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2

I think, Kyle's link is a really good answer. For they layman (like me), you can run some numbers here: http://xaonon.dyndns.org/hawking/ a 1 earth mass black-hole (about the size of a ping-pong ball) would radiate so little energy that it would easily destroy the earth, even from a low orbit. It might take some time to devour it, but no question, it ...


7

I'd like to add a bit more to Adam Davis's concise answer. Newton's Sphere Production Method Firstly, note that the lathing process shown in The video that Adam Davis's Answer links to with the sphere rotated and an oscillating abrasive cup sliding over its surface about pseudo-random rotation axes is the standard technique invented by Isaac Newton that ...


4

Another reason that nobody has mentioned might be durability. The corner of a cube would be easy to accidentally break because a small force on it would equate to a large pressure (its area gets smaller as the shape of the cube gets more accurate). In fact I think it would be likely to break during machining, resulting in truncated corners (even if only ...


49

If you know the diameter of the sphere, you know everything you need to know about the dimensions. It all comes down to one single value. Any other shape requires multiple dimensions and thus multiple values. Further, measuring a cube or another shape for accuracy is harder than measuring a sphere. Making very accurate spheres is not as difficult as you ...


28

There is a nice article in the New Scientist that describes how these spheres were made. Sphere can be made very precisely (and their shape measured accurately) simply because of their symmetry - and therefore their volume can be determined most accurately. The video clip in the above article shows this in detail. Of course a sphere also has the lowest ...


4

A sphere might be harder to machine, but easiest to verify the accuracy of, especially when accounting for slight changes due to temperature. It should be noted that any standard like this must not only have a mass of 1 kg (or whatever) but also have some secondary method of verifying mass, in this case, being able to count (to some accuracy) the number of ...


6

I like to explain this using a figure from a talk by Marco Limongi some years ago. Based on a given set of models, the $x$-axis shows the initial mass of the models and the $y$-axis the final mass. The different coloured layers show the composition of the star at the moment of collapse. The mass ejected in the supernova is the difference between the curve ...


0

m includes all kinds of energies for the mass at rest, including the thermal energy, and those from the other degrees of freedom, such as the ones internal to the nucleus, as AnnaV mentioned. So you only have to compute mc^2, the HTM term is redundant.


4

To deal with your last point first, the coupling of anti-matter to the gravitational field of the earth is discussed in Has the gravitational interaction of antimatter ever been examined experimentally? It's exceedingly unlikely we'll ever be able to amass enough anti-matter to measure the gravitational field created by the anti-matter, but if the ...


1

A small note on the definition. It should be $$ M(S) := \sqrt{ \frac{\text{Area}(S)}{16 \pi}} \left(1- \frac 1 {16 \pi}\int_S \theta^-\theta^+ d_{\sigma_S}\right), $$ where $\theta^\pm$ are the divergences along the two null directions. It is equal to what you have written if the 2-surface $S$ lies in a space-like 3D submanifold with vanishing extrinsic ...


1

I don't know if you have already done integration in high school but at least here is a solution without using multivariable calc. The kinetic energy of the spring with length $L$ is the following $$E_{kin}=\int_m \frac{u^2}{2} \, \text{d}m$$ where $u$ is the speed of a little mass element d$m$ and the integration is over the mass of the spring. Assuming ...


3

Dark matter does not readily "accumulate". If(?) it exists then it interacts very weakly with normal matter and is primarily influenced by gravity. The Earth's gravity is far too small to make a local concentration of dark matter. The local dark matter would be moving in the Galactic potential at speeds similar to that of the Sun around the Galaxy ($\sim ...


0

There is no absolute values in a nature. So, there is no absolute zero gravity between any (gravitating) particle in the universe. Assuming gravity is quantized, we could conclude that there is could be two particles in the universe that never exchanged any quant of gravity, but the probability of this event will never fall to absolute zero.


1

"Masked by fluctuations" is philosophically different from "is exactly zero." There would be a nonzero gravitational force which would be impossible to measure except in an empty universe.


-1

Newton's inverse square law of gravitation is similar to Coulomb's law for static electric forces. It is natural therefore to ask if there are negative masses like negative charges. The difference here would be that same charges repel whereas same masses attract. Thus one would expect that negative masses repel resulting in them being expelling each other- ...


1

Mostly it's to throw physics at the wall and see what sticks. You see, there are several important energy conditions in GR. These energy conditions allow you to derive several important theorems, see here for a list. Many of these theorems are about the topology and causal structure of spacetime. For example, assuming the averaged null energy condition ...


-1

Force carriers have a range told by the equation: $$ R≈h/4\pi mc $$ h is Planck's constant, m is mass and c is the speed of light This can be derived from the Heisenberg uncertainty principal: $$ \Delta E \Delta t ≥ h/4\pi $$ E is energy and t is time. This states that an uncertainty of energy can only be sustained over a limited amount of time depending on ...


