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Weight of a person on earth vs. weight on a different planet

  1. Jul 9, 2014 #1
    1. The problem statement, all variables and given/known data
    A person on earth weighs 750N. What would the person weigh on a planet with 1.2 times the radius and 1.9 times the mass?

    2. Relevant equations
    The two ms cancel, leaving:

    3. The attempt at a solution

    Since the two ms cancel out, I thought that the most logical way to approach the problem would be to take the equation g=Gme/re2 and the values I was previously given for the earth's radius and mass, multiply by the factors given,and solve the problem.

    g=(6.67*10-11)(1.1362*1024) /7656000

    g=98981121.21 N

    The answer seems too large. Did I miss a step, was my logic flawed, or am I just overthinking the answer?

    Thanks so much.
  2. jcsd
  3. Jul 10, 2014 #2
    Heya :smile:

    It looks like you didn't multiply the factors correctly. Try again and see what you get.

    Note: You can also do it without the numerical values for the Earth's mass and radius. You can do it by getting the ratio of g of the new planet to the one of earth's)
  4. Jul 10, 2014 #3
    ow :redface:

    You are correct actually. You just forgot to square the radius and you missed an extra power for the mass (24 should be 25). Happens :smile:
  5. Jul 10, 2014 #4
  6. Jul 10, 2014 #5


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    That's great!

    But just for clarity, let me rewrite that as

    [tex] g_{Earth} = G \frac{m_{Earth}}{\left( r_{Earth} \right)^2} [/tex]

    Of course you know that [itex] g_{Earth} = \mathrm{9.81 \ m/s^2}[/itex], so you don't actually need to solve that if you don't want to. But you know how to derive that value if you wanted do. :smile:

    You can also solve for the person's mass, m, using simple algebra, with W = mgEarth, where W = 750 N.

    I'm sorry, you lost me at this point. I'm not sure where those values are coming from.

    Something is certainly not right here. g is a measure of acceleration, not force/weight. However you have it in units of Newtons, a measure of force/weight. Something is not quite right there.

    Now try

    [tex] g_{other} = G \frac{m_{other}}{\left( r_{other} \right)^2} [/tex]

    Once you find [itex] g_{other} [/itex], the weight of that person on the other planet is [itex] m \left( g_{other} \right) [/itex]
    Last edited: Jul 10, 2014
  7. Jul 12, 2014 #6

    Hm, okay. I'm not sure if you're saying that my answer is correct and I made a typo here, or that my logic was correct and my answer was slightly off.
  8. Jul 12, 2014 #7
  9. Jul 12, 2014 #8
    Thanks! I got the values from my previous physics lessons. They are constants, as far as I know.

    My first instinct was to calculate the mass using the equation for weight. However, at least as I understand, the person's' original mass cancels out, leaving me with the other information to solve the problem. If I'm wrong about the mass canceling out, I'm open to other ideas! Thanks again! :)
  10. Jul 13, 2014 #9
    It doesn't matter if you cancel the m or not. If you cancelled it, then you will multiply the g with m again to get his/her weight. Whatever makes it easier for you. Just check your numbers and algebra again as I said earlier.

    Did you square the radius?
    Check the powers
    Last edited: Jul 13, 2014
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