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Time of satellite orbit around moon?

  1. Nov 19, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the time of an orbit of a satellite around the moon close to the moon's surface? What quantities do you have to look up to be able to answer this question?

    Radius of moon is 1736482 meters

    2. Relevant equations

    F = Gm1m2/r^2
    and maybe
    F=mv^2/r

    2pir

    3. The attempt at a solution

    1. 2 pi (1736482 meters) = 10910638.19 m = distance around moon
    .....and thats all I have. I don't know how to achieve time just given a radius. I do know that centripetal force has to come into play somewhere.

    Would the answer for the second question just be the radius?

    I'm so lost!
     
  2. jcsd
  3. Nov 19, 2015 #2

    Bandersnatch

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    Hi Dillion.

    Let's take it step by step.

    For an object to move in circles (e.g. an orbit), what magnitude must the net force acting on it be?
     
  4. Nov 19, 2015 #3

    PeroK

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    Gold Member

    You have the two key equations. Do you understand them? What does ##F=mv^2/r## mean?
     
  5. Nov 19, 2015 #4
    I hate to say this but I don't even know what you're asking.....
     
  6. Nov 19, 2015 #5
    Is it the force something creates while going around in a circle? When I see this equation I think about tension force and a ball on a string being swung around.

    F = Gm1m2/r^2 is the gravitational force of an object a certain distance away from the moon.

    I think a light bulb just clicked...

    G = 6.67 x 10 ^-11
    M1 = I will make it up to be a mass of an object lets say 1 kg
    M2 = mass of moon which is 7.35 × 10^22 kg
    r = 1736482 meters

    When I plug those value into F = Gm1m2/r^2, I get F = 2823208.072. I can set this as the F in F = mv^2/r.

    2823208.072 = (1kg) (v^2) / 1736482

    solving for v^2, v = 2214147.7 m/s

    Velocity = distance / time..... so 2214147.7 = 10910638.19 / time.

    time = 4.9 seconds.


    is this right?
     
  7. Nov 19, 2015 #6

    Bandersnatch

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    You've got the procedure figured out right, but:
    You've made a mistake in calculations here. Try again, and make sure you plug in everything into the equation. See what result you get then.

    Once you've got that done, we can talk about what the equation means, and why it was even a good idea to use it here.
     
  8. Nov 19, 2015 #7

    Okay I forgot the square the radius :(

    but after I redid my calculations I got 1.63 for F

    1.63 = (1kg) (v^2) / 1736482

    solving for v^2, v = 1682.4 m/s

    Velocity = distance / time..... so 1682.4 = 10910638.19 / time.

    time = 6485.16 seconds.


    second time the charm?
     
  9. Nov 19, 2015 #8

    Bandersnatch

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    Looks about right. How many hours is that?


    About the equations used: a general fact of nature is that every body moving in circles, no matter what or where it is, is under the influence of a force (or a net result of many forces added together), whose magnitude must be exactly equal to the centripetal force: ##F=mV^2/r##.

    The important bit about the centripetal force is that it's just a statement about how strong some other, actual force is. When dealing with circular motion you must find what force is acting on the object (gravity? tension in a string? floor of a space station?), and then you can say with confidence that this force is equal to the centripetal force.

    Here, you knew that the object was moving in circles (orbit near the moon surface, which is round), you knew what force was acting (gravity), so all that remained was to say that the only force acting is equal to the centripetal force.
     
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