Einstein's cosmologicas constant divided by energy density x c^4= Newtons gravity constant.?
965826
L = 8pi G rho/c^4 is the definition of the cosmo constant L (1/m^2); where c = light speed and rho is the energy density of the known universe in a vacuum.\n\nF = GmM/r^2 is the equation for the attraction force F between two masses m and M, r distance apart.\n\nL was created by Einstein to make his equations come out right.  By that, he meant the universe should be spherical (not flat or saddle shaped).  His original results showed a flat universe, and he just "knew" that was wrong; so on came L to fix the shape problem.\n\nF was the result of experiments by Newton.  Through observation, he found that the attractive force between two bodies was proportional to the inverse square of the distance between them and to the product of the magnitudes of the two masses.  That is, F varies as (~) (mV/r^2).  \n\nWhenever there is a proportionality, as in F ~ mM/r^2, there has to be a constant of proportionality.  In this case Newton called it G; so that F ~ mM/r^2 becomes F = GmM/r^2 when the constant is inserted.\n\nThe G in both equations is the same constant.  But Sir Newton came upon it long before Einstein found it could be used in his equations and the calculation of L.  So there is no way that Newton (1642-1727) could know about Einstein's (1879-1955) use of it much later.
879076
According to Newtonian gravity, gravitational force is -GMm/r^2.  Any mass experiences a gravitational force toward every other mass in the universe, proportional to the product of their masses, and inversely proportional to the square of their distance.  This means my body experiences a certain gravitational force from the sun, no matter what else is happening in the universe.  That says it would be impossible to shield me from the sun's gravity.\n\nTo find the net gravitational force on me, simply do a vector sum of all the gravitational forces on me from all other masses in the universe.  Simple to say, not so simple to do.\n\nThat sum is zero at the center of mass of the universe, wherever that is.  Unless the expansion of the universe is perfectly uniform, that point will shift with time.  I could find other points where the sum is zero.  I move away from that center of mass but toward a large nearby mass, just close enough that the pull of that nearby mass exactly balances the pull of the rest of the universe.  These will be single points, not regions.  At these points, I still can't say there's no gravitational force acting on me.  But I can say there are countless gravitational forces acting on me, but their sum is zero.\n\nLet's say I tell you I found such a zero-sum spot, about 1/3 of the way between our sun and Alpha Centauri, and I take you there.  How will you know I am telling you the truth?  The principle of equivalence says that there's no test you could make to tell whether the net gravitational force there is zero, or whether you are just accelerating according to the effect of the nonzero net force on your mass.\n\nThe relativistic warping of spacetime in the presence of gravitational mass is more a way of representing or measuring that force, rather than explaining it.  Neither Newton nor Einstein could tell us why gravity exists or how it works.  If spacetime is not warped in your region, it doesn't mean the gravitational field is zero, it means it is uniform.  And again, how would you tell whether spacetime is warped in your region?\n\nI've tried for decades to think of a gravity shield or a gravity insulator.  If I do find one, watch for my Nobel Prize in physics, and for me to quickly surpass Bill Gates on the wealth scale.
895733
Energy=Mass * (speed of light)^2
110744
Short answer: usam da inverse square law Jar Jar!\n\nLong answer:\n\nNewton simplified things by picturing all of a planets mass gathered in a single point in the center of the planet and the surface as a point on a certain distance from that center of mass. So the gravitationall pull (F) on a body (m) at a distance (r) from the mass (M) can be calculated using this equation:\n\nF=G*((M*m)/r^2)\n\nwhere G is the gravitational constant (6.67 * 10^-11 Nm^2kg^-2)\nEarths (equatorial) radius is 6.378 * 10^6 meters and its mass is\n5.974 * 10^24 kilogrammes\nSo lets test this on a dude weighing 100 kg:\n\n6.668 * 10^-11*((5.974*10^24 * 100)/4.068*10^13)\n==>\n6.668 * 10^-11*(5.974*10^26 /4.068*10^13)\n==>\n6.67 * 10^-11* 1.424*10^13 \n=\n979 N (what I got in my calculator)\n\nSo earth pulls on a guy, with the mass 100 kg, with the force of 979 Newtons. That means 9.79 N/kg, earths gravitational force (g). This should also mean that a mass dropped from a small height should accelerate at the rate of 9.79 m/s^2. It´s a simple experiment to do but where I am the value of g is actually 9.82. This is because I am closer to the earths center. Earth bulges out at the equator because of its rotation. I am farther north. But still a simple experiment can be done to validate this calculation. \nSo if the earths mass was 3 times greater than g shoud be 3 times greater. Right? You betcha. And I have already shown that moving closer to earths center increases g. Moving twice as far from the center should decrease g by half. Right? Wrong! Now what? We where so close. Oh man...\ng is reduced to a quarter if we increase the radius by 2. This is because of the inverse square law, a very important law in astronomy. It means that the apparent brightness of a light source is inversely proportional to the square of its distance. This also aplies to gravitational fields. Double the distance, divide by 4.\nSo lets put this all together. You should be able to do it yourself by now. I got 734 N or 75 % of the dudes weight on earth. And the reason he doesn´t weigh more than on earth is because of the inverse square law. Eventhough the new world is three times as massive we still moved twice as far away from the center. So take the weight of your astronaut, multiply by 3 and divide by 4.
