New answers tagged

1

The chain looks pretty long, only a meter or so is actually in the air. The gravitational force on the runner by the chain is not that big. The runner is bulled back/down in the angle that the chain attaches to the back. The chain cannot convey any torque or transversal force. So there are only a few kilograms of weight directly pulling on the runner. A lot ...


6

The temperature limit for laser cooling is not related to gravity but to the always-present momentum kick during absoprtion/emission of photons. Ultracold atom experiments typically use laser cooling at an initial stage and afterwards evaporative cooling is used to reach the lowest temperatures. In evaporative cooling the most energetic atoms are discarded ...


0

What is more fundamental: Geometry and Topology or physical matter? None of the above. Like CuriousOne said, geometry and topology are mathematical disciplines rather than real objective things. And whilst people talk about electrons and quarks as "fundamental" particles, they aren't really fundamental because after annihilation they aren't there any ...


1

Well when there is no answer available, I do not think it hurts to try to come up with one. As far as existence is concerned, Geometry and Topology can exist without physical matter, but not vice versa. In fact, there is so much empty space without matter, but no matter without being in space. Empty space has some Geometry and Topology. However, to ...


1

That's fairly small for an object. It wouldn't have any significant gravitational effect on the moon or the earth. Tidal effects go as the cube of the distance. So the sun has about half the tidal effects of the moon. If this object were in low earth orbit (400km altitude), then the relative tidal effects on the surface when it is overhead would be about ...


3

You can extract geothermal energy from the interior, but it's not an infinite source of energy. Extracting geothermal energy will make the core cool down a bit faster than it otherwise would (so, in case of a perfectly insulated planet this would be the only source of heat loss), the drop in pressure would make the planet shrink a bit more, so you then get ...


2

It is the "infinite source" that is giving the downvotes. As the answer by Ramchandra Apte states, gravitational compression is a one way street as far as energy goes. But of course a complicated state like a planet has energy exchanges with its sun, and one can get transformation of sun energy to gravitational energy as in hydroelectric power.. One can ...


1

The heating is one-time as once the planet reaches equilibrium with gravity (i.e. compression force from gravity equals normal force), no more compression with occur. Basically the planet will compress one-time and that's it. (In reality, this occurs when the planet forms and its various pieces join together)


1

The following interpretations are taken from Thorne [2014]. Chapter 17, entitled Miller's Planet, discusses the issue of the large waves on the water planet in the movie Interstellar. There Kip mentions that the waves are due to tidal bore waves with height of ~1.2 km. In the appendix entitled Some Technical Notes, Kip estimates the density of Miller's ...


4

For a given mass the gravitational attraction remains the same -- but only if you are far away. For example, the surface gravity of Sol, our sun, is $274$ $ m/s^2$, about 28 times the surface gravity of Terra, which is $9.8$ $ m/s^2$. But as the material is compacted, the surface gravity increases: this is because the effective mass can be treated as ...


2

If cosmological expansion applies on the scale of the earth moon system, then in some short period of time $\delta t$ the distance between the earth and moon increases from $r$ to $r+\delta r$. So the force of gravity between the bodies changes to: $F+\delta F=\frac{GMm}{(r+\delta r)^2}\approx\frac{GMm}{r^2}\left(1+\frac{\delta ...


0

You can indeed balance the horizontal forces at each point, and the sum of vertical components should equal the weight. That does seem to leave you with an over constrained problem (four equations with three unknowns) which will only have a solution when the angles are chosen "just so". If one of the angles was not given you could solve. Pick one and prove ...


0

Edit I was wrong. Look at Floris's answer. If you split it into sections it should be easier than combining it into one. If you have $F_1$ , $F_2$, $F_3$ as the tensions of the strings $$ F_{1x} = F_{2x} $$ $$ F_{2x} = F_{3x} $$ $$ F_{1y} + F_{2y} = 12g $$ $$ F_{2y} + 7g = F_{3y} $$ Then you can use trigonometry and then eliminate. Edit: Eliminate ...


0

Somebody correct me if I'm wrong but I do believe you would have five tensions. There would be the tension in the one rope joining the two ropes holding the masses, the tension in the two ropes from the intersection to the mass, then the tension in the two rope from the top of the diagram to the intersection of all the ropes. The tension in the rope from ...


0

I know what you say, that's How I did initially, and it gave me a wrong solution. I will post the exercise to you solve by yourself. If I got the same system, without the rope, considering there is friction between the blocks, and A and the ground, is there friction between the blocks while I'm pulling block B but the system doesnt start moving?


