Tag Info

New answers tagged

0

The difference between Caltech's and Crowell's solution to this problem is that Caltech does not take into account tidal effects, which are excluded in the problem statement explicitly: "Also, neglect earth-tide effects." Tidal effects: the influence of the gradient of moon's gravity over the volume of planet Earth, which can be characterized by the ...


1

I posted a link to a summary paper on tides in a comment yesterday. That paper is Agnew, D. C. (2007), "Earth Tides", pp. 163-195 in Treatise on Geophysics: Geodesy, T. A. Herring, ed., Elsevier. That paper contains the answer to your question. I don't know how long that link will last, so I'll summarize some of what Agnew described. This is a summary ...


2

There is no known analytical way to predict this, because the equations are chaotic. So just set up the equations of motion carefully, integrate them properly, and watch what happens. There will be some near misses, and then there will be a collision. You might need to "speed up time" in your model to have it happen in your life time - in a real planetary ...


3

I believe the explanation can be found in Manual of Harmonic Analysis and Prediction of Tides : In deriving mathematical expressions for the tide-producing forces of the moon and sun, the principal factors to be taken into consideration are the rotation of the earth, the revolution of the moon around the earth, the revolution of the earth around ...


0

An overview of how to do the tidal corrections is at http://gravmag.ou.edu/reduce/reduce.html (from Googling 'gravity tide correction'). You might also follow up looking at http://www.applied-gravity.com/gb/html/tides.html. From http://eclipse.gsfc.nasa.gov/phase/phases1901.html one can see that May 4th of 1981 (the Zumberge paper) was a new moon. For the ...


5

The question refers to "laboratory" measurements, i.e., local ones. Such a measurement can only be sensitive to the gravitational acceleration of a test mass relative to the laboratory. For example, if one drops a mass in a vacuum column and measures the time it takes to hit the floor, the acceleration inferred is the acceleration of the mass relative to the ...


13

There is a sort of analog called gravitomagnetism (or gravitoelectromagnetism), but it is not discussed that often because it applies only in a special case. It is an approximation of general relativity (i.e. the Einstein Field Equations) in the case where: The weak field limit applies. The correct reference frame is chosen (it's not entirely clear to me ...


4

There is a gravitational analogue of the magnetic field. See gravitoelectromagnetism and frame dragging on Wikipedia.


0

You might be better off posing this question at a site geared towards Ufology. The entire concept of these beings is inconsistent with the laws of physics and is not in any way grounded in science, let alone a topic suitable for discussion on this site. First, any such ships would indeed be influenced by the gravitational field of objects around them and the ...


2

The reason why we don't see dust particles orbiting comparably tiny rocks within the solar system is just that the rest of the solar system interferes due to how slow such an orbit has to be: the presence of other massive bodies perturbs the orbits of dust particles. the solar radiation deorbits particles smaller than about 1mm. the solar radiation ...


3

This is in addition to LuboŇ° Motl's answer. Due to the friction, the rain drops move with the wind while falling. So, on windy days, with the wind in the back, the cloud will always be behind you. When we say, the falling speed of $10 \frac{\mathrm m}{\mathrm s}$ is fixed, the angle $\vartheta$ corresponds directly to the wind speed over $$\vartheta = ...


3

You can actually see rain falling from distant clouds. The best situation to see it is a hot summer day that gives birth to a relatively isolated cloudburst a few miles away, where you can watch from the top of a hill. Before the rain, the bottom of the cloud is more or less well-defined and cloudy-looking, with same sort of "tufty" texture that you're ...


5

Suppose a rather simple case where the wind is completely uniform, so it has the same strength and direction at all altitudes and all times (highly unlikely)! Then the horizontal motion of the raindrop will be ambivalent to the wind, moving with it from the moment it leaves the cloud until the moment it falls on your head. The consequence is that the cloud ...


3

The droplets are falling pretty much exactly vertically down from the viewpoint of the moving air mass at the given place because friction would almost immediately eliminate any relative motion of the droplets and the air (except for its vertical portion which is driven by gravity). In other words, the horizontal speed in the $x,y$ directions is equal to ...


0

Maybe a qualitative answer motivated from thermodynamics: If you let your black hole rotate, you reduce the number of symmetries of your system, this will decrease your entropy $S$ which is proportional to the surface area. The surface area however is for sure monotonic increasing with your Schwarzschild radius, therefore, if your break symmetries, $r$ will ...


-2

Escape velocity is much easier to understand using the following definition. For the example of earth: Escape velocity is the velocity at which a body at rest at the edge of the universe would impact the earth given only the action of gravitational force between the earth and the body. Since the action is reversible it stands that a body reaching escape ...


