They do not “accelerate”, but there is still an acceleration due to gravity or the curvature of space time or whatever we choose to call it these days. Therefore, they “feel” a force when something (like the ground) is opposing the acceleration. The force is the effect, not the cause.
If we try to mix Newtonian mechanics with general relativity we end up with a hodgepodge.
In Newtonian mechanics, free-fall is the motion that results when gravity is the only force acting on an object.
However, in the curved spacetime of general relativity, an object in free-fall has no force acting on it.
The second statement raises the question, "Why does a free object which is initially at rest in curved spacetime begin to accelerate?".
I read that general relativity textbooks are not clear on the answer to the above question.
I simply have to accept that an object in free-fall has a spatial trajectory which is explained in terms of spacetime geodesics and the energy of curved spacetime - come back Newton, all is forgiven! 😀
P.S. Bonsai may be interested in reading this article on "Gravity - from weightlessness to curvature":
https://www.einstein-online.info/en/spotlight/geometry_force/
In Newtonian mechanics, free-fall is the motion that results when gravity is the only force acting on an object.
However, in the curved spacetime of general relativity, an object in free-fall has no force acting on it.
The second statement raises the question, "Why does a free object which is initially at rest in curved spacetime begin to accelerate?".
I read that general relativity textbooks are not clear on the answer to the above question.
I simply have to accept that an object in free-fall has a spatial trajectory which is explained in terms of spacetime geodesics and the energy of curved spacetime - come back Newton, all is forgiven! 😀
P.S. Bonsai may be interested in reading this article on "Gravity - from weightlessness to curvature":
https://www.einstein-online.info/en/spotlight/geometry_force/
There is some deformation in the flesh, yes... but ALL of the flesh is being acted upon gravity.. pulled downwards.
Again, the ground is creating a resisting force and there is a transfer of energy from gravity pulling the body down and the ground resisting it.
Remember that the ground is NOT solid either when you look at its molecular and atomic structure.
HEAT.
Indeed, the crust of the Earth is also being pulled downwards by gravity... so that heats up the center. (Actually, with plate tectonics it gets more complicated ).
Again, the ground is creating a resisting force and there is a transfer of energy from gravity pulling the body down and the ground resisting it.
Remember that the ground is NOT solid either when you look at its molecular and atomic structure.
HEAT.
Indeed, the crust of the Earth is also being pulled downwards by gravity... so that heats up the center. (Actually, with plate tectonics it gets more complicated ).
The atoms at or near the surface of the supporting structure could be said to act like little springs due to the electromagnetic interactions between them.
A very simple analogy of what's happening at an atomic level is when you lie on a bed.
The force of gravity acting on you squashes the bed springs and they react, providing an upward force to support you.
And squashing the springs will result in a temperature rise, however slight.
Interestingly... at the mass near the center of the Earth.... there should be no gravity there, huh?
So all the drawings about 3d representations of a "gravity well" are wrong... because the maximum gravity is found on the surface of the Earth. Once you go below the surface, mass starts to experience a gravity pull from all sides, including from above. Hence, at the center of the Earth's mass, the pull of gravity from Earth's mass cancels itself.
So all the drawings about 3d representations of a "gravity well" are wrong... because the maximum gravity is found on the surface of the Earth. Once you go below the surface, mass starts to experience a gravity pull from all sides, including from above. Hence, at the center of the Earth's mass, the pull of gravity from Earth's mass cancels itself.
What if the Earth were hollow?
I assume there would be gravity on the inner surface of the Earth.
Or how else would Tarzan be able to have so much fun there? 😀
I assume there would be gravity on the inner surface of the Earth.
Or how else would Tarzan be able to have so much fun there? 😀
Well, if the Earth had a big hole in the center, then Tarzan would be upside down, his feet towards the surface.
But, depending on how big the hole is, in relation to the size of the Earth, he would be much lighter than on the surface. The bests he fought would be tall and spindly and walk slowly. Waterfall would fall slowly and the mists of water would form long lasting fog banks.
I don't think those science fiction writers were physicists.... not like Arthur C. Clark, Asimov, David Brin, Neil Stephenson, etc.... at least those guys follow the Physics or warp it in somewhat logical ways.
But, depending on how big the hole is, in relation to the size of the Earth, he would be much lighter than on the surface. The bests he fought would be tall and spindly and walk slowly. Waterfall would fall slowly and the mists of water would form long lasting fog banks.
I don't think those science fiction writers were physicists.... not like Arthur C. Clark, Asimov, David Brin, Neil Stephenson, etc.... at least those guys follow the Physics or warp it in somewhat logical ways.
