Gravity and Gravitational Vectors

[allrighty]

Hi everyone, im new here but i need some help

 

Ive been asked to find the angle between the gravity and gravitational vectors as a function of latitude where gravity is pointing to the center of the earth and the gravitational vector is the sum of the gravity vector and centripetal vector. Can anyone help? Thanks

[swansont]

Hi everyone, im new here but i need some help

 

Ive been asked to find the angle between the gravity and gravitational vectors as a function of latitude where gravity is pointing to the center of the earth and the gravitational vector is the sum of the gravity vector and centripetal vector. Can anyone help? Thanks

 

What direction does the certripetal vector point?

[allrighty]

The centripetal vector points away from the axis of rotation of the earth, it remains horzontal throughout earth's latitude

[allrighty]

Here is a picture to show vectors and things that i drew

where the vectors F_centripetal + F_gravity = F_gravitation

[insane_alien]

umm, just an observation from 1am, but aren't centripetal forces(in this case gravity) always acting towards the axis of rotation by definition?

[allrighty]

Yes your right... my mistake... anyways there is a fictional force pulling us away from the earths surface, that is what i tried to represent

[Rocket Man]

fictional? what? centripetal forces are very real.

what holds the moon up?

 

your diagram is correct, all you need is to put numbers to it.

for simplicities sake, take gravity as a constant 9.8m/s^2 at the surface, and centripetal force as a function of the smallest distance to the axis of rotation and the velocity at the surface.

[math]a=\frac{v^2}{r}[/math]

 

the earth rotates at a rate of 1/(24*3600) rotations per second

v = rotation rate * pi * radius

[math]a=\frac{(86400^{-1}\pi r)^2}{r}[/math]

 

[math]f_x = 9.8cos\theta - a_{centripetal}[/math]

[math]f_y = 9.8sin\theta [/math]

that's your total force vector, determine the angle between it and the surface.

[allrighty]

Thanks everyone for their help especially Rocket Man I really appreciate all the advice you give me

 

There are a couple things i dont understand however...

 

1. Doesnt the centripetal force depend on the latitude. It seems that it would be greatest at the equator, with no centripetal force at the pole....

Maybe if r represented the distance to the earth's vertical axis... (horizontal in the diagram from the earth's surface at some point.) If r is a function of latitude, how would I relate the two?

 

2. Could you explain the step getting V in the centripetal equation relation ship a little clearer for me? Im just not sure where the expression v = rotation rate * pi * radius comes from....

 

3. How would the angle phi be obtained, adding fx and fy through a vector system i and j and then what?

 

 

Overall I am just overwhelmed by this problem relating phi to theta to gravitational vector... sorry if some of my questions seem redundant

[timo]

I think you´re missing how a representation of vectors in (x,y) coordinates relates to a representation via the magnitude and an angle. If the x-axis is what you´ve called 0° and the y-axis is what you called 90°, then for any vector [math] \vec v [/math]: [math] \vec v = \left( \begin{array}{c} v_x \\ v_y \end{array} \right) = | \vec v | \left( \begin{array}{c} \cos \theta \\ \sin \theta \end{array}\right) [/math], where [math] \theta [/math] is the angle to the x-axis increasing towards the y-axis (same as in your diagram). Using this and the definitions of sine, cosine and tangens, you can transform from one representation to another and back.

 

As for Rocket Man's post: Note that he completely omitted any units, which is usually a bad idea.

[swansont]

Yes, the centripetal force will depend on the latitude, as will the direction of the gravitational force.

 

a = v^2/r and both v and r depend on the latitude.

 

The distance travelled around a circle is 2*pi*r, so the speed is that distance multiplied by the rotation rate (or divided by the period), i.e. 1 revolution in 24 hours

[swansont]

fictional? what? centripetal forces are very real.

what holds the moon up?

 

your diagram is correct, all you need is to put numbers to it.

 

The diagram has the centripetal force pointing out, rather than in. That would imply it is a centrifugal force, which is not real (in an inertial coordinate system)

 

Nothing "holds the moon up." Like any object in orbit, it is in freefall, but has a trajectory that does not hit the earth.

[allrighty]

So here is where I am now

 

Sorry i dont know how to make those equations so bear with me...

 

This is the coordinate system I have been using

"I" represents a vector from origion going horizontally from left to right

"J" represents a vector from origion going vertically up

 

I have the vector a_c = ((86400^-1{sec}*pi*r)^2)/r I

I found a relationship for r in the horizontal plane as

r = 6,388,000{earth's radius in m}*Cos(theta)

 

then substituted r into the vector a_c and simplified it to

a_c = .00845{m/s^2}Cos(theta) I

 

and a_g = -9.81{m/s^2}Cos(theta) I - 9.81{m/s^2}Sin(theta) J

 

So i summed up the vectors in the I and J directions to get:

 

SUM a_x = 0.00845Cos(theta) I - 9.81Cos(theta) I

SUM a_y = -9.81Cos(theta) J

 

and can just combine the vectors in one expression for a_xy now right?

 

But ive forgotten how to find angle between vectors (cross product? dot product? or something else?)

[Rocket Man]

for simplicities sake, keep the components. the dot product is the one you want.

a.b = |a| |b| cos(phi)