0

The majority of the sources that LIGO (and other gravitational wave detectors) are aiming for are astrophysical (e.g. neutron stars, black holes, supernovae, pulsars). The expected cosmological gravitational radiation from standard inflationary models (see this recent article from P. Steinhardt) would be very weak in the LIGO band (10-1000 Hz). There are a ...


1

According to General Relativity, energy is equivalent to inertial mass, and all inertial mass generates gravity. Since electrons have measureable inertial mass, they should have a small influence 0.1% contribution to the gravitational force in neutral matter. That being said however, the question cannot be answered experimentally because the electrical ...


0

It is a supermassive black hole but its mass is nothing compared to the mass of the whole galaxy. So if you remove the black hole, we will not feel any difference because we are so far away. It is not the central black hole that keeps the galaxy together, our sun is orbiting around the center of mass of the galaxy and the supermassive black hole just happens ...


0

You're just comparing forces to forces, Newtons to Newtons. I guess if you want to be really nitpicky, you're talking about force per charge vs force per mass. But when one says that the electric force is much stronger than the gravitational force, one means that in dealing with the usual objects, like electron, the effects of gravity on the electron's ...


3

Mathematically, that's due to superposition. Both masses produce some gravitational field, which add together to give the "net field". (The same goes for electromagnetism, where one may add electric/magnetic field strengths,electric potentials etcetera for every point in space.) Ever so slightly changing the field strength at the well. As gravity gets weaker ...


1

Anyway, if I absorb gravitons, why does the planet care? It acts as though it absorbed equal and opposite gravitons in terms of its resultant force experience. Not only the planet is producing gravitons, so are you, albeit a lot less than a complete planet. The amount of gravitons is directly related to the mass of the object in question, so is the ...


2

You 'feel' gravity due to the gravitational field of everything around you. Near earth, the earth is the biggest source. You are also a source, and the earth is affected by your gravity. The law governing the strength of these fields is symmetric in that whatever force you feel from the earth's field, the same force is felt by the earth due to your field. ...


2

You can't jump very quickly. The average vertical jump of NBA players is 28 inches. How fast are they going at max? $$v^2 = 2 a x$$ $$v = \sqrt{2 g (0.71m)}$$ $$ v = \sqrt{13.95m^2/s^2}$$ $$ v = 3.7 m/s$$ That's not very fast. It means that at a maximum you can remove less than 4m/s of your impact speed. Since a plane crash may be 100m/s, that doesn't ...


-1

You cannot jump from an object in free fall. The reason is this: If you are standing on an object in free fall this means you are yourself in free fall. In that case the Normal force experienced by you is $0$ N. If the Normal force is $0$ N then you cannot exert a force on the object (hence the object cannot exert a force on you via Newtons 3rd Law) ...


1

Spacetime curvature and gravity are not two distinct concepts, they are one and the same. You can say that one "translates into" (can be represented and looked at as) the other, they do not have a cause-and-effect type of relationship. Gravity simply is the curvature of spacetime. To do a quick analogy - it's like someone saying either "Au revoir" or ...


-1

Space-time is affected by energy forms, like gravity, electro-magnetic flux-but in yet unknown ways. Space-time shifts under influence of these fields, but you will agree that the mass and electro-magnetic field too shifts mass and electro-magnetic fields! Hence, we may notice immeasurable change in mass under the influence of the energy-for mass and ...


-1

I found many explanations for this type of questions http://settheory.net/cosmology http://settheory.net/general-relativity It's better than "The Meaning of Einstein's Equation" (John Baez). In particular - It is directly applied to an important example (universal expansion) - The expression is simpler (relating 1 component of the energy tensor to 3 ...


0

There are several effect that could explain retrograde captures being easier than prograde. As Nickolai noted, aside from atmospheric effects and non-uniform gravitational fields, there's not way for an orbiting object to know about the spin of the primary. However, in general, planet's do have non-uniform gravitational fields- they bulge due to rotation, ...


1

It might not be best to say you'd feel it, as the oscillations of the sun's gravity (as due to rotations and revolutions) are relatively subtle here on earth. The moon's gravity has twice as much influence in terms of tidal force, which is probably the most prominent consequence of gravitational "stacking" in the macroscopic sense you had in mind.* Here's an ...


1

Well given Newtons famous equation the force of gravity you feel is equal to $$F_{\textrm{gravity}}=G\frac{mm'}{r^2}.$$ It increases with the mass of the two objects being measured and is inversely proportional to the distance between the two objects in question. Strangely enough however, this technically stretches out nearly infinitely so the ...


1

Technically, yes, however in the case of you standing on a planet: You feel the gravity of the planet and star, so you accelerate accordingly. However the planet also feels the gravity of the star, so accelerates towards it. So you only notice your acceleration towards the planet.


