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Quick tangential speed question

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Two satellites A and B of the same mass are going around the earth in concentric orbits. The distance of satellite B from the earth’s center is four times that of satellite A. What is the ratio of the tangential speed of satellite B to that of satellite A?

    2. Relevant equations


    3. The attempt at a solution
    So if v=rw and B is 4 times the distance of A, wouldn't the ratio (V_B/V_A) just be:
    4rw = rw
  2. jcsd
  3. Feb 26, 2008 #2

    D H

    Staff: Mentor

    Think again, this time using Kepler's laws.
  4. Feb 27, 2008 #3
    I'm sorr, i don't see wher eyou are coming from.

    Kepler's Laws are:
    1. The path of the planets about the sun are elliptical in shape, with the center of the sun being located at one focus. (The Law of Ellipses)
    2. An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time. (The Law of Equal Areas)
    3. The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. (The Law of Harmonies)

    But I don't see how these apply if I do not have a period and am only trying to calculate the ratio? Doesn't v=rw? Or am i totally off base?
  5. Feb 27, 2008 #4

    D H

    Staff: Mentor

    The satellites are in orbit. Of course they have a period, and because the satellites are at different orbital altitudes their periods are not the same per Kepler's third law. You don't know exactly what the period is, but that doesn't matter here. You only need to know the ratio of the periods.

    What is the relationship between orbital period T and orbital angular velocity w?
  6. Feb 27, 2008 #5
    The angular velocity specifies the angular speed at which an object is rotating along with the direction in which it is rotating so it is a portion so....when i looked it up the period equals (w/2*pi)^-1 where w is the angular velocity...where would I go from there?
  7. Feb 27, 2008 #6

    D H

    Staff: Mentor

    You wrote Kepler's third law in English. Write it as a mathematical equation instead. Next, write an alternative version of the law that uses angular velocity in lieu of period. Finally, apply this to the problem at hand.
  8. Feb 27, 2008 #7
    T^2=r^3 so since t=w/2*pi^-1:

    I'm sorry but i feel so confused I appreciate you being patient with me but where would I go from here. The thing is that we haven't used angular velocity in my class yet and i didn;t know that the period equaled w/2*pi^-1 until i looked about 3 chapters ahead. So, does this look right? How would I apply this to the problem at hand?
  9. Feb 27, 2008 #8

    D H

    Staff: Mentor

    First, you can get rid of the pi. Kepler's third law is a proportionality, not an equality.

    Second, if you haven't learned about angular velocity yet, why did you bring it up in your original post? All you need is Kepler's third law and the formula for the circumference of a circle. You know the formulation; you even used it in a different thread, quoted here:

  10. Feb 27, 2008 #9
    So, if v=2*pi*R/T and T^2=r^3, wouldn't v/2*pi*r =r^3 and one r would cancel out making it v=2*pi*r^2 so for the ratio, 2*pi *(4)r^2 =2*pi * r^2 and everything would cancel out except the 4 making the answer 4 like I originally thought?
  11. Feb 27, 2008 #10

    D H

    Staff: Mentor

    No. Think about it this way. Saturn is about 10 AU from the Sun (the Earth is 1 AU). By your logic, Saturn should have about 10 times the Earth's orbital velocity. However, Saturn has a period of about 30 years. Coupling that with the distance means that has about 1/3 of the Earth's orbital velocity, not 10 times the Earth's velocity.
  12. Feb 27, 2008 #11
    So the ratio would be 1/4 or would you have to square the r term making it 1/16?
  13. Feb 27, 2008 #12

    D H

    Staff: Mentor

    No. Look at Saturn again. The ratio is not 1/100, or even 1/10. It is about 1/3.

    Kepler's third law:

    [tex]T^2 \sim r^3[/tex]

    Note that I did not use the equal sign in the above expression. Kepler said the square of the period is proportional to the cube of the average distance. This is a proportionality, not an equality. Taking the square root of both sides,

    [itex] T \sim r^{3/2}[/itex]

    You already know [itex]v=2\pi r/T[/itex], or, as a proportionality,

    [tex]v \sim \frac r T[/tex]

    You have an expression for T in terms of r. Apply that to the above.
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