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Help with Circular Motion

  1. Mar 18, 2010 #1
    Hello guys, good morning. :)

    I need your help regarding Circular Motion. I'm hoping that someone will help me heartily with this.

    These are the following problems:
    1. When the radius and mass are held constant, what do you expect to happen to the centripetal force if the frequency of rotation is increased?

    2. Determine the speed of the satellite orbiting at a height of 700km above the Earth's surface.

    mass of earth = 5.98 x 10^24kg
    radius of earth = 6.38 x 10^6m

    3. How much centripetal acceleration is produced when the object rotates 40 times a second with 4.89m radius? Also compute for the tangential velocity.

    That's all guys. Hope you can help me with these. Thanks advance.
     
  2. jcsd
  3. Mar 18, 2010 #2

    rock.freak667

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    What are your attempts on the questions? What equations do you need to use?

    For the first one, how does centripetal force relate to the frequency of rotation?

    For the second one, what is providing the centripetal force needed to keep the satellite in orbit?
     
  4. Mar 18, 2010 #3
    Hello rock.freak667, good morning. :)

    Thanks for replying my post Sir. Actually, I really don't know Sir. I don't know how will I answer you questions about my homework. Because I'm kinda late with this topic because I'm on the tour when they discussed this. I'm a 3rd year student and my classmates are 1st year. So it's a bit favor with them because I'm on the tour with my other 3rd year batch here in the Philippines.
     
  5. Mar 18, 2010 #4

    rock.freak667

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    Wikipedia explains it well

    Read it, see if you can grasp the concepts and are able to get started on the questions

    http://en.wikipedia.org/wiki/Uniform_circular_motion" [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Mar 19, 2010 #5
    Hello rock.freak667.

    Thanks again for replying my post. Here's what I got:

    1. When the radius and mass are held constant, what do you expect to happen to the centripetal force if the frequency of rotation is increased?

    Answer:
    The centripetal force will also increase (force = m x (v^2/radius))

    2. Determine the speed of the satellite orbiting at a height of 700km above the Earth's surface. The mass of earth is 5.98 x 10^24kg and the radius of earth is 6.38 x 10^6m.

    Answer:
    g(h)=g(0)*(R/(R+h))^2 where g(0)=9.81
    g(700)=7.966
    v=radical(rg)=7510 m/s

    3. How much centripetal acceleration is produced when the object rotates 40 times a second with 4.89m radius? Also compute for the tangential velocity.

    Answer:
    Compute for Tangential Velocity:
    Tangential Velocity = frequency x 2pi x radius
    Tangential Velocity = 40 x 2pi x 4.89 = 251.327
    Tangential Velocity = 251.327 x 4.89 = 1228.99 m/s

    Compute for Centripetal Acceleration:
    Centripetal Acceleration = tangential velocity^2 / radius
    Centripetal Acceleration = 1228.99 m/s^2 / 4.89
    Centripetal Acceleration = 308878.6135 m/s^2 or 308879 m/s^2


    I think number 1 and 3 are already OK. But I'm bit confuse with number 2 (red color). I just copy the answer of number 2 from Yahoo Answers. I don't understand what is g(0), R, and h. What are these? Hope you can help me with this.

    Thanks advance.
     
  7. Mar 19, 2010 #6

    rock.freak667

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    While this answer is correct, the asked for the frequency of rotation ω, so remember that v=rω, so put that back into your equation for centripetal force.

    Do you know Newton's law of gravitation? If you do, the gravitational force between the Earth and the satellite provides the centripetal force such that

    Fgravitational=Fcentripetal

    What is the gravitational force between the Earth and the satellite.

    this looks correct to me.
     
  8. Mar 19, 2010 #7
    I don't know about the Newton's Law of Gravitation. Can you help me solve with this?
     
  9. Mar 19, 2010 #8

    rock.freak667

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    The gravitational force between two point masses M and m separated by a distance r is given by

    [tex]F= \frac{GMm}{r^2}[/tex]


    Where G is the universal gravitational constant 6.67x10-11 Nm2/kg2
     
  10. Mar 19, 2010 #9
    Here's my answer in number 2. Kindly check if the equation is correct.

    Answer:
    The force on the satellite due to the Earth is given by:
    F = G(me * ms)/r^2

    me = 5.98 x10^24 kg = 5,980,000,000,000,000,000,000,000kg
    ms = mass of satellite
    G = Gravitational Constant = 6.67 x 10^-11 N(m/kg)^2 = 0.0000000000667 N(m/kg)^2
    r = 6.38 x10^6m + 700km x 1000m / 1km = 7,080,000m

    Object moving in a circular motion will have force given by:
    F = (ms * v^2)/r

    Equating the equations:
    “G(me*ms)/r^2 = (ms*v^2)/r” to “v = square root (G*me/r)”

    V = sqrt (0.0000000000667N(m/kg)^2 x 5,980,000,000,000,000,000,000,000kg / 7,080,000m)
    V = sqrt (56337005.64971751412429)
    V = 7505.79813542287547472866
    V = 7505.80m/s
     
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