0

Gravity can be seen as a gauge theory of the Lorentz group (which acts on the tangent space). These was pointed out by Kibble and Sciama during the 50s and 60s. As John said before, it's better seen in terms of differential forms. Another reference you might find interesting is the Lecture notes on Chern-Simons gravity by Jorge Zanelli (available in ...


3

Just to add to John Rennie's answer, the objects where we expect to see the largest frame dragging effects are spinning black holes. There, there is actually a surface called the ergosphere (outside of the event horizon), where it is impossible for observers to stay stationary with respect to observers far from the black hole. In a sense, their reference ...


5

I disagree with your premise that fields and curvature are different at all. The gravitational field strength tensor is (or, can be seen as, but usually isn't) the Riemann curvature tensor of spacetime. Likewise, the electromagnetic field strength tensor is the curvature tensor of a gauge principal bundle. The fields and the curvatures are not distinct ...


3

The spacetime outside a spinning mass is described by the Kerr metric. To explain how the Kerr metric produces frame dragging is hard, because it's not something for which there's an easy intuitive model. Frame dragging arises because the spacetime geometry links the angle measured around the spinning object to time, and this means the angle changes with ...


2

You have a chain of action and reaction. There is twice the weight on our head because the forces felt by the lower block are its weight, plus the action of the top block (minus the reaction force of your head). Then how is it on a molecular scale? Well, just the same: if you imagine a crystalline solid with horizontal layers, one layer feels the action of ...


0

You're mixing up different things. Two blocks of iron press your head more than one block because the Earth's gravity pulls two blocks stronger than one. This is why two blocks can tear through the paper where one can not. The molecular bit comes into consideration if you ask why the top block doesn't pass through the bottom one. This is because the ...


1

As i can remember, for the case of classical electrodynamic the d.o.f counting start after the assumption of the Bianchi Identity, and at the end the desired result came all from the gauge freedom. In fact $\partial_{[\mu}F_{\nu\alpha]}=0$ only choose a suitable form for $F_{\mu\nu}$, for example $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.$$ After ...


1

You can think of diff as bianchi id. The additional 4 dof is killed by the fact that 4 of the 4 of the EFE are constraints.


2

The force you describe is a capillary force. The following diagram illustrates where this force comes from: The diagram is supposed to show the thin film of water between the tea back and the wall of the mug. Any air/water interface has a surface tension that makes it behave like an elastic membrane. The air water surface is trying to shrink, and that ...


0

Let us clearly draw the line between two things here, since the question can easily involve opinion based answers, which may also be dubbed non-mainstream (which isn't welcome on this site). 1) The existence of dark matter is generally believed by a majority of the Physics community, since astronomical observations, notably by the Planck space ...


6

Yes, there have been suggestions that such particles exist, and an example is the sterile neutrino. But your question is a little more involved than you might think at first sight. For example if the sterile neutrino only interacts through gravity what interaction caused it to be created in the first place? There is nothing in the Standard Model that could ...


2

This answer is within the current physics and theoretical understanding, which has developed a successful formalism that includes all the experimentally seen particles in the Standard Model. The model has been very successful in predicting several new particles using its symmetry and mathematics, the experimental observation of the Higgs boson serving as ...


9

Based on your comment, I think you are indeed asking a more profund question than your teabag suggests: Why is it that gravity is so weak compared to the other forces? The answer is: We don't know. Seriously, that is one of the holy grails: To first find the Grand Unified Theory of nature in which all forces except gravity are explained as coming from one ...


1

Besides the adhesive force of water it might be good old friction, caused by the air pressure. As you mention the bag is wet. A vacuum could be created between the wall of the cup and the bag, causing the air to push the bag tight to the wall. This in turn causes a friction force between the wall and the bag.


0

Neither, however... If you are standing on such a board, there is a simple way to propel yourself, assuming you can change direction (i.e. steer) the board. You cannot propel yourself forward, but you can propel yourself sideways, by pushing the board to one side. This gives you some sideways velocity. Then (before you fall over) turn the board so it is ...


0

You can't actually propel yourself forwards or backwards in this way (unless you are taking advantage of significant friction in the bearings). Moving your body forward or backward would cause the platform to move in the opposite direction but only so much as to leave the person+platform system's center of mass unchanged. Another consideration. Where would ...


2

Dark energy is responsible for the acceleration of the universe at large scale, that is, it causes the second derivative of the cosmic scale factor $a(t)$ to be positive. But a smaller scales we found agglomerates of particles where the effect of the four forces (not only gravity) is much stronger that the repulsive effect of dark energy. A useful analogy ...


6

Is it because the acceleration is too weak? It is too weak with respect to the four forces we measure. The fact that the four known forces are so much stronger means that agglomerates of particles, up to the scale of galaxies are not internally affected, they keep their structure intact, like the famous raisins in the rising bread. It is only at the ...



Top 50 recent answers are included