New answers tagged

-1

For a very simple answer, the sun is bigger but the average mass density of the sun from its center to the surface is smaller compared to the average mass density of the moon from its center to its surface, it is the reason that the major causes of tides from the moon than from the sun.


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Are there any numerical predictions for G from modern theories What are physics theories? They are mathematical models which fit data, measurements in a laboratory or observations with real numbers. If they only fit data, then they are just mathematical maps of reality. To be a theory they also have to predict new setups, they have to be predictive to be a ...


3

Newton would not say that an individual standing on the Earth represents an inertial frame of reference. An inertial frame is one in which Newton first law applies i.e. an object moves in a straight line at constant speed. Since the observer dropping the apple observers the apple to accelerate that observer's frame is non-inertial. However you are correct, ...


0

That's because the Centrifugal force that is created by the rolling counteracts gravity. If the pilot would do the roll over slower, the water would spill.


1

Because this is exactly the same as swinging a bucket of water in a vertical circle. The water will not fall out, if it is swung fast enough. https://www.youtube.com/watch?v=Zjqrx7wrpJc source When you swing things around in circular orbits, it wants to fly "out of" the orbit. That is why water in the bucket is squeezing against the bucket which is the ...


2

Since you mention the following in one of your comments I'm less interested in Einsteins historical struggles and would love a more modern perspective on how to get to this insight. I hereby unashamedly ignore history, and offer instead a quick plausibility argument. Let's start with the equivalence principle which, loosely speaking, says that a (...


0

The holes you are used to seeing most likely occur on Earth. Since Earth's surface is a 2-dimensional object, holes in it are also 2-dimensional. Given that we live in a 3-dimensional world (spatially/geometrically), these holes would then be traversable; leading to a "pit" below. Since it isn't realistic to have a pit that goes completely through Earth, you ...


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The Quora quote is not useful because it does not tell you how to do any calculations. You need websites which explain how to calculate radius and period of orbit. eg http://www.physicsclassroom.com/calcpad/circgrav http://www.physicsclassroom.com/class/circles/Lesson-4/Mathematics-of-Satellite-Motion A more advanced webpage for calculating general (non-...


9

The zig-zag strategy seems trivially obvious - but it might not be the better strategy in a particular situation. I suggest rather than asking under what conditions this strategy is preferable, you ask under what conditions the counter-intuitive straight-line strategy is preferable. The advantage of zig-zagging is that it presents a smaller "collision ...


0

On an ideal spherical rotating Earth an mass $m$ on the Earth is subjected to the gravitational attractive force $\frac{GMm}{R^2}$ which provides the force required to make the mass move in a circular path $mR\omega^2$ and the force the object exerts on the surface of the Earth $mg$. The force the mass exerts on the Earth has a direction which is not ...


0

The rotation curve exists because of he presence of dark matter and baryonic matter. And we can estimate it using the orbital velocity of the Sun and its distance from the Galactic Centre. The fact that we can do that demonstrates that we can estimate the interaction generated by Dark Matter. That is how its existence was initially suggested. We knew the ...


3

The answer is that we do not know. It is a working assumption that it does, and it is this assumption that leads to an estimate of the total mass and how that mass is distributed - essentially by the application of Poisson's equation for gravitation. $$ \nabla^2 \Phi(r) = 4 \pi G \rho(r) $$ If for some reason dark matter did not couple gravitationally in ...


2

We assume that gravity couples to the stress-energy tensor i.e. the Einstein equation relates the curvature to the stress-energy tensor. In cosmology the only important terms in the stress-energy tensor are the diagonal terms $T_{00}$, $T_{11}$, $T_{22}$ and $T_{33}$. The $T_{00}$ term is the energy density i.e. how much stuff there is per unit volume - note ...


7

They actually don't separate in distance, but Laplace himself asked the same question and made the exact same mistaken assumption that you did, so don't feel bad - you're in good company. The best way to approach the problem is through an expansion in powers of $v/c$, where $v$ is the velocity at which the particles move and $c$ is the speed of light (or, ...


2

This isn't a complete answer, but you might find it helpful anyway. Any corrections are appreciated. Whether this happens in theory I can't say, but I'm pretty sure it doesn't matter in practice. First notice that the effect is going to be negligible unless the bodies are moving close to the speed of light; in other words, the delay in gravity has to be ...


0

Huh, people? :-) He says the main tank is closed! That means it can be pressurised, and the pressure will ofc push out water, depending on pump efficiency. Nothing else needed than a valve that hinders the water from flowing back to the open reservoir when the pump is off. In the pic below a self-adjusting system. - If there is too little water in the ...


