Variation of gravity along latitude

[wilsonlye]

Factor contributes to variation of gravity along latitude is:
1. shape of the earth
2. rotation of the earth

gravitational field strength is resolved into two components, (R cos$$\theta$$)$$\omega$$ square, and g'
at the poles,$$\theta$$ =90 degree, therefore, g' = g which is 9.81
at the equator $$\theta$$=0, therefore, g' = g- R omega square

what I couldn't understand is that why the components of g are not perpendicular to each other.
can we resolve a force into components which are not perpendicular to each other ?

 

[D H]

The components of g are not perpendicular because they come from two distinct sources.

Think of it this way: Consider a mass at rest on an inclined frictionless plane. A cable connects the mass to the top of a fixed pole which rises above the top of the plane. The forces on the mass include gravity (down), the normal force (normal to the plane), and tension (toward the top of the pole). None of these forces are perpendicular to one another.

 

[wilsonlye]

then what are the two distinct sources?
one of them is R cos($$\theta$$)$$\omega$$square

the other source?

 

[D H]

$R\omega^2\cos\theta$ is not a source. It is a mathematical expression. What is the source of that term?

The other source is of course gravity itself. Where does it point?

 

[wilsonlye]

g points towards the centre of the earth

 

[wilsonlye]

that term is centripetal force

 

[Creator]

what I couldn't understand is that why the components of g are not perpendicular to each other.
?

These are the directions of the two forces...
The force of gravity acts from the center of mass of earth and so intersects normal to the Earth surface (approximately) at any latitude, BUT...
Centrifugal force (from Earth rotation) acts perpendicular to the axis of rotation, and so it intersects at various angles at the (curved) surface depending upon latitude.

...

 

[wilsonlye]

In another word, mg and centripetal force are two different forces act on a particle and mg' is the resultant force experienced by the particle?

 

[Creator]

In another word, mg and centripetal force are two different forces act on a particle and mg' is the resultant force experienced by the particle?

Yes, basically; one arises from the gravitational mass (of earth) and the other from the rotating reference frame.



Creator

 

[Aer-ki]

Factor contributes to variation of gravity along latitude is:
1. shape of the earth
2. rotation of the earth

gravitational field strength is resolved into two components, (R cos$$\theta$$)$$\omega$$ square, and g'
at the poles,$$\theta$$ =90 degree, therefore, g' = g which is 9.81
at the equator $$\theta$$=0, therefore, g' = g- R omega square

what I couldn't understand is that why the components of g are not perpendicular to each other.
can we resolve a force into components which are not perpendicular to each other ?

A side question...I've always wondered about the high jump in track being affected by the position of the moon.

Shouldn't the total gravitational affect on an individual be slightly less when the moon is directly overhead?

 

[Charlie Costa]

"The components of g are not perpendicular because they come from two distinct sources."

I'm confused by the above statement. "g" is gravitational field strength and comes from wherever gravity comes from! Single source if you will.

Also, I understand that at the equator Fn= Fg- mw^2 R, where I am taking g as a positive value, but what about at at a latitude of $$\theta$$ not equal to zero or 90 degrees?