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Gravitation and Newton's Synthesis

  1. Dec 23, 2009 #1
    1. The problem statement, all variables and given/known data

    13) (II) At what distance from the Earth will a spacecraft on the way to the Moon experiance zero net force due to these two bodies becasue the Earth and Moon pull with equal and opposite forces?

    2. Relevant equations

    NET F = ma
    G = 6.67 E-11 (Nm^2)/kg^2
    Fg = (GmM)/r^2
    Mass Moon = 7.35 E 22 kg
    Mass Earth = 5.98 E 24 Kg
    r Earth to Moon = 384,403,000 m

    3. The attempt at a solution

    Apply Newton's s second law in the radial direction

    NET F = m_craft( a_radial) = Fg moon = Fg Earth = 0
    = (G m_craft m_moon)/(384,403,000 m - r)^2 = (G m_craft m_Earth)/r^2

    m_craft cancels
    G cancels

    m_moon/(384,403,000 m - r)^2 = m_Earth/r^2

    m_moon/((384,403,000 m)^2- r^2) = m_Earth/r^2

    raise both sides to negative one power

    ((384,403,000 m)^2- r^2)/m_moon = r^2/m_Earth

    multiply both sides by m_Earth

    m_Earth( (384,403,000 m)^2 - r^2 )/m_moon = r^2


    ( m_Earth(384,403,000 m)^2 - m_Earth(r^2) )/m_moon = r^2


    ( m_Earth(384,403,00 m)^2 - 2(m_Earth)(384,403,000 m)m_Earth(r^2) - m_Earth(r^2) )/m_moon = r^2

    simplify further

    ( m_Earth(384,403,00 m)^2 - 2(m_Earth)^2(384,403,000 m)(r^2) - m_Earth(r^2) )/m_moon = r^2

    multiply both sides by 1/m_Earth and m_moon to clean up

    (384,403,00 m)^2 - 2(m_Earth)(384,403,000 m)(r^2) - (r^2) = (m_moon (r^2))/m_Earth
    Last edited: Dec 23, 2009
  2. jcsd
  3. Dec 23, 2009 #2
    Do you know how to use the quadratic equation?
  4. Dec 23, 2009 #3
    yes i do
  5. Dec 23, 2009 #4


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    Staff Emeritus
    Science Advisor
    Gold Member


    Recall that: (a - b)2 ≠ (a2 - b2)

    By definition:

    (a - b)2 = (a - b)(a - b)

    = aa - ab - ba + bb

    = a2 - 2ab + b2

    Another tip:

    Don't bother plugging in numbers and carrying them through all of those intermediate steps. It just clutters things up! Work through the entire problem algebraically (i.e. in symbols) until the end, when you have solved for an expression for r, the distance where the forces are equal. The nice thing is that this expression will be general. It will give the distance between any two masses m1 and m2 (with a given separation R) at which the gravitational forces cancel. Then you can plug in specific numbers for the case of Earth and its moon.
  6. Dec 23, 2009 #5
    Good, because that is what I think will finish the problem: but before going there check your work basically you have A*(k-r)^2=B*r^2 Seems you might have dropped some terms: the left side would be A(k^2-2akr+r^2).

    Edit: just as Cepheid points out, and the tip should be taken to heart: keep the numbers out of it until the end--lots of wasted writing, and makes it MUCH harder to follow.
  7. Dec 23, 2009 #6
    Ok I'm here what do I do now

    0 = (m_M r^2)/m_E + r^2 + 2(3.84403 E 8 m)r - (3.84403 E8 m)^2

    sorry about that
  8. Dec 23, 2009 #7
    There that should be correct sorry
  9. Dec 23, 2009 #8
    I was wrong

    0 = (m_M r^2)/m_E - r^2 + 2(3.84403 E 8 m)r - (3.84403 E8 m)^2

    i believe this is correct
  10. Dec 23, 2009 #9
    Assuming you have done the math right, which I'll check in a minute--

    you have r^2(m/M-1) + 2d*r - c^2. Use the quad, and see if your answer makes sense--you can check the work yourself by plugging in the distance in your original formula--forces should balance.

    That looks better, I thought you had flipped a sign, remember it is so much easier keeping the numbers out of it!
  11. Dec 23, 2009 #10
    I think I did it right
  12. Dec 23, 2009 #11
    excellent--again your answer can be put back in the original eqn as a check. Really do give some effort at keeping the numbers out of it until you're ready to use your calculator. Errors are so much easier to spot and your prof will be appreciative as well!
  13. Dec 23, 2009 #12
    were do i go from here
  14. Dec 23, 2009 #13
    To the next problem, right? I'm assuming you used the quadratic formula to solve for r.

    Recall (-b +/- sqrt(b^2-4ac))/2a Now the real fun begins.
  15. Dec 23, 2009 #14
    I don't know which one is a, b or c here becasue there are four terms...
  16. Dec 23, 2009 #15
    please!!! i'm almost done
  17. Dec 23, 2009 #16
    Collect the terms around r^2 This is a. The term in front of r is b and the pure number is c.
  18. Dec 23, 2009 #17
    and how do I do that

    0 = (m_M r^2)/m_E - r^2 + 2(3.84403 E 8 m)r - (3.84403 E8 m)^2

    there are two terms around r^-2

    m_M r^2/m_E and
    - r^2
    all contain an r
    how do i just simply collect this
  19. Dec 23, 2009 #18
    Notice what i did way back when -- I reposted your work as r^2(m/M-1) where m is the little mass, and M is the big mass. The quantity (m/M-1) is A. BTW thats ((m/M) - 1)
  20. Dec 23, 2009 #19
    how did you go from this

    0 = (m_M r^2)/m_E - r^2 + 2(3.84403 E 8 m)r - (3.84403 E8 m)^2

    to this

    ((m/M) - 1)

    for the first term and getting rid of the second by combining it with the first to get

    ((m/M) - 1)

    i don't see were that comes from

  21. Dec 23, 2009 #20


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    Homework Helper

    (m_M r^2)/m_E - r^2
    =r^2 (m_M/m_E-1)
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