unison with the moons gravitational field and acceleration working in two opposing directions at the same place and time at the near sides
If you where to graph the gravitational potential of the EarthMoon system, it would (very roughly) look something like this
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The larger dip is the Earth and the smaller dip is the Moon. Note that while the "bump" in between Earth and Moon is higher than the low points of either the Earth or Moon, it is still lower than what it would to either side of the Earth and Moon.
In this graph, down in towards lower gravitational potential (it takes energy to move from lower to higher potential. The strength of gravity at any given point is the "steepness" of the Slope with horizontal being zero net gravity force.
while the at the peak of the Bump between Earth and Moon is horizontal, representing zero net g forces at that point, it would still take energy to move from that point to some point far removed from the Moon Earth pair where gravity also is at or near zero.
The gravitational field of every mass extends for an infinite direction. The net result of gravitational force at any point is due to the combination of all the fields at that point. In most cases, the vast majority of objects contributing to the local gravity are so far away that they don't make enough of a difference to even bother about, and you really only have to concern yourself with objects which are relatively nearby.

This was all put to bed hundreds of years ago by Newton, Kepler et al.
The gravitational force between any two bodies is F = Gm_{1}m_{2}/r^{2} where the symbols have their obvious meaning and G is a universal constant.
F is a vector, that is, it has a magnitude as given by the equation, and a direction, between the centers of mass of the bodies.
If you introduce more bodies, you can solve the equation between any pair, so if m_{1} is your "test mass" you will get a set of vectors F_{12}, F_{13}, F_{14}.....all diverging from m_{1}. You can add them graphically or algebraically to produce one resultant vector, the net force F_{1n} acting on m_{1}.

tidal forces can be calculated as a vector subtraction of gravitational forces between the center and surface of the Earth (two different points).
The center of the Earth and the surface of the Earth are joined by the fairly incompressible rock of the Earth.
I have heard that daily tides stretch the Earth's rocks by about 0.3m in 6,000km, or a stretch of 0.000005%.
So the forces felt by the center of the Earth (in free fall around the Sun) are transmitted to your feet, so you can do the vector sum at a single point  you.
In this sum, the only force that is detectable by the human body is the gravitational attraction of the Earth (and even that we frequently forget about).
The other tidal effects are imperceptible.

tidal forces can be calculated as a vector subtraction of gravitational forces between the center and surface of the Earth (two different points).
The center of the Earth and the surface of the Earth are joined by the fairly incompressible rock of the Earth.
I have heard that daily tides stretch the Earth's rocks by about 0.3m in 6,000km, or a stretch of 0.000005%.
So the forces felt by the center of the Earth (in free fall around the Sun) are transmitted to your feet, so you can do the vector sum at a single point  you.
In this sum, the only force that is detectable by the human body is the gravitational attraction of the Earth (and even that we frequently forget about).The other tidal effects are imperceptible.
I reckon that the wt of an extra say 300mm of water at high tide would make the 6000 km compress more than that there expansion of 0.3mm, say a compression of 0.3mm (or praps a compression of 3mm or 33mm or 333mm even) (not important)(or is it).

tidal forces can be calculated as a vector subtraction of gravitational forces between the center and surface of the Earth (two different points).
The center of the Earth and the surface of the Earth are joined by the fairly incompressible rock of the Earth.
I have heard that daily tides stretch the Earth's rocks by about 0.3m in 6,000km, or a stretch of 0.000005%.
So the forces felt by the center of the Earth (in free fall around the Sun) are transmitted to your feet, so you can do the vector sum at a single point  you.
In this sum, the only force that is detectable by the human body is the gravitational attraction of the Earth (and even that we frequently forget about).The other tidal effects are imperceptible.
I reckon that the wt of an extra say 300mm of water at high tide would make the 6000 km compress more than that there expansion of 0.3mm, say a compression of 0.3mm (or praps a compression of 3mm or 33mm or 333mm even) (not important)(or is it).
You are both wrong ! (I believe)
The earth is liquid core with a crispy coating. These tidal effects will aff3ct that surely. And the gravitational tidal efffect causes heat and expantion.

tidal forces can be calculated as a vector subtraction of gravitational forces between the center and surface of the Earth (two different points).
The center of the Earth and the surface of the Earth are joined by the fairly incompressible rock of the Earth.
I have heard that daily tides stretch the Earth's rocks by about 0.3m in 6,000km, or a stretch of 0.000005%.
So the forces felt by the center of the Earth (in free fall around the Sun) are transmitted to your feet, so you can do the vector sum at a single point  you.
In this sum, the only force that is detectable by the human body is the gravitational attraction of the Earth (and even that we frequently forget about).The other tidal effects are imperceptible.
I reckon that the wt of an extra say 300mm of water at high tide would make the 6000 km compress more than that there expansion of 0.3mm, say a compression of 0.3mm (or praps a compression of 3mm or 33mm or 333mm even) (not important)(or is it).
You are both wrong ! (I believe)
The earth is liquid core with a crispy coating. These tidal effects will aff3ct that surely. And the gravitational tidal efffect causes heat and expantion.
@Petrochemicals  you are right  surprisingly ;)  the effect does cause some heating, however, @evan_au is also correct, because although the core is liquid it is also very incompressible. @mad aetherist is forgetting that the water or rocks are lifted by a tidal force opposing gravity.