# Tag Info

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 ...

43

Yes, the particle would continue to accelerate and would never reach a terminal velocity. But that is not what this equation tells you. This equation tells you what the terminal velocity is, given the parameters of the function. When in a vacuum, there is no terminal velocity. It is not zero, it is not infinity. A terminal velocity literally does not exist ...

36

The answer lies in something called the virial theorem. You are correct, a cloud that is in equilibrium will have a relationship between the temperature and pressure in its interior and the gravitational "weight" pressing inwards. This relationship is encapsulated in the virial theorem, which says (ignoring complications like rotation and magnetic fields) ...

16

Let's assume you mean that Earth now has the mass of Jupiter (as opposed to actually launching from the literal planet Jupiter - whole different question...). Then: radius of Earth = $6.4 \times 10^6~\text{m}$ mass of Jupiter = $1.9 \times 10^{27}~\text{kg}$ Escape velocity, $v_\text{escape} = \sqrt{\frac{2GM}{r}}$ This gives a value for ...

13

As gas clouds collapse, they increase in internal energy (measured by temperature). This is part of what causes their pressure to increase. As they increase in temperature, though, they also increase the amount of radiation they emit. As they emit radiation, their internal energy decreases and thus their pressure also decreases, allowing for further ...

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 / ...

10

Hey! The question keeps getting edited! Make up your mind! You asked about Mars originally, then edited the question. Actual, real Jupiter is flat out impossible. Does it have a surface to launch from? Who knows? What's the pressure at that depth? Can our probes even survive at that depth? Probably not? What if Earth had the mass of Jupiter? More ...

9

You have to think of the symmetry of the situation. For each point A and B, there is a point A' and B' opposite the center line that has an attractive force pulling in such a way that the components perpendicular to the center line are canceled. Therefore, the net force is along the center line.

9

As in math, you get the physical meaning via the limit: it's not "undefined", it's "going to infinity" when $\rho$ goes to 0 from above. (which is the only way in the physical world, that why its well defined there).

8

In all cases, the two objects move towards one another. In fact they experience exactly the same gravitational force. However, because acceleration equals force over mass $$\mathbf{a} = \frac{\mathbf{F}}{m}$$ that equal forces causes the heavier object to accelerate much less than the lighter one. But technically, the Earth does move towards you very ...

6

The issue here is that the existence of terminal velocity depends on the assumption that the particle is moving through a fluid. As real fluids have only positive non-zero densities, simply plugging in 0 breaks this assumption. You could say that the classical limit of terminal velocity as the fluid approaches vacuum (i.e., as the density approaches 0) ...

6

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 ...

5

As a general rule (regardless of the definition of V) we have: $$\frac{\mathrm{d}}{\mathrm{d}t}(\bf{V}\cdot\bf{V})=\frac{\mathrm{d}\bf{V}}{\mathrm{d}t}\cdot \bf{V}+\bf{V} \cdot \frac{\mathrm{d}\bf{V}}{\mathrm{d}t}$$ $$\frac{\mathrm{d}}{\mathrm{d}t}V^2=2\left(\bf{V}\cdot \frac{\mathrm{d}\bf{V}}{\mathrm{d}t}\right)$$ ...

5

Obviously this is one of many examples of what can happen in physics problems, but what does undefined actually mean in terms of physics? I hope I am explaining myself clearly. The mathematical formulas used to model physical observables are valid in specific frameworks, where they have fitted observations and predicted new observations successfully. An ...

4

Your question indeed very clear,a good example, and cuts to the heart of physics - being able to define, predict the world using mathematical models. But you have to make sure your models are correct - and complete. Fabrice and Robert are right - velocity will go to infinity without constraints. And there is the clue as to what's going on as a physics ...

4

Let's look at each answer in turn: a) The force of gravity acting on the astronaut and spacecraft is negligible Wrong. To be in orbit, gravity needs to be acting. If it were negligible, they would just head off in a straight line in space. b) The spacecraft and the astronaut are in orbit around the Sun with the Earth This is true, but irrelevant. ...

