Variation in g with latitude

In summary, the conversation discusses the concept of circular motion and the force required for it. The participant initially suggests that the force providing the necessary centripetal force is mgcos(lamda), but is corrected by another participant who states that mgcos(lamda) is actually much greater than the needed centripetal force due to the constraint force of the ground pushing up. The conversation then explores the effects of Earth's rotation on weight and discusses the calculation of apparent g as a function of latitude.
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upload_2016-12-2_23-28-4.png

Here the particle is performing circular motion due to the rotation of the earth. And for the circular motion it requires centripetal force then which force provides the necessary centripetal force. I think it is mgcos(lamda). Am i right?
 
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  • #2
Hardik Batra said:
I think it is mgcos(lamda). Am i right?
Yes, that's correct.
 
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  • #3
Can I do this?

mgcos(lamda) =
upload_2016-12-3_22-51-26.png
(centrifugal force) (In PQ direction)

PR = mgcos^2(lamda)

and resultant force = mg - mgcos^2(lamda).
 
  • #4
No, that is incorrect. mg cos(Q) is much larger than the required centrepital force. You are forgetting about the constraint force of the ground pushing up.

For example the radius of the Earth is about 6.4e6 m. The rotational rate is once per day making the velocity about 460m/s. So V^2/R at the equator is an out 0.034 m/s^2 which is 300 X smaller than g. The normal force which would otherwise have to be = mg now is reduced to 99.7% of mg. So you weigh 0.3% less at the equator. Actually it's more than that because of the Earth's centrifugal bulge, but that's another story.
 
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  • #5
Hardik Batra said:
Can I do this

mgcos(lamda) = View attachment 109856 (centrifugal force) (In PQ direction)

PR = mgcos^2(lamda)

and resultant force = mg - mgcos^2(lamda).
No. As Cutter Ketch stated, mgcosθ is much greater than the needed centripetal force.

I was sloppy before: I just meant to say that it is the component of the weight perpendicular to the axis of rotation that provides the centripetal force, not that it equals the centripetal force.

To find the apparent g as a function of latitude (pretending a spherically symmetric earth) you'd add the centrifugal force to the force of gravity.
 
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What is "variation in g with latitude"?

Variation in g with latitude refers to the change in the acceleration due to gravity (g) as one moves from the equator to the poles on Earth's surface. It is caused by various factors such as the Earth's shape, rotation, and distribution of mass.

Why does g vary with latitude?

G varies with latitude due to the oblate shape of the Earth, which is slightly flattened at the poles and bulging at the equator. This uneven distribution of mass causes a variation in the distance between the Earth's center and a point on its surface, leading to a change in the gravitational force at different latitudes.

How does g change as one moves from the equator to the poles?

As one moves from the equator to the poles, g decreases due to the increasing distance from the Earth's center. At the equator, g is approximately 9.78 m/s², while at the poles it is around 9.83 m/s². This difference may seem small, but it can have significant effects on things like the weight of objects and the measurement of time.

What other factors can affect the variation in g with latitude?

Apart from the Earth's shape and rotation, the distribution of landmasses and changes in elevation can also contribute to the variation in g with latitude. For example, areas with high elevations will experience a slightly stronger gravitational pull than those at sea level.

How is the variation in g with latitude measured?

The variation in g with latitude is typically measured using instruments such as gravimeters or pendulums. These devices can detect small changes in g and are often used in geophysical surveys to map out the Earth's gravitational field and understand its variations at different latitudes.

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