EK PHYSICS QUESTION

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kfcman289

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I have no clue about the following question:

The earth spins on its axis, flattening its spherical shape from pole to pole and bowing out at the equator. An object is placed on a scale at the equator. How does the centrifugal force and distance from the center of gravity affect the weight (as measured by the scale) at the equator?

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It will weigh less at the equator. Mostly because the object is farther from the center of the earth (center of gravity).
The centrifugal force is less significant compared to the gravitational force but also would cause you to weigh less.
 
It will weigh less at the equator. Mostly because the object is farther from the center of the earth (center of gravity).
The centrifugal force is less significant compared to the gravitational force but also would cause you to weigh less.

Can you explain why its farther from the equator and why the centrifugal force even matters/ why it is less? I am really confused about what is going on.
 
Can you explain why its farther from the equator and why the centrifugal force even matters/ why it is less? I am really confused about what is going on.
Sure.
This -> "flattening its spherical shape from pole to pole and bowing out at the equator." Tells you that it's wider at the equator. Take the earth and squish it at the poles so that it bulges around the middle. That's the image they are trying to convey.
Based on that it's safe to say if you squish a ball then the squished part is closer to the center (center of gravity).
The universal gravitation equation is: Force = GM1m2 / r^2
So you can see as radius (distance from object to center of earth) gets smaller the force gets larger. So your heavier the closer you are to a gravitational source.

Centripetal/centrifugal forces are only marginally important at the surface of earth, versus something much farther away like a geosynchronous satellite where it is a big factor.
It's saying that because the earth is spinning at the equator, an object will be sort of "flying off" the earth the way something would fly off a spinning merry-go-round near the edge.
If you weigh the same object at the poles it won't be experiencing the same centripetal force so it won't be being pushed away by the rotation of earth.

The centrifugal force mostly applies to when you spin something so fast that it cannot hold itself together. Fast rotation produces a great deal of inertia the farther you get from the center of mass.
If you spin something like the earth (or an asteroid) fast enough, it will be torn apart. Hard to explain, sorry if that's still unclear.
 
Nit picking only, so don't pay too much attention to me:
The acceleration due to the rotation of the Earth at the equator is: ω^2*r=(2π/T)^2*r=4π^2*6.4x10^6/(24*60*60)^2=0.03 m/s^2 or about 0.3%g
The difference in radius of the Earth at the equator and at the pole is about 21 km. That makes for about 1-((6400-21)/6400)^2=0.007 difference in g or about 0.7%g
Overall, both effects contributions are within the same magnitude. They both work in the same direction, so the relation of their magnitudes does not really matter that much.
 
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