Weight is less at equator than poles -- Acceleration vectors involved?

In summary, neglecting equatorial bulge and using 'g' (9.81m/s^2) as standard before calculations, the speaker wanted to calculate the change in 'g' at the equator based on Newton's 3rd law. However, the speaker was confused about the direction of the acceleration vector responsible for this effect and how something can weigh less at the equator without a component of force in the opposite direction. The expert explains that in a rotational frame of reference, the pseudo forces generated due to the motion, such as centrifugal force, decrease the effect of acceleration at the equator. This is similar to the effect of acceleration on a lift going up or down. The expert also clarifies that when measuring weight
  • #1
papazulu
3
0
Neglecting equatorial bulge and using 'g' (9.81m/s^2) as standard before calculations, I wanted to see if I could calculate the change in 'g' before looking it up. Obviously this not a difficult sort of problem but I am finding myself a tad confused about something. It seemed intuitive to me that based on Newton's 3rd law I could find radial accel. at equator A-rad=(465.1^2 m/s) / (6378137 m) = (.033912 m/s^2) then subtract it from 9.81m/s^2.

'g' at equator = (approx.) 9.776 m/s^2

My question is exactly which direction is the acceleration vector responsible for this effect pointing? I've read that there is no 'true' force acting outward in rotational motion. How can something weigh less at the equator if there is no component of force that is in the opposite direction of the radial acceleration?

I'm not doing this for a class so please (if you care to) just go right to the basic explanation of it.

thank you
 
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  • #2
papazulu said:
My question is exactly which direction is the acceleration vector responsible for this effect pointing? I've read that there is no 'true' force acting outward in rotational motion. How can something weigh less at the equator if there is no component of force that is in the opposite direction of the radial acceleration?

I'm not doing this for a class so please (if you care to) just go right to the basic explanation of it.

thank you

When you are in a rotational frame of reference the pseudo forces which are generated due to the motion is a force being experienced

so its a true force and its called centrifugal force . in an inertial frame its not present or its counterpart centripetal force is present.
as Earth is rotating and you are also rotating with it the centrifugal force will act radially outward and necessarily decrease the effect of acceleration g. g is radially inward.
as you move from equator to the pole the radius of circle of rotation decreases so the centrifugal acceleration which is proportional to radius will also decrease and ideally at the poles its effect will vanish.
this effect is same as in linear accelerated motion like in a lift going down with acceleration one weighs less than the lift going up.
 
  • #3
It has to do with the fact that the centripetal force required to keep you moving in a circle depends on both the radius from the center and the speed at which you are traveling. If you increase the radius but keep your speed the same, you need less force to keep you at the same distance (because you need less centripetal acceleration). If you speed up but keep the radius the same, the amount of force required increases (because you need a larger centripetal acceleration).

Gravity supplies more than enough force required to hold you to the Earth's surface, provided you don't attach yourself to a rocket and accelerate up to a few dozen mach. Part of that force is used solely to keep you moving in a circular path while the rest is simply excess and must be opposed by the upward force from the ground (weight). If the Earth rotated twice as fast, your tangential velocity would increase and you would need more centripetal acceleration to stay on that same circular path. Again, gravity would still be more than enough to hold you to the surface, but more of it would be used to as centripetal force, leaving less for the surface to counteract.
 
  • #4
papazulu said:
Obviously this not a difficult sort of problem but I am finding myself a tad confused about something. It seemed intuitive to me...
Write out the symbolic equations, instead of juggling numbers by intuition.
 
  • #5
papazulu said:
How can something weigh less at the equator if there is no component of force that is in the opposite direction of the radial acceleration?

Because when you are at the equator you are not in an inertial reference frame. Imagine the Earth as a giant rotating sphere far, far from any other body or gravitational sources. Imagine further that you are observing the Earth spin from an inertial frame placed along the axis of rotation, say above the North pole. If you look down at a point mass located at the north pole it will look to you like it is not moving. It is just sitting peacefully on top of the globe (point masses don't "rotate"). But when you look at a point mass located at the equator, it will be seen to move in your inertial frame. It is going around the axis of rotation. Both are being pulled by the same gravitational force, but their motions are very different in the inertial frame, and thus so are their accelerations.

If a point mass at the equator measured its weight and then went to the north pole to do so again, it would find a difference as you suggest. If the point mass were foolish enough to think it had measured its weight in the same inertial reference frame both times, then it might posit the existence of another force in addition to gravity as a way to explain the discrepancy between measurements. But there is no other force. There is a difference in the measured weight because the same force is coupled to different accelerations which the point mass neglected when it assumed both measurements were taken in the same inertial reference frame.

