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During a solar eclipse calculate forces.etc

  1. Nov 9, 2007 #1
    During a solar eclipse....calculate forces.......etc

    1. The problem statement, all variables and given/known data

    During a solar eclipse, the moon, Earth, and sun lie on the same line, with the moon between Earth and the sun.
    (a) What force is exerted on the moon by the sun?
    __________N
    (b) What force is exerted on the moon by Earth?
    __________N
    (c) What force is exerted on Earth by the sun?
    __________N

    2. Relevant equations

    F=(G*M1*M2)/R^2

    3. The attempt at a solution

    scroll down.
     
    Last edited: Nov 9, 2007
  2. jcsd
  3. Nov 9, 2007 #2

    mgb_phys

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    R is the distance between the centres of the objects -- not the radius.
    (it's R because it's the radius of the orbit)
     
  4. Nov 9, 2007 #3
    ok so for part A i did:

    G=6.67*10^-11
    Mass of moon: 7.36*10^22 kg
    Mass of sun: 1.99*10^30 kg
    dist between sun and moon: 1.5*10^11 m

    (6.67*10^-11)(7.36*10^22)(1.99*10^30) ALL OVER (1.5*10^11) and i got 6.51e31

    but it says it's incorrect...
    should i be converting to newtons?
     
  5. Nov 9, 2007 #4

    D H

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    G=6.67*10-11 -- you are missing the units! G is not a number like pi. It is a physical constant, which means its value depends on the units in which it is represented.
     
  6. Nov 9, 2007 #5
    alright then.. i don't know what the units would be in this case for G. i just know it's a universal constant. i never knew i would have to convert it somehow, and either way i still dont know how right now :(
     
  7. Nov 9, 2007 #6

    D H

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    It is 6.673*10-11 m3/kg/s2. So of course you got the wrong answer when you mixed meters and kilometers.
     
  8. Nov 9, 2007 #7
    thank you, but...that is very confusing, and my teacher failed to mention those units in class. now i REALLY am confused as to how to incorporate it into the equation!
     
  9. Nov 9, 2007 #8

    D H

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    Sorry. For some reason I saw you using kilometers. You did use meters.

    You're problem is here:

    What is the equation for the gravitational force?
     
  10. Nov 10, 2007 #9
    for which planet? i know that for earth we should multiply the number by 10 to get it in Newtons from Kilograms
     
  11. Nov 10, 2007 #10

    D H

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    I am talking about Newton's universal law of gravitation. You applied it incorrectly.
     
  12. Nov 10, 2007 #11
    it's Fnet=ma
    and F of gravity ~ (m1*m2)/d^2

    how did i apply it incorrectly
     
  13. Nov 10, 2007 #12

    D H

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    You divided by the distance, not the square of the distance.
     
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