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A thought regarding angular acceleration, gravity, and weight.

  1. Feb 17, 2009 #1
    One of the first things we learn is that the Earth rotates very quickly (about once a day). We, as very imaginative 4/5/6 year old children, quickly realize the dangerous implications of this. However, as quickly as the fear beset us, we are calmed by our teacher/parent/television.

    "Gravity holds us to the Earth so we don't fly off."

    Whew! Crisis averted.

    What we feared as children was that our bodies would resist the angular acceleration demanded of us by the rotation of the Earth and that we would fly off into space. We then learn that the gravitational pull on our bodies easily counters the "centrifugal force".

    But I've been thinking lately...

    The angular acceleration we feel and resist isn't always the same. Relative to the Earth's axis, someone close to the north pole wouldn't feel nearly as much centrifugal force as someone at the equator. And it's not only the Earth! The Earth revolves around the Sun, the Sun around the center of the galaxy, and the galaxy around god-knows-what. Surely, our bodies resist all of these accelerations. If you think about it even more, these orbits could compound or diminish the centrifugal force we feel if they are synchronized in a particular way. Like constructive/destructive interference in waves.

    For example, the Earth rotates counter-clockwise and orbits the sun counter-clockwise (as viewed from the North Star). Therefore, person A standing on Earth and facing the sun will feel the centrifugal force from being hurled around the Sun, but the centrifugal force from the Earth rotating will be in felt the opposite direction, so the net centrifugal force is diminished. However, person B standing on the opposite side of the Earth, away from the Sun, will resist the rotation of the Earth and the Earth's orbit around the Sun in the same direction (out away from the sun), so the net centrifugal force is compounded. The angular acceleration we undergo, and therefore resist, is always changing, depending on how all the Earth's/Sun's/Galaxy's orbits are interacting, and different points on the Earth's surface (Relative to the Earth's axis, someone standing on the equator should feel twice the centrifugal force as someone at +/- 60 degrees latitude).

    If you factor into a person's free body diagram the resistance to all of these celestial spins, the normal force acted on us by the Earth isn't always the same. Factor in the gravitational pull by these masses on a person and the change is even more dramatic.

    Our weight isn't always the same! It changes depending on the time of day, the day of the year, even how far you live from the equator!

    Out of curiosity, I did some numbers. I live in Ohio, which is at 40 degrees latitude, so I revolve about the Earth's axis at about .766 times the radius of the equator and therefore at .766 times the speed. I found that I feel a centrifugal force of .026 N due to Earth's rotation. Keep in mind that I feel that force pushing me normal to the Earth's axis, not away from Earth's center. Gravity, however, does pull me to the Earth's center with a force of 1067 N. So... not really comparable... but still interesting! After adding the vectors, I find that The normal force of the earth on me is actually 1066.98 N. What's more interesting is that, because the centrifugal force isn't normal to Earth's surface, I also constantly feel a force of .017 N to the South! Can't really say I've ever noticed it, but cool! What about forces from the Sun? I found that the Sun and I pull on each other with a force of .6 N and I feel a force of .005 N directly away from the sun because of the revolution of Earth around it. Let's say that it's 12:00 noon in the middle of summer in Ohio (wishful thinking). This means that all 23.4 degrees of tilt of the Earth's axis are toward the Sun and that Ohio is facing the Sun more directly than at any other point in the 24 hour day. At this moment, the orbital plane of the Earth is offset from Ohio's latitude by 16.6 degrees. I'm going to ignore the angle between the orbital plane and the Ohio-Sun line, because it makes things a lot more complicated and it's negligible (.0006 degrees). Earth is pulling me toward it's center with 1067 N. I feel a force of .02 N away from Earth because of it's spin. The Sun is pulling me away from the Earth with .6 N and I feel a force of .005 N into the Earth because of Earth's orbit. After adding all the vectors, I find that the normal force of the Earth on me is 1066.385 N. Now, let's say the opposite. It's the 12:00 midnight in the middle of winter in Ohio (that's more like it). In this situation, the normal force of the Earth on me is 1067.035 N. That's a .65 N difference! That's .146 pounds! Okay, so it's not that much, but I think it's cool. That's just from gravity from the Sun and Earth and our motion through space, relative to the Sun! I haven't even begun to incorporate all of the bodies we revolve around and all the masses that pull on us!

