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Universal Gravitation and Satelite Motion

  1. Oct 10, 2006 #1
    A couple of questions that Im struggling with again...that I really need to figure out ASAP

    On the way to the Moon the Apollo astronauts reached a point where the Moon's gravitational pull became stronger than the Earth's.

    (a) Determine the distance of this point from the center of the Earth.

    (b) What is the acceleration due to the Earth's gravity at this point?

    I have this set up (7.36e22)/(.38e9-x)^2 = (5.98e24)/x^2

    and i get 1.11x = .38e9 and yea im lost from there....any help there? and once i get the radius i can get part b with GMe/R^2

    and a couple more on Satelite Motion...


    Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This means that the masses of the two stars are equal (Fig. P14.15). If the orbital velocity of each star is 220 km/s and the orbital period of each is 20.4 days, find the mass M of each star. (For comparison, the mass of our Sun is 1.99e30 kg.)

    v = 2pir/T

    220000 m/s = 2pir/ (1762560)

    r = 6171443003 m? and mv^2/r = G2m/R

    and thats where i'm lost...

    and my last question....

    Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. Suppose that the mass of a certain spherical neutron star is twice the mass of the Sun and its radius is 9.0 km. Determine the greatest possible angular speed it can have for the matter at the surface of the star on its equator to be just held in orbit by the gravitational force.

    Im completely drawing blanks on this one....
    r = 9000 m, m = 4e30....and how exactly do i get angular speed? really lost...


    thanksssssss a lot for any and all help
     
  2. jcsd
  3. Oct 10, 2006 #2

    OlderDan

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    Your first set-up looks good (assuming you have the correct values for the known quantities; I don't have those numbers memorized) I don't know where the next line comes from. You should have a quadratic equation in x to solve.

    In your second problem, where did you get mv^2/r = G2m/R

    Get these two before you worry about the last one.
     
  4. Oct 10, 2006 #3

    Andrew Mason

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    Better to do the problem algebraically and then plug in numbers at the end.

    [tex]GM_e/x^2 = GM_m/(R-x)^2[/tex]

    where R = 3.8e5 km or .38e9 m. The G's cancel out. So far this is what you have done.

    You then have to rearrange into a quadratic equation form and solve for x (using the quadratic formula).

    AM
     
  5. Oct 10, 2006 #4

    Andrew Mason

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    Forget about plugging in numbers until the end. Show the solution algebraically. As Dan has pointed out, your equation is incorrect. You are equating centripetal force to gravitational acceleration - (the m on the left side should be deleted or one added on the right).

    AM.
     
  6. Oct 10, 2006 #5
    Algebraically I got R^2 - 2Rx + x^2 = Mm/Me x^2

    is this correct? but im getting two positive values for the quadratic?

    and for the 2nd one...

    v^2/r = G2M/ R^2 ?

    where R is 2r?
     
  7. Oct 10, 2006 #6

    OlderDan

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    You should get two answers for x. One is a point between the earth and the moon. Where is the other one?

    Yes R is 2r, but why 2M?
     
  8. Oct 10, 2006 #7
    for the first one i got 431305921.4 m/ 337924847.8 m

    arent those both between the earth and the moon? i dont get which one would be the correct answer in this case...

    and for the 2nd question...would it be m^2

    because u have would have the masses of each being equal?

    so v^2/r = Gm^2/4r^2 ?
     
  9. Oct 10, 2006 #8

    OlderDan

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    If your .38e9m for the distance from the earth to the moon is right, one of these x values is beyond the moon. There is a point out there where the forces are again equal. You want the one in between.

    For the second, either set force = force or acceleration = acceleration. You are still mixing the two.
     
  10. Oct 11, 2006 #9
    hmm im still gettin two #'s that dont make sense for the first one...

    okay so acc = acc

    v^2/r = GM/4r^2<<?
     
  11. Oct 11, 2006 #10
    well i got part a...instead of a quadratic i just took the root of everything....now part b seems to be confusing me...g = GMe/R^2?
     
  12. Oct 11, 2006 #11

    OlderDan

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    This is good. What is the R in this equation?
     
  13. Oct 11, 2006 #12

    Andrew Mason

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    The first number (assuming you have done the arithmetic correctly), .43e9 m is greater than .38e9 m. (the figure you are using for the distance between the centres of the earth and moon) so it is not between the earth and moon. You want the position where the forces of gravity are equal and opposite. If you used x as the distance from the moon, x would work out to a negative number.
    This is correct or:

    [tex]\frac{v^2}{R/2} = \frac{2v^2}{R} = \frac{GM}{R^2}[/tex]

    where R = distance between the centres of the stars.

    AM
     
    Last edited: Oct 11, 2006
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