Suppose that there is only two point masses in the universe. They are positioned far from each other at rest and at distance d and are identical. How much does it take two masses to meet together?\nWhat is the relative velocity when they meet?\nDon’t make any extra simplification on problem.
123450
Your experimental model is corrupt in terms of current known information about matter and energy in a space-time continuum.  No prediction or calculation can be applied.\nWith two point masses, the only force in the system would be do to gravity.  This is an assumption based upon multiple moving masses in what we would call the normal physical universe.\nIf the distance between their centers (d) was sufficiently small, the points would be touching one another and no force would be exerted on the masses.  If the distance was sufficiently large, then it is possible that the force would be so small that the masses would not move.  Only in a finite range of distances would any gravity possibly exist.\nOne can make no assumptions about G as a gravitational constant, since G assumes that one of the masses is the earth.  One would need additional information about the masses to solve the problem.
482427
There may be two answers to this question.\n   First;The universe is spherical.expanding from  a single point.\n   It is definitely finite and will someday become extinct.\n   Seems like a big ball!!\n\n   Second;Time acts as just more than an arrow barreling inexorably in one direction.\n   Every entity in the universe has a zero  time point.\n   You,me and an atom.\n   In any direction we look into the past.\n   This would imply a two dimensional expanding sphere.\n   This would not be much fun!\n   We know the universe works.It incorporates a sort of holographic feature that allows entities to interact,seemingly in a solid sphere.\n   Two people in a room see each other as they were in the past.\n   Maybe three-hundred millionths of a second ago.\n   Because our reaction time is about one-tenth of a second,we have no awareness of this time lag.\n   To be significant to two people they would have to be about 18,000 miles apart.They would be absolutely unaware of each other.\n   This holographic trick of the universe extends to fundamental particles[The uncertainty principle] and to celestial bodies,stars,galaxies and large structures.\n  Philosophers may want to jump in here to try and hone a comprehensive configuration to this phenomenon.\n   Well I hope I did not confuse you to much. I am going to have sit down and figure out just what I said.
311781
'Outside the universe' seem to me to belong to the universe!\nYou think the universe is that small???\nI do not know the answer on your question.\nLet me ask you: can the space time be curved without mass or only thanks to mass?
256130
Hey, man, if a rock is heavy, that means it is attracted to another body of mass.  So the question is: Can God create two bodies of mass that are so attracted that he cannot keep them apart?  Well, if he could compress the whole universe into the size of a pin-head to create the big bang, then I would have to say....maybe.
556817
F = -G(m1*m2)/r^2\nF = force\nG = gravitational constant = (6.6742 +/- 0.001)*10^(-11) N * m^2 / kg^2\nm1 = mass of object 1 = 47 kg\nm2 = mass of object 2 = 6.00×10^24 kg\nr = distance between objects\n\nF = -(6.6742 +/- 0.001)*10^(-11) * (47 kg * 6.00*10^24 kg) / r^2\nF = -4.225 * 10^14 N*m^2 / r^2\nSince the distanct between the Earth and the person weren't given this is all the farther the force can be calculated to.  The force will be equal on both the objects (person and Earth) only that the acceleration of the Earth will be almost nothing due to its size and the force will be in the opposite direction.\n\nBut based on the acceleration of the person, you can calculate a force, but it will only be an estimate\nF = m*a = 47 kg * 9.8*10 m / s^2 = 460.6 N\nForce on Earth is 460.6 N\n\nNow the acceleration of the Earth is\n460.6 = 6.00*10^24 * a\na = 7.68 * 10^(-23) m/s^2
387755
We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.
424569
Oscar. You have written too much. This 125m tower stuff does not matter compared to r=6.38*10^6. Why is this talk about 1.00 km and angle theta? See what “zipper” above says. Besides I answer your last question << Why is the trajectory of the cannonball not parabolic?>>. Why should it be if both you and Newton demand it to be circular? Assume that V1 of cannonball is the 1st cosmic speed corresponding to a circular track. Let V2 be > than V1, then you receive an oval track. Take V3 > than V2, then the track may become parabolic, so that cannonball would never return back to the cannon to linger out among planets and stars. If you are intrigued calculate a set of velocities and corresponding orbits of the cannonball. If cannon is not available consider a football! I like your inventive wits – to be honest!
347974
The idea is that all forces at the nodes (connecting points) are balanced. Draw the two free body diagrams.\n\nLeft point:\nFx\n0=-T1cos53+T2cos11\nFy\n0=T1sin53-T2sin11-57\nRight point:\nFx\n0=-T2cos11+T3cos42\nFy\n0=T2sin11+T3sin42-W2\n\nBy inspection we see only the first 2 eqns are required. Solve the system:\n0=-T1cos53+T2cos11\n0=T1sin53-T2sin11-57\n\nT1=83.62 N\n{=}8.5 kg
509937
What is " a bunch of m's"?\nThe 6.67 refers to the Gravitational constant: 6.67x10-8 dyne cm2 g-2.  \nAccording to the law of universal gravitation, the attractive force between two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them.\nThe formula is F=G (m1m2)/r2, where F is the force, G is the 6.67 business above, m1 and m2 are their masses (in grams) and r2 is the distance between them squared (in cms).
