Tag Info

Hot answers tagged

35

Actually, it doesn't have the same mass, it has significantly less mass than its precursor star. Something like 90% of the star is blown off in the supernova event (Type II) that causes the black holes. The Schwarzschild radius is the radius at which, if an object's mass where compressed to a sphere of that size, the escape velocity at the surface would be ...


30

When you watch a pop-sci TV show, you need to take everything you see with a very healthy grain of salt. This is particularly the case if the show's host isn't a scientist, but even when a scientist is the host, you need to be suspicious. Stellar black holes do not turn into monsters that reach out and pluck objects from the heavens. From far away, a black ...


21

If you look closely at the crocodiles' tails you'll see that they wave their tails from side to side to provide propulsion for the jump. Compare this to a fish swimming: The side to side motion of the fish's tail propels it forward, and the crocodiles are using exactly the same sort of side to side motion to propel themselves upwards.


12

I've recently started trying to swim the butterfly. Unlike other swimming strokes, there doesn't seem to be any way to "go easy" and do a relaxing length of the pool: if I want to get my face out of the water to breathe, I essentially have to use the water to do a push-up. Now if a scrawny guy like me can lift his chin a few inches out of the water using ...


9

It actually goes the other way around: when a star collapses to form a black hole, its planets (if it has any) will become unbound and fly away to infinity. Simple reason: when the star explodes to form a compact object (neutron star or black hole), it releases most of its mass in the form of a SuperNova explosion, so that the central object around which ...


6

The upwards force comes from the rather violent tail movement. When the rest of the body is out of the water, the tail still acts sort of like a hydrofoil pushing the crocodile upwards, only not with a linear but oscillating motion, and obviously it's rather instable but enough to get the whole animal up in the air for a short while.


6

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 ...


5

The acceleration of the expansion is currently observed to be happening. This observation is one of the pieces of data we use to infer the amount of dark matter. It tells us that there can't be more than a certain amount of dark matter, because that would be incompatible with the observed acceleration.


4

The short answer is yes, the presence of dark matter would act to counter the expansion of the universe. And in fact it does--but not enough to stop the expansion. Dark matter has gravity just like normal matter. In fact, that's pretty much the only reason we know dark mater exists at all: we can observe dark matter's gravitation effects in the rotation ...


4

If you measure the large-distance strength of the gravitational acceleration $g\approx \frac{GM}{r^2}$ of a star / black hole with the assumption that your distance $r$ is much further out than the various mass parts, shock wave, and ejected material; then $g\approx \frac{GM}{r^2}$ is (within a percent or so) the same before and after the supernova. This is ...


4

I'll take the question to be referring to solid rock. In reality, I think small asteroids are loose jumbles of rubble with a lot of vacuum between the rocks, and larger bodies like Ceres may have been liquid when they formed. Googling turned up [Scheuer 1981], which can be found online for free by googling. S/he estimates the maximum height of a mountain to ...


4

The masses can't repel each other because gravity is mediated by a spin 2 field, and for spin 2 the force between charges of equal signs is attractive. See the question Why is gravitation force always attractive? for an explanation of this. But it's impossible to say why the force can't be zero. Experiment shows that masses do attract each other, and ...


4

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 ...


4

General relativity is only conformally invariant in two dimensions. This can be proven by making the transformation $g_{ab} \rightarrow \phi g_{ab}$, and seeing what transformation Einstein's equation${}^{1}$ makes. What you will find is that Einstein's equation will MOSTLY transform, but you will get terms proportional to $(d-2)(d-1)$ and derivatives of ...


3

g is telling you the constant rate at which the velocity is changing. So initially the velocity is 0 and after 1 second the velocity is 10 m/s. The average velocity during that first second is 5 m/sec so the mass has fallen 5 m. In the second second the initial velocity is 10 m/s and at the end of that second it is 20 m/s. The average velocity over that ...


3

You are not making a fundamental error, and your approach is in principle correct. Basically, what you get, is a discretization error. Basically, what you are doing is evaluating the integral $$d=\int_{t_0}^{t_1}v(t)dt=\int_{t_0}^{t_1} gt dt.$$ Which you approximated with a Riemann sum (maybe unintentionally?), i.e $$d=\sum_{i=1}^N g t_i \Delta t.$$ You ...


