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Answer:

By using mathematical models from what we can measure and observe from the surface.

Explanation:

www.ice-age-ahead-iaa.ca
Physicists can't look inside the sun to find out what's there. Instead, they have to do a bit more thinking. By observing the particles and radiation emitted, the motions of gas and something called solar oscillations, we can make models using physics, maths and what we already know about how matter behaves.

en.wikipedia.org

Solar Oscillations - The sun's surface oscillates (moves up and down) due to sound waves that are generated and trapped within the sun's interior. The way that these sound waves are reflected within the sun's atmosphere can tell us a lot about the materials within in, and their nature.

earthsky.org

Solar neutrinos - The reactions within the sun emit tiny particles called neutrinos. Unlike photons (light) these can travel from the sun's core to its surface very quickly. By observing these neutrinos, we can tell a lot about the rate of reaction within the sun as well as its pressure, temperature and density.

Then, once you have some data, it's a matter of applying laws and formulae that have been confirmed through measurable experiments on earth to form models.

I hope this helps; let me know if I can do anything else:)

Answer:

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 #1/4# as strong at distance 2, #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 #(1/2000)^2=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 #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.

Answer:

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
#color(white)(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.

Answer:

The energy required to release a photon depends on its frequency.

Explanation:

The energy of a photon is related to its frequency or wavelength. The energy #E# of a photon with frequency #nu# is given by the equation:

#E=h nu#

Where #h = 6.62607004 × 10^-34 kg\ m^2 \/ s# is Planck's constant.

Photons are commonly emitted by electrons in atoms. Electrons can only occupy certain discrete energy levels. When an atom is in an excited state, one or more of its electrons are pushed into a higher energy level. This requires energy. When an electron drops into a lower energy level it releases the energy by releasing a photon at a precise frequency which corresponds to the different between the two energy levels.

Photons can have any frequency from the very low frequency radio waves, through visible light up to the very high frequency gamma rays.

Lower frequency photons are emitted by electrons losing energy. Gamma rays are emitted when an atomic nucleus needs to lose energy.

Hence a photon can be released by any amount of energy.

Answer:

Gravity keeps a planet in orbit.

Explanation:

A planet's orbit is governed by gravity. Orbits were first described by Johannes Kepler. His three laws of planetary motion are:

  1. The orbit of every planet is an ellipse with the Sun at one of the two foci.
  2. A line joining a planet and the Sun sweeps out equal areas during equal intervals of time
  3. The square of the orbital period is directly proportional to the cube of the semi-major axis of its orbit.

Isaac Newton described gravity as an attractive force between bodies using the equation:

#F=(GMm)/r^2#

Where #G# is the gravitation constant, #M# is the mass of the Sun, #m# is the mass of the planet and #r# is the radius of the orbit.

Albert Einstein showed that gravity is not in fact a force. It the effect of curvature of 4 dimensional spacetime caused by the masses of the Sun and planets. Each planet follows a geodesic which is the 4 dimensional equivalent of a straight line.

Newton's model of a force of gravity is a good approximation as long as the masses and speeds are not too big. In fact Newtonian gravity can't accurately predict the orbit of Mercury because there are relativistic effects.

So, there are no forces keeping a planet in orbit Each planet is simply following a geodesic in curved spacetime.

Answer:

The gravitational parameter for a body #GM# is the gravitational constant #G# multiplied by the mass of the body.

Explanation:

When doing calculations involving gravity, the gravitational constant #G# is required. It is however difficult to measure the value of #G# to high degrees of accuracy.

The known value of #G=6.674 08 \times 10^-11 m^3 kg^-1 s^-2# but the uncertainty in the value is #0.000 31 \times 10^-11#. So, effectively we only know the value of #G# to four decimal places.

Calculating the mass #M# of a body, such as a planet or the Sun, is also difficult. To do so accurately would require knowledge of the body's volume and density which we can't know accurately.

Fortunately, in gravitational equations the quantities #G# and #M# are multiplied together to form the gravitational parameter #\mu=GM#.

Using data from measurements of the orbits of planets, moons and satellites it is possible to measure the value of the gravitational parameter for bodies with a high degree of accuracy. The values of the gravitational parameter for the Sun, Earth and other planets have been carefully measured.

So, if you want to calculate the orbital parameters of a satellite orbiting the Earth, you don't get the values for #G# and #M# and multiply them together. Instead you get the much more accurate gravitational parameter #\mu# for the Earth.

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