0

The bodies don't 'know' anything. If at t=0, the bodies had sufficient velocities relative to each other, the bodies would orbit their centre of mass and if at t=0 the bodies were still with respect to each other, the bodies would fall towards each other. There's still no absolute frame of reference and if you pick any inertial frame of reference, you would ...


2

You should have a look at this paper by the LIGO team, which describes in some detail the fitting process which is used to estimate the system parameters from the observed data. It also gives a brief plausibility argument for why a pair of merging black holes are expected to give up $\sim 5$% of their rest mass as gravitational waves. The parameters are ...


1

I think userLTK's answer pretty much covers your question, but I would add a footnote for clarification. Your question is a bit unclear because you don't say whether you're asking: does the process of contraction lose mass? or is a contracted object lighter once it has reached equilibrium? As userLTK says, the process of contraction does not cause a ...


1

The act of contracting converts potential energy into kinetic energy or heat, so the answer in this case is no difference, though the somewhat hotter contracted planet does lose heat (and mass) faster, but the actual contracting should be zero change in system energy. This is consistent with Newton's energy can't be destroyed theorem. The only energy ...


0

If you randomly assign velocities independently of distance from centre of mass, the simulation will take much longer to settle down to a near-stable configuration. The outer particles which are moving outward will have relatively high kinetic energy, so the configuration will expand further from the CoM, leading to weaker interactions and fewer collisions, ...


0

I was taught that the first Newton's law should be seen as a definition of an inertial frame and a postulate that there exists such a frame. If you were observing the two bodies mentioned in your post and believed in "our gravity", you would think that this "theory of gravity" is probably not true, since the bodies behave much differently than described. ...


0

One different aspect to this in case we are really talking about a living planet. Because gravity really kicks in at that scale, anything the size of a planet is so smoothly round that a polished billard ball feels ashamed for its own imperfection. So if this being is really planet-sized, it better be of really low density ...


0

The only time you cannot define a velocity is if you only have one body in the universe. Two and more bodies define a center of mass and a velocity vector with respect to the center of mass. Two bodies will be rotating about the center of mass if there exists an angular momentum. Conservation of angular momentum makes the orbits stable . If no initial ...


0

You are right to be astonished by, and question, the claim that 1 solar mass equivalent of pure energy was radiated by gravitational waves. I was in a similar position myself in February when they announced the first black hole merger they detected had radiated 3 solar masses (or more than the entire observable universe in that split second). But having ...


2

This is from the Wikipedia article mentioned. I think this is really worth reading, as it not only describes some of the physics associated with the accepted answer and it's citation of Feynman's argument about a bead a stick, but perhaps gives a little insight into the development of the field as well. Feynman's argument Later in the Chapel Hill ...


9

Feynman gave an argument of beads on a string or rod. The passage of a gravitational wave would cause the beads to move in a way similar to the arms of the LIGO interferometer. He argued that the motion might have friction on the string. We might think of this as magnets on a solonoid. If there are magnets at different places on the solonoid their motion ...


0

In order to be detected, the gravity waves have to be converted to some other form of energy (eg mechanical, electrical). So the answer to question 1 is Yes. The answer to question 2 is any article about gravity-wave detectors/detection.


0

Another suggestion: close valve 1 and 2 and turn on pump to make 'fountain' effect. Close valve 1 and open valve 2 and turn on pump to transfer water from reservoir to tank. Close valve 2, open valve 1 and turn off pump, to transfer water from tank to reservoir.


0

The answer is no, simply because the pressure on the check valve on the side of the open pond must be greater than that within the main tank before the valve can open and let water flow from the open pond and the main tank. This will only happen if either: The open pond's water level is higher than that of the open tank; as drawn, the system will simply ...


0

@phillip_0008 Pretty sure this one would work! Basically, no need for valve 3. And wouldn't be able to access valve 2. So split pump inlet between bottom of tank and bottom of reservoir, directly outlet to fountain. Water coming out of fountain sits in reservoir until it gets sucked back in thru pump. makes sense in my mind but let me know if you see ...


0

How about: open valve 2 to transfer water from the reseroir to the tank. Close valve 2 to create the 'fountain' feature. Open valve 3 to transfer water from tank to reservoir.


0

The water level in the pond must be the same as in the tank, so: :-)


11

Thanks to Michael Seifert's answer, I found a paper he referenced: Satellite Relocation by Tether Deployment by G. A. Landis and F. J. Hrach, 1989. By extending a tether radially, a satellite can increase or decrease its orbital speed (pictures below copied from the paper): The principle then can be used to pump an eccentric orbit: Similarly a planet ...


2

In the presence of a gravitational field the hot air raises because it feels a buoyancy force. There is a difference of pressure between the bottom and the top of the hot air portion resulting in an upward force. If there is no gravity, there is no difference of pressure (in principle) so the hot air does not raise. It just expands uniformly. If you are in ...


