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## Do black holes defy the laws of physics?

Phillip E.
Featured 1 year ago

Black holes challenge the laws of physics as we know them.

#### Explanation:

Nothing should be able to defy the laws of physics. If something is inconsistent with the laws of physics then they need to be modified to accommodate the inconsistency.

Black holes are extreme objects. They were predicted from the Schwarzschild solution to Einstein's General Theory of Relativity. Many people didn't think that they existed until evidence was found. A black hole could explain the galactic X-ray source at Cygnus X-1. It is now believed that most large galaxies have a supermassive black hole at their centres.

One issue with black holes is that the theories suggest that there is a singularity inside them. A singularity is a point of infinite density and infinite curvature of spacetime. The physicist Kip Thorne described the singularity as the point where all laws of physics break down.

Another problem with black holes is the black hole information paradox. The issue is that if a particle falls into a black hole information about its state is lost. This is forbidden by the laws of physics as we know them. Stephen Hawking is working on a new theory by which the information is somehow retained at the event horizon.

Clearly we need new laws of physics if we are to completely understand black holes. As nothing which goes past the event horizon can ever escape. This makes it impossible to see inside a black hole.

So, yes, black holes defy the laws of physics as we know them. This means that our laws of physics are incomplete.

## How does star formation begin?

Hunaid L. Hanfee
Featured 12 months ago

Stars formation begin inside when dense concentrations of molecular clouds core collapse under their own gravity.

#### Explanation:

Stars begins as vast clouds of cold molecular hydrogen and helium left over from the Big Bang.

These clouds is held in balance between their inward force of gravity and the outward pressure of the molecules but when some external force disturbs this, it causes cloud to begin collapsing.

These clouds are extremely cold (temperature about $10$ to $20$K, just above absolute zero). The deep cold also causes the gas to clump to high densities.

These cloud core have masses around ${10}^{4}$ solar masses in the form of gas and dust. The cores are denser than the outer cloud, so they collapse first. As the cores collapse they fragment into clumps around $0.1$ parsecs in size and $10$ to $50$ solar masses in mass. .(credit)

Each clump eventually becomes so hot and dense that nuclear reactions begin. When the temperature reaches $10$ million degrees Celsius, the clump becomes a new star and the whole process takes about 10 million years.

After the cloud become star in its core, proton-proton chain reaction starts which gives the stars energy.

The proton–proton chain reaction is one of the two (known) sets of fusion reactions by which stars convert hydrogen to helium. It dominates in stars the size of the Sun or smaller.

## The Martian satellite Phobos travels in an approximately circular orbit of radius #9.4 * 10^6# #m# with a period of #7# #h# #39# #min#. What is the mass of Mars?

Parzival S.
Featured 11 months ago

${M}_{m a r s} = 6.479 \times {10}^{23}$ kg (I used Google calculator so you may get a slightly different number if you approximate $\pi$).

#### Explanation:

The equation that relates the mass of Mars with the orbital information provided about Phobos is the following:

${T}^{2} / {R}^{3} = \frac{4 {\pi}^{2}}{G \cdot {M}_{m a r s}}$

T is the period. Let's express that in seconds:

$7 \cdot 3600 + 39 \cdot 60 = 25200 + 2340 = 27540$ seconds

Let's plug in and solve for M:

$M = 4 {\pi}^{2} {R}^{3} / \left({T}^{2} G\right) = 4 {\pi}^{2} {\left(9.4 \times {10}^{6}\right)}^{3} / \left({\left(27540\right)}^{2} \cdot 6.673 \times {10}^{-} 11\right)$

$M = 6.479 \times {10}^{23}$ kg (I used Google calculator so you may get a slightly different number if you approximate $\pi$).

A quick check of "the internet" tells me that the actual number for the mass of Mars is $6.39 \times {10}^{23}$ kg, so we're pretty close.

Thanks to the following link for the equation: http://www.physicsclassroom.com/class/circles/Lesson-4/Mathematics-of-Satellite-Motion

## How does the composition between irregular, spiral, and elliptical galaxies compare?

What a wonderful world
Featured 8 months ago

They vary in a number of things, types of stars they contain, the amounts of metal, dust, and gas they contain, and the ages of the stars within them.

#### Explanation:

Typically, elliptical galaxies tend to be composed of older, 'lighter' stars, and are thought to be old galaxies, so there is little inter-stellar matter, and hence little star-formation taking place. They also tend to have a number of globular clusters around them.

Spirals, as you might guess, have a spiral structure (about 60% of them seem to be barred spirals, which means they have a bar across them). The stars in the arms especially tend to lie more or less in a rough plane, and orbit the galactic bulge (which is in the centre).