1

You are on the right track. Mathematically, the center of mass $\vec{c}$ is defined as $$ \vec{c} \sum_i m_i = \sum_i m_i \vec{r}_i $$ where $m_i$ are the individual masses and $\vec{r}_i$ the individual position vectors.


1

The answer is, you'll have to be more specific. When a fist hits a bag, the force varies a lot in terms of time. If you imagine trying to punch a bag as slowly as possible, there is a slow onset as your fingers just graze the leather, followed by something more "meaty" as your bones start to press against your skin which is pressing against the actual ...


1

Firstly, the ADM mass of an end is in no way the sum of the masses of the things in any portion of (or entirety of) the spacetime. Even the ADM mass of a black hole or even a regular star is in no way the sum of the masses of the parts. For instance if you put 10^20 cold iron atoms into a spherical asteroid in an otherwise empty universe, it will have an ...


1

"the mass of an isolated proton plus the mass of an isolated electron is slightly different than the mass of the two combined, because their interaction energy affects the mass of the system" Yes, this is true, but unfortunately, when we bring together some particles, and a stable configuration is formed, the system emits what is called binding energy. ...


0

When an object (ball) of mass $m$ at rest is struck elastically by another object of mass $M$ traveling with initial velocity $v$, then the velocity after impact is given by $$v_{ball}=\frac{2M}{m+M}$$ Two limiting cases: $m=M$, maximum transfer of energy (the racket stops and the ball travels with the velocity of the racket); when $m<<M$, the final ...


1

It could have been beautiful, and indeed it would have been the case if the Universe consisted of "normal" matter — i.e. baryons and dark matter – above a certain critical density threshold (which you can calculate to $\sim10^{-29}$g/cm$^3$). Unfortunately, as it was realized in 1998, a mysterious "energy" labeled dark energy seems to permeate empty space, ...


6

The rotational energy of a body is given by: $$ E = \tfrac{1}{2}I\omega^2 $$ where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a uniform sphere the moment of inertia is related to the mass of the sphere, $m$, and the radius of the sphere, $r$, by: $$ I = \frac{2}{5}mr^2 $$ You already have the mass, and you can Google for the ...


1

In standard model, the mass of a particle can be explain by either Dirac or Weyl equation. The first thing is that neutrinos are can't be described by any of the above equations (Dirac equation or Weyl equation) in the standard model because no right handed neutrinos are observed. Dirac equation needs four spinors to explain the mass of any particle. But in ...


0

In theory, a massive particle can be accelerated asymptotically close to the speed of light. But one can never actually reach the speed of light because the kinetic energy of such a (massive) particle would be infinite. This is because the mass of the particle itself become heavier at higher speeds due to relativistic effects. Specifically: $$\text{Kinetic ...


1

Well, by massive, I assume you mean objects that have non-zero rest mass. In that case, it would take infinite energy for that object to reach the speed of light. However, their speed would get closer and closer to the speed of light as more energy is put in, until their speed was practically (but not exactly) the speed of light. Additionally, the smaller ...


1

No, that's not possible. Even if the two bodies could be compressed to be just larger than their Schwarzschild radius (they can't really, without collapsing further to black holes), their combined Schwarzschild radius, which grows linearly with mass, is twice their individual Schwarzschild radii. That means that even if they effectively rolled on each other ...


3

Far away from a black hole, spacetime is curved only a little bit, and many different things could curve it like that out there. It's like if you had a dollar in your pocket, and it's been there for a long time, and you can't remember if you got it from your boss or from your friend. But a dollar is a dollar. So you could have a massive star, or a black ...


11

Although we don't have a quantum theory of gravity, we think we have some reliable knowledge about the properties of black holes from general relativity. One thing we think we know is the so-called "No-hair conjecture", which says that black holes can be described by just three numbers: mass, charge, and angular momentum (i.e. how much they are spinning). ...


0

The singularity probably does not exist, as GR likely breaks down at those size / energy scales. When we have a full quantum description of gravity we may know what's really there. By the way, the part of the black hole we fully understand is actually the vacuum solution - the Schwarzschild metric - which includes the event horizon but not the source mass. ...


0

It's almost certainly incorrect that the center of a black hole is a singularity as this would be at odds with quantum mechanics. Just how exactly it looks like would be something to ask of a theory of quantum gravity! Regardless of being a singularity or not, the mass is determined by how much mass you stuff into your black hole. Hence black holes of ...


0

The density of black holes isn't infinite. Some black holes have the billionfold density of our sun (like the black holes in center of galaxies). There are big and small black holes.


1

Unfortunately, no, your calculation does not seem to be correct. Your calculation is based on Einsteins famous equation $E=mc^2$; however, this equation is actually only valid for objects at rest, while all experiments confirm that photons in a vacuum move with a constant speed of $2.99...\times10^8$~m/s. The equation Einstein gave for moving particles is ...



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