2

You can use these free body diagrams:


2

In general relativity, the field equation relates the metric (through the associated curvature tensor) to the stress energy tensor $T^{\mu\nu}$. This can be interpreted as a flux of energy and momentum in spacetime (i.e. integrating $T^{\mu\nu}$ over a spacetime hypersurface, like a three dimensional hypersurface of constant time, tells you the rate at which ...


2

Why does a mass attract all the masses around it? Because this is what has been observed . No apples falling up have been observed. Why should't it repel Because no repulsion of masses has been observed up to now. There exist experiments at CERN where the gravitational behavior of antimatter is probed and maybe in the future there will be an ...


-1

In refference to: http://physics.stackexchange.com/a/250821/84895 The reason you are sticking to the floor right now is that the shortest distance between today and tomorrow is through the center of the Earth.


3

This is actually a famous theorem known as the Einstein Equivalence Postulate (sometimes Equivalence Principle). It's true that since Earth is spinning, acceleration in a spacecraft isn't quite the same situation we experience daily, but in general, yes, gravity is indistinguishable from uniform acceleration. Specifically, if you are in a box with no windows ...


1

The time dilation is not dependent on the gravitational force, but rather on the gravitational potential energy. As long as we are well away from black holes we can use an approximation called the weak field limit. In this approximation the time dilation relative to an observer at infinity is given by: $$ \frac{dt}{dt_0} = \sqrt{1 + \frac{2\Phi}{c^2}} $$ ...


6

That quote requires some modification for it to make sense: "General Relativity basically says that the reason I am sticking to the floor is that the path of maximal aging between 'here now' and 'here tomorrow' is through Earth's center."


-1

Which comes first for gravity: mass or space-time? Neither I'm afraid. As Einstein said, the mass of a body is a measure of its energy-content, and it's actually a concentration of energy that causes gravity. Hence a massless photon has a (very weak) gravitational field. On top of that there's a bit of an issue with spacetime as per CuriousOne's ...


41

That is awesome! And it makes complete sense too! (other than a possible misusage of the word "distance"). Let's have a look at the equations of motion of you in Earth's curved spacetime, assuming that your feet are not touching the ground: $$ \frac{\mathrm d^{2}x^{\mu}}{\mathrm ds^{2}}+\Gamma^{\mu}_{\nu\sigma}(x(s))\ \frac{\mathrm dx^{\nu}}{\mathrm ...


11

What GR says is correct: the straight line between, say, London today and London tomorrow is not the curve that spends all the time between in London: whether it actually passes through the centre of the Earth I'm not sure, and it depends on how fast you are moving as well as where you are. The caveat is that the straight line (geodesic) not the shortest ...


7

It makes sense as a "visual" description. In GR, free particles with mass move on time-like geodesics. A common description of geodesics are such curves that locally minimalize path length, but this desciption comes from Riemannian geometry, not Lorentzian geometry, which GR is. In Lorentzian geometry, timelike geodesics are those that locally maximalize ...


1

Let's ignore the planet in the middle for now and just consider a big drop of water. The mass of water is proportional to the radius cubed: $$ M = \tfrac{4}{3} \pi r^3 \rho $$ and the surface gravitational acceleration is given by: $$ a = \frac{GM}{r^2} $$ so substituting for $M$ we get: $$ a = G\tfrac{4}{3} \pi \rho r $$ So the surface gravity ...


0

$f(a)$ is not an independent degree of freedom from $R$ or from $a$. You have a system of equations here, and if you solve for the exact form of $a(t)$, you are not likely to admit solutions that have $f(R(a(t))) = 0$ for finite $t$. If you do, however, it's likely an indication of geodesic incompleteness of the solution.


0

If we were discussing electromagnetic forces, you might be asking whether electromagnetism causes the forces or vice versa. To make this less of a chicken-egg question, I'd assume "electromagnetism" here means the existence of nonzero values for electric charges and the electric and magnetic fields. The electromagnetic force on a particle of charge $q$ of ...


1

As general relativity says, gravity is nothing but a geometry of space time. The bending of spacetime is caused by mass and energy distributions, which we see as gravity. Both accounts to the same thing. The spacetime is affected by mass as well as condensed energy. It could successfully explain why there is gravity and almost everything related to that. But ...


2

Per Einstein's theory, they both are one and the same thing. There is some, yet not known property of space, and mass, that causes this phenomenon. Even if the property becomes known today, the question will be why that property exists?


0

Keep in mind that pressure is not force. Also, pressure is not mass per area. Pressure expresses the ratio of differential force exerted normal to a surface per differential area, and is expressed in units of force per area. A pressure of 100 kPa exerted on a surface means that on 1 cm$^2$ of that surface there is a total force of 10 N normal to the ...


