How to simply solve for E=MC^2?

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  • Thread starter Zarich12
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I was wondering how to simply solve for E=MC^2. I have basic idea but I just want to check it. You take the mass times the speed of light (in miles per second?) and square that. Over one second that is equal to the number of watts the object could produce. Is that right. If there are any mistakes there please let me know. Thanks!
 

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  • #2
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Solve for what?
You can divide or multiply the equations by the speed of light or any other non-zero variables as often as you want. It does not matter which system of units you use, as long as you keep them consistent.

Energy divided by time is power, right.
 
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  • #3
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Solve for E in that equation, but yes, that answers my question. I just wanted to make sure i was running the computations correctly. Thanks!
 
  • #4
Nugatory
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You take the mass times the speed of light (in miles per second?) and square that. Over one second that is equal to the number of watts the object could produce.
If you're measuring energy in joules and power in watts, you'll need to use meters for distances and meters per second for speeds.
 
  • #5
Orodruin
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And no, it is not the power you can produce per second with an object. It is the total energy content of the object if it is at rest.
 
  • #6
Bandersnatch
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Also, it's not clear if you're squaring the result of mass times speed of light (which is incorrect) or just the speed of light (which is correct).
 
  • #7
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Solve for E in that equation, but yes, that answers my question. I just wanted to make sure i was running the computations correctly. Thanks!
The equation is solved for E already: it is "E=something".
 
  • #8
jtbell
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Also, it's not clear if you're squaring the result of mass times speed of light (which is incorrect) or just the speed of light (which is correct).
In more explicit mathematical terms, ##mc^2## means ##m(c^2)## not ##(mc)^2##.
 
  • #9
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If you're measuring energy in joules and power in watts, you'll need to use meters for distances and meters per second for speeds.
Right, sorry about that. I meant meters.
 
  • #10
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Also, it's not clear if you're squaring the result of mass times speed of light (which is incorrect) or just the speed of light (which is correct).
Just the speed of light.
 
  • #11
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In more explicit mathematical terms, ##mc^2## means ##m(c^2)## not ##(mc)^2##.
Right, and given that you know only to square the speed of light, not the entire equation.
 
  • #12
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Why are you interested, by the way? As Orodruin notes, there are relatively few circumstances where this calculation would give you a meaningful power output. You may be using one of them, but you may not. A spot of context would enable us to comment.
 
  • #13
ChrisVer
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Right, and given that you know only to square the speed of light, not the entire equation.
then you have 1 equation with 2 unknowns as a result it does not have 1 solution. It's exactly the same as [itex]y= ax[/itex] if you know [itex]a[/itex]. If you know neither [itex]y,x[/itex] then this equation is solved by an infinite set of [itex]y,x[/itex] which lie on a line with slope [itex]a[/itex].
As things are you are only confusing yourself; what's the question you want to answer?
If you know the rest mass you input it in [itex]m[/itex] (in kg) and multiplying with the [itex]c^2[/itex] (in m/s) you can obtain the energy [itex]E[/itex] (in Joule) of that mass at rest. It's not an equation that needs a lot to be solved.
In order to speak for radiated away power, you must have an energy difference [itex]\Delta E = E_{fin}-E_{init}[/itex] in some time interval [itex]\Delta t = t_{fin} - t_{init}[/itex]:
[itex]Power= \frac{\Delta E}{\Delta t}[/itex]
The only way to have some [itex]\Delta E[/itex] is if your mass is changing with respect to time: ([itex]\Delta m= m_{fin} - m_{init} \ne 0[/itex].)
 
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  • #14
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I know it doesn't have much to be solved. I just wanted to make sure I was doing it right. I'm thirteen, so I just wanted to make sure I understood the logic behind the equation.
 

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