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B How much mass with the centrifugal forces of earth to break gravity?

  1. Oct 28, 2015 #1

    d w

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    I'll start with this question, how much "mass" does something have to have before centrifugal force exceeds gravity on earth?

    I used basic physics of centrifugal force and gravity, used force vectors, and the math doesn't jive. This is what I mean.

    There are 2 forces on earth that everything experiences if the earth rotates about its axis and orbits around the sun.

    Gravity and centrifugal force are quantified algebraically by acceleration (unit of distance)/(unit of time)^2 and thus they interact as force vectors.

    When you look at the values for objects with a mass of 4250kg or more, you will see the centrifugal acceleration begins to exceed the acceleration of gravity.

    I used these figures [when calculating centrifugal acceleration]:

    Earth's orbit values:

    Velocity 29,722ms−1

    Radius 149,668,992,000m

    Mass ≥22,679.6kg

    Earth's rotation values:

    Velocity 464.922ms−1 @ the equator (this value decreases down to zero as it approaches and reaches the "poles")

    Radius 6,371,390m

    Mass ≥4,250kg

    Moon orbit values:

    Velocity 1023.1ms−1

    Radius 384,400,000m

    Mass 7.34767309×10^22kg

    With these values and their products in mind, notice that the force vectors begin to exceed gravity and counter any claim that the earth is spinning or that the moon could orbit the earth. Is this right?

    Value I used for constant of gravity is 143 N

    Any value that exceeds 143 N from the centrifugal equation, is force away from earth exceeding gravity towards earth.
     
    Last edited: Oct 28, 2015
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  3. Oct 28, 2015 #2

    Orodruin

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    No.

    The gravitational pull on the Moon is pretty much exactly what it needs to be to provide the correct centripetal acceleration.

    It is also not clear why you think the mass has anything to do with things. Both the gravitational force and the required centripetal force are directly proportional to the mass and therefore cancels out in the comparison.
     
  4. Oct 28, 2015 #3

    d w

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    Hi and thanks for responding, I was looking for someone who knew the math and physics to respond with an explanation though because I want them identify the error in the math, not the conclusion. The conclusion is stated in the math, so that is why I want to see the math error. Thank you :)
     
  5. Oct 28, 2015 #4

    Janus

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    Centripetal acceleration is independent of mass, and is found by [itex]\frac{v^2}{r}[/itex]

    At the equator this works out to 0.0339 m/s^2 vs 9.8m/s^2 squared for the acceleration due to gravity.

    For the Moon orbiting the Earth, the centripetal acceleration is ~0.0027 m/sec^2 and the acceleration due to gravity is ~0.0027, which is what you expect for an orbiting object.
     
  6. Oct 28, 2015 #5

    Janus

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    In order for us to show the math/physics error you made, you need to show the math you used to get your answers, not just the numbers you plugged in.
     
  7. Oct 28, 2015 #6

    d w

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    Hi, thank you also.

    According to Newton's second law the centripetal force can be expressed as

    Fc = m ac

    = m v2 / r

    Are you saying that the centrifugal force is not derived from the same equation and centripetal force? Or are you saying that the result of centrifugal force is not the actual acceleration multiplier on the mass? I get that F has the mass/acceleration mixed in, but even "mass" has mass/acceleration mixed in because the "mass" is a result of gravity on the mass. Gravity is a constant so they use that constant to form weights. I am saying the energy expressed in this equation is a rate of energy transfer. Newtons are really a rate of energy transfer, but it has to have time or distance for it to express the rate.

    So far this is what I think, centrifugal force is on mass, so it isn't "independent" of mass. Centrifugal force expresses the acceleration on the weight in Newtons. Gravity force is expressed as weight in newtons as 143 N. Since I am expressing gravity this way, I expressed centrifugal force the same way. Both my equations are expressing the effects of acceleration as force, no?
     
  8. Oct 28, 2015 #7

    Orodruin

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    You still have not specified how you have compared this with the gravitational force. You need to show what you have actually done and not assume that everybody is a mind reader, in particular as it is clear that you have misunderstood something along the way.
     
  9. Oct 28, 2015 #8

    d w

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    Since it is all basic math and physics, I was hoping to discuss it with someone who already knew how to work the problem out. If I explain it, I have to explain centrifugal force, units, gravity, force vectors, and work that all out on a keyboard. I rather discuss it with someone who already knows that part of the problem. If you can't see how I arrived at my conclusion in the first place, you can't help.
     