501038
This is a simple ratio problem. The equation would be Dgalaxy/Dearth=Dearth/x so x=(Dearth)^2/Dgalaxy. Asssuming a parsec is 3.6 light years, and a light year is 5.87 trillion miles and the diameter of the Earth is 8,000 miles, then the Earth would be 0.001 millimeter or one micron in size, which is the size of an average bacterium, maybe? Very cool. Oops! I re-did my calculation and I see that I made a factor of ten error. The correct answer is about 0.1 microns, or the size of some viruses. But tell me how you would make such a model of the Earth?
14672
yikes, everyone is giving you the wrong formula. it's not e=mc^2. that would be converting energy into mass and back which is done like when subatomic particles crash into each other really fast and such... the mass dialation formula you are looking for is \nm'= m/sqrt(1-v^2/c^2)\nwhere m' is the mass you would be not moving at the high velocities, m is the mass moving at the high velocities, v is the velocity you're traveling at, c is the speed of light, sqrt means we're square rooting everything in those parenthesis, and ^2 means squared. if you were to plug numbers in for velocity, you can see that this effect is absolutely neglagible unless you are moving very very close to the speed of light. and it is impossible to actually move at the speed of light, because then v^2/c^2 would be 1, and 1-1=0 so you would have a 0 in the denominator of that equation, which is impossible.  that equation is actually very interesting in that it is part of the lorentz factor which also effects time and length. in fact, as you increase speeds close to the speed of light, not only does mass increase, but time moves slower for you and length shrinks.  I've had to derive these equations in physics class, it's very interesting to do it yourself and come up with the equation proving that things like velocity, mass, and length are affected by velocity.  but again, you need to be going very very close to the speed of light for the velocity to have any effect on these things.\n\nas far as practicality, instances of the lorentz factor have been proven. there are these particles that have a certain lifespan before they disintigrate.  so now, we know how fast they are going and we know how long before they disintegrate, so theoretically, we should easily be able to figure out how much distance they cover before they disintigrate, right? but somehow, the particles go a lot farther than they should... if we plug their velocity into the lorentz transformation formula however, and put in how far they should travel (but don't), they we come up with how far they actually do travel.  this proves the validity of the lorentz factor.\n\nthere is no practical application of the mass increase since objects of great mass would require way too much energy to accelerate to speeds great enough to see this effect, so therefore it can't really happen.. if ever it did though, the only thing this mass expansion would serve to do would be to limit time travel. i mean, if time is slowing down for us at high speeds, technically we could make it slow down so much so that it would be going slower for you than everyone else and so you could basically travel forward in time. but if your mass was expanding and length shrinking, then either the pressure would kill you, or your blood would get so dense it couldnt carry oxygen to your cells and that would kill you, so there goes the time travel thing.
43767
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.
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Physicists like to lump time with space. The distance between A and B can be expressed as distance or time. \n\nBut, what is time? That's a bit like "what is mass?" or "what is charge?"\n\nThis is what we know about time. \n1. It is relative.\n2. It has a forward bias with respect to our perspective.\n3. The faster your friend goes, the slower he moves.\n\nI think you just have to accept it for how it is.
471898
If m1 is the mass of the moving particle and m2 is the mass of the particle at rest, then \n\nm2 / (m1+m2) of the kinetic energy of the moving particle will be lost after inelastic collision.\n\n\nBy law of conservation of momentum the final velocity of the two together is \n\nV = m1 v1/ (m1 +m2) becasuse v2=0.\n\nAfter collision the kinetice energy is 0.5 (m1+m2) V^2.\n\nBefore collision the kinetic energy was only 0.5 m1 v^2.\n\nUsing these relations we get the above result.
300228
V = (v + u) / (1 + uv/c²)\n\nTry that. The velocities are quite high so relativistic. Haven't tried it yet though.\n\nNevermind, the velocites aren't nearly high enough to be relativistic. If it was km/s it would work.
26292
In Newtonian space, F = ma or m = F / a (mass is equal to the force it has divided by the acceleration applied to it).\nIn an space-time continuum (relativity aspects as per Einstein) the mass is relative to its location and velocity with respect to the observer.
675473
The force of gravity of one object on another is given by the equation\nF = (GMm)/r^2\n\nWhere G is the universal gravitational constant, M is the mass of one object, m is the mass of the other, and r is the distance between the two.\n\nYou can see that the greater the mass of the two objects, the stronger the force of one on the other is.\n\nIn short, things that are more massive get a better tug.