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To answer your specific question: absolutely none. The Millenium run is a "dark matter-only" simulation. In this sort of simulation gas physics is taken to play a negligible role. All the gas (and stars, indeed all "baryonic matter" as it's called in the jargon) is removed and replaced with additional dark matter. The extra dark matter is added just to keep ...


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I will try to answer as many questions as I can. I won't presume to give you complete exhaustive answers, but maybe they will be nonetheless useful to you. What variables determine the range of temperatures over which matter is liquid? My understanding of thermodynamics is that matter changes from a solid to a gas when some thermal vibrations create an ...


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Liquid: molecules form bonds with neighboring molecules for most of the time, but there are enough energy for the bonds to break momentarily and be formed again with another molecule "These explanations seem hand-wavy to me." What is the level of your knowledge of physics? Are you aware of the quantum mechanical nature of atoms and molecules, at a ...


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But my problem is that I don't understand how a force can be a vector, in my head I see it as a direction vector and some power number Right. If the direction is a "unit" vector, then you can compare the magnitudes of different forces to compare the strengths. But you can multiply the magnitude and the direction to get a new vector that contains ...


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Force has both a magnitude and direction, which are the properties of a vector. The magnitude can be given by $\vec{F} = G\frac{M_1M_2}{\vec{r^2}}$ where $G$ is the gravitational constant and $M_1,M_2$ are the masses. the distance from each other is represented by the vector $\vec{r}$ which will be the displacement from the origin.


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Well the difficulty is when the ball in contact with wall. Few things you can consider. elastic impact. In this case, you can treat the ball as an elastic spring. When it impacts the wall with a speed, the kinetic energy will be converted to the spring's potential energy. And then, after it reaches the maximum deformation, the ball will spring back into ...


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Visual Solutions (Now Altair) makes VisSim Software. Here is a demo block diagram that they have used to simulate a bouncing ball: The $1/s$ blocks are integrator blocks from the VisSim library. The plot of the bouncing ball with these parameters is shown below. If you are interested in running this, the demo comes with the install and you can download ...


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This is really a difficult problem, but possibly not for the reason you imagine. The following naif criterion seems at first highly appropriate: take any pair of stars, subtract the center-of-mass motion, compute the kinetic ($T$) and gravitational ($W$) energies, and check whether $T + W < 0$. If so, the pair is bound, otherwise it is not. How can ...


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The solution is the same as the elliptic case, except: $$ t = \sqrt\frac{a^3}{GM} (\theta - e\sin(\theta))$$ becomes: $$ t = \sqrt\frac{a^3}{GM} (e\sinh(\theta)-\theta)$$ and $$ \nu = 2 \arctan\left( \sqrt{ \frac{1-e}{1+e} } \tan\left( \frac \theta 2 \right) \right) $$ becomes: $$ \nu = 2 \arctan\left( \sqrt{ \frac{e+1}{e-1} } \tanh\left( \frac \theta 2 ...


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The basic equations, assuming no air drag($^*$) are as follows. At $t=0$ we drop the object from height $H$, we assume its initial speed is zero ($v_0=0$). Only gravity ($mg$) is acting on it. If the object free falls to height $h_1$, energy conservation then shows us that its speed has now become: $$v_1=\sqrt{2g(H-h_1)}$$ At height $h_1$ a braking ...


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The notation $k^2$ for vector $\boldsymbol{k}$ means the square length/magnitude of the vector; if $\boldsymbol{k} = (k_x,k_y,k_z)$ in Cartesian coordinates then $$ k^2 = \left|\boldsymbol{k}\right|^2 = k_x^2+k_y^2+k_z^2 $$ so that $$ \widetilde{g}(\boldsymbol{k}) = \frac{4\pi}{k_x^2+k_y^2+k_z^2} $$ (using Pythagoras)


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It is not clear at which level you want to simulate this. Fine grained If you want to do this on a very fine grained level, you just need so simulate Newtonian mechanics and gravity and it should emerge from itself when you have a trajectory set up. This means that you compute the gravitational force to be $$ \vec F = - G \frac{Mm}{r^3} \, \vec r \,, $$ ...



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