And plz if you know what is self gravitational potential energy discuss it.\nand how do we calculate it.
210095
Gravitational potential = G *Mass of the earth/radius of the earth.\nIn other words its equal to Gravitational energy per unit mass.
427458
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.
32003
ja :), that's a good one. E=mc^2 talks about the equivalence between matter and energy; c is just the constant that joins them both ..you won't attain a c ¨velocity¨out of it ; you'll just have to use \n E=mc^2/(sqrt(1-(v/c)^2) ( Is it Lorenz transf?) and I think Xav was pretty clear with his explanation! :)
561727
Use Newton's universal gravitation equation, then sum the forces.\n\nF = G*m1*m2/r^2\n\nF = force (newtons)\nG = gravitational constant (6.6742 x 10^-11)\nm1 = mass of object 1 (kg)\nm2 = mass of object 2 (kg)\nr = distance between objects (meters) ==> don't forget to convert the units\n\nIn this case, you do it twice:\nF1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left\nF2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right \n\nNet Force = 0.84 - 0.067 = 0.772mN to the left
154644
What is this gravity that you mention?  :P\n\nGood point tho, funny how it didn't exist until very recently in our history.  *hugs*
50140
This problem is solved with the application of work and energy principle which states that total work done on a moving particle is equal to the change of its kinematics energy:\nW=T2-T1\nHere the only work done on the particle is dot to gravitation force and is equal to -m*g*h:\n\n-m*10*0.45 = (1/2)*m*(1)^2-(1/2)*m*(v)^2 --->\nm*v^2=(4.5+0.5)*m*2=10*m --->\nv=sqrt(10) m/s\n\nv is equal to sqare root of 10.(approximately 3.162 m/s)
543379
According to energy conservation law: m2*g*h=m1*g*h+m1*v^2/2+m2*v^2/2+I*w^2/2.\nm2*g*h is initial potential energy of the system;\nm1*g*h is final potential energy of m1;\nm1*v^2/2 is kinetic energy of m1;\nm2*v^2/2 is kinetic energy of m2;\nI*w^2/2 is kinetic energy of pulley with inertia moment of I & final rotary speed w rad/s^2.\nNow m1=0.46kg, m2=0.5kg, h=0.75m, r=0.05m, t=5s, g=9.8m/s^2.\nAs h=a*t^2/2, then a=2*h/t^2 – linear acceleration of m1 and m2. \nAs v=a*t, then v=2*h/t – linear speed of m1 and m2.\nAngular acceleration of pulley w’=a/r; angular speed w=v/r.\nThus 2*(m2-m1)*g*h=(m1+m2)*v^2+I*(v/r)^2, so I=(extract it please and plug values in).\nT1=m1*(g+a), T2=m2*(g-a). Don’t worry, be happy!
489517
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
215346
I'll give the framework of the answer, but without doing any of the actual work (math), since I'm on my own free time.\n   Define the key variable to be theta, the angle relative to vertical of the line from the center of the disk to the center of the half-circle.\n   Now, specify the formula for gravitational potential energy of the disk as a function of theta (OK, I'll do that: PE = mgh = mg (R+r) cos(theta) )\n   Now, specify the formula for kinetic energy of the disk as a function of d(theta)/dt.  This has two components, (1)energy associated with the velocity of the center of mass, v = d(theta)/dt * (R+r), [use KE=mv^2/2] and\n(2) energy associated with the spinning of the disk, omega = d(theta)/dt * (R/r + 1) [use KE= I(omega)^2 / 2] (the +1 in that equation is the "gotcha" in this problem)\n  Now, use conservation of energy to equate total kinetic energy to loss of gravitational potential energy.  This equation thus relates theta to d(theta)/dt.  By solving this differential equation with the proper initial conditions, you will get the function of theta versus time.\n   Determine the vertical acceleration by taking second time derivative of (r+R)cos(theta).  When this vertical acceleration equals g, the rolling disk loses touch with the large disk.
303330
Hmmmm......... could it be energy?????
84826
Black holes are a theory, we suspect that some anomalies in space are black holes. White holes are desired by our nature to balance the negative energy of black holes. However the Begy concept of gravity as an external force acting on our universe more than compensates for the negative E of black holes and even predicts a source of energy that is constantly flowing into the universe which can only be detected as the force of gravity. According to the Begy theory, black holes would be anchors in time and space where the gravitational force is neutralized.
173496
There is potential energy and kinetic energy.  Kinetic energy is the energy that something has from motion.  It's equation is E=1/2*m*v(squared) where m=mass and v=velocity.  For potential energy there are many forms.  Gravity E=mass*height*gravity, Electric and magnetic fields are potential energy.  And nuclear energy E=m*c(squared) which is the rest potential energy that anything with mass will have.  There is also energy from light E=h*nu, nu=frequency h=planck's constant.  There is also chemical potential energy which is the energy contained within molecules.
121932
Why would you think either a black hole or Earth would be made up of "pure energy" just because they have mass?\nMass is mass, energy is energy, only in very specfic circumstances does the line blur (E=mc^2).
583378
Total energy at the start is equal to the gravitonal potential energy plus the initial kinetic energy. That equals 320000j + 14400j. At the bottom of the hill, the kinetic energy is 160000j and there is no more potential energy. The amount of energy lost to friction, in the form of heat, is 174400 joules.
258669
if the centrifical forces are as great as you say it is. then the probablity of this force creating a dimensional vortex because of its immense gravitational energy is very probable...but how you gonna test something like this.\n\nCan I cheat and ask stephen hawkins?
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.
198895
because according to general relativity E=mc^2, so you can say that energy is a form of mass, but when you include this to the uncertainty principle it's incorrect to say that void has no energy (x & v = 0) so you have to go the other way, since there's a lot of void out there you should have the equivalent mass, and since mass is gravity we should be seeing the BIG CRUNCH right now, obviously that ain't happening, so one of the two theories must be wrong, but scientists prefer to say that both of 'em r parcially or entirely wrong