    I don't claim to be an astronomer, and I know Newtonian physics gets a little funky on certain scales, so I don't doubt that I could be wrong. What do you think?
  2. jcsd
  3. Feb 17, 2009 #2


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    By this reasoning, shouldn't a person at the north or south pole move infinitely fast? Clearly, this cannot be right. The problem isn't really with the infinity either, because 1m away from the north pole you will still move at ridiculous velocity if this analysis were correct.
  4. Feb 17, 2009 #3


    Staff: Mentor

    I didn't check your work so I don't know if your numbers are correct, but your general idea is correct.
  5. Feb 17, 2009 #4
    A very good first go. The centrifugal force in Ohio points "south." This is a question we normally give our undergrads in intro physics. Usually our question also asks "Why are space shuttles and rockets launched close to the equator?"
    If you plot out the centrifugal force felt by an object from 0 deg to 90 deg latitude, you should get something that goes like a sine. You can obtain this if you plot out the vector diagram.

    Nabeshin, the equation you are using is too simple. I would bet it is mv^2/r. When you are computing this for Earth, you must keep in mind that you are in a rotating reference frame. This means you will have to pull out a full dp/dt calculation (ugh, cross products). Specifically, this will be a cross product between angular velocity and the radial vector.
  6. Feb 17, 2009 #5
    No, not at all! The speed with which someone revolves around the Earth's axis is 465.1 m/s (the Earth's rotation at the equator) multiplied by the absolute value of the cosine of their latitude.

    A person on the equator is revolving at 465.1cos(0)=465.1*1=465.1 m/s
    A person at the North or South pole is revolving at 465.1cos(90)=465.1*0=0 m/s
    A person in Ohio is revolving at 465.1cos(40)=465.1*.766=356.267 m/s

    I apologize for not stating the equation I was using explicitly. I hope you see what I'm saying now.
  7. Feb 17, 2009 #6

    D H

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    Centrifugal force is a pseudo force. It arises solely from your perspective of being in a non-inertial frame. In the sense that you can build a device to measure real forces, centrifugal force is not real. Hence another name for pseudo forces is fictional forces.

    Doing physics from the perspective of a non-inertial frame can at times be much easier than from the perspective of an inertial frame. The way to do this is by inventing pseudo forces that allow you to pretend that Newton's laws of motion still apply. These apparent forces that arise from trying to pretend Newton's laws still apply are not real. They're a fiction.

    Depends on what you call weight. You are talking about what a spring scale will register: apparent weight. You are correct; apparent weight varies with latitude. There are two primary effects. One is directly due to your motion around the Earth's axis, the other is due to the non-spherical nature of the Earth (which also results from the Earth's rotation).

    Let's say your mass is 109 kg. Most spring scales display mass (pounds or kilograms) rather than force (pounds-force or newtons). Let's assume a somewhat non-standard spring scale that displays force. Ignoring that the Earth is not spherical, that spring scale would register 1068.21 newtons at the north pole, 1065.38 newtons in Ohio, and 1064.52 newtons at the equator. Account for the Earth's oblateness and the difference between the readings will be even greater.
  8. Feb 17, 2009 #7


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    Yes sorry, my silly mistake :grumpy:
  9. Feb 17, 2009 #8
    Don't know how much vector formalism you post-ers are into but writing down "centrifugal force" as
    F = -m*OMEGA x (OMEGA x R) gives you the forces of interest.
    Here OMEGA is you angular velocity vector, and R is the radial vector.

