New answers tagged

4

Yes, the sun's mass isn't constant causing disturbances of orbits, but should you care. Mass of sun now: $\approx 2 * 10^{30}$ kg so $E \approx 1.8 *10^{47}$ J. Radiation per year is about $10^{34}$ J. You've got a lot bigger problems that affect your calculation of orbits. Especially like where Jupiter is relative to where you thought it was. Namely the ...


0

When we talk about a (non-spinning) black hole we normally mean the spacetime geometry called the Schwarzschild metric. This is one of the simpler solutions to the Einstein equations, but it's important to understand that the solution is based on a number of assumptions and as a result it is only an approximation to a real black hole. The assumption that is ...


1

The idea of a "Zero-Energy Universe" is a theory held by a limited number of scientists. There are several stackexchange question that expand on the theory and may help you. Zero energy universe Total energy of the Universe


1

I think userLTK's answer pretty much covers your question, but I would add a footnote for clarification. Your question is a bit unclear because you don't say whether you're asking: does the process of contraction lose mass? or is a contracted object lighter once it has reached equilibrium? As userLTK says, the process of contraction does not cause a ...


1

The act of contracting converts potential energy into kinetic energy or heat, so the answer in this case is no difference, though the somewhat hotter contracted planet does lose heat (and mass) faster, but the actual contracting should be zero change in system energy. This is consistent with Newton's energy can't be destroyed theorem. The only energy ...


1

Well, we should not forget about the dirt machine millions/billions of year old where stuff grows in dirt, you get wood, burn it and get energy. Although there are other ingredients like seeds, water and sunlight. This can be considered a machine because, there is a biochemical mechanism at work. Therefore, if this process can be fine tuned, and sped up, ...


4

The 'rule' that "matter can not be [created or] destroyed" simply isn't true. Matter can be created and destroyed under the right circumstances, it's just that those circumstances are not met for macroscopic quantities of matter in places where humans live. This means that the conservation of matter can be taken as true in almost all of chemistry and large ...


1

Well, uranium ore is a form of dirt, and with a suitable processing does give out energy at the cost of its weight in mass. The machines invented are called reactors, quite complicated though, one needs: Neutron moderator to slow down the neutrons. In light water reactors and heavy water reactors it doubles as the nuclear reactor coolant. Nuclear ...


1

Yes, relative to a photon, all objects of mass $m$ are moving at the speed of light in empty space. (Note however no observer can really travel at the speed of light, but as they approach the speed of light their observations approach what I've described)


0

For an indivisible particle, probably the simplest operational definition of "mass" in your context is $\sqrt{E^2 - (p c)^2}/c^2$, where $E$ is its energy and $p$ its momentum. Equivalently, the mass of an indivisible particle at rest is $E/c^2$ in that frame. A lump of gold and a proton are not indivisible particles, because their constituent parts can ...


3

Suppose you start with your (stationary) 1kg block of gold. If you raise its temperature you have to add energy to it, and that means it's different after you've raised its temperature. For example you could shine an infrared lamp on it, in which case you've added the energy from the IR lamp. The mass changes because if you add an energy $E$ the mass goes up ...


2

No. It means that the energy content of a particle at rest whose mass is $m$ is equal to $mc^2$. In general, $E=\dfrac{mc^2}{\sqrt{1-\dfrac{v^2}{c^2}}}$ represents the energy content of a particle of mass $m$ moving with a speed $v$. If it is at rest, i.e. $v=0$, then $E=mc^2$.


2

No. It means that when one is accounting for the total energy of the system one must account for the energy contribution of the mass. It also means that the mass isn't necessarily conserved in a system, while energy and momentum will be. In special cases, energy and/or momentum may even be constant in an defined subsystem. The quantity $E^2- p^2c^2$ of a ...


2

No, it doesn't. It just means that particle of mass $m$ can be converted to $mc^2$ of energy. You can find a good popular article on that topic here: https://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence


-1

Yes, spring mass will increase but in very very small amount. As per Mass energy equivalence the amount of mass gain will be(E=mc2): m= E/C2 for example if spring have store 1 Joule of potential energy when it is extended then the increased mass will be: m =1/299792458= 3.33564095198152e-9


0

So far gluons appear massless, and if they have mass this would mean there is some sort of symmetry breaking or Higgs mechanism involved with QCD. So far there is no evidence of this, and theory does not make predictions of gluon masses. The coupling parameter of QCD decreases with transverse momentum or $\sqrt{s}$. This higher energy limit means that ...



Top 50 recent answers are included