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I know that this distance would be many light years in length

also know that gravity would vary over this distance (but what else was calculus inventend for)
would two masses attract each other at this distance?
if the two masses are attracted to each other at this distance or more importantly greater than this distance than the potential energy could be greater than the energy of its own mass?

2006-08-21 03:26:12 · 10 answers · asked by treb67 2 in Science & Mathematics Physics

10 answers

First. e=mc^2 is an equation relating how much energy is stored in the "rest mass" of an object. "Rest mass" means it is stationary. So this has nothing to do with "Potential Energy" of an object within a gravitational field. But let's try to answer your question:

If the water is 1 Kg, then E=C^2

PE = mgh = gh, where g is the acceleration due to gravity (it's about 9.8 m/s^2, so let's just round it to 10m/s^2)

Therefore PE = E -> 10h = C^2 -> h=C^2/10

since C = 300,000 km/s, then h = 9 x 10^14 m

So you can see it is some ridiculously big number, but that's because you're trying to equate the rest mass energy to it's potential energy within Earth's gravitational field, which by the way at that distance from Earth, there would be absolutely no influence at all from Earth's gravitational field.

2006-08-21 04:16:39 · answer #1 · answered by PhysicsDude 7 · 0 0

You want the KE = mc^2

KE = mv^2/2
so v = 1.414 * c

First Hurdle. Here Is it possible any particle can exceed the light speed. Assuming yes then you have to calculate the height.

This is not a simple calculation. It involves calculus integral equations. Since the velocity is very high the distance it has to travel is also very high to reach that velocity. The gravity for long distance is not constant it is proportional to 1/r^2. where is r is the distance between the objects. I can assure you that this can not happen at the earth. So it has to be a black hole where the gravity is very high. So you can assume some height and calculate the gravity at the surface of the black hole.

So do the math and let us know the result.

2006-08-21 04:26:10 · answer #2 · answered by Dr M 5 · 0 0

First of all, u need to understand that as a body falls, its potential energy decreases. Its kinetic energy increases. So i think you should have asked about Kinetic Energy.
Now, since you are refering to E=mc^2, your talking about 'Modern Physics'. So u have to realise that according to Theory of Relativity if material object (particle) is moving with respect to a observer, then the mass of that material object (particle) as measured by that observer increases with increase in speed (or velocity). What this means is that as the Kinetic Energy increases, so does mass! Infact it is the Kinetic Energy of a body that contributes to increase in its mass. Energy and Mass are interconvertible. So its not like u can reach a point where kinetic energy exceeds energy equivalent of mass.
Now, if ur talking about earlier classical theories (which are approximations), then Energy and Mass are totally different. U cant talk about Kinetic Energy exceeding energy of the mass since they are totally different. Either way, the question you asked is not properly framed.
I hope i have been of some help.

2006-08-21 04:14:36 · answer #3 · answered by Maverick 2 · 0 0

It's only potential energy if it's standing still.

Once gravity starts doing work on it (that's force through a distance, remember) the potential energy starts becoming kinetic energy.


Doug

2006-08-21 03:56:05 · answer #4 · answered by doug_donaghue 7 · 0 0

E=mc² does not apply here.

That equation only applies to REST mass.

As speed increases, its total energy increases. The energies you are concerned with will never be equal for any moving mass.

2006-08-21 06:20:03 · answer #5 · answered by Jay T 3 · 0 0

It has to do with e=mc^2 and the strength of gravity on earth

2006-08-26 20:11:24 · answer #6 · answered by Anonymous · 0 0

Your question has nothing to do with "e=mc^2". It is related to law of conservation of energy in a gravitational field.

K.E. = 0.5 mv^2
P.E. = mgh

Rest of the calculation is easy knowing what is the velocity of the water and the initial height.

2006-08-21 03:35:54 · answer #7 · answered by GreenNGreen 1 · 0 0

assume, the ability released in the course of the flood is A. Now we can get the precise wide type through multiplying A with the type of situations the flood handed off. for that reason, which will be 0. A x 0=0 this is the mathematics for you. i'm grateful i did not ought to study that in college.

2016-11-30 22:34:22 · answer #8 · answered by scharfschwerdt 3 · 0 0

This is not a good place to ask a physics question!

2006-08-21 03:35:36 · answer #9 · answered by Anonymous · 0 0

?????

2006-08-26 00:08:35 · answer #10 · answered by marco v 2 · 0 0

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