# Tag Info

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 ...

8

I assume a steady-state universe and that the bodies have no velocity relative to each other. Yes, they will eventually collide. Gravity has an effect over any distance, including the ~46 billion light-year radius that constitutes the spherical observable universe (the actual size of the universe may be much larger). Of course, the force will not be very ...

7

The answer to your questions is very nuanced for the most part. I'll start with the easy answers: Light does not experience time, neither does it experience no time, the most accurate statement I could probably make off-hand is that it experiences null time. Null time does not mean time has stopped, that would be zero time; null time means null time, null is ...

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.

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.

5

Yes, they will collide given the initial conditions. The speed at collision can be calculated. We can presume them to begin with virtually zero gravitational potential energy. We need an assumption of size when they collide. Let's assume a size of 50cm. That way when they collide, the centers will be 1m apart. $$U = -\frac{GMm}{d}$$ When they collide, ...

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

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 ...

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 ...

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 ...

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

If the bodies are initially at rest, then the orbit will be a degenerate ellipsis of finite semi-major axis and eccentricity 1, i.e., a line segment. The semi-major axis $a$ is half the initial distance. Time to collision is half the period $T$. This can be directly derived from Kepler's Third Law. $$\frac{T^2}{a^3} = \frac{4\pi^2}{GM}$$ $$T = ... 3 Your figure for being below water is not correct. As you descend in the ocean the ambient pressure increases by about 1 atm/10 meters. In a uniform sphere, the gravitational field is linear in the radius, zero at the center. Making the incorrect assumption that the earth is a uniform sphere, being down 1 km would decrease the gravitational acceleration by ... 3 A few thoughts to help you on your way. When an elevator is moving, you have to do work against gravity. You are changing the potential energy of the system. The faster the elevator moves, the more work per unit time is needed (because power = work times velocity). If you are changing the velocity of an object, you are changing its kinetic energy: if it's ... 3 Correct. For a sphere of uniform density, the acceleration drops off linearly.$$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 ...

2

I believe you have stumbled upon an important property of General Relativity (GR) that gravity is geometry. As you say, the gravitational force is not an ordinary force, but a property of space-time. Objects that appear to be under the influence of gravity are in fact in free-fall, simply following the (geodesics of) the curved space-time. You can ...

2

Assuming you are using C code: #include <stdio.h> int main(void) { float position = 0, velocity = 8, time, timeStep = 0.01; float g = 9.8; for(time = 0; velocity > -10; time+=timeStep) { velocity = velocity - timeStep * g; position = position + velocity * timeStep; printf("time %f; velocity %f; position %f\n", time, velocity, ...

2

The standard approach in numerical simulations is to do a discrete time stepping. If the motion is limited to the vertical direction, you just need to keep track of vertical position $z$ and vertical velocity $v$ (with positive $v$ denoting upward speeds). Starting with given values for $z$ and $v$ (the initial position and initial velocity) you update the ...

2

I'd put this as a comment, but I don't have enough reputation. That being said, it sounds like nonsense to me as well. I found some attempts to debunk it on rational wiki, http://rationalwiki.org/wiki/Scalar_wave It's not the most neutral article, but I think it may help you sort out fact from fiction. I think the treatment in the "In real physics" section ...

2

As for the helicopter problem, theoretically, arbitrarily low power can be sufficient to float a load, if you push down a lot of air with very low speed. However, you need longer and lighter (and maybe wider) blades for that, so the problem you'll have to solve is that of structural strength. Let me note that recently a muscle-driven helicopter ...

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