We started with me asking how F=ma would describe the force on an object at rest on the surface of the Earth and you then went on to talk about an 'upward force' and a 'downward force' represented by your diagram with the vectors and I rejected the notion that there was an upward force but agreed there was a contact force. We now seem to have progressed to W=ma and a downward force exerted by an object caused by gravity. Wasn't this what I said earlier?
BTW, the idea of fictitious forces appears to come out of Newtonian physics and it is discussed in the book I linked to earlier. Einstein did not like the idea and rejected it during the development of GR, but made sure GR produced the same results as Newtonian physics where it mattered.
It would appear that I have a problem understanding your statements because your nomenclature differs from that which I am familiar.
Indeed you are correct, as is confirmed by Wikipedia:
"A fictitious force is a force that appears to act on a mass whose motion is described using a non-inertial frame of reference, such as a linearly accelerating frame or rotating reference frame. Fictitious forces are invoked to maintain the validity and thus use of Newton' second law of motion in frames of reference which are not inertial."
Fictitious forces are no longer necessary in general relativity since the physics is explained by the geodesics of spacetime.
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Centrifugal force is one of the reasons why you appear to weigh less at the Earth's equator than at its poles.
At the equator, you will experience a centrifugal force because you are in a rotating reference frame.
Since centrifugal force acts outwards from the centre of rotation, it tends to cancel out a little bit of Earth's gravity acting on you.
This makes your perceived weight at the equator slightly smaller than at the poles.
How much smaller? By about 0.4%. This would mean that a 150 lb person will appear to be a more slender 140 lb 10 oz!
And that's before we take into account the other reason which is that the rotation of the Earth makes it bulge out at the equator. This contributes a further weight reduction of around 0.1% due to your greater distance from the centre of the Earth.
At the equator, you will experience a centrifugal force because you are in a rotating reference frame.
Since centrifugal force acts outwards from the centre of rotation, it tends to cancel out a little bit of Earth's gravity acting on you.
This makes your perceived weight at the equator slightly smaller than at the poles.
How much smaller? By about 0.4%. This would mean that a 150 lb person will appear to be a more slender 140 lb 10 oz!
And that's before we take into account the other reason which is that the rotation of the Earth makes it bulge out at the equator. This contributes a further weight reduction of around 0.1% due to your greater distance from the centre of the Earth.
Good exercice: The gravity inside an hollow sphere.
Answer: Zero gravity everywhere inside the cavity. If i remember correcty from 1960 studying.
Tarzan would be floating in zero gravity.
Well remembered, mchambin!
The mathematics will be a doddle to you and is available to interested parties in this (safe) link:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html
However, I shall add that the field inside a hollow Earth would be zero only if there are no other masses present. The hollow Earth would not shield Tarzan from the gravitational pull of the Moon for instance, so he might not float quite as freely as you suggest.
P.S. Thanks for spoiling one of my favourite recurring fantasy themes!
The mathematics will be a doddle to you and is available to interested parties in this (safe) link:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html
However, I shall add that the field inside a hollow Earth would be zero only if there are no other masses present. The hollow Earth would not shield Tarzan from the gravitational pull of the Moon for instance, so he might not float quite as freely as you suggest.
P.S. Thanks for spoiling one of my favourite recurring fantasy themes!
You are right about the Moon pulling Tarzan.
But, what about Tarzan pulling the Moon ?
This reminds me a home work my eldest son had, while attending highschool:
Given à mass of 1 kilo gramme at the surface of the Earth, find the pull by the Moon.
He found zillions of Newtons......And could not see, it was obviously wrong.
But, what about Tarzan pulling the Moon ?
This reminds me a home work my eldest son had, while attending highschool:
Given à mass of 1 kilo gramme at the surface of the Earth, find the pull by the Moon.
He found zillions of Newtons......And could not see, it was obviously wrong.
That must illustrate the perils of using a pocket calculator!
If we use F = GMm/R² to calculate the pull of gravity between a mass of 1 kg and the Moon's mass of 7.35 x !0^22 kg, separated by a distance of 384,400 km, the result is 0.000032 N.
If we estimate Tarzan's mass as 80 kg, the pull of the Moon on him would be 0.002656 N.
If we use F = GMm/R² to calculate the pull of gravity between a mass of 1 kg and the Moon's mass of 7.35 x !0^22 kg, separated by a distance of 384,400 km, the result is 0.000032 N.
If we estimate Tarzan's mass as 80 kg, the pull of the Moon on him would be 0.002656 N.
Interestingly, in one of my Motion Mountain books, Christoph Schiller talks about what would happen if the Earth stopped rotating. Apparently the whole of Europe would be under about 4000 metres of water. Although the bulge at the equator is not huge, it stores up an enormous amount of water purely due to centrifugal force!
(no word from @system7 - I hope all is ok with him)