A spiral's arms tend to contain a number of hotter, younger, blue stars, whereas the galactic bulge (the bit in the middle) may be more like an elliptical galaxy, and may contain more older, metal-poor, redder stars, this seems to vary though, depending on how tightly wrapped the arms are. The more tightly wrapped ones (Sa) contain older stars, and the more open ones (Sc) contain younger, bluer, more metal-rich stars.

The bulge may well also contain a very massive black hole, though this is still not fully confirmed. It has also been speculated that Spirals may contain a dark matter halo of some sort.

As for irregular galaxies, some may once have had a structure, which got deformed by some external gravitational force. A lot of them contain great amounts of dust and gas (unless they are dwarf irregular galaxies).

The irregular galaxies fall into a 3 basic classes: Irr-I, Irr-II, and dlrrs. Irr-I has some structure, but not enough to be classified as a spiral, barred-spiral, or elliptical; the Irr-II has no structure at all; and the dlrrs are dwarf galaxies with low levels of metals and high levels of gas. They may be very similar to the first galaxies to form in the Universe.

## What is the spectroscopic method of analysis?

reudhreghs
Featured 2 months ago

Spectroscopy is measuring light to find out what something is made of on an atomic scale.

#### Explanation:

Spectroscopy is measuring the light emitted by stars to tell what sort of atoms the star is making, or what a substance is made up of.

Each atom is a nucleus surrounded by a load of electrons. The electrons have specific energy levels, like stairs, and can move between the levels by absorbing or emitting photons of certain frequencies.

When an electron absorbs a photon, it gains energy and moves up the stairs. When it emits a photon, it loses energy and moves down the stairs again. Luckily for us, photons are also light, so we can see it when electrons lose energy in atoms.

Also, each atom has very specific energy levels that we can measure precisely here on earth, so we know from the frequency of light emitted what energy is lost and so what atom it is in the star, or any other substance.

[The above image is an example of spectroscopy.]

Energy of a photon is given by:

$E = h f$ or $E = h \nu$

where $h$ is called Planck's constant and is approximately $6.626 \cdot {10}^{-} 34$, and $f$ is the frequency of the light emitted in Hertz (oscillations per second).

We may not be able to directly measure the frequency, but we know that the speed of light is constant and that it is given by:

$c = f \lambda$

where $c$ is the speed of light, $\lambda$ is wavelength, and $f$ is, again, frequency. The speed of light is about $3 \cdot {10}^{8} \frac{m}{s}$ approx. The real value is taken as $2.99792458 \cdot {10}^{8} \frac{m}{s}$. These values are speed of light in a vacuum.

Therefore,

$f = \frac{c}{\lambda}$, so

$E = h f = \frac{h c}{\lambda}$

This is great because the colour of light is how humans see wavelength, so just from the colour of something we can determine the energy of light emitted, and the energy of light emitted is specific to each atom.

We can also go the other way, rather than measuring the light that something emits, measure the light that it absorbs, which, although it is exactly the opposite, gives us exactly the same result. We can do this by shining white light on a gas in the lab and seeing what light comes through the other side. The light that doesn't come through is absorbed, and from this we can tell precisely what the gas is.

## How many neutrinos are in the universe?

Phillip E.
Featured 4 weeks ago

There should be in excess of ${10}^{79}$ neutrinos in the Universe.

#### Explanation:

We can't be sure for certain how many neutrinos there are in the Universe, but we can make an estimate.

First of all let's assume that there were relatively few or no neutrinos created in the big bang. The only particles being considered are protons and electrons. There should be equal amounts of antiprotons and electrons, but we have no explanation as to why matter dominated over antimatter.

Estimates put the number of atoms in the Universe at ${10}^{80}$ in the observable Universe. Actually estimates vary from ${10}^{78}$ to ${10}^{82}$.

About 75% of the mass of matter is Hydrogen and most of the remaining 25% of the mass is Helium. Heavier elements are relatively rare. Given that Helium has two protons and two neutrons, we can estimate the number of neutrons to be about 10% of the number of protons or a number of ${10}^{79}$.

All of the neutrons in Helium and heavier elements were created as part of the fusion process where a proton decays by the weak force into a neutron, a positron and an electron neutrino.

So, as creating a neutron also creates a neutrino, they should occur in similar numbers at about ${10}^{79}$.

This figure is probably on the low side as there may have been neutrinos created at the big bang. Also, antineutrinos are created by some radioactive and fusion processes, where a neutron decays into a proton, an electron and an electron antineutrino.

If an neutrino and an antineutrino of the same flavour collide they will annihilate each other. As neutrinos don't interact much such collisions will be quite infrequent. Also, the neutrino-antineutrino annihilation is a weak force process where they become a ${Z}^{0}$ boson. The Z decays rapidly. Depending on the neutrino energy the Z can decay into two photons, but it is more likely to decay back into a neutrino-antineutrino pair.

So, in summary the number of neutrinos is in the order of ${10}^{79}$ to within two orders of magnitude.

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