HTH
 
  • #6
So very hypothetically if gravity and the effects of atmosphere ceased for just me alone, me and the Earth would begin to move away from each other, I in a straight line tangent to the Earth at the point of separation, but the point on Earth heading away in the path of an arc the shape of the earth. The POINT or PLOT OF DIRT upon which I was standing would be experiencing the force, NOT me. It would be the Earth trying to get away from me, not me from the earth. I am moving at constant velocity in a straight line. ∑F=0 now for me. But for the plot of dirt the force is not zero. Thus work is being done TO THE PLOT in moving it away from me, no work is being done on me. Now to someone standing near the plot where I was standing observing me move away from the Earth it would appear as if some force were catching me away, but, in fact, they would be being caught away from me by the Earth's torque. They are being accelerated away from me, not me from them. And it would be all the more illusory because we would be moving along with each other at the same apparent velocity magnitude while I am moving away from the earth. Throughout the course of the first few hours it would appear to someone on Earth as though some force were lifting me straight up. I wonder how long it would take before I would disappear from view into the sky. What kind of interests me now is what angle would I be from the point of departure at different points of time. At one quarter of the day (6 hours) I would be (2π⋅r)/4, or 6220 miles away from where I was standing at its current point of rotation and at an angle of almost exactly 30 degrees from that point's tangent line.

So anyway, the value .033912 m/s^2 is the acceleration at which Earth is trying to get away from me and it is this that shows up on the weight scale when someone is weighed at the equator vs at the poles.

Here is a question I want to try to answer in a succeeding post. At what rate would the plot of ground's velocity be changing as it accelerated away from me?
 
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  • #7
Most of that looks correct, papazulu.

papazulu said:
Here is a question I want to try to answer in a succeeding post. At what rate would the plot of ground's velocity be changing as it accelerated away from me?

I believe the change in rate would take the shape of a sine wave.
 
  • #8
you are correct in that the acceleration of the surface of the Earth at the equator is 3.39 x 10-2ms-2
Imagine that you are standing in a lift descending with this acceleration and the force acting on you (from the Earth) is 9.776 N/kg can you visualise/calculate your motion?
What would bathroom scales read ?
 
  • #9
Drakkith said:
Most of that looks correct, papazulu.
I believe the change in rate would take the shape of a sine wave.

yeah probably some cycloid type equation that could model the x, v, & a. Over the long term (assuming I continued in a straight line unaffected by the gravity of the sun or other planets) the Earth's cycloidal motion relative to me would begin to be affected of course by its rotation around the sun. That is something I want to delve into actually. A line drawn from the surface of a wheel to an origin but the wheel is rolling within another wheel. That is what would be needed to model the point I left on Earth and its line to me as we move apart, or rather model its x, v, & a relative to me.

I just began two Hibbeler texts Dynamics and Statics. I've been teaching myself math and physics off and on for a few years. I learned through Calc III from Stewart's 1320 page Early Transcendentals. I've gotten about halfway through a couple of ODE texts, Zill and a Schaum's DE. And the physics 1 and 2 I learned from a Halliday/Reznik and a Young and Freedman. From what I've been able to tell, the Hibbeler Statics and Dynamics are the next logical step, obviously while still pressing forward in the DE. I also just got a 300/400 level Vector Analysis text that is basically just the vector parts of Calc 2 and 3 but deeper and more intense.

Anyway the x, v, and 'a' with cycloidal rolling inside of a circle is definitely something I'll probably run into at some point moving forward.
 

1. Why is weight less at the equator compared to the poles?

Weight is less at the equator than the poles because the Earth's shape is not a perfect sphere. It is flattened at the poles and bulging at the equator, causing a decrease in gravitational force at the equator.

2. How does the Earth's rotation affect weight at the equator?

The Earth's rotation causes a centrifugal force at the equator, counteracting some of the gravitational force, resulting in a lower weight compared to the poles.

3. Is weight affected by latitude?

Yes, weight is affected by latitude due to the Earth's shape and rotation. The closer one is to the equator, the lower their weight will be compared to those at the poles.

4. How does the acceleration vector at the equator differ from the poles?

The acceleration vector at the equator is larger than at the poles due to the centrifugal force caused by the Earth's rotation at the equator.

5. Does weight vary at different points on the equator?

Yes, weight can vary slightly at different points on the equator due to variations in the Earth's shape and rotation, but the difference is minimal.

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