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It proved the relationship #T^2 prop R^3# (Kepler's third law).


Kepler's laws were based completely on observation, but Newton managed to prove them using mathematics.

If we assume that an object is orbiting another object in a perfect circle, we can use #F=(mv^2)/R#.

We know that #v = omega*R#, so #F=(m*omega^2*R^2)/R=m*omega^2*R#

We also know that #omega = (2*pi)/T#, so #F=m*R*((2*pi)/T)^2= m*R*((4*pi^2)/T^2)#

Using the law of universal gravitation, #F=(GMm)/R^2#

If we equate these two, we get #(GMm)/R^2=m*R*((4*pi^2)/T^2)#

The two #m#'s cancel out, to give: #(GM)/R^2=R*((4*pi^2)/T^2)#

Dividing both sides by #R# gives us: #(GM)/R^3=(4*pi^2)/T^2#

Flipping these equations upside down gives us #R^3/(GM)=T^2/(4*pi^2)#

Multiplying both sides by #4*pi^2# gives us #(GM)/(4*pi^2)R^3=T^2#

Therefore, #T^2=kR^3# where #k=(GM)/(4*pi^2)#

Which shows that #T^2 prop R^3#, which is Kepler's third law.


Nuclear fusion is the process of two small atomic nuclei coming together to make a larger nucleus which is stable.*


Nuclear fusion is the reverse process of nuclear fission. This process also results in the conversion of mass into energy.

Nuclear fusion is the energy source which causes stars to "shine", and hydrogen bombs to explode.

Nuclear fusion in star
The enormous luminous energy of the stars comes from nuclear fusion processes in their centers. Depending upon the age and mass of a star, the energy may come from proton-proton fusion, helium fusion, or the carbon cycle

(Credit- http://hyperphysics.phy-astr.gsu.edu/hbase/astro/astfus.html)

Fusion releases energy. The energy released is related to Einstein's famous equation, #E=mc^2.#

To start nuclear fusion, protons and neutrons must be exposed to temperatures approaching 10 MK Celsius (180 million degrees Fahrenheit). In these reactions too much energy is released.

When nuclei fuse together, part of their mass is converted into energy. The sun and other stars convert this energy into light.

#=># For example - Fusion of deuterium with tritium creating helium-4, freeing a neutron, and releasing 17.59 MeV of energy, as an appropriate amount of mass changing forms to appear as the kinetic energy of the products, in agreement with kinetic E = Δmc2, where Δm is the change in rest mass of particles. (Source : Wikipedia)



As #r#, the distance between the astronaut and the Earth increases in #F_g=G(m_1m_2)/r^2#, the gravitational force between the two masses. The weightlessness of an astronaut is actually a continual falling towards but never landing on Earth.


There are a couple of ways of answering this question: one is to answer your specific question about the force of gravity experienced by an astronaut, and the other is to talk about the weightlessness experienced in space.

Let's first talk gravity:

Gravity is a force that one mass exerts on another mass. The equation for that force is:


where #G# is the gravitational constant, the #2color(white)(0)m# terms are the #2# masses exerting the force on each other, and #r# is the distance between them.

The question being asked is about the experience of gravity by an astronaut in space. So let's work this out: as a person rises up into the sky, what in the equation is changing? #G# is the same, as are the mass of the person and the mass of the Earth. It's #r# that is increasing and it increases at an exponential rate.

So as the astronaut rises up above the Earth, the denominator in the equation gets bigger, resulting in less gravitational force the astronaut experiences from the Earth (and coincidentally the Earth experiences less gravity from the astronaut - but the Earth is so massive it hardly notices!)

Ok - so now let's talk about weightlessness:


We've seen on TV and youtube and movies how an astronaut in space, say for instance one living aboard the ISS (International Space Station), seemingly floats in midair. They can take a water droplet and play with the little spherical drop and can add more and more water to make the sphere bigger. So - how is this possible? Is it a result of experiencing a lesser amount of gravity? The answer is - partly.

Remember the story about Sir Issac Newton and the apple - how the apple hit him on the head and suddenly he had the framework for the theory of gravity? What's far more likely in that story is that after the apple hit him on the head, he wondered why the Moon didn't do the same thing - why it keeps moving around the Earth and not plummeting onto it.

The answer here is what @SCooke is getting to in his answer: the movement of someone (or something) in orbit is one part falling towards Earth and one part moving tangentially away from Earth, with the two vectors adding up to being a sort of continuous falling - you fall but never land. Everything in orbit, whether the astronaut or the water drop or the Moon, they are all falling towards the Earth but will never hit it. And it's that continual falling that is the experience of weightlessness.


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


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.


Please see this link for more explanation.


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


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

#1.# Theory of Relativity by Albert Einstein
#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.




The layer chromosphere is #2000# km thick.

The radius of the sun is #0.7 * 10^6# km.

To calculate this you have to just do same as we do in calculation of our exam's marks percentage.

Just imagine,

Your exam is of total marks of #0.7 * 10^6#(not possible but you have #color(white)0000000000000000000000000000000#to just imagine this )
out of this you got #2000# marks.

To calculate the percentage you have to do something like this:

#=2000/(0.7 * 10^6) * 100#


Just do same calculation for this question,
Total radius of sun is #0.7 * 10^6# km.
out of this #2000# km is chromosphere.

#=2000/(0.7 * 10^6) * 100#



I have added image for your help by which you can know where is chromosphere of the sun.


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