2
Active contributors today

## During the fusion process, how is mass converted into energy?

Waleed A.
Featured 9 months ago

$E = m {c}^{2}$

#### Explanation:

This is calculated using the famous equation of Einstein,

$E = m {c}^{2}$

In Fusion reaction like the ones taking place in the core of a Star, there is enough pressure to fuse hydrogen nuclei to form one helium nucleus.

So, 4 hydrogen nuclei are fused together to form one Helium nucleus. But, where does the energy come from that keeps the Sun from collapsing?.

When 4 Hydrogen nuclei are merged together they show a certain discrepancy in the mass when a Helium atom is formed, i.e the mass of 4 Hydrogen atoms before Fusion is less than the mass of the Helium atom after the reaction this mass defect is converted into energy by $E = m {c}^{2}$.

$\text{mass of a hydrogen atom " = " 1.00794 u}$

$\text{mass of one helium atom " = " 4.002602 u}$

Where $u = 1.6605 \times {10}^{- 27} \text{kg}$.

The mass defect, $\Delta m$, will be

$\Delta m = 4 \times {m}_{H} - {m}_{H e}$

$= 4 \times \text{1.00794 u" - "4.002602 u}$

$= \text{0.029158 u }$ or $\text{ "4.8416859 * 10^(-29)"kg}$

In the equation $E = m {c}^{2}$, $c$ is the speed of light in a vacuum, approximately equal to $3 \cdot {10}^{8} {\text{m s}}^{- 1}$.

This means that you have

$E = 4.8416859 \cdot {10}^{- 29} {\text{kg" * (3 * 10^8)^2"m"^2"s}}^{- 2}$

$E = 4.35751731 \cdot {10}^{- 12} \text{ J per reaction}$

## What is the Aurora Borealis?

Hunaid L. Hanfee
Featured 9 months ago

Aurora Borealis are lights caused due to collision between charged particle & the air molecule in Earth's magnetic field.

#### Explanation:

Aurora Borealis are the dancing light which can be seen at sky in night. This light can be seen above the magnetic poles of the northern and southern hemispheres. They are known as 'Aurora Borealis' in north & 'Aurora Australis' in south.

For more picture please visit this site - Aurora near Akureyri

This light is actually the collision of charged particle from sun between air molecule in the Earth's magnetic field. When they strike the molecules to become excited. This molecules gives off the light as they calm down.

This lights are also know as Northern Lights.

Aurora displays appear in many colors although pale green and pink are the most common. But sometimes red, yellow, green, blue, are also visible. People call it dancing light because it appears in many form like arc, cloud, shooting rays etc.

Aurora shows different colors because of different gases present in the air.
Oxygen - Pale green or orange red.
Nitrogen - Blue or red.

## Does the Oort cloud exist? If so, why can't we see it?

Parzival S.
Featured 8 months ago

The Oort Cloud is theorized to exist but due to many factors we can't see it.

#### Explanation:

First let's talk about what the Oort Cloud is (and what it isn't).

The first thing to know is that it is theorized that the cloud exists. While definitive proof will be difficult to come by, the theory helps explain many questions in astronomy (where do long-period comets come from and how to describe their orbits, etc)

From http://space-facts.com/oort-cloud/, it's a region of space that surrounds the solar system where ice, rock, and the occasional larger body (sometimes called Dwarf Planets) exist. The region is a spherical shell and starts about 2,000 AU (Astronomical Units) from the Sun (and to put that into perspective, 1 AU is the average distance from Earth to the Sun. Pluto is 40 AU from the Sun.) and extends well out towards the closest star (perhaps as much as 100,000AU away from the Sun or roughly 1/4 of the distance between the Sun and the closest star).

What it isn't is a cloud in the sense we think about clouds on Earth - big puffy things made up of water vapour that are easy to spot in the sky. Clouds in the astronomical sense are far less dense but are much more vast so that we see them when the light of stars and galaxies is interfered with.

So what would make the cloud hard to see?

Things that we observe in space are observable only if we can see them.

So what does it take to be able to see something? Light.

And where does the light come from? It either is produced by the object (ex. the Sun, a lamp, etc) or the light is reflected off the object (ex. the Moon).

When light is reflected, all the light that strikes the object doesn't bounce back - it gets scattered, absorbed, and the like, so only a tiny fraction of what strikes the object is reflected back (which is why the Moon, even though it can be quite bright at night, isn't anywhere near as bright as the Sun. The Moon absorbs some light and what is reflected is scattered off in all sorts of directions - only a fraction reaches Earth).

Clouds in the Earth's sky are visible because there is a strong light source (the Sun) that allows the clouds to scatter the light and yet still have plenty left for us to look up and see them. There is no light source like that for the Oort Cloud, so what light it gets, when scattered, doesn't come back towards Earth but instead flies off in random directions.

How much light does the Oort Cloud receive? Light diminishes exponentially with distance. A strong light source at distance 1 will only be $\frac{1}{4}$ as strong at distance 2, $\frac{1}{9}$ at distance 3, etc. And this is for both the journey of the light to the object and the reflection back! What that means is that for an object at the closest edge of the Cloud, it receives ${\left(\frac{1}{2000}\right)}^{2} = \frac{1}{4 , 000 , 000}$ the amount of light we get (remember that Earth is at distance 1 AU and the edge of the Cloud is at distance 2000AU)! And then light that is reflected back towards Earth and so we see $\frac{1}{4 , 000 , 000}$ of that!

