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## What is the difference between the strong and weak nuclear forces of the universe?

Phillip E.
Featured 1 year ago

The strong force holds atomic nuclei together and the weak force causes radioactive decay.

#### Explanation:

The strong nuclear force is responsible for binding protons and neutrons together in an atomic nucleus. It is strong and short ranged and has to overcome the electromagnetic force which is pushing positively charged protons apart.

A good example of the strong force is the fusion process which happens in smaller stars such as our sun. Positively charge protons repel each other. At the extreme temperatures and pressures in the sun's core, two protons can get close enough together for the strong nuclear force to bind them into a bi-proton or Helium-2 nucleus.

A bi-proton is very unstable and most of them fly apart. For the fusion process to continue to produce Deuterium the weak nuclear force is required.

The weak nuclear force is responsible for radioactive decay by being able to convert a proton into a neutron of vice versa. To be more precise it converts an up quark to a down quark or vice versa by means of the W boson. In the case of fusion a proton is converted into a neutron, a positron and an electron neutrino.

$u \to d + {W}^{+}$
${W}^{+} \to {e}^{+} + {\nu}_{e}$

In fact the strong nuclear force doesn't really exist. Early theories described the strong nuclear force as binding protons and neutrons using the pion as the force transmitting boson. We now now that protons, neutrons and pions are composite particles consisting of quarks bound by the colour force transmitted by gluons. So, the strong force is actually a residual effect of the colour force extending beyond the inside of protons and neutrons to bind them together.

## How long ago was Earth formed and how does anyone know that for sure?

George C.
Featured 1 year ago

At least $4.404$ billion years, probably $4.54$ billion years.

#### Explanation:

Date estimates are obtained by radiometrically dating very old rocks.

OK so far, but:

• How can you find a very old rock when rocks are typically recycled by tectonic processes?

• How do you know that a rock is actually part of the Earth and not part of a meteorite?

• What is this radiometric dating anyway and how do we know that it is accurate?

Firstly, though rocks are subsumed in the mantle and new rock formed from volcanic eruptions, there are some samples of very old zircon found in Austrailia that managed not to be recycled.

Meteorites have specific characteristics that allow us to differentiate them from other rocks. They may have fusion crust - though that does tend to break off after time. They may differ in density from terrestrial rocks, or exhibit particular kinds of fractures, etc.

Radiometric dating measures the relative proportion of different isotopes. For example, over time uranium turns into lead. So by measuring the relative proportions of lead and uranium in a sample you can date it. That's a very crude description of what is a precise and carefully executed technique.

By means of radiometric dating we have measured the age of the old zircon in Australia at $4.404$ billion years and meteorite material at $4.54 \pm 0.05$ billion years.

So what does the age of meteorites tell us about the age of the Earth? According to our understanding, they are kind of left over material from the formation of the solar system. Unlike rocks on Earth they have not been subject to volcanic activity, but were formed at about the same time as the Earth. So it is generally reckoned that their age is the same.

How sure are we?

Radiometric dating techniques are well corroborated. Uranium based dating has the advantage of the ability to cross check between ${\text{^235U -> }}^{207} P b$ and ${\text{^238U -> }}^{206} P b$ decay processes for extra confidence.

So I would say we are pretty sure.

$\textcolor{w h i t e}{}$
Footnote

Beware of some internet sites which claim the radiometric dating is faulty. They often have unscientific agendas.

## A star has an absolute magnitude of -5 and an apparent magnitude of 10. How many parsecs away is this star?

A08
Featured 1 year ago

${10}^{4} \text{ pc}$

#### Explanation:

We know that the apparent magnitude ${m}_{\text{app}}$ of an object tells us how bright that object will appear as observed from Earth.

And the absolute magnitude ${M}_{\text{abs}}$ of an object tells us how bright that object would appear when it is observed from a standard distance of 10 parsecs.

Difference between apparent and absolute magnitudes of an object,${m}_{\text{app" – M_"abs}}$, is called the distance modulus.

Magnitude system is based on the response of the human eye which shows a logarithmic response. Also the magnitude system, a logarithmic scale, assumes that a factor of $100$ in intensity corresponds exactly to a difference of $5$ magnitudes.
Therefore, we have this scale of base ${100}^{\frac{1}{5}} = 2.512$.

For two stars $A \mathmr{and} B$ if there magnitudes and intensities are denoted by $m \mathmr{and} I$ respectively, we have the expression connecting both as
${I}_{A} / {I}_{B} = {\left(2.512\right)}^{{m}_{B} - {m}_{A}}$

Taking the log of both sides and using ${\log}_{10} {M}^{p} = p {\log}_{10} M$ we get
${\log}_{10} \left({I}_{A} / {I}_{B}\right) = \left({m}_{B} - {m}_{A}\right) {\log}_{10} 2.512$

This is commonly expressed in the form
${m}_{B} - {m}_{A} = 2.5 {\log}_{10} \left({I}_{A} / {I}_{B}\right)$ ......(1)

We know that intensity of a light source follows inverse square law of distances. After comparing intensities and magnitudes of two different stars as in equation (1) let us compare the intensities and magnitudes of the same star at two different distances.
Substituting ${\left({d}_{B} / {d}_{A}\right)}^{2}$ for $\left({I}_{A} / {I}_{B}\right)$, (1) becomes