4

Lets just start with the cosmic velocity assuming you dont want to escape fully. $v \le \sqrt{\frac{2GM}{r}}$, where M is the mass of the planet and r is the distance to the center of Mass. We assume a spherical planet. We know the average density $\rho$ of our planet earth and since you will probably want to live on the new planet we will assume it has ...

4

Actually you can go to the orbit of Jupiter with a $\approx 2500$ tonnes rocket and a $3$ tonnes payload. From there you can use an ionic engine. A rocket launched from the equator of Jupiter that turns at $12.6~\text{km}/\text{s}$ needs just an increase in speed $v = 29.5~\text{km}/{\text{s}}$. $$v_{rj}:= 12.6~{\text{km}}/{\text{s}} \;\;\; R_j := ... 4 To two significant figures, the acceleration due to gravity is g=9.8\:\mathrm{m/s^2} everywhere on Earth (at sea level). That is to say, if you use e.g. a pendulum to measure g to two sig figs, you will get this value no matter where you are. In a sense, this is the precision to which the Earth is well-approximated by a uniform sphere of matter. The ... 4 This video by Richard P. Feynman might explain how hard it is to answer such a 'why' question. An excerpt: But the problem, you see, when you ask why something happens, how does a person answer why something happens? ... When you explain a why, you have to be in some framework that you allow something to be true. You have to know what it is that you’re ... 3 There is a mutual attraction from gravity, and we generally only consider the smaller object here on earth because the earth is so massive, the acceleration of the earth is negligible. This is because a = F/m, and with equal F between the two objects, the acceleration will scale as a\propto 1/m. For the earth, this leaves a ridiculously small, but ... 3 Yes, Newtonian Physics works on a galactic scale. Still, for long distance interactions on fast objects you might want to take into account the finite speed of gravity, but I don't think it is necessary for ordinary galaxies simulations. Conversly a lot of phenomena occur that impact the galactic material: writting a decent simulation is not easy. 3 There are three classic tests of general relativity: the anomalous precession of Mercury's orbit the deflection of light by the Sun the gravitational redshift of light Newton's theory predicts zero precession in test (1) and zero redshift in test (3). For test (2) Newton's theory predicts a deflection half the size of the prediction in general ... 3 The problem is in your assumption that "we fall into open space" unless the planet is large enough. Even if there were no planet at all, we would not fall. In open space you just stay where you are - unless you are affected by some star or planet. In other words you will always drift slowly towards something or other. Now the earth's gravity is so large ... 3 For a uniform, spherical distribution of mass (cloud of gas and dust) of radius R and mass M in absence of magnetic, radiation fields etc, we have dm = 4\pi \rho r^2 dr and the potential energy of a spherical shell of inner radius r and outer r + d r is dU = -G\frac{m(r)dm}{r}, m(r) = \frac{4}{3}\rho r^3, and a simple integration yields, ... 3 According to conservation of momentum, the center of mass of a system cannot accelerate without external forces. In other words, if the center of mass starts out at rest (which is generally a good procedure in simulations), then it should always stay at rest. It is normal for numerical errors to introduce deviations, but the motion you are seeing looks ... 2 The spacetime outside a spherically symmetric arrangement of mass is described by the Schwarzschild metric. This is a consequence of Birkhoff's theorem. So the changes in the interior structure of your object make absolutely no difference to an object outside it. The orbit will be exactly the same as if the object was unchanging, or indeed if it was a black ... 2 What is gravitational potential? Usually a potential is defined as the potential energy per mass or per charge or similar. This is most often seen in relation to electricity or chemistry and less often to gravity. GPE is gravitational potential energy. GP is gravitational potential energy per mass:$$GP =\frac{GPE}{m}  Is it defined for the system ...

2

Does an object decelerate when reaching terminal velocity No, it ceases to accellerate. So while the force of air resistance increases to equal the force of gravity is the sky diver decelerating? Consider $F = ma$. You say because $F_r = -F_g$, net $F$ is zero. What does that imply about $a$?

2

Let's discuss a whole set of cases. You're standing on the planet. I'm guessing that you would describe the sensation as 'feeling' gravity pulling you down. You're on a skydiving trip, standing in the plane. You still 'feel' gravity. You're on the skydiving trip and you've just step out of the plane, but not had time to pick up speed. Here you don't 'feel' ...

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