Question of latitude of Earth and measured field strength

[Clara Chung]

Homework Statement

My book said:
As the Earth is rotating about its axis, objects on the Earth's surface are performing circular motion. Therefore, objects on the Earth are accelerating and they need a centripetal force. The weight contributes a small portion to the centripetal force needed. Therefore, the measured g is smaller. The decrease in g is largest at the equator. At higher latitude, the radius of the circle is smaller while w is the same. Therefore, the centripetal force needed is smaller. The measured g is larger at higher latitude.

Homework Equations

The Attempt at a Solution

However from my calculation, let the latitude be θ, W=weight, N=normal force, R=radius of the earth
Wcosθ - Ncosθ = mw^2 Rcosθ
W-N=mw^2(R)...(1)
Wsinθ=Nsinθ
W=N...(2)
Which is totally the same as considering an object on the equator.
Why am I wrong, please help

 

[cnh1995]

Homework Statement

My book said:
As the Earth is rotating about its axis, objects on the Earth's surface are performing circular motion. Therefore, objects on the Earth are accelerating and they need a centripetal force. The weight contributes a small portion to the centripetal force needed. Therefore, the measured g is smaller. The decrease in g is largest at the equator. At higher latitude, the radius of the circle is smaller while w is the same. Therefore, the centripetal force needed is smaller. The measured g is larger at higher latitude.

Homework Equations

The Attempt at a Solution

However from my calculation, let the latitude be θ, W=weight, N=normal force, R=radius of the earth
Wcosθ - Ncosθ = mw^2 Rcosθ
W-N=mw^2(R)...(1)
Wsinθ=Nsinθ
W=N...(2)
Which is totally the same as considering an object on the equator.
Why am I wrong, please help

https://www.physicsforums.com/posts/5643868/
This recent thread might help.

 

[CWatters]

My book said:
As the Earth is rotating about its axis, objects on the Earth's surface are performing circular motion. Therefore, objects on the Earth are accelerating and they need a centripetal force. The weight contributes a small portion to the centripetal force needed.

That last sentence is badly worded in my opinion.

Gravity provides ALL of the centripetal force needed and a lot more. Your apparent weight is the force left over after some has been used to provide the needed centripetal force.

 

[Clara Chung]

That last sentence is badly worded in my opinion.

Gravity provides ALL of the centripetal force needed and a lot more. Your apparent weight is the force left over after some has been used to provide the needed centripetal force.

I know it. The book actually means that a small portion of weight is used to contribute all the centripetal force. It is because the centripetal force is the net force by W - N. However by my calculations, I don't think the apparent weight ( Normal force ) will decrease when the latitude change (except for the poles) .

 

[haruspex]

I know it. The book actually means that a small portion of weight is used to contribute all the centripetal force. It is because the centripetal force is the net force by W - N. However by my calculations, I don't think the apparent weight ( Normal force ) will decrease when the latitude change (except for the poles) .

W and N cannot be in the same straight line.

By the way, there is another reason N is greater at the poles than at the equator.

 

[Cutter Ketch]

The Attempt at a Solution

However from my calculation, let the latitude be θ, W=weight, N=normal force, R=radius of the earth
Wcosθ - Ncosθ = mw^2 Rcosθ

You have applied a cos theta on every term. Why? Have you drawn a picture and figured out which forces point which way and how they vary with latitude?

 

[Clara Chung]

W and N cannot be in the same straight line.

By the way, there is another reason N is greater at the poles than at the equator.

 

[haruspex]

View attachment 110822

Since the body is static, the centripetal force is the resultant of all applied forces. If W and N are in the same straight line, how can their resultant lie in a plane of latitude? Which of these forces might not be as you have shown it? (Hint: is the Earth really spherical?)

 

[Clara Chung]

 

[haruspex]

View attachment 110824

Umm... that would give a resultant pointing away from the Earth's axis.

 

[Clara Chung]

Umm... that would give a resultant pointing away from the Earth's axis.

 

[haruspex]

View attachment 110825

Good.
What equations can you write down now?

 

[Clara Chung]

W cosa - Rcosb = mw^2 r...(1)
Wsina = Rsinb ...(2) where a<b
So by (2),
W/R = sin(b) / sin(a)
I don't know how to analyze it, does sinb/sina increases when a increases?

 

[haruspex]

W cosa - Rcosb = mw^2 r...(1)
Wsina = Rsinb ...(2) where a<b
So by (2),
W/R = sin(b) / sin(a)
I don't know how to analyze it, does sinb/sina increases when a increases?

I assume R is the same as N.
Can you derive a single equation that does not mention angle b?

 

[Clara Chung]

I assume R is the same as N.
Can you derive a single equation that does not mention angle b?

That is a bit long...
sinb = Wsina / R
cosb = root (1 - (Wsina / R)^2)
Put this into (1)
Wcosa - R root (1 - (Wsina / R)^2) =mw^2(r)
Wcosa - √[ R^2 - W^2 sin^2(a) ] = mw^2(r)
However R and r are still variables?..

 

[haruspex]

That is a bit long...
sinb = Wsina / R
cosb = root (1 - (Wsina / R)^2)
Put this into (1)
Wcosa - R root (1 - (Wsina / R)^2) =mw^2(r)
Wcosa - √[ R^2 - W^2 sin^2(a) ] = mw^2(r)
However R and r are still variables?..

OK, but R is what you are trying to find, so rearrange it as R=...
You can determine r.

It might be better to go back to using N instead of R so that you can use R for Earth's radius.

 

[Clara Chung]

I get something like
R^2 = W^2 - 2Wcosa(m w^2 r) + m^2 w^4 r^2
when I did some research r is something like √[ A^2 cos^2(a) + B^2 sin^2(a) ], A should be larger than B in the sphere I drew, so can I say
r = √[ (A^2-B^2) cos^2(a) + 1 ], where A^2 - B^2 is positive and r's value depends on cos^2 (a) ?

However, the second term and the last term seems to be contradictory

 

[haruspex]

when I did some research r is something like

For the purposes of estimating r in this question, you can treat the Earth as spherical, so just Earth radius cos (latitude).
And yes, you can neglect the m² term.

 

[Clara Chung]

R^2 = W^2 - 2Wcosa(m w^2 r) + m^2 w^4 r^2
= W^2 - 2Wcosa(m w^2 R cosa) + m^2 w^4 R^2 cos^2 (a)
= W^2 + cos^2 (a) m w^2 R ( m w^2 R - 2W )
Therefore, R increases with latitude (a), thank you so much