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Earth's varying centripetal acceleration

  1. Dec 11, 2004 #1
    Hi,

    I've got a problem that I'm working on that is stated as follows:

    Earth's elliptical orbit around the sun brings it closest to the sun in January of each year and farthest in July. During what month would Earth's (a) speed-and (b) centripetal acceleration be greatest?

    Now, using Kepler's second law, I stated that a greater arc length must be made when the Earth is closer to the sun, and so the speed there must be greater. I was about to put down that acceleration must also increase since [tex]a_{c} = v^2{}/r[/tex], but then I realized that the radius has also decreased, and so I wasn't sure how I could determine where on the Earth's elliptical rotation around the sun the centripetal acceleration is greatest. Can someone help me out?
     
  2. jcsd
  3. Dec 11, 2004 #2
    If the radius has decreased, wouldn't that mean that the acceleration has increased since r is on the bottom (of the formula)?
     
  4. Dec 11, 2004 #3
    But v has also increased, as was deduced by looking at Kepler's second law...
     
  5. Dec 12, 2004 #4
    Come on, someone must be able to explain how to find centripetal acceleration in an elliptical orbit to me. I'd imagine this is a fairly widespread topic considering every planet in the solar system revolves the sun in an elliptical orbit to varying degrees.
     
  6. Dec 12, 2004 #5

    Ivan Seeking

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    I'm not sure what level you're working at but I think you have what you're looking for. If v is a maximum when r is a minimum, what has happened to ac?
     
  7. Dec 12, 2004 #6
    There's no telling, since the magnitude of both r and v is unknown. If r is sufficiently small, the acceleration will have increased, and if it is sufficiently large, it will have decreased. Or am I missing something?
     
  8. Dec 12, 2004 #7

    Ivan Seeking

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    Are you thinking that ac must be constant? We only need the equal areas in equal time, which led you to conclude that v is largest when r is smallest, right? . At this point we know v is large where r is small, so what must be true for a?

    What level are you at in school? It's hard to provide the proper level of help without knowing.
     
  9. Dec 12, 2004 #8
    Grade 12 Physics right now. It's almost 1AM and I can't believe it's taken me so long to grasp that when you divide by a small number, it gets bigger. God I'm brain-dead sometimes. Thanks very much for the help.

    However, I do have another problem that's probably worthy of this forum.

    At what distance from Earth, on a line between the centre of Earth and the centre of the moon, will a spacecraft experience zero net gravitational force?

    Now, I start out by stating [tex]r_e{} + r_m{} = 3.84 \times 10^8{}[/tex], 3.84 x 10^8 being the distance between the Earth and the moon.
    I then solve for re: [tex]r_e{} = 3.84 \times 10^8{} - r_m{}[/tex]

    Now I use the gravity equation, [tex]g = \frac{GM}{r^2{}}[/tex]

    Since g must be equal from both planets, I equate the two: [tex]\frac{GM_m{}}{r_m{}^2{}} = \frac{GM_e{}}{(3.84 \times 10^8{} - r_m{})^2{}}[/tex], where [tex]M_m{}[/tex] is the mass of the moon and [tex]M_e{}[/tex] is the mass of the Earth.

    Over the course of the last hour, I have found myself utterly unable to solve the equation for rm. I'm pretty sure my equation is correct, and I think it's simply due to my poor algebraic skills that I have been unable to isolate rm. Could someone please tell me how I can do this?

    The numbers I'm using are 6.67 x 10^-11 for the gravitational constant, 5.98 x 10^24kg for the mass of the Earth, and 7.35 x 10^22kg for the mass of the moon.
     
  10. Dec 12, 2004 #9

    Ivan Seeking

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    Also, you are implicitly using two different r's here. On one hand you mean r as the distance from the sun, and in the other, when you look at the acceleration, you mean r as in uniform circular motion.
     
  11. Dec 12, 2004 #10

    Ivan Seeking

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    I posted late, hang on...
     
  12. Dec 12, 2004 #11
    You're right, that equation wouldn't work for the problem I'm given, and I'm completely restricted to using Kepler's laws. I see where I went wrong, and I'm pretty sure I've got the right answer now. Thanks for your help on that one.

