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A few little questions about force

  1. Aug 6, 2009 #1
    If I were in a completely empty space, I wouldn't notice if I were moving at a constant speed, but would I notice acceleration/force, since there is no reference point?

    In a uniform circular motion is the acceleration = mass*(angular speed)², how can I derivate this formula without calculus?

    Why is there high and low water in a period of twelve hours and not 24 hours, since the moon attracts the water two times in 24 hours (when the earth is turned towards the moon and when it's turned towards the sun)?
  2. jcsd
  3. Aug 6, 2009 #2


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    Without a reference point you cannot define a coordinate acceleration. But you can measure your proper acceleration. However forces which accelerate everything equally (like gravity) do not cause proper acceleration.
    There are so many sites with nice pictures on this:
    Simple answer: In a non-uniform gravitational field everything gets stretched, along the field gradient direction. Here explained by Feynman (chapter 10):
    Last edited: Aug 6, 2009
  4. Aug 6, 2009 #3


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  5. Aug 6, 2009 #4
    Yes indeed my bad, thanks a lot for the help :smile:
    By the way, looking out of my window I wonder, do we only see one side of the moon because one side is heavier and thus more attracted?
    And how can I measure my acceleration if I have no reference?
  6. Aug 6, 2009 #5
    I liked this http://lmgtfy.com/?q=why+we+only+see+one+side+of+the+moon" [Broken] :tongue:
    Last edited by a moderator: May 4, 2017
  7. Aug 7, 2009 #6
    Ok you have a point, but I like to get a general idea of the concept, before I look something up, so I can see what's important and what not in an explanation. :smile:
    Last edited: Aug 7, 2009
  8. Aug 7, 2009 #7


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    It is not heavier, just closer and therefore more attracted. And yes, tidal deformation slows down spinning until the orientation is fixed along the gravitational gradient.
  9. Aug 7, 2009 #8
    Well, it isn't coincidence that the side of the moon that is tidal locked with the Earth is also the side with greatest amount of basalt... that side has more mass and is therefore more likely to be the side that locks with the Earth -- it would have locked either way in the end to do the simple loss of angular momentum, but it locked with that side more likely because of that side's greater mass.

    Also, wouldn't an accelerometer be a point of reference? The mass that lies inside of it seems to act as a point of reference for acceleration calculations.
  10. Aug 7, 2009 #9


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    My point was that is not necessary to have a heavier side to lock. But yes, a non uniform mass distribution makes it easier.
    I assume the OP meant a point of reference that defines a reference frame in which you measure coordinate acceleration (dv/dt).
  11. Aug 8, 2009 #10
    i'm sorry.... but i believe it to v2/r

    If you think about it logically, considering that the velocity remain the same, increasing the radius using v2*r would result in a larger acceleration. This is contradictory to what it really is.

    if you increase the radius, the acceleration is seen to decrease indicating an inversely proportional relationship, hence v2/r rather than v2*r
  12. Aug 8, 2009 #11
    It depends on the scale of the acceleration. if you close your eyes in a car (i.e. have no reference point) and you take off like a shower of **** at the traffic lights, would you be able to tell?

    the answer is yes because you can feel the seat exerting a force upon you (you are not in fact being pushed back into the chair).

    However the acceleration can be small enough for you not to notice.
  13. Aug 8, 2009 #12

    Doc Al

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    Don't confuse angular speed (ω, in radians/sec) with linear speed (v, in m/s). The centripetal acceleration can be written as ω²r or v²/r. Since v = ωr for circular motion, these two expressions are equivalent.
  14. Aug 8, 2009 #13
    Oh ok... sorry about that.

    So as v = 4pi2R/T

    2pi/T = your angular velocity

    wow...that is cool
  15. Aug 8, 2009 #14
    I think that there is a high and low tide because the moon revolves around the earth in an ellipse thus there are 2 points that are close to earth and 2 points that are far from earth... when the moon revolves till the closer points, there is a high tide, when it moves further from the 2 close points the tides drop

    just my two cents, hope they are right
  16. Aug 8, 2009 #15

    D H

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    Nope. As mentioned previously, the tides result from the fact the Moon's gravitational field is not uniform. The Earth as a whole is accelerating toward the Moon at

    [tex]a_{\text{earth}} = \frac {GM_{\text{moon}}}{R^2}[/tex]

    where [itex]R[/itex] is the distance between the centers of the Earth and the Moon.

    The point on the surface of the Earth directly between the Earth and Moon is a bit closer to the Moon than is the center of the Earth the while the point on the surface of the Earth exactly opposite the Moon is a bit further. The accelerations at these points toward the Moon are

    [tex]a_{\text{node}} = \frac {GM_{\text{moon}}}{(R\mp r_e)^2}[/tex]

    where [itex]r_e[/itex] is the radius of the Earth, [itex]R-r_e[/itex] is used for the sub-Moon point, and [itex]R+r_e[/itex] for its antipode. The relative acceleration is

    a_{\text{rel},\text{node}} &= a_{\text{node}} - a_{\text{earth}} \\
    &= \frac {GM_{\text{moon}}}{(R\mp r_e)^2} - \frac {GM_{\text{moon}}}{R^2} \\
    &= \frac{GM_{\text{moon}}}{R^2}
    \left(\frac 1 {(1\mp r_e/R)^2)} - 1\right) \\
    &\approx \pm 2\frac{GM_{\text{moon}}}{R^3}r_e

    Note that the relative acceleration is directed away from the center of the Earth in both cases.
  17. Aug 8, 2009 #16
    are you referring to me ?:tongue2: just want to know where im wrong
  18. Aug 8, 2009 #17

    D H

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    While the magnitude of the tides depends on the Moon's orbit, the timing of the tides does not. Think of it this way: Most places have two high tides and two low tides every day (24.84 hours, to be precise). In comparison, the Moon orbits the Earth but once per month.
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