2

As you points very well, in the lack of gravity there is no hydrostatical separation. I.e. it doesn't matter that the warmer air has a smaller density, there is no gravity which would separate them. But, there are many other effects which would still work, because they aren't caused by the gravity. Most important is the diffusion: Further, in the case of ...


1

Fun question! Try this very simple answer (Newtonian as you asked).. If a planet changes its shape from a round ball into a twine spool like shape (i.e. with a thicker center and elongated thinner ends) and assuming the elongation is done exactly along the radial line to the star (Sun), then for simplicity sake, the amount of mass that gets closer to the ...


6

You can always use the process of tidal acceleration/deceleration. In nature this process might be very slow, such as for the system Earth/Moon. However, you can alway speed it up, by artificially increasing the frequency of the shape oscillations. In a natural system tidal acceleration will stop when the two objects are in tidal locking (both object ...


14

A different mechanism: On a long timescale, by increasing the surface area exposed to the sun (flattening the planet), the radiation pressure would increase, boosting to a higher orbit. Changing the albedo would be a more effective means to the same end but could allow assymetric force as well Either way it would be simpler in a tidally-locked planet. ...


67

If you allow for non-Newtonian gravity (i.e., general relativity), then an extended body can "swim" through spacetime using cyclic deformations. See the 2003 paper "Swimming in Spacetime: Motion by Cyclic Changes in Body Shape" (Science, vol. 299, p. 1865) and the 2007 paper "Extended-body effects in cosmological spacetimes" (Classical and Quantum Gravity, ...


16

This is a really nice question. Conservation of angular momentum tells us that in an isolated system, total angular momentum remains constant in both magnitude and direction. The key here is that the conserved quantity is the total angular momentum: spin+orbital angular momentum. An example: For a planet, angular momentum is distributed between the ...


4

By conservation of momentum and energy, the only possible way to change a planet's trajectory is to eject some (large) mass at high velocity in specific direction, like rockets do. But you are also correct that by increasing the moment of inertia, the rotational speed can be changed. But this cannot influence the movement of center of mass. Edit2: Other ...


0

Ever after Maxwell unified electricity and magnetism, the goal of unification is a holy grail for theoretical physics, for all forces. The macroscopic forces of electricity and magnetism emerge from the underlying quantized level of nature, photons, electrons and all the elementary particles in the standard model table enter with fields in quantum field ...


0

You can, in theory, explore this scenario with the Kerr metric, which will model the spacetime outside your spherical mass. However, the Kerr metric's main difference from the Schwarzschild metric on test particles is on their motions normal to the radius; these effects are much more pronounced than any effect on the motions in the radial direction, the ...


2

Negative mass and by corollary negative energy have some strange consequences. If you have a set of masses in a region of space with a volume $V$ the density of energy is $\rho = \sum_im_ic^2/V$ This defines $T^{00} = \rho$ component of the stress energy tensor. The Hawking-Penrose energy conditions are that $T^{00} \ge 0$. Violations of this creates various ...


0

One has to differentiate between negative inertial mass, which is very unlikely, and a negative gravitational mass charge, which is what you are looking at. The latter has not been ruled out for antimatter. More generally, we have not confirmed, yet, that the equivalence principle holds for antimatter.This leaves room for such hypotheses and it requires ...


0

I suppose you are asking about locally inertial frames? Chapter 2.4: Postulate (2) of general relativity implies that at each point of spacetime it is possible to choose locally inertial coordinates: $\xi^m$ Say you have coordinates $x^\mu$ and you want to transform to inertial coordinates $\xi^m$ which are in locally inertial frame $ds^2=\eta_{m ...


-2

The quantum jumps happen in random directions, as long as there is no additional field. If you have a gravitational field, the probabilities of jumps could change, but still the jumps themselves are random (but distributed differently). Quantum jumps have been shown experimentally by the group of Serge Haroche: Quantum jumps of light recording the birth and ...


3

There are no quantum leaps. This is an errant notion from early theories that was latched onto by the public and continues to be popularized by 'pop science' programming. It is a misunderstanding of quantum mechanics. There are discrete states for negative energy systems, but any Hamiltonian term that allows transitions between them does so continuously. ...


1

In the context of GR and the equivalence principle, given a Lorentzian manifold $(M,g)$, the following comments seem relevant: If the (Levi-Civita) Riemann curvature tensor does not vanish in a point $p\in M$, then there does not exist a neighborhood $U \subseteq M$ of $p$ (and a coordinate system defined on $U$) such that the metric $g_{\mu\nu}$ becomes ...



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