704788
According to Mark McCutcheon and his Final Theory (of everything), gravity is the result of everything in the universe expanding at a universal constant atomic expansion rate (XA) of 0.00000077 per second squared. \n\nFrom his book:\n\n"This numeric result for the universal atomic expansion rate, XA, means that we can now calculate the amount of overall expansion each second of any atom, object, or planet simply by multiplying their radii by XA. Multiplying the radius of the Earth by XA gives 4.9 meters, which is the distance an object "falls due to gravity" on Earth in its first second. So, in essence, we can now calculate the effective gravity of every atom, object, and planet simply by knowing their size. This is radically different from Newton's idea of gravity, which states that a gravitational force emanates from all objects in proportion to their mass."\n\nSo, as all atoms composing the Earth and all the atoms composing yourself are growing at a certain expansion rate, and since the Earth proportionally grows faster than you, in reality you are never falling towards the Earth, on the contrary the Earth is pushing you constantly as it grows. \n\nYou would not feel falling in a free fall (in a vacuum), the Earth would be growing towards you very fast at a speed of 4.9 meters per second. As you also grow in size yourself at the same expansion rate, the distance between you and the Earth shrinks at a speed of 9.8 meters per second squared.\n\nNo strange force emanating from objects, attracting each other. Simply shrinking distances between them due to everything expanding at a constant rate. That is gravity according to Mark McCutcheon.\n\nSee the link below. It is not very clear on that website what the new gravity theory is all about, but if you contact me I will give you all the details. This is the kind of thing that could change the future, and yet, could virtually pass unnoticed in the history of humanity. Check it out... Mark McCutcheon probably does not have all the answers, but I'm sure he is up to something important which could answer many of your questions you have asked so far.
1130053
As mentioned in other replies, physicists don't yet have all of the "how" questions answered. Einstein, in the General Theory of Relativity, pretty much nailed the underpinnings of the current theoretical understanding of gravity. Here is my attempt at the "nutshell." I'll start with an analogy. Let's say you are driving somewhere in the midwest or plains states--essentially flat ground, with a few rolling hills. Let's say you stop your car at the top of a slight rise. As you look around you, you see flat land in every direction. You're sitting on essentially a two-dimensional plane. Taking your foot off the brake, you notice the car starting to roll forward. Your initial observation about the 2-dimensional "space" you were sitting on isn't quite accurate. The 3rd dimension (altitude) has come into play, and gravity is pulling your car into the swale. What Einstein said (and how he came up with this back in the early 1900's is incredible) was that our THREE dimensions are really a slightly inaccurate description of the world (and universe) we inhabit. A fourth dimension, time, is part of the "fabric" of our surroundings. As in the case of that slight "warp" in the 2-dimensional farmland that led to a noticeable effect on your car, Einstein said that warps in the space-time continuum are experienced as gravity. As you know, Einstein was into the "energy = mass" thing big time, and you know that mass is tied up with gravity as well. If you want to get more cozy with these related concepts, I strongly recommend David Bodanis' book, "E=mc2: A Biography of the World's Most Famous Equation" -- available as a "BooksOnTape." It's exciting, full of cool anecdotes about the heroes of 20th century physics, and... NO MATH, just concepts. Best of luck!
225813
The constant in Newton's law of gravitation that yields the force one body exerts on another when multiplied by the product of the masses of the two bodies and divided by the square of the distance between them. It equals 6.67 × 10-11 m3kg-1s-2.