  10. Oct 28, 2015 #9

    d w

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    Okay, I added some information. If you are looking for key numbers, sure, but if you want me to work it all out, it is really just 2 forces (expressed as acceleration in Newtons), 2 force vectors, 1 mass. Easy.
     
  11. Oct 28, 2015 #10

    Orodruin

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    Since what you are doing is very basic and you are obviously doing it wrong, the task is with you to actually provide your computations. Most people here are familiar with basic gravity and mechanics, which you do not seem to be. You have already been provided with the correct arguments by Janus. It is up to you to either show your work or where you think he is wrong.
     
  12. Oct 28, 2015 #11

    d w

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    Let me ask you a question. Have you worked out the math yet? IF you have, did you come to a different answer for the amount of weight that it would take for the centrifugal forces to equal the force of gravity?
     
  13. Oct 28, 2015 #12

    Orodruin

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    This is a value taken from nowhere and has nothing to do with how gravitation works on a general object. The gravitational force on a body depends on its mass, you cannot just assume it to be a particular value for all objects.
     
  14. Oct 28, 2015 #13

    d w

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    From Wikipedia, "
    One slug has a mass of 32.174049 lbm or 14.593903 kg based on standard gravity, the international foot, and the avoirdupois pound.[1] At the surface of the Earth, an object with a mass of 1 slug exerts a force of approximately 32.2 lbF or 143 N.[2][3]"

    Notice that 143 N is not out of nowhere.
     
  15. Oct 28, 2015 #14

    Orodruin

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    No, because it is specified which mass it belongs to. It is the force on 1 slug in a gravitational field equal to that at the Earth's surface. It is therefore completely arbitrary and will depend on what mass you are using as well as on the actual gravitational field, which changes with the distance from the gravitational source.

    You cannot use it as the gravitational force on an arbitrary mass at an arbitrary distance.
     
  16. Oct 28, 2015 #15

    d w

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    The constant of gravity is considered 9.8 m/s^2, no? IF that is a constant, the force of acceleration can be expressed as a force by converting to units using 1kg m/s^2 = 1 Newton. Since that is a measure of energy and acceleration is a demonstration of energy, they translate. Perhaps you have to see through the units a little, but the units represent ideas. The idea represented in acceleration and force is strength of energy force does it by saying "on mass" is all, and then over time it become rate of energy.
     
  17. Oct 28, 2015 #16

    Orodruin

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    No, you simply cannot do this. Force has physical dimension of mass*length/time^2 and the gravitational constant has physical dimensions of length/time^2 (just as acceleration). You will notice that m/s^2 is not the same as kg m/s^2... To find the gravitational force, you need to multiply by the mass of the object the gravitational field acts on, the result of which will have the correct physical dimension.

    In addition, the gravitational field is 9.8 m/s^2 only at the Earth's surface. It varies with distance.

    I am sorry, but these are just ramblings which have nothing to do with actual physics.
     
  18. Oct 28, 2015 #17

    d w

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    I disagree with you, but, let us solve it this way. With the known earth orbit and rotation values, how much mass is needed to equal the value of gravity under centrifugal force?
     
  19. Oct 28, 2015 #18

    Bandersnatch

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    @d w when you multiply the centripetal (or -fugal) acceleration by mass you end up with centripetal force. Similarly, when you multiply gravitational acceleration by mass you end up with gravitational force. Both are proportional to the mass you're multiplying by, so their ratio doesn't change in the process.
    If one's greater than the other, it stays greater no matter the mass (unless it's really huge and messes up the orbit).

    You can not compare acceleration with force and expect it to mean anything.
     
  20. Oct 28, 2015 #19

    Orodruin

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    There really is not anything to disagree with here, this has been well known since the time of Newton. If you have a different idea about how Newtonian gravity works, it has long since been disproved by observations.
     
  21. Oct 28, 2015 #20

    d w

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    Hi, the question I pose to you then is, what is the critical mass for when centrifugal force takes over gravity with the known values in our universe?

    Earth's orbit values.

    Velocity 29,722 m/s

    Radius 149,668,992,000 m

    Mass ? (I got 22,679.6 kg or greater)


    Earth's rotation values.

    Velocity 464.922 m/s @ the equator (this value decreases down to zero as it approaches and reaches the "poles")

    Radius 6,371,390 m

    Mass ? (I got 4,250 kg or greater)
     
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