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Velocity of satellite at specific altitude

  1. Feb 1, 2013 #1
    1. The problem statement, all variables and given/known data

    How fast must a satellite 10,000 miles above Earth's surface travel, and how long does it take to complete one orbit of Earth?

    2. Relevant equations

    [itex]v^2 = g*r[/itex]


    3. The attempt at a solution

    10,000 miles + radius of Earth = 14,000mi =

    [itex]v^2 = (9.8)(2.253*10^7m)[/itex]

    v = 14859.13 m/s

    The answer says 9400 miles per hour which is about 4202m/s.
    Is there something I'm doing wrong or overlooking?
     
  2. jcsd
  3. Feb 1, 2013 #2

    jhae2.718

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    Is gravitational acceleration constant?
     
  4. Feb 1, 2013 #3
    [itex]\frac{g_1}{g_2}[/itex] = ([itex]\frac{r_2}{r_1}[/itex])^2

    [itex]\frac{22}{g_2}[/itex] = ([itex]\frac{14000mi}{4000mi}[/itex])^2

    [itex]g_2[/itex] = 1.796mi/h/s

    [itex]v^2[/itex] = (1.796)(14000)

    v = 158.56 mi/h ?
     
  5. Feb 1, 2013 #4

    jhae2.718

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    Check your units. Also, what units are your value for ##g_1##? Recall ##g_0 \approx 32.17\text{ ft/s}^2##.
     
  6. Feb 1, 2013 #5
    [itex]g_2 = 1.7959 mph [/itex]

    (1.795mph)*(14000mi) = v^2

    v = 158.56 mi/√h


    Edit: For [itex]g_1[/itex] I used 22mph because the other units are in miles.
     
  7. Feb 1, 2013 #6

    cepheid

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    Sorry to say it, but this isn't making any sense. The quantity g is supposed to be an acceleration, but you have units of speed instead.

    My approach to solving this problem would be to look at Newton's Universal Law of Gravitation. I would also ask myself: considering that the object is moving in a circle at a constant speed, what kind of force must be present?
     
  8. Feb 1, 2013 #7
    Okay sorry, I meant 22 mi/h/s...

    and g_2 = 1.7959 mi/h/s at 10,000 miles above Earth's surface.
     
  9. Feb 1, 2013 #8

    jhae2.718

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    Check what you're doing with ##g_2##...the unit you should have is mi/(h*s)...so, what conversion do you need to do to your result?

    You should maintain units throughout your calculations to avoid these types of errors.

    Also, you should carry more precision for ##g_1## and ##R_\oplus##.
     
  10. Feb 1, 2013 #9
    Okay, I'm getting confused. In the beginning I converted everything to meters.

    altitude = 1.609*10^7m + 6.3675*10^6m
    radius of Earth = 6.3675*10^6m
    gravitational acceleration = 9.8m/s^2
    v = ?

    v^2 = g*r
    v = sqrt(g*r)

    v = sqrt( 9.8 * (1.609*10^7m + 6.3675*10^6m))
    v = 14835.2m/s

    But this doesn't equal 9400 mph.

    I was using this website for help: http://www.ux1.eiu.edu/~cfadd/1150/05UCMGrav/Sat.html
     
  11. Feb 1, 2013 #10

    cepheid

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    Suddenly we're back to square one. It has already been explained that you're using the wrong value for g, the one that applies at the Earth's surface, not 10,000 miles up. EDIT: This is why I like my approach of deriving the equation for the speed from Newton's Law of Gravitation. It's conceptually clearer, because stuff isn't hidden in "g."
     
  12. Feb 1, 2013 #11

    jhae2.718

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    ##g \not = 9.8\text{ m/s}^2## that far from Earth. Recall that ##F_g \propto \frac{1}{r^2}##.

    Use Newton's gravitational law to find ##g## at the altitude specified in the problem.

    (Generally, SI units are almost exclusively used in the space world.)
     
  13. Feb 1, 2013 #12
    Okay, [itex]g_2[/itex] = (6.6742*10^(-11))*((5.9736*10^24)/(16371011.61))

    [itex]g_2[/itex] = 1.4875 m*s^-2
     
  14. Feb 1, 2013 #13

    jhae2.718

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    You're using the wrong radius. ##R = R_\oplus + r##, where ##R## is the total radius, ##R_\oplus## is the radius of Earth, and ##r## is the geometric altitude of the satellite.

    Your value is off by an order of magnitude.

    My advice would be to solve the problem in terms of symbolic variables until you have an expression for ##v## in terms of the known variables, and only then substitute numerical values.
     
  15. Feb 1, 2013 #14
    Okay.

    10,000 miles = 1.609*10^7 meters

    R = (6.37101*10^6 + 1.609*10^7)
    R= (22461010)
    R^2 = 5.044*10^14m

    g_2 = (6.6742*10^-11)*(5.044*10^14)
    g_2 = 0.790424

    v^2 = 0.790424 * 22461010
    v = 4213.51 m/s

    Edit: Which is approximately 9400 mph. Thanks for helping me! :)
     
  16. Feb 1, 2013 #15

    jhae2.718

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    No problem!
     
  17. Feb 1, 2013 #16
    Actually there's one more question I have which is finding the time it takes the satellite to orbit the Earth once.

    Can you please verify this?

    T = 2*pi*sqrt((r^3)/(G*m))

    T = 2*pi* Sqrt([itex]\frac{22461010^3}{G*m_E}[/itex])


    T = 33627.87 seconds
    T = 9 hours 20 minutes
     
  18. Feb 1, 2013 #17

    cepheid

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    This is fine, but there's an easier way to calculate it, which is to say speed = distance/time, where the distance is the circumference of the orbit.

    Since you *already* calculated the speed, you can just plug in into this equation. If you use the equation that has r , G, and M_E in it, then you are implicitly calculating the speed *again*, which is a waste of time, since you already know it.
     
  19. Feb 1, 2013 #18
    Oh, right.
    Well, the answer says it takes the satellite 6 hours to complete one orbit around Earth which is about 3 hours shorter than what I came up with (9 hours). Am I wrong?
     
  20. Feb 1, 2013 #19

    cepheid

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    No, you just used the wrong radius for the orbit. The orbit is a circle. The radius of a circle is the distance between the *centre* of that circle and each point on the circle. In this case, what is the radius of the circle? (Hint: it is not 10,000 miles).
     
  21. Feb 1, 2013 #20
    It's the radius of Earth + the distance of the satellite from the surface of Earth. This is what I used though...

    Edit: I used 22461010 meters.

    6.3675*10^6 + 1.609*10^7 = 22457500 meters

    ^Using this radius, I still get around 9 hours.
     
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