The Oort Cloud does not make its own light, so we need reflected light to see it. The Cloud is made up of small rocks and pieces of ice with huge distances between them, so most of the light that does manage to make it there passes right through and never reflects. Space rocks and ice are notoriously "dirty" - or covered in a dark coloured "space dust", so most of the light is absorbed by the rocks in the Cloud. And the distances involved are massive, so much so that to see Pluto and a few of the other large objects in the Cloud, it takes very powerful telescopes and a lot of luck in finding them.

The only other way of seeing the Cloud is with the use of a deep space probe. However, because of the vast distances, it's difficult to build a spacecraft that can make the journey. According to https://en.wikipedia.org/wiki/Oort_cloud, Voyager 1, the farthest and fastest moving probe we have and launched in 1977, will be another 300 years before it reaches the Cloud and by then will have no power to explore or send back data. Other probes also making their way out of the solar system will also run out of power long before making contact with the Oort Cloud.

## 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 8 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

## What is the string theory? How is it related to space time?

Azhar A.
Featured 7 months ago

String-Theory is the theory that replace the particles(atoms, electrons, photons) to vibrating strings.

#### Explanation:

In the early Twentieth century, two frameworks of physics were published:

$1.$ Theory of Relativity by Albert Einstein
$\textcolor{w h i t e}{0000000000000000000}$and
$2.$ Quantum Mechanics

Both these theories try to explain the formulated laws of Physics that we use now. But the problem arises when we try to reconcile Quantum Mechanics with Theory of Relativity because of Gravity.

String Theory try to solve this problem by replacing particles with strings. That's why it is a member for Theory of Everything.

On large scale distances, the string have normal properties like mass, charge, etc. It is a vibrating string that vibrates in all different ways.

When we get electromagnetic field in space-time, it is called as Quantum Field Theory. As we know that, quantum mechanics is based on probabilities, we can compute that with the help of Perturbation Theory.

But how string theory relates to space-time? If a closed string is traveling in a curved space-time, then the coordinates of the string in spacetime feel this curvature as the string propagates. The answer lies on the string worldsheet or stringy space-time. In order for their to be a consistent quantum theory in this case, the curved space in which the string travels must be a solution to the Einstein equations.

This was a very convincing result for string theorists. Not only does string theory predict the graviton (A hypothetical gravitatioal force carrier) from flat spacetime physics alone, but string theory also predicts the Einstein equation will be obeyed by a curved spacetime in which strings propagate.(Credit)

The problem with string theory is that it is not experimentally discovered yet. If we will get results, we can be able to reconcile gravity with quantum mechanics.

## Mars has an average surface temperature of about 200K. Pluto has an average surface temperature of about 40K. Which planet emits more energy per square meter of surface area per second? By a factor of how much?

JMicheli
Featured 6 months ago

Mars emits $625$ times more energy per unit of surface area than Pluto does.

#### Explanation:

It is obvious that a hotter object will emit more black body radiation. Thus, we already know that Mars will emit more energy than Pluto. The only question is by how much.

This problem requires evaluating the energy of the black body radiation emitted by both planets. This energy is described as a function of temperature and the frequency being emitted:

$E \left(\nu , T\right) = \frac{2 {\pi}^{2} \nu}{c} \frac{h \nu}{{e}^{\frac{h \nu}{k T}} - 1}$

Integrating over frequency gives the total power per unit area as a function of temperature:
${\int}_{0}^{\infty} E \left(\nu , T\right) = \frac{{\pi}^{2} c {\left(k T\right)}^{4}}{60 {\left(\overline{h} c\right)}^{3}}$

(note that the above equation uses $\overline{h}$, the reduced Planck's constant, rather than $h$. It is difficult to read in Socratic's notation)

Solving for the ratio between the two, then, the result is incredibly simple. If ${T}_{p}$ is Pluto's temperature and ${T}_{m}$ is Mars' temperature then the factor $a$ can be calculated with:

$\frac{{\pi}^{2} c {\left(k {T}_{m}\right)}^{4}}{60 {\left(\overline{h} c\right)}^{3}} = a \frac{{\pi}^{2} c {\left(k {T}_{p}\right)}^{4}}{60 {\left(\overline{h} c\right)}^{3}}$
$\frac{\cancel{{\pi}^{2} c {k}^{4}}}{\cancel{60 {\left(\overline{h} c\right)}^{3}}} {T}_{m}^{4} = a \frac{\cancel{{\pi}^{2} c {k}^{4}}}{\cancel{60 {\left(\overline{h} c\right)}^{3}}} {T}_{p}^{4}$
${\left({T}_{m} / {T}_{p}\right)}^{4} = a = {\left(\frac{200}{40}\right)}^{4} = {5}^{4} = 625$ times as much

##### Questions
Ask a question Filters
• An hour ago · in Gravity Waves
• · 2 hours ago
• 2 hours ago · in Stellar Clusters
• 6 hours ago
• 7 hours ago · in Gravity Waves
• 7 hours ago · in Gravity Waves
• 10 hours ago · in Relativistic Orbits
• 11 hours ago · in Blackbody Radiation
• 13 hours ago · in Planet Composition
• 18 hours ago · in Galaxy Composition
• 20 hours ago · in Relativistic Orbits
• 20 hours ago · in Galaxy Composition
• 20 hours ago · in Gravity Waves
• · Yesterday · in Space-Time
• · Yesterday · in Black Holes
This filter has no results, see all questions.
×
##### Question type

Use these controls to find questions to answer