3

A few thoughts to help you on your way. When an elevator is moving, you have to do work against gravity. You are changing the potential energy of the system. The faster the elevator moves, the more work per unit time is needed (because power = work times velocity). If you are changing the velocity of an object, you are changing its kinetic energy: if it's ...


3

Correct. For a sphere of uniform density, the acceleration drops off linearly. $$g = g_{surface} \frac{r}{R}$$ where $r$ is the location under consideration, $R$ is the radius of the sphere and $r < R$. Under such a scheme, gravity would be one half that at the surface. The earth is not a uniform sphere though. The outer crust is much less dense than ...


3

Your figure for being below water is not correct. As you descend in the ocean the ambient pressure increases by about 1 atm/10 meters. In a uniform sphere, the gravitational field is linear in the radius, zero at the center. Making the incorrect assumption that the earth is a uniform sphere, being down 1 km would decrease the gravitational acceleration by ...


3

Gravity may be treated as a quantum field theory. In this kind of theory, interactions are represented by field correlations, more known as "virtual particles", "virtual gravitons" in the case of gravity. The fact that two charges (more precisely, in the case of the gravitation, $2$ positive energy densities) attract each other is due to the sign ...


3

Newton's law does predict the bending of light. However it predicts a value that is a factor of two smaller than actually observed. The Newtonian equation for gravity produces a force: $$ F = \frac{GMm}{r^2} $$ so the acceleration of the smaller mass, $m$, is: $$ a = \frac{F}{m} = \frac{GM}{r^2}\frac{m}{m} $$ If the particle is massless then $m/m = 0/0$ ...


3

UPDATED: See below. Your NDSolve inputs seem to be doing what I would expect for a mass around a gravitational center. Using: a = 0; b = 0; traj = Table[ s = NDSolve[{x''[t] == -x[t]/((x[t] - a)^2 + (y[t] - b)^2)^(3/2), y''[t] == -y[t]/((x[t] - a)^2 + (y[t] - b)^2)^(3/2), x[0] == 1, y[0] == 0, x'[0] == 0, y'[0] == v}, {x, y}, {t, -20, ...


3

I am rather surprised that neither link posted above gives a simple discussion of the effect, so here it goes. Let us consider many asteroids of cubic shape, of constant density $\rho$, and of varying side $l$. We ask when, roughly, self-gravity will be able to perturb this shape into a spherical one. A cube of side $l$ has the same volume as a sphere of ...


2

Some simple scaling relations suffice to determine the size beyond which gravity prevents non-spherical rocks from forming: A molecule of mass $m$ is bound to a mass $M$ of linear size $R$ with gravitational binding energy approximately equal to $G M m / R$. If this gravitational binding energy far exceeds the molecular binding energy $E_b$, gravity will ...


2

You may define a conserved total stress-energy tensor (matter + gravitation). The main problem is that a conserved total stress-energy tensor is not covariant, and that a covariant stress-energy tensor is not conserved. Said differently, $\nabla^\mu T_{\mu\nu}=0$, which is a covariant equation, does not represent a conservation law, while $\partial^\mu( ...


2

If your potential is $\propto 1/r$, you're effectively simulating gravity. If it gives you better intuition, imagine it as the earth around the sun. As long as your numerical solver is doing a decent job, you shouldn't expect the ball to spiral in, the correct solution would be a conic section, that is it would orbit the origin in an elliptical path if the ...


2

There is a difference between "feeling the force" and "being stretched". If you imagine two balls connected by a spring, and falling towards a massive object, then the closer ball will experience a greater force and therefore "accelerate away" from the ball that is further away - the spring between them will stretch, and thus provide a force balance. A ...


2

Yes, it an extremely small effect but it exists in Einsteins general relativity. There is one case of a double star where there rotation around each other seems to lose energy at rate that this phenomena should give according to general relativity


2

There are more spins than just 2. There are particles with spin zero (Higgs particle). Spin half (electrons, positrons, neutrions, quarks, muons, etc.), spin 1 (photons, gauge bosons of weak interaction), spin 3/2, spin 2 (hypothetical gravitons). During attraction / repulsion there are 2 things that come into play: 1) The particles that get attracted / ...


2

The mathematical transformation from particle to antiparticle reverses the sign of the charge and the sign of the sign of the intrinsic parity. An antiparticle has the same (positive) mass and same spin as the corresponding particle. In the theory of supersymmetry, the known particles have "superpartners" with different spin. Supersymmetry is a very ...



Only top voted, non community-wiki answers of a minimum length are eligible