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Law of Universal Gravitation

  1. Jul 16, 2010 #1
    1. The problem statement, all variables and given/known data
    The average distance separating Earth and the moon is 384000km. What is the net gravitational force exerted by Earth and the moon on a 3.00 x 10^4kg spaceship located halfway between them.


    2. Relevant equations
    F=Gm1m2/r^2

    G = 6.673x10^-11
    earth mass = 5.98x10^24
    moon mass = 7.36x10^22


    3. The attempt at a solution

    I found the force exerted by the earth on the spaceship and then the force exerted by the moon on the spaceship. Did I do that right?

    Please help, this is one of my non-favorite topics.

    thanks!
     
  2. jcsd
  3. Jul 16, 2010 #2

    rock.freak667

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    Homework Helper

    Yes that is what you had to do. Those two forces act in opposite directions, so the net force is?
     
  4. Jul 16, 2010 #3
    I didn't get the right answer. :confused:

    For the force exerted by Earth I got 6.24x10^10 using half the distance between earth and moon as r. For the force exerted by the moon I got 7.67x10^8.

    The answer is 321N toward Earth.
     
  5. Jul 17, 2010 #4

    Ush

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    The "r" in the law of universal gravitation is from the center of mass. Meaning, you have to consider the radius of the earth and the radius of the moon. (if they have not already done so).
    edit: don't forget to convert km -> m
     
    Last edited: Jul 17, 2010
  6. Jul 17, 2010 #5

    Ush

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    that force is also pretty large o_o
    might want to try again.
    Force of gravity decreases with distance and you're in space!
    meaning.. your "force by earth" should be less than Ma(earth) and your "force by moon" should be less than "Ma(moon)"

    I haven't actually done the calculation- I'm just letting you know simple checks you can do on the way-
     
  7. Jul 17, 2010 #6
    k. I did the calculation over. Since the spaceship is between earth and moon, for r, I took the radius of the planet and added half the distance between earth and the moon. is that right?

    If so, i still didn't get the answer. But I did get a smaller number. For earth i got 305N and for moon I got 3.92N.

    Please help me find my mistake. Thanks.
     
  8. Jul 17, 2010 #7

    Ush

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    type what you're doing here
     
  9. Jul 17, 2010 #8
    Distance between earth and moon = 384000km/2 = 192000 --> 1.92x10^8m

    Earth:
    F = Gm1m2/r^2

    = (6.673x10^-11)(5.98x10^24)(3.00x10^4)/ 3.93x10^16
    = 305N

    where, r is equal to the radius of the earth plus the distance from earth to the spaceship.

    Moon:
    F = Gm1m2/r^2

    = (6.673x10^-11)(7.36x10^22)(3.00x10^4)/ 3.75x10^16
    = 3.93N

    where, r is equal to the radius of the moon plus the distance from moon to the spaceship.


    Can you please tell me where I went wrong?

    THANKS.
     
  10. Jul 18, 2010 #9

    Ush

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    your answer looks correct.
    final answer: Fnet = 305 - 3.93 = 301.07N [toward earth]
    I'm not sure why your book says it's wrong.
    =/
     
  11. Jul 18, 2010 #10
    The answer is 321N toward the earth.
     
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