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Homework Help: Cicular Orbit Question

  1. Mar 9, 2005 #1
    A satellite moves in a circular orbit around
    the Earth at a speed of 6.7 km/s.
    Determine the satellite's altitude above the
    surface of the Earth. Assume the Earth
    is a homogeneous sphere of radius Rearth =
    6370 km and mass Mearth = 5.98 x 10^24 kg .
    You will need G = 6:67259 x 10^-11 N m^2/kg2
  2. jcsd
  3. Mar 9, 2005 #2
    This is a simple question. How long did you try to do it?
    Use mv^2/(R+h) =GMm/(R+h)^2
  4. Mar 9, 2005 #3


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    Be careful with your units. Convert to mks and use:

    [tex]altitude = \frac{GM}{v^2}-radius[/tex]

    to get altitude in meters
  5. Mar 9, 2005 #4
    what is mks??

    what is mks?
  6. Mar 9, 2005 #5


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    mks stands for meters, kilograms, and seconds. Most formulas that you use in physics require that the numbers be in these units. For example, distance = velocity * time. If your velocity were 20 meters/second and your time was 2 hours, it would not be correct to say that distance = 20 m /s * 2 hours = 40. (Actually, you could do it this way if you were satisfied with a unit that no one used. It would be 40 meter hours/second. But try telling someone that the distance from their house to yours is 40 meter hours/second, and you'll gain an appreciation for mks). But if your time was in seconds then you would have distance = 20 m/s * 7200s= 14400 m. Now if you want, you can stray from mks and express the answer as 14.4 km, which in everyday talk is the best way to say it. But as long as you remain in mks while working the problem, you'll avoid lots of mistakes.

    Look at the units in your problem:

    G: N m^2/kg^2 : you've got meters and kilograms
    Mass: kg
    Radius: km

    Everything looks good except for radius. It's expressed in km, and the formula wants meters. So you must convert it to meters first by multiplying it by 1000.

    When you get an answer, your answer will be in meters, but it would be easier to comprehend if it were in kilometers. So you would want to divide it by 1000.
    In your problem, notice the units after G. There's a "m" in there and a "kg".
  7. Mar 9, 2005 #6

    i used this formula but i still keep getting the wrong answer. i changed radius from km to meters and km/s to m/s but i still keep getting the wrong answer
  8. Mar 9, 2005 #7

    i forgot to change the answer back to km's
  9. Mar 9, 2005 #8

    i have a final question, which is how do i find this. i know what the question is askin i just dont understand how to get the answer

    Two planets A and B, where B has twice the
    mass of A, orbit the Sun in elliptical orbits.
    The semi-major axis of the elliptical orbit of
    planet B is two times larger than the semi-
    major axis of the elliptical orbit of planet A.
    What is the ratio of the orbital period of
    planet B to that of planet A?
  10. Mar 9, 2005 #9


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    The fact that the planets are different masses are irrelavant. But the ratio formula is P^2 = A^3.

    Period^2 = semi-major axis ^3
  11. Mar 10, 2005 #10


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    It's NOT P2= A3!

    P2 is PROPORTIONAL to X3. That is P2/X3 is the same for both planets (that's one of Kepler's laws). (I've changed to X rather than A since, in the original question, A is used as the label of one planet.)

    If we let PA and PB be the periods of planets A and B respectively and XA and XB be the semi-axes, then
    [tex]\frac{P_A^2}{X_A^3}= \frac{P_B^2}{X_B^3}[/tex]
    and so
    [tex]\frac{X_B^3}{X_A^3}= \frac{P_B^2}{P_A^2}[/tex].

    Now, a question: does the problem actually say "two times larger" rather than "two times as large as"? The latter, "two times as large as" would mean
    [tex]\frac{X_B}{X_A}= 2[/tex]
    while, strictly speaking "two times larger" means "three times as large":
    [tex]\frac{X_B}{X_A}= 3[/tex]
    but is often used incorrectly!

    If it is "two times as large", then
    [tex]\frac{X_B^3}{X_A^3}= {\frac{X_B}{X_A}}^3= 8[/tex]

    If it is "two times larger" then
    [tex]\frac{X_B^3}{X_A^3}= {\frac{X_B}{X_A}}^3= 27[/tex]
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