    [edit]
    dontcha hate when that happens?
    [/edit]
     
  13. Dec 12, 2004 #12

    Ivan Seeking

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    Since g must be equal from both planets, I equate the two: [tex]\frac{GM_m{}}{r_m{}^2{}} = \frac{GM_e{}}{(3.84 \times 10^8{} - r_m{})^2{}}[/tex], where [tex]M_m{}[/tex] is the mass of the moon and [tex]M_e{}[/tex] is the mass of the Earth.

    Multiply both sides of the equation by the denominators and solve the quadratic equation for rm
     
  14. Dec 12, 2004 #13

    Ivan Seeking

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    You will do well in physics! :biggrin:
     
  15. Dec 12, 2004 #14
    lol, thanks Ivan.

    Attempting to solve... please hold...

    [edit]
    gah, this may take awhile. If you want to leave, could you leave the answer in spoiler tags or something for me?
    [/edit]
     
    Last edited: Dec 12, 2004
  16. Dec 12, 2004 #15

    Ivan Seeking

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    Are you comfortable solving quadratic equations?
     
  17. Dec 12, 2004 #16
    Ok, I'm stuck again. I apologize for my ineptitude.

    Multiplying both sides by their respective denominators yields:
    [tex]GM_m{}(3.84 \times 10^8{} - r_m{})^2{} = GM_e{}r_m{}^2{}[/tex]

    Getting rid of G and expanding [tex](3.84 \times 10^8{} - r_m{})^2[/tex], I get:

    [tex]M_m{}(3.84 \times 10^8{} - r_m{})^2 - M_m{}(3.84 \times 10^8{})r_m{} + r_m{}^2{} - M_e{}r_m{} = 0[/tex]

    Plugging in all the largish numbers, I get values that I try to plug into the quadratic formula, but the answer I get that isn't negative has an exponent of 44. I'll try it again, but if you can see something wrong with what I'm doing, please tell me.

    [edit]I got 40277514, which sounds feasible. Checking it now...[/edit]

    [edit2]No, doesn't work. Making third attempt...[/edit2]
     
    Last edited: Dec 12, 2004
  18. Dec 12, 2004 #17

    Ivan Seeking

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    I'll be on for a little while yet. No hurry. Separate the terms according to the powers of r and reduce to the standard quadratic form.


    edit: wait, again I posted just as you did. :yuck:
     
  19. Dec 12, 2004 #18

    Ivan Seeking

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    okay, you need to multiply out the (3.84E8 - rm)2 and get r2, r, and the constant alone. Remember the form of the quadratic equation: aX2 + bX + c = 0
     
  20. Dec 12, 2004 #19
    Alright, I'll post exactly what I've got on my sheet right now:
    [tex]-5.98 \times 10^{24}r_m{}^2{} + r_m{}^2{} - 7.35 \times 10^{22}(3.84 \times 10^8{})r_m{} + 7.35 \times 10^{22}(3.84 \times 10^8{})^2{} = 0[/tex]

    I calculate all the numbers, and it doesn't seem to work.

    Sorry that took so long, LaTeX is cool, but takes me awhile to type out.

    [edit]
    Wow. This is starting to freak me out :tongue2:

    Looking at my second equation up there, I get the impression that it's utterly borked. I did actually do what you're saying, but the way I've got it entered is totally messed. I'll try to redo it, one sec.
    [/edit]

    [edit2]
    [tex]M_m{}(3.84 \times 10^8{})^2 - M_m{}(3.84 \times 10^8{})r_m{} + r_m{}^2{} - M_e{}r_m{}^2{} = 0[/tex]

    All I did was forget to take the rm out of the constant part. AND forget to square the last rm.
    [/edit2]

    [edit3]
    Must go sleep, bye-bye!
    [/edit3]
     
    Last edited: Dec 12, 2004
  21. Dec 12, 2004 #20

    Ivan Seeking

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    [tex]M_m{}(3.84 \times 10^8{})^2 - 2M_m{}(3.84 \times 10^8{})r_m{} + (M_m- M_e{})r_m{}^2{} = 0[/tex] :smile:
     
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