${m}_{B} - {m}_{A} = 2.5 {\log}_{10} {\left({d}_{B} / {d}_{A}\right)}^{2}$

$\implies {m}_{B} - {m}_{A} = 5 {\log}_{10} \left({d}_{B} / {d}_{A}\right)$

When ${d}_{A} = 10 \text{ pc}$, so that ${m}_{A} = {M}_{\text{abs}}$, and ${d}_{B} = d$ be specified in pc, above equation reduces to
${m}_{\text{app" - M_"abs}} = 5 {\log}_{10} \left[\frac{d}{10}\right]$

Above can be rewritten as
${m}_{\text{app" - M_"abs}} = - 5 + 5 {\log}_{10} d$
$\implies d = {10}^{\frac{{m}_{\text{app" - M_"abs}} + 5}{5}}$ .....(2)

Inserting given values in (2) above we get
$d = {10}^{\frac{10 - \left(- 5\right) + 5}{5}}$
$\implies d = {10}^{4} \text{ pc}$

## How do we know that galaxies farther away from us are moving faster than nearby galaxies?

Dwight
Featured 12 months ago

Astronomers use the Doppler Shift to determine the speed at whih a source of light is moving. The faster the source moves, the greater the shift in the colour of the light we observe (compared to the colour of a stationary source).

#### Explanation:

Anyone who has listened to a train approaching and passing on a track has been aware of the change in sound pitch that occurs. This effect is known as the Doppler Shift, and is well understood for all types of waves, including sound and light.

The image is meant to show that if the wave source is moving toward us, the waves will be compressed ahead of the source, and a high frequency is heard. Behind the source, the observed pitch is lower than that of a stationary source. The faster the source moves, the greater the change in pitch that is observed.

With light, the effect is not in pitch, but colour. If a source of light moves toward an observer, the waves will be shorter in wavelength and higher in frequency. The light will appear to be shifted toward the blue end of the spectrum. If the source is moving awat from the observer, the shift in frequency causes the colour to appear more red. As in the case of sound, the faster the source moves, the greater the change in colour that is observed. This is the red shift that is seen in all galaxies.

So, to finally answer your question, when we measure the light from more distant galaxies, we routinely note larger shifts in the colour of the light they emit. This tells us they are moving at greater speed than the galaxies that are "close by".

## Estimate the approximate (minimum) mass of an exoplanet in a 3.312 day orbit around a 1.3 solar-mass star, where it is known that the radial velocity caused by orbital motion is 471 metres per second?

Phillip E.
Featured 8 months ago

#### Explanation:

${a}^{3} = \frac{G M}{4 {\pi}^{2}} {p}^{2}$

Where $a$ is the semi-major axis, $G$ is the gravitational constant, $M$ is the mass of the star and $p$ is the orbital period.

Now for the Sun $G M = 1.327 \cdot {10}^{20}$, so for a 1.3 solar mass star $G M = 1.725 \cdot {10}^{20}$.

The period is $p = 3.312 \cdot 86400 = 886156.8$ seconds.

Substituting the values gives the semi major $a = 7.099 \cdot {10}^{9}$ metres.

The radial velocity of the planet is ${v}_{p} = \frac{2 \pi a}{p}$.

Rearranging the Kepler equation:

$\frac{4 {\pi}^{2} {a}^{2}}{p} _ 2 = \frac{G M}{a}$

Taking the square root gives:

$\frac{2 \pi a}{p} = {v}_{p} = \sqrt{\frac{G M}{a}}$

Substituting values gives ${v}_{p} = 1.55 \cdot {10}^{5}$ m/s.

The momentum equation relates the sun star mass and velocity to the planet mass and velocity.

$M {v}_{s} = {m}_{p} {v}_{p}$

A solar mass is $1.99 \cdot {10}^{30}$kg. So:

${m}_{p} = \frac{1.3 \cdot 1.99 \cdot {10}^{30} \cdot 471}{1.55 \cdot {10}^{5}} = 7.86 \cdot {10}^{27}$kg.

Jupiter's mass is $1.898 \cdot {10}^{27}$ kg.#

This makes the exoplanet about 4 Jupiter masses.

## What is a solar day?

Phillip E.
Featured 4 months ago

A solar day is the period between successive solar noons.

#### Explanation:

The problem with defining the length of a day is that there are several definitions.

The length of a day as measured by clocks is the mean solar day. It is exactly 24 hours long.

The sidereal day is the time it takes for the Earth to complete one rotation with respect to the fixed stars. It is about 23 hours 56 minutes.

The solar day is the period between two successive high noons. High noon being the moment the Sun is at its highest in the sky. It varies continuously from day to day due to the Earth's orbit being elliptical and due to the ${23}^{\circ}$ axial tilt. Successive days vary in length by up to 20-30 seconds.

Solar days were traditionally measured using sun dials. When clocks were invented a correction factor needed to be added to sundial time to make it agree with clocks. The time difference is called the equation of time and the difference between clock noon and solar noon can be as much as 18 minutes.

The graph shows the equation of time. The orange curve is the difference due to the Earth's orbital eccentricity. The green curve shows the difference due to the axial tilt. The blue curve is the sum of the other two curves which is the equation of time.

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