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Another gravitation problem

  1. Sep 7, 2016 #1
    1. The problem statement, all variables and given/known data
    If the gravitational accleration at the Earths surface is 9.81 m/s^2 what is its value at a height equal to the diameter of the earth from its surface?

    2. The attempt at a solution
    I have heard that it becomes 1/4th the value as the center of mass is moving away form each other.
    But I really can't understand the idea. Why is it 1/4?
     
    Last edited: Sep 7, 2016
  2. jcsd
  3. Sep 7, 2016 #2

    PeroK

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    What do you know about the force gravity?
     
  4. Sep 7, 2016 #3
    I know that it is a force acting between 2 masses with a distence between them. It is represented by F= G m1m2/ d^2.
    I also know that it is a concervative force and a central force.
     
  5. Sep 7, 2016 #4

    PeroK

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    That's good. So, what can you do with that formula for ##F##?
     
  6. Sep 7, 2016 #5
    Well I don't understand how I can apply it here. How can we use the formula of force to find gravitational accelereration? Why not find the gravitational acceleration at a hight h from surface of Earth?
     
  7. Sep 7, 2016 #6

    PeroK

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    Are force and acceleration not related?
     
  8. Sep 7, 2016 #7
    Yes but I still don't understand how that solve the problem..... Say I use the formula G=9.81m/s^2. Earths radius is 6400km so its diameter is x2. what about mass?
    should it be volume x density? Won't that make the problem unnecessarily large and complicated?
     
  9. Sep 7, 2016 #8

    PeroK

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    One thing at a time! Let me help. You have a general formual for the gravitational force. You want the gravitational acceleration due to the Earth:

    1) On the Earth's surface

    2) At a height equal to the diameter of the Earth above its surface.

    So, you need:

    Let ##M## be the mass of the Earth, let ##R## be the radius of the Earth. Now, what can you can about the accleration of gravity at the Earth's surface?
     
  10. Sep 7, 2016 #9
    Acceleration due to gravity at the earths surface = GM/R^2
     
  11. Sep 7, 2016 #10

    PeroK

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    What about above the Earth's surface, at a height equal to the Earth's diameter?
     
  12. Sep 7, 2016 #11
    Well, using Gp= GM/(R+h)^2. So....9.81 x M/ (6400+12800)^2 ?
    Isn't M= volume x density. How can I find M withought density?
     
  13. Sep 7, 2016 #12

    BvU

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    Work with symbols. Leave it in as M. It's the same for h = 0 as for any other value of h
     
  14. Sep 7, 2016 #13
    So...... 9.81 x M /(6400)^2+h^2?
     
  15. Sep 7, 2016 #14

    BvU

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    No. And I still see numbers, so I haven't been clear enough.
    You have
    1) at ground level $$g = {GM\over R^2}$$
    2) at height h $$g^* = {GM\over (R+h)^2}$$All you want to do is pick a good value for h and express ##g^*## in terms of ##g##
     
  16. Sep 7, 2016 #15
    Already pointed out both equations but I don't get the value part....earths radius=6400km so shouldn't its diameter be 6400 x 2.
    Is it the convertion? should I make it into meter?
     
  17. Sep 7, 2016 #16
    Also what is the value of M?
     
  18. Sep 7, 2016 #17
    Ahhhh I see G is the gravitational constent 6.674x 10^11 and M = gr^2/G.Right?I though G was g.This was my own mistake.
    But still can't get the right answer. Maybe I need to understand the concept more deeply.
     
    Last edited: Sep 7, 2016
  19. Sep 7, 2016 #18
    In BvU's post # 14, algebraically, what is g*/g equal to? Do you need to know the mass of the earth M to get this ratio?
     
  20. Sep 7, 2016 #19
    Okay. I have to find the ratio..so GM cancel out...

    so g*/g= R^2/(R+h)?
     
  21. Sep 7, 2016 #20
    Yes, except the R+h is also squared. Does this give you your desired answer?
     
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