    Do the calc (pad and paper) and you will get the nice result DH and I refer to.
    If you are so amped by these results, you can grab your nearest wikipedia and determine how forces behave when observed from the Sun. Keep in mind the pseudo centrifugal force will be WEAK.
  10. Feb 17, 2009 #9
    Not a problem! It happens to everyone (especially me, haha)

    It's amazing how many people will never own up to it though!
  11. Feb 17, 2009 #10
    Yea, I tried to be careful to not talk about it like I would a real force. I tried to say things like "I feel a centrifugal force of" instead of something like "the Earth acts on me with a centrifugal force". I also tried to aknowlege that it is just an illusion caused by matter's resistence to angular acceleration.

    Apparent weight, yes, thank you. Again, I should have been more specific.

    True, all my calculations were based off of a spherical Earth. Your results sure are more dramatic than mine! It's probably because of the constants we used and how much we rounded. I used the mean radius of Earth, which I got from wikipedia, as the radius. I got the gravitational constant from wikipedia, too. If I had been sending a bus-load of children to the moon, I would have been more careful. I just wanted something ball park so I could illustrate my point. Thanks for the numbers!
  12. Feb 17, 2009 #11


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    No one addressed this (that I saw):
    No. The earth is in orbit around the sun and one way of looking at an orbit is to say that the centrifugal force equals the centripetal force....in other words, you feel no force. That's why astronauts are weightless. If what you say were true, astronauts would feel weight that varied at different points in their orbits and the orbits of the earth and sun, etc.
  13. Feb 17, 2009 #12


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    It does not, however, vary with time of year.
  14. Feb 17, 2009 #13

    D H

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    So were mine. I made a mistake on the 40o latitude value. It should be 1066.05 newtons, not 1065.38. Assuming a spherical Earth with radius r, the magnitude of the force on the scale is

    [tex]F=m\sqrt{g^2-cos^2\phi\, w^2r(2g-w^2r)}[/tex]


    [tex]g=\frac{G M_e}{r^2}[/tex]
  15. Feb 17, 2009 #14
    True, the Earth wouldn't stay in orbit if they didn't.

    But is there anything to be said for the fact that we aren't orbiting the sun, but riding on something that is? My thoughts are that if we were floating in space, our mass would determine an appropriate orbit and the forces would balance, but since we are forced to share an orbit with Earth, the forces don't quite balance. There is a larger centripetal force on the Earth than on a person, right?
  16. Feb 17, 2009 #15


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    If you're interested in adding additional accuracy to your consideration of weight, you might look at buoyant forces. I may not be correct though. Even if I were, it would be quite small.
  17. Feb 17, 2009 #16

    D H

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    You do feel a force. The Earth is in orbit around the Sun, not you. At local midnight, the acceleration due to the Sun's gravitational force on you is [itex]GM_s/(R_e+r_e)^2[/itex] where [itex]R_e[/itex] is the distance between the Earth and Sun and [itex]r_e[/itex] is the Earth's radius. Meanwhile, the Earth (and hence you) are accelerating Sunward at [itex]GM_s/R_e^2[/itex]. That gradient in the gravitational field means you feel a force, the tidal force. The acceleration from this force (at local midnight) is about [itex]2GM_sr_e/R_e^3[/itex] anti-Sunward. For someone who masses 109 kg, this force is about 28 micronewtons.

    That is not why astronauts are weightless. Think about it this way. The trajectories followed by the Apollo astronauts to and from the Moon were quite elliptical. Centrifugal force and gravitational force do not cancel in an elliptical orbit. Yet the Apollo astronauts felt weightless on those trips to and from the Moon.

    Astronauts feel weightless because (Newtonian view) gravity is the one force that cannot be felt or sensed or (general relativistic view) gravity can't be felt because it to is a pseudo force.

    It does, by tiny, tiny amount. The Earth's orbit isn't circular. The solar tidal forces on our 109 kg person are about 2.8 micronewtons stronger at perihelion than at apohelion.
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