9301
ok, well the formula you need is:\n\nFg=Gm1m2/r^2\n\nwhere Fg is the force of gravitation, G is the universal gravitational constant(just a constant, you dont have to worry about it) m1 and m2 are the masses of the 2 objects, and r^2 is the distance between the two objects squared.\n(a) ok, so if r increases by 10 times, since r is squared, the denominator would actually increase by 100 times.  Since the denominator increases by 100 times, and the numerator stays the same, you're basically multiplying it by (1/100) so that means that you have to do the same thing to both sides, so you just multiply Fg by 1/100 so thats 500/100 so the answer is 5 newtons.  This makes sense because the farther objects get from each other, the less each other's gravity will affect them.\n\n(b) ok, now one of the m's, lets say m1 doubles.  The rest stays the same again, and this time its in the numerator, so you're just multipying the whole thing by 2, so the new force would be 1000 newtons.  This makes sense, because something with more mass would experience more gravitational force, for example, a person with more mass experiences a greater gravitational force with the earth, and therefore weighs more.
803150
check out this site for the answer\nhttp://math.ucr.edu/home/baez/physics/Relativity/GR/centre.html
201357
This is an awesome question!  Because it relates Newton's law of gravity to real life applications.  The simple answer is yes.\n\ng = Gm1m2/r^2 is Newton’s equation for gravity.  It basically says that any two objects exert an equal and opposite force on each other.  So with that lets apply this to your question.\n\nFirst lets find out how dense coal is.... between 1.25 and 1.5 (depending on the specific kind of coal your looking for) with coke coal which is the most commonly use coal (I think) its about 1.25.  1.25 g/mL isn't very dense; in fact liquid water at 25 C is 1 g/mL so that gives you an idea of how dense coal is.  \n\nWell density is a measure of how much mass is packed into a particular volume or area, so the less mass in a particular area the "lighter" that area is (this is not very specific but its the only word I can think of at 6 am).  Anyways, moving on, when satellites orbit the earth they actually speed up and slow down depending on where they are over the earth.  For instance, when a satellite travels over the Rocky Mountains it actually slows down.... Why?  Because of Newton’s law of gravity!  There is more dense material in the Rockies, and so there is a greater attractive force exerted on the satellite (of course it is very minute, obviously the satellite doesn't drop out of the sky) but nevertheless the satellite does experience a slight change in velocity.  So, if you get two satellites following each other, this change in velocity over certain areas of the earth can help a scientist determine the relative density of that portion of the earth.  So, if these two satellites are traveling over a part of the land, and detect a pretty low density for solid matter, say around 1.5 or even as much as 2.5... etc. Scientists can deduce pretty accurately the composition of matter in that particular location and this ability has enabled humans to locate coal underground without even having to start digging.\n\n\nPretty cool huh?\n\nSo in short, satellites velocity helps determine the density of specific areas of the earth.  This density can help us determine the composition of matter in these specific areas.  Areas of land with low densities may contain coal.\n\nOn last disclaimer, this is how I was taught in my college physics course as the way we find the specific densities of the earth.  However, there may be a more technologically advanced way that I don't know of, that is used to locate coal.\n\nHope this helps!
1033843
e=mc^2\n\ne (energy)=m (mass) times c (the speed of light) squared
70889
convert to Newtons from mN, then divide by gravity because F=mg. so you get .00000780/9.8
834129
You are confusing energy and gravitational force. The two are\nnot equivalent. The property of gravity is based upon the mass of an object, the more mass an object has, the more gravity it posesses. The amount of gravity an object has does not increase or decrease with time.
433896
Yes, energy can cause a gravitational attraction (or repulsion - cosmological constant not equal to zero), via a distortion in spacetime. This is because of the energy tensor in the gravitational field equations, namely:\nG(uv)=8*pi*T(uv).\n\nThe most obvious example of energy distorting spacetime would be that which is carried off by gravy waves, due to for example, a neutron star orbiting a massive black hole. The LIGO and other detectors are trying to detect these waves. Hope this answers your question! ;-)
649603
To say its an equation involving the relation between energy possesed by a mass
834818
Yes.  The cosmological constant represents the energy density of empty space.  Einstein introduced it in order to adjust his equations to allow for a static universe, but then disowned it after Hubble's measurements suggested the universe was expanding.  However, recently it has been picked up again, as an elegant way to account for evidence that the universe is not only expanding, but accelerating (i.e., the rate of expansion is increasing).  A positive cosmological constant means that empty space has positive energy, which would cause space to expand at an increasing rate.  No adjustment of the value of the constant is necessary; think of it as a measure of force or pressure, not mass or mass density.
814715
no. \nthe stress-energy tensor tells us how the energy is spread, and moves, across spacetime.\nthe gravitational constant is a parameter that determines to what degree 1 kg of mass (for instance) will "warp" the spacetime. \n\nso, you can't derive it from the einstein equation.\n\nin fact, relativists usually set the gravitational constant to = 1, so that they don't have to deal with it. although it's essential when you're trying to determine exactly what your physical predictions will be, it's not really an active element of einstein's theory. it just kind of sits there.