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Gravitational motion usingly earths radius

  1. Oct 29, 2007 #1
    1. The problem statement, all variables and given/known data
    A communication satellite is in circular orbit around the Earth at 36,000 km above the earth. Starting with newton's law of gravitation, determine the velocity of the satellite. The only numerical values you are allowed to substitute in solving this problem are the radius of the earth and the acceleration due to gravity at the surface of the earth

    2. Relevant equations
    (mv^2)/(r)

    F = ((Gm1m2)/(r^2))


    3. The attempt at a solution

    I set them equal and got the equation

    v^2 = (Gm2)/(r)

    i dont know what to do, from here.
     
  2. jcsd
  3. Oct 29, 2007 #2

    learningphysics

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    That's correct. Now plug in the right values for

    m2 and r into:

    v^2 = (Gm2)/(r)

    and solve for v.

    m2 is the mass of the earth. look it up online or in your text.

    r is the distance from the center of the earth.

    ie r = radius of the earth + 36,000*10^3

    Look up the radius of the earth.

    EDIT:

    Sorry zelphics... I didn't read the whole post.

    you can use g = Gm2/r^2

    so m2 = g*r^2/G where r is the radius of the earth...

    then plug this in for m2 in:

    v^2 = (Gm2)/(r)
     
    Last edited: Oct 29, 2007
  4. Oct 29, 2007 #3

    D H

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    You are allowed to use a numberical value for the acceleration due to gravity at the surface of the Earth, which is of course one Earth radius from the center of the Earth. What does Newton's law of gravitation (along with Newton's second law of motion) tell you the acceleration due to gravity is at a distance of one Earth radius from the center of the Earth? What does that mean in terms of the satellite's motion?
     
  5. Oct 29, 2007 #4

    D H

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    He can't use the mass of the Earth or the value for G in this problem. He has to use the given satellite orbital radius, the radius of the Earth, the acceleration due to gravity at the surface of the Earth, and nothing else.
     
  6. Oct 29, 2007 #5
    Learningphysics: I cant just plug in the mass of the earth...

    D H: newtons second law is F = ma, I have no idea - im really confused.
     
  7. Oct 29, 2007 #6

    learningphysics

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    Yes, sorry about that. :redface:
     
  8. Oct 29, 2007 #7

    learningphysics

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    you can use g = Gm2/r_earth^2

    where r_earth is the radius of the earth... solve for m2 here... plug it into your equation.
     
  9. Oct 29, 2007 #8

    D H

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    The law of gravitation describes the force due to gravity. The force due to gravity translates into the acceleration due to gravity after dividing the force by the mass, which you can do by virtue of Newton's second law.
     
  10. Oct 29, 2007 #9

    D H

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    Solve for the product Gm2. He can't use G.
     
  11. Oct 29, 2007 #10
    Edit
     
    Last edited: Oct 29, 2007
  12. Oct 29, 2007 #11

    learningphysics

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    Thanks D H. I'm not thinking straight today.
     
  13. Oct 29, 2007 #12
    aaron???

    from my physics class at FM?

    NO WAY...it's zach!!!!!!!!!!
     
  14. Oct 29, 2007 #13
    hahahaha!

    I knew it was someone from my class

    it was the same exact problem.

    Gm2= g at Re Re^2

    thats what you are looking for

    Me and Kagan just figured that out we hope its right.
     
  15. Oct 29, 2007 #14
    do you guys have aol or something?

    i am soo confused - im not even going to sleep tonight =[
     
  16. Oct 29, 2007 #15
    Yeah we were just on AOL trying to do them all.

    We think we got all of them except the very last part(c) of #6.

    We gave up and got off but I'm on just re-writing all of them over on nice paper.
     
  17. Oct 29, 2007 #16
    im still on #3, are there any pages in the book that are helpful?

    i read it a million times...
     
  18. Oct 29, 2007 #17
    Umm not that I could find...

    I re-read it tons too.

    But I think I can do most of these mayybe.

    I can try to help you but I won't guarantee they are all perfect.

    On #3 you are really close you have v^2=Gm2/r

    GMe=g at Re Re^2 plug that in

    That me and Kagan found in our notes from Oct. 22
     
    Last edited: Oct 29, 2007
  19. Oct 29, 2007 #18
    that would be really helpful if you could help me, but i dont want to keep you up - maybe we shouldnt use this forum as a chat room though =p
     
  20. Oct 29, 2007 #19

    D H

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    Aaron, please do not give out answers to homework problems. Doing so doesn't help the person who asked the question learn anything.

    Even more important, please do not give out wrong answers. Doing so hinders learning.
     
  21. Oct 29, 2007 #20
    Well maybe I did do it wrong but i had to find a way to do it without the 36,000 since we aren't even allowed to use that.
     
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