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Gravity of shrunken Earth

  1. Sep 14, 2016 #1
    1. The problem statement, all variables and given/known data
    If the radius of of the earth were to shrink by one percent, its mass remains the same,g would decrease by nearly.....%

    2. Relevant equations
    g=GM/ R^2

    3. The attempt at a solution
    Well, I took g1= GM/R^2

    And gravity after shrinking as g2= GM/(R-(R/100))^2
    Then I did g1-g2 and I didn't get the answer.
     
  2. jcsd
  3. Sep 14, 2016 #2

    Bystander

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    Start with a calculation of the new volume.
     
  4. Sep 14, 2016 #3
    V1= 4/3 pi r^3

    V2= 4/3 pi (r-r/100)^3
     
  5. Sep 14, 2016 #4

    Bystander

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    Next, is this the exact statement?
     
  6. Sep 14, 2016 #5
    Yes, doesn't gravitation decrease to zero when we go to the core?
     
  7. Sep 14, 2016 #6

    Bystander

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    Yes, but we're talking about surface gravity.
     
  8. Sep 14, 2016 #7
    I understand but if earth shrinks then isn't it like moving towards the core?
    Also the question does say decrease it was not an error.
     
  9. Sep 14, 2016 #8

    jbriggs444

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    What does volume have to do with it when density is not remaining constant? Post #1 already has the correct formula.
    As you move toward the core, you dig down, leaving a portion of the earth above your head. The portion of the earth that is farther from the core than you are ceases to have any net gravitational effect on you.

    If the Earth shrinks beneath your feet, the entire mass of the earth remains beneath your feet and continues to affect you.
     
  10. Sep 14, 2016 #9

    Bystander

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    Where is the mass located relative to the new surface, interior, or exterior?
     
  11. Sep 14, 2016 #10
    So if the mass remains the same gravity does not change?
     
  12. Sep 14, 2016 #11

    jbriggs444

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    If you consider Newton's universal law of gravity:

    ##F = \frac{G m_1 m_2}{r^2}##

    indeed gravity does not change. But if you want to keep your feet on the surface of the shrunken earth than, as Bystander suggests, it is surface gravity that you need to consider. Your original post correctly considered the effect of this in reducing the r term.

    Edit: The problem statement asks for a percentage decrease. What percentage decrease did you compute?
     
  13. Sep 14, 2016 #12
    I got g1 - g2 =98GM/99

    How can I convert this to percentage?

    Also if you think about it in terms of

    g is inversely proportional to R^2 then shouldn't gravity increase as radius decrease? So won't that make the question wrong?
     
  14. Sep 14, 2016 #13

    jbriggs444

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    I do not understand that answer at all. Where did the 98 come from, where did the 99 come from and what are the units on that answer?

    Edit: I have a suspicion that there is a rounding error, but the R went away as well, so it's more egregious than that.
    It makes the question "interesting".
     
  15. Sep 14, 2016 #14
    g1= GM/R^2

    g2= GM/ (R-R/100)^2.

    Using (a-b)^2 rule

    g2= GM/ 100R^2-2R^2+R^2 = GM/99R^2

    g1- g2 = GM/R^2 - GM/99R^2
     
  16. Sep 14, 2016 #15

    Bystander

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    This could be a "re-tread" problem; it "expands" one year, "contracts" the next, and the subsequent parts aren't always kept "in sync" with the rest of the statement.
     
  17. Sep 14, 2016 #16
    So the question is wrong?
     
  18. Sep 14, 2016 #17
    Well I guess you can say it's interesting when it poses no threat. But imagine sitting in an exam hall and this shows up and you hear a voice in your head saying "your God, where is he now?"...
     
  19. Sep 14, 2016 #18

    Bystander

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    No. It is out of synchronism with itself; it has to be interpreted carefully.
     
  20. Sep 14, 2016 #19

    jbriggs444

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    That is incorrect. Please learn to use parentheses and decimal points.

    A correct result is ##g_2 = \frac{GM}{(0.99R)^2}##
     
  21. Sep 14, 2016 #20
    Is the right answer 1 percent?
     
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