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Homework Help: Issues with gravity!

  1. Feb 23, 2005 #1
    I'm encouraged by the amount of people I see viewing this website... I'm in serious need of equation help.

    I'm currently taking an online physics class because of my very full workload of 5 jobs, one being full time, another being very close to that, and the rest part time. (a few hours a week each - don't panic!) I'm currently about 2.5 weeks ahead in my homework! (Yay!!!)

    Ahh, can you tell I'm a workaholic?

    I've come to a real issue, and I need someone out there to maybe give me some helpful advice on where I can find out how to apply the correct equation to my problems.

    How do I find out what force of earths gravity would have on me as a percentage of mg? I'm trying to solve that question for 1 earth radius up, 2 earth radii up, and if orbiting in a shuttle around 200 mi up. I'm MORE than willing to do the math - I love to learn! - but I need to know where to start. I feel a little lost, and have nobody I can ask, since it's an online class.

    Thank you everyone!! I really appreciate it.

    Amanda Blackwood.
    Last edited: Feb 23, 2005
  2. jcsd
  3. Feb 23, 2005 #2
    I would like to help you, but I don't understand your question, at all.... can you state it more clearly pls.....
  4. Feb 23, 2005 #3

    Doc Al

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    Staff: Mentor

    The force of the earth's gravity can be found using Newton's law of gravity:
    [tex]F = G \frac{M_e m}{R^2}[/tex]
    Where R is the distance from the center of the earth. (For R > = earth's radius)

    So, at the earth's surface:
    [tex]F = mg = G \frac{M_e m}{R_e^2}[/tex]
    so... [tex]g = G \frac{M_e}{R_e^2}[/tex]
    where [tex]R_e[/tex] is the earth's radius.
  5. Feb 23, 2005 #4

    Doc Al

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    So, the question is: How does the weight change when the distance goes from 1 earth radius to 2 earth radii ? Substitute [itex]R = 2 R_e[/itex] and then take the ratio of that force compared to the force when [itex]R = R_e[/itex]. (No fancy math; things will cancel!)
  6. Feb 23, 2005 #5
    Honey, to me, this IS fancy math! LOL Thanks so much for your help. I'm working on the problem now, though I'm sure my mind is about 10 to the 5th slower than yours!

    Thank you SO much!
  7. Feb 23, 2005 #6
    So are my calculations correct for this being the answer?
    9.8[tex]{m/s_^2}[/tex] at earths surface,
    19.6[tex]{m/s_^2}[/tex] one radis up,
    39.2 [tex]{m/s_^2}[/tex] two radii up, or am I going in completely the wrong direction?
  8. Feb 23, 2005 #7
    Oh, MAN! Here's another....

    How do I apply the same equasion to find the [tex]M_jupiter[/tex] , if Europa orbits a nearly perfect circle in 3.55 days to go around once. It's
    [tex]6.71 x {10^5} km [/tex] from Jupiters center....

    I feel so lost...
  9. Feb 24, 2005 #8
    Do you know Kepler's third law??

    If not then think about it this way: (If you know what centripetal force is)
    What forces does Europa feel from Jupiter? Also because it revolves around jupiter it feels another force simply because of it's cuircular motion.
    Since europa isnt going anywhere (that is not escaping from Jupiter) the sum of thse two forces must be zero
    Thus they must equal each other
  10. Feb 24, 2005 #9
    yes, you are completely wrong....

    [tex]g' = G \frac{M_e}{R^2}[/tex] as Doc Al suggested
    you should able to "see" whenever you double the R, the g' will decrease 4 times, the relation goes like [tex] g\alpha\frac{1}{r^2}[/tex]therefore, the right one should be:
    9.8m/s^2. at earths surface,
    2.45m/s^2. one radis up,
    1.09m/s^2. two radii up,
  11. Feb 24, 2005 #10

    Doc Al

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    Right. Now to put things in the form that the question asked (force as a percentage of mg) I would say that:
    At the earth surface, F = 100% mg (of course!)
    At 1 earth radius up (R = two earth radii from the center), F is 1/4 as much, so F = 25% mg;
    At 2 earth radius up (R = three earth radii from the center), F = 1/9 mg = 11% mg, etc.
  12. Feb 26, 2005 #11

    I dont know Keplers third law... and I still feel a little lost. PLEASE tell me this is as bad as the math is going to get! The sylabus actually states that all I need to know how to do is be able to "count my money" to get all the math correct. I've been a bartender and server for 8 years. Counting MONEY isnt a problem!!!

    Is there some simple equasion I can use and memorize to put this all together?

    Thanks again.

  13. Feb 26, 2005 #12
    ok, lets see how much i remember from physics.

    if im not mistaken, this is how you can solve for the mass of jupiter.

    define velocity as a function of period and radius:

    [tex]v = \frac{2\pi r}{T}[/tex]
    acceleration is equal to the velocity squared, so:
    [tex]a = (\frac{2\pi r}{T})^2=\frac{4 \pi^2 r^2}{T^2}[/tex]

    now define force:
    [tex]F=M_{satelite}\frac{4 \pi^2 r^2}{T^2}=\frac{GM_{satelite}M_{jupiter}}{r^2}[/tex]

    just canel out and solve for [itex]M_{jupiter}[/itex]. i hope i still remember physics correctly. those equations may be wrong. :redface: hope it works out fine.
  14. Feb 26, 2005 #13
    oh, make sure you have the period (T) in the correct units.
  15. Feb 26, 2005 #14

    Doc Al

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    This one's a little trickier; you'll need to know about circular motion. You'll need the period (T) of the orbit: convert 3.55 days to seconds. You'll need this to calculate the angular speed of Europa: [itex]\omega = 2 \pi / T[/itex].

    The only force on Europa is the gravitational attraction, so:
    [tex]F = m_E a[/tex]
    The force is gravity; the acceleration is centripetal:
    [tex]\frac{G M_J m_E}{R^2} = m_E \omega^2 R[/tex]
    solve for the mass of Jupiter:
    [tex]M_J = \frac{\omega^2 R^3}{G}[/tex]
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