# Tag Info

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There are plenty of satellite galaxies orbiting larger galaxies. The question is how long are you willing to wait for an orbit? The Milky Way has a mass $M$ of something like $6\times10^{11}$ solar masses, or $10^{42}\ \mathrm{kg}$. The small Magellanic Cloud is at a distance $R$ of $2\times10^5$ light years, or $2\times10^{21}\ \mathrm{m}$. A test mass ...

61

Great question. Observations show that Dark Matter (DM) only noticeably interacts gravitationally, although it's possible that it may interact in other ways "weakly" (e.g. in the 'WIMP' model --- linked). Everything following has no dependence on whether DM interacts purely/only gravitationally, or just predominantly gravitationally --- so I'll treat it as ...

27

They do! There's an entire class of galaxy, called a 'satellite galaxy' which is defined entirely based on them orbiting a larger galaxy (which would be called a 'central galaxy'). Our own milky-way is known to have many orbiting satellite galaxies, or at least 'dwarf-galaxies'. If dwarf-galaxies aren't enough, the milky-way itself is gravitationally ...

11

Because the dark matter does not interact a lot, there is no mechanism that would slow it down quickly. When a dark matter particle is falling towards some gravitational center, it is speeding up, then it flies through the periapsis and continues away into the distance. Normal matter clumps into planets, because it is slowed down by interactions / ...

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At this point we know a lot more about what dark matter is not, than what it is. It does not interact via the electromagnetic force, and interaction via the strong force is also unlikely. Interaction via the weak force is still an active area of research (See here). To understand why dark matter does not form clumps, imagine two particles of dust whizzing ...

3

Comments to the question (v1): In Newtonian mechanics with Newtonian gravity, a body can have orbital angular momentum wrt. a reference frame. A non-point-mass can also have spin angular momentum. Bodies can exchange angular momentum via tidal forces. In GR, it possible to assign angular momentum to certain space-time regions (such as e.g. the Kerr ...

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Q1. The Earth spins because it formed in the accretion disk of a cloud of hydrogen that collapsed down from mutual gravity and needed to conserve its angular momentum. It continues to spin because of inertia. Q2. No, Venus is quite a famous example.

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There is no force that causes the planets to rotate. Most of the rotation comes about from the conservation of angular momentum. Angular momentum is given by $L=m\omega{r^2}$ where $m$ is the mass, $\omega$ is the angular velocity in radians per second, and $r$ is the radius of the circular motion. Due to conservation of angular momentum, if the radius of ...

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The possible answer can be this: During the big bang, a large amount of energy was given out. That energy till makes the planets rotate( as there is no friction in space).

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You can think of the spin state of an electron as represented by a vector $(\alpha,\beta)$. Depending on how you set things up, "Up" might be represented by $(1,0)$, "Down" by $(0,1)$, "Left" by $(1,1)$, and "Right" by $(-1,1)$. Up is orthogonal to Down, and Left is orthogonal to Right, but Up is not orthogonal to Left.

1

You need to use vectors. Since $L \neq r \times p$, you need to use $\vec L = \vec r \times \vec p$ instead, where the $\times$ is the vector cross product of vectors, not the scalar multiplication of scalars. So you have $$\vec L= \left[(v_o t \cos \theta) \hat x+ (v_o t \sin \theta - \frac{1}{2}gt^2)\hat y\right] \times m\left[(v_o \cos \theta)\hat x+ ... 1 Yes, the rotational kinetic energy decreases. The extra energy is converted to thermal energy in the wheel and environment. If you imagine letting the weight go, it will slide across the surface of the wheel as it moves towards the edge. This sliding is motion against friction, so energy is lost there. Then the weight might bang into whatever holds it at ... 1 A quick Google search turns up a couple of additional resources besides the free paper abstract. Stanford has a summary of the paper, and the University of Delaware has slides associated with a talk on the paper. Offshore wind turbines currently exist and do act the opposite of an aircraft propellor: wind energy is converted into electricity, slowing the ... 1 You are quite correct that if you have items floating freely inside your space station they won't experience any artifical gravity as the station starts spinning. The artificial gravitational acceleration of an object is a consequence of its tangential velocity v and is given by:$$ g = \frac{v^2}{r}  where $r$ is the distance to the axis. The freely ...

1

In quantum mechanics and particle physics, spin is an intrinsic form of angular momentum carried by elementary particles, composite particles (hadrons), and atomic nuclei. Spin is one of two types of angular momentum in quantum mechanics, the other being orbital angular momentum. The orbital angular momentum operator is the quantum-mechanical counterpart to ...

1

Spin and orbital angular momentum are two different things, as already pointed out in Aniket's answer, but there is a good reason why we still call spin a "spin". This is because the Einstein-de Haas-Richardson experiment shows that electron spin is indeed of the nature of an angular momentum, although not exactly due to a "spinning electron". In fact, ...

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It depends on what you want how you choose the polarization of the light. The polarization of your light determines the recoil of your electron and your ion. In photoelectron vmi you would like to see the angular distribution of how the electron detaches from the molecule, so you should select the polarization of your light such that the velocity vector of ...

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(sorry, I couldn't write this in the comment section) Have you met the postulates of quantum mechanics? Here is a summary of them http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html Postulate 3 says if an observable has associate a (hermitian) operator, the only values we would observe for one photon the spin-angular momentum are the eigenvalues ...

1

Interesting question which even made me laugh. As already pointed out the Power of an hurricane is too high to be connected to any grid. My laugh came about this; "Giant heaters out to sea which dump heat into sea water?" -Why? Because it's basically the heat of the sea which feeds the energy to the Hurricane, and thus this kind of system would not do ...

1

$m^2\leq l^2$ implies that $\lvert m \rvert \leq l$ (for positive $l$, which it is in this case). Therefore, both the ascending and the descending chain have to terminate at $m=l$ and $m=-l$, respectively.

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Angular momentum in quantum mechanics in general works like this: the total is measured by $L^2 = \hbar^2 \ell (\ell+1)$ whereas the projection along any axis is measured by $L_z = \hbar~m$ between $-\ell \le m \le \ell.$ Both $\ell$ and $m$ are simultaneously measurable (i.e. the $L^2$ and $L_z$ operators commute), and they must be spaced by integers but ...

1

Its not that tough. You can work it out by using just two equations. But the one thing you should keep in mind is that when the comet is at the minimum distance from the sun, its velocity must be perpendicular to the radial vector (sun to comet). So the minimum distance is itself the minimum perpendicular distance used in the angular momentum formula at ...

1

For these kinds of system we often define a pair of quantities, one which is characteristic of objects or systems and one which is characteristic of interactions. Examples of these pairs are work (interaction) and energy (system) or impulse (interaction) and momentum (system). There is no commonly applied name for the interaction quantity that pairs with ...

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