2

An example of the sort of system you describe would be the Earth and the Moon, with the Earth playing the part of your mass $m_1$ and the Moon $m_2$. Neither of these are point masses, but courtesy of Gauss' law we know that the gravitational field of a sphere is the same as the gravitational field of a point mass provided you are farther away than the ...


2

The field of quantum black holes is an hot topic of research right now, and the firewall proposal is still being debated. I have the feeling that no one really take the proposal seriously. By saying this I don't mean that it was a bad paper, on the contrary it's a nice thought experiment that forced us to think even more about the black hole information ...


3

Consider the virial theorem, which says (ignoring complications like rotation and magnetic fields) that twice the summed kinetic energy of particles ($K$) in a collection of particles (could be the gas in a star, could be stars in a star cluster) plus the (negative) gravitational potential energy ($\Omega$) equals zero. $$ 2K + \Omega = 0$$ Now you can ...


3

This is an active (hot?) topic of research, in fact I attended a workshop on the subject just last week. In brief, no one has found a dark matter (DM) halo yet that does not host a galaxy, though we would very very much like to! The first reason it's so difficult to find a DM halo that does not have a galaxy is that a common working definition of a galaxy ...


3

Basically, it's a consequence of negative heat capacity. Gravitationally bound systems can (often do) behave such that adding energy results in reducing temperature, and vice versa. You can understand this intuitively if you consider a simple two-body system: adding energy to the system causes the orbits to expand, and bodies on larger orbits move at lower ...


0

The drag force due to the atmosphere would be larger for the larger raindrop, resulting in a reduced acceleration. Therefore, the smaller raindrop would start moving faster in time. Eventually, you will have two rain drops falling at terminal velocity. The faster rain drop will be the one with less surface area perpendicular to the motion. edit: Sorry, ...


2

I'm not sure what you mean about "assuming there is no rain" but larger raindrops do have a faster terminal speed than smaller drops. EDIT: Here's a whiteboard video using Stoke's law to explain why.


18

Interstellar space is an excellent vacuum, but it's not a perfect vacuum. For example Earth is constantly bombarded with protons from the solar wind, which stream outward uninterrupted until the heliopause when matter from other stars becomes more dominant. If there were, say, an antimatter star nearby, the place where its stellar wind of antiparticles met ...


2

In its superconducting phase, it'll have exactly the same effect on magnetic fields as any other superconducting object like the Meissner-effect. It won't affect gravity other than by its mass - because electricity and gravity are separate things.


0

The weight of the person on the swing is causing gravity to pull them down you could swing in a full circle if you had enough force


2

You mean go all the way around? It could if you had enough force to overcome gravity and like a tether ball swing all the way although most humans do not have the strength to apply the force needed to push another or them selves to a full revolution around the bar of a swing with out a jerk, but if the chain was replaced with a solid bar to prevent jerking ...


0

why it doesn't it swing fully ? I think the question does point to the fact that swings which are usually available in the park does not provide free swinging , low amplitude of the swing and needs constant pushing. All the above is related to energy dissipation of the initial potential energy provided to the swing- and the dissipation is at the hinges ...


2

I would guess that the argument applies to the case where the frequency of the gravitational wave is low compared to the natural frequency of the stick, or put another way, the rate of change of length due to the gravitational wave is small compared to the speed of sound in the stick. If you take some line normal to the GW and of length $\ell_0$ then the GW ...


3

In an expanding universe, the farther away an object is, the faster is it's recessional velocity relative to an other object at distance $r$: $$v_{rec}=H(t) \cdot r$$ with $H(t)$ as the current Hubble parameter. On the other hand, for every given distance there is also an escape velocity $$v_{esc}=\sqrt{\frac{2\cdot G\cdot M}{r}}$$ If you want your 2 ...


0

Well, in your very ideal system and from the classical point of view, yes the dices (considering they as an electromagnetic neutral system) will be in touch at some time independent of where are they at the beginning.


-4

I was considering this question as well. Neither matter/energy can be created or destroyed, only converted from one form to the other. Consumption of star stuff by a singularity would convert that star stuff into star energy, but with no other way of expressing that energy there seems no choice but for nature to convert with 100% efficiency that matter into ...


-3

The constant is only $8\pi$ when you're working in units where $G=1$. The number $G$ was originally defined as the constant of proportionality in Newton's $$F=G\frac{m_1m_2}{r}.$$ So in some sense the only thing we can do is to pass to the Newtonian limit and calculate the force between two point masses.


1

For each step in walking; In a horizontal road: F=ma, And in an inclined road: F-mgsinθ=ma. So you need more power (greater F) when you are walking up a hill. It may be useful that we know maximum force F we can use when we are walking is the limiting friction force.



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