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Thread: Determining magnitude of centripetal force

  1. #1 Determining magnitude of centripetal force 
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    Hi everyone! Here is my question!

    The orbit of Uranus around the sun is nearly a circle radius 2.87 x 10^12 metres.
    The speed of Uranus is approximately constant at 6.80 x 10^3 m/s. The mass of Uranus is 8.80 x 10^25 kg.

    a) Name the force that causes centripetal acceleration. (My answer, which I think is correct, is Fg)
    b) Determine the magnitude of this force. (Unsure as to which equation I am supposed to use?)
    c) Calculate the orbital period of Uranus, both in seconds, and in Earth years.

    Merci to all who can help :-D


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  3. #2 Re: Determining magnitude of centripetal force 
    Forum Ph.D. Leszek Luchowski's Avatar
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    Quote Originally Posted by Jprojectrunway
    a) Name the force that causes centripetal acceleration. (My answer, which I think is correct, is Fg)
    Yes and no. Fg might be the "correct" symbol for the force (what's a correct symbol? Technically, anybody is free to use d for length and p for time), but its name is a meaningful word. Such as "viscosity" or "air pressure" (but these are wrong answers - I am sure you can find the right one yourself).

    Quote Originally Posted by Jprojectrunway
    b) Determine the magnitude of this force. (Unsure as to which equation I am supposed to use?)
    Before you start using equations, do some plain logical thinking.
    Answer these questions:
    - How long will Uranus take to complete, say, 0.001 radian of arc on its orbit?
    - How much will its vector of speed change during that time? (hint: the length of the vector is constant but its direction changes all the time, so the difference is not zero)
    - How much force does it take to accelerate so much mass by so many meters per second in so many seconds?

    Quote Originally Posted by Jprojectrunway
    c) Calculate the orbital period of Uranus, both in seconds, and in Earth years.
    See above, but forget the 0.001 radian bit and just ask how long to complete the circle. I suppose you know how to convert gazillions of seconds into years, don't you.

    Good luck,
    Leszek.


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  4. #3  
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    Thank you for your relpy.
    However, I am still confused.
    This is highschool physics, and I was only given 3 equations:

    Force centripetal= m4pie^2r/ T^2
    Force centripetal= m4pie^2rf^2
    Force centripetal= mv^2/r

    There is no eqn that deals with gravity, which I think is the correct force acting in this eqn.
    Thanks for any other help anybody can provide!
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  5. #4  
    Forum Ph.D. Leszek Luchowski's Avatar
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    Quote Originally Posted by Jprojectrunway
    Thank you for your relpy.
    However, I am still confused.
    This is highschool physics, and I was only given 3 equations:

    Force centripetal= m4pie^2r/ T^2
    Force centripetal= m4pie^2rf^2
    Force centripetal= mv^2/r

    There is no eqn that deals with gravity, which I think is the correct force acting in this eqn.
    Thanks for any other help anybody can provide!
    Yes gravity is indeed the force.

    One way of computing the value of this force is to use the laws of gravity; take the mass of Uranus, of the Sun, and the distance, and hey presto! You get the force. Computed, so to speak, from its source: two large masses at a certain distance.

    But there is another way: you can compute a force by analyzing its effect. If you have the mass of Uranus alone, and the speed, and the radius of a circular orbit, you can compute the centripetal force which is needed to keep that body in that orbit (and prevent it from departing along a tangent, into the emptiness of outer space). You don't have to know if it's gravity or rocket engines, or if someone is pulling a string, or if Saturn is rolling on a gigantic circular track. The nature of the force would be different, but its value has to be the same for the same effect.

    This second way is what I suggested when I asked you to see how fast (or how slowly) Uranus' speed vector is changing. If you follow this path, you should get a formula which doesn't mention gravity. It will probably be one of those three formulas in your book. But I find them hard to read - aren't you having trouble with your keyboard? You typed the number 4 where asterixes should be.

    All the best,
    Leszek.
    Leszek. Pronounced [LEH-sheck]. The wondering Slav.
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  6. #5  
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    Quote Originally Posted by Jprojectrunway
    Force centripetal= m4pie^2r/ T^2
    Force centripetal= m4pie^2rf^2
    Force centripetal= mv^2/r
    Puzzling and misleading. None of those equations actually describes the centripetal force, but one of those is the centrifugal force. The centripetal force is the real physical central force that causes the deviation from a straight movement, hence in this case the gravitational force. For a stable circular orbit, the centripetal and centrifugal force have the same absolute value.

    Did you know, you can use "TEX" code to produce equations? For example "F_c=\frac{mv^2}{r}" produces .
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  7. #6  
    Forum Ph.D. Leszek Luchowski's Avatar
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    Quote Originally Posted by Dishmaster
    Puzzling and misleading. None of those equations actually describes the centripetal force, but one of those is the centrifugal force.
    It's not misleading. If the body is stably in orbit, as Uranus has been for awhile, the centripetal force equals the centrifugal force. What's wrong by determining one value by calculating another which is known to be equal to it?
    Leszek. Pronounced [LEH-sheck]. The wondering Slav.
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  8. #7  
    Moderator Moderator Dishmaster's Avatar
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    Yes, the values are the same in this case of equilibrium. And one can use this boundary condition to calculate its value. But this is only true for this assumption. By definition, the centripetal force is the physical force, in this case the gravitational force. So, the answer to the question "What is the centripetal force?" is correctly answered with "gravitational force". But in general, one cannot calculate the gravitational force with any of the listed formulae.
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  9. #8  
    Forum Ph.D. Leszek Luchowski's Avatar
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    Quote Originally Posted by Dishmaster
    Yes, the values are the same in this case of equilibrium. And one can use this boundary condition to calculate its value. But this is only true for this assumption.
    Well, if there is no equilibrium, say for a stray high-velocity meteor passing by and flying away towards infinity, then there is no circular orbit. But even then, the formula would be correct at the adhelium, if you take the radius deduced from the local curvature of the (hyperbolic) trajectory.

    As a matter of principle, you can determine a force by looking either at its cause (for gravity: the masses and distance) or its effect (for bodies in space: change in momentum over time). Both approaches are valid; choose the one you have the data for.
    Leszek. Pronounced [LEH-sheck]. The wondering Slav.
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