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Would the energy be in if you used grams for mass and km/sec for c?

  1. Apr 30, 2005 #1
    What unit would the energy be in if you used grams for mass and km/sec for c?

    And how many lights or something would 1 gram of the energy fuel (for comparision).

    Last edited: Apr 30, 2005
  2. jcsd
  3. Apr 30, 2005 #2
    for kilograms, meters, and seconds, you would get joules. if you use kilometers instead of meters you would be getting kJ, since a kilometer is 1000m. If you use kilometers and grams, you would (I believe) be getting joules, since the "1000" conversion facor for meters -> kilometers and kilograms -> grams cancels.

    Your second question can be answered by plugging in the mass in kilograms.
  4. Apr 30, 2005 #3
    Look at the Joule, the standard unit of energy and see how your units relate.

    [tex]Joule = \frac{m^2*kg}{s^2}[/tex]
  5. Apr 30, 2005 #4


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    It depends on how long you want to fuel them for. Lights require a certain amount of energy per second (in mks units, that's "watts") to stay alight. If 1 gram of mass energy were converted with 100% efficiency to power the light, then it could keep a 100-Watt light bulb going for 30,000 years (or 300 for 100 years)!
  6. Apr 30, 2005 #5


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    not quite. there is a square term in [tex] E = m c^2 [/tex] for the kilometers that isn't there for kilograms or for Joules.

    r b-j
  7. May 1, 2005 #6


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    Why change it? The equation was formulated in the gram/centimetre/second system. Just convert your kilometres to metres first.
  8. May 1, 2005 #7
    The questioner appears to be asking for the purpose of understanding what happens when certain things are done. In this case it pertains to units.
    The equation was not formulated in terms of any units whatsoever.

  9. May 1, 2005 #8


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    Sorry; I misunderstood the question.

    Perhaps a poor choice of words on my part; when Einstein first used it, those were the units that he employed. You couldn't arbitrarily change it to energy in dynes, mass in metric tonnes, and speed in rods per hour and still get a reasonable result.
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