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Question about Universal Gravitation

  1. Feb 12, 2005 #1
    How much energy is required to move a 1070 kg object from the Earth's surface to an altitude four times the Earth's radius?

    Having problems with this question.

    I figure it would be GMm(1/r(final) - 1/r(initial)), with the final being 4x earth's radius, and initial being earth's radius. I plugged in the constants for G, and M(earth), but the answer i'm getting is wrong. Please help.
     
  2. jcsd
  3. Feb 12, 2005 #2
    [tex]W=\Delta{E_{P_{g}}}[/tex]

    [tex]E_{P_{g}}=-\frac{GMm}{r}[/tex]

    Watch the signs.
     
  4. Feb 12, 2005 #3
    hmm, I tried with the signs, however, looks like theres something up with the number i'm getting
     
  5. Feb 12, 2005 #4
    What answer are you getting? The number should be positive, and the formula you're using will fetch you a negative number.
     
  6. Feb 13, 2005 #5
    Well, I used (6.37x10^6)(4), to get Rfinal, since that is the earth's radius times 4, and I use 6.37E6 for Rinitial, since the particle starts on the earth's surface. Of course, G and M are constants, and I used 1070 for the mass of the object. Plugging everything in, I get

    - (6.67E-11)(5.98E24)(1070)(1/25480000 - 1/6370000) = 5.025E10
     
  7. Feb 13, 2005 #6

    Curious3141

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    You're misreading the problem : altitude of 4X the Earth's radius. They did not say to a distance from the Earth's center of mass of 4X the Earth's radius. Do you see now ?
     
  8. Feb 13, 2005 #7
    I see what you mean, so what would Rfinal be?

    Its 2AM, the reason why i'm asking stupid questions :smile:
     
  9. Feb 13, 2005 #8

    Curious3141

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    Five times the Earth's radius. If you can't see why this is so, please post.
     
  10. Feb 13, 2005 #9

    dextercioby

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    Just a piece of advice:it's not indicated to (try to ) solve physics problems at 2AM,especially in the weekend,when you have enough time (i think) during the day...

    Daniel.
     
  11. Feb 13, 2005 #10
    Wait, but I did set that as Rfinal, four times the earth's radius, and came out with the answer that I got before. Am I supposed to do it based on 5x the radius??? If that is so, I'm kinda confused :confused:
     
  12. Feb 13, 2005 #11

    dextercioby

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    Yes."move a 1070 kg object from the Earth's surface to an altitude four times the Earth's radius".I think the word "altitude" is referred wrt the Earth's surface,therefore the distance between the orbit and the Earth's center is the altitude+Earth's radius which is 5 times the Earth's radius...So [itex] 5R_{E} [/itex] is your [itex] R_{fin} [/itex].

    Daniel.
     
  13. Feb 13, 2005 #12
    Yes, that worked out. Thanks a lot for the help. I actually have one more question:

    At the Earth's surface a projectile is launched straight up at a speed of 10.3 km/s. To what height will it rise? Ignore air resistance and the rotation of the Earth.

    Now this question is listed under Universal gravitation, but looking through the formula's I can't seem to find one that can fit with this problem.
     
  14. Feb 13, 2005 #13

    dextercioby

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    Apply the law of conservation of mechanical energy.Again,pay attention with the signs,as the PE is negative (by convention,it's chosen 0 at infinity).

    Daniel.
     
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