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Homework Help: What is the satellite's instantaneous velocity

  1. Mar 8, 2005 #1
    Could someone tell me what formula I should use here?

    A satellite travels around Earth in uniform circular motion at altitude 35836 km above Earth's surface. The satellite is in geosynchronous orbit (that is, the time for it to complete one orbit is exactly one day). In the figure below , the satellite moves counterclockwise (ABCDA). State directions in terms of the x- and y-axes.

    (a) What is the satellite's instantaneous velocity at point C?
    km/s in the -y-direction
    (b) What is the satellite's average velocity for one quarter of an orbit, starting at A and ending at B?
    km/s at 45° above the -x-axis
    (c) What is the satellite's average acceleration for one quarter of an orbit, starting at A and ending at B?
    m/s2 at 45° below the -x-axis


  2. jcsd
  3. Mar 8, 2005 #2
    for uniform circular motion [tex]a=\frac{v^2}{r}[/tex], you know what r is and you know what a is, so you should be able to find out what V is. I would draw free body diagrams and then just wright the equations you have out. Atleast that is what I would think you would do. :tongue2:
  4. Mar 8, 2005 #3
    There is no one formula that will answer all of those questions. Physics is a class all about solving problems, so think about the big picture for a while. I have written what I would do while looking at the problem if you need a little assistance but I seriously urge you to try it yourself first.

    A) Find the radius of motion for the satellite (center of the earth to the satellite) and then find the circumfrence using that radius. Divide this distance by seconds in a day and you will find velocity. At point C, there is no x component of velocity so do not worry about using trigonometry.

    B) The only acceleration I see is radial acceleration (which changes the direction of velocity), so the magnitude of velocity is constant.

    C) Again, I only see radial acceleration in this problem so you can use the equation mewmew gave.
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