Gravity Gradient: Southward Displacement at $\phi = 30^{\circ}$

[ognik]

Q: "The dependence of fee fall acceleration g on geographical latitude $\phi$ at sea level is given by $g=g_0\left(1+0.0053 Sin^2\phi\right)$. What is the southward displacement near $\phi = 30 ^{\circ}$ that changes g by 1 part in $10^8$?"

This is in a section on gradient ($\nabla$) but I can't visualise the directions involved. I assume the equator is $\phi = 0^{\circ}$ so is the gradient of g pointing at the centre of the Earth at the equator? and $f\left(\phi\right)$ is along the surface of the earth? Then is $\phi = 30 ^{\circ}$ pointing parallel to the gradient at the centre?

Maybe I should be using $df(\phi) = C_1 - C_2 = \Delta C = \left(\nabla f(\phi) \right).dr$ with $df(\phi)$ the '1 part in $10^8$' and dr is the southward displacement?

 

[I like Serena]

Q: "The dependence of fee fall acceleration g on geographical latitude $\phi$ at sea level is given by $g=g_0\left(1+0.0053 Sin^2\phi\right)$. What is the southward displacement near $\phi = 30 ^{\circ}$ that changes g by 1 part in $10^8$?"

This is in a section on gradient ($\nabla$) but I can't visualise the directions involved. I assume the equator is $\phi = 0^{\circ}$ so is the gradient of g pointing at the centre of the Earth at the equator? and $f\left(\phi\right)$ is along the surface of the earth? Then is $\phi = 30 ^{\circ}$ pointing parallel to the gradient at the centre?

Maybe I should be using $df(\phi) = C_1 - C_2 = \Delta C = \left(\nabla f(\phi) \right).dr$ with $df(\phi)$ the '1 part in $10^8$' and dr is the southward displacement?

The gradient in polar coordinates is:
$$\nabla g = \pd g r \boldsymbol {\hat r} + \frac 1 r \pd g \phi \boldsymbol {\hat \phi}$$

Latitude is indeed measured from the equator.

A bit of background information: due to centrifugal forces the Earth is not quite spherical, but a bit flattened.
As a result the gravity at the poles is greater than the gravity at the equator.
Moreover, someone standing on the surface also experiences centrifugal forces, causing a slight change in magnitude and direction.

 

[ognik]

Thanks ILS

The gradient in polar coordinates is:
$$\nabla g = \pd g r \boldsymbol {\hat r} + \frac 1 r \pd g \phi \boldsymbol {\hat \phi}$$

But from the eqtn I have for g, there is no $\partial/\partial r$ component? So is the southward displacement value from 1/r?

And directions, g must point into the centre, so does $\nabla g$ point south across the Earth's surface?

 

[I like Serena]

Thanks ILS

But from the eqtn I have for g, there is no $\partial/\partial r$ component? So is the southward displacement value from 1/r?

And directions, g must point into the centre, so does $\nabla g$ point south across the Earth's surface?

We can consider $g$ an undirected scalar that varies over the surface of the earth.

The dependency on $r$ is indeed not given, but then, we're only interested in the 'southward displacement'.
With only the coordinate $\phi$, the gradient simplifies to:
$$\nabla g = \frac 1 r \pd g \phi \boldsymbol{\hat\phi}$$
where $\boldsymbol{\hat\phi}$ is the unit vector in the direction of increasing $\phi$, which is northward, parallel to the surface of the earth.

 

[ognik]

Thanks ILS, allow me to ramble a bit please - I know I have a weakness in visualising fields, just not much practice...

You've helped me see nicely the scalar gravity field itself - different values at diff. places on the surface of the earth, (I must remember to remember these have no direction!). The field itself is g, a scalar function of $\phi$ - which is directly the latitude coord, 0 at the equator, +/- 90 at the poles? If so, and we are south of the equator, wouldn't $\hat{\phi}$ increase southward?

Isn't r the radius coord of the Earth (which we don't know)?

My book says the gradient is $\perp$ to the surface of the field, but the eqtn has it parallel to the surface? I don't just want to apply formula and hope :-)

Attached is the best image I could find, not very useful ...View attachment 4948

 

[I like Serena]

The field itself is g, a scalar function of $\phi$ - which is directly the latitude coord, 0 at the equator, +/- 90 at the poles? If so, and we are south of the equator, wouldn't $\hat{\phi}$ increase southward?

$\phi$ keeps decreasing to more and more negative values south of the equator.
So $\boldsymbol{\hat{\phi}}$ still points northward, even south of the equator.

Isn't r the radius coord of the Earth (which we don't know)?

Yes.
It's $6,371\textrm{ km}$ on average, which we can look up.

My book says the gradient is $\perp$ to the surface of the field, but the eqtn has it parallel to the surface? I don't just want to apply formula and hope :-)

Since we only have 1-dimension, perpendicular is somewhat meaningless, since it means perpendicular to a point.

If we do introduce the second dimension $r$, we can draw ovals of equal $g$ values.
Then we'll find indeed that the gradient is perpendicular to any of those ovals.
However, the direction of that gradient will not correspond to the direction of $\vec g$.
Instead the gradient will point in the direction where $g$ will increase the most, which is a combination of going to the center and the direction where the centrifugal force becomes less (northward).

As for the formula, we can explain it in words, which should help to avoid the feeling to helplessly apply a formula.
The word gradient means the ratio that a scalar function changes if we change the position an infinitesimal amount.
In this case it means the ratio of $dg$ to $r\,d\phi$, where the latter is the distance we travel when changing $\phi$ by $d\phi$.

Attached is the best image I could find, not very useful ...

It's a nice picture that shows how we vector $\vec g$ is determined, which as you can see, points slightly southward, while the gradient of the scalar $g$ points slightly northward.

 

[ognik]

Thanks.
That Radius is approximate though, I was hoping to find a way to ratio it out of the eqtn...
Anyway, getting down to the nitty-gritty,

$ \nabla g = \frac{1}{r}\pd{g}{\phi} \hat{\phi} = \frac{g_0}{r}\left(0.0053)\right) 2 Sin \phi Cos \phi$

$ \therefore dg =\frac{g_0}{r} \left(0.0053)\right) (2 Sin \phi Cos \phi) d\phi \hat{\phi} = \frac{1}{10^8}$

with $g_0 = 9.8 ms^{-2} $ and $ r = 6,371,000 m $ I get $d\phi = 2.8$ degrees?

 

[I like Serena]

Thanks.
That Radius is approximate though, I was hoping to find a way to ratio it out of the eqtn...
Anyway, getting down to the nitty-gritty,

$ \nabla g = \frac{1}{r}\pd{g}{\phi} \hat{\phi} = \frac{g_0}{r}\left(0.0053)\right) 2 Sin \phi Cos \phi$

$ \therefore dg =\frac{g_0}{r} \left(0.0053)\right) (2 Sin \phi Cos \phi) d\phi \hat{\phi} = \frac{1}{10^8}$

with $g_0 = 9.8 ms^{-2} $ and $ r = 6,371,000 m $ I get $d\phi = 2.8$ degrees?

The $1$ part in $10^8$ would correspond to the ratio between $dg$ and $g_0$.
That is, $g_0$ should be on the other side of the equation.

Furthermore, the problem statement asks for a 'displacement'.
$d\phi$ is not a displacement but an angle. The displacement is $ds = r\,d\phi$, which actually rids you of $r$ nicely.
I guess we should use $\nabla g = \frac{dg}{ds}$.

 

[ognik]

I think I went down the wrong track anyway, I wrote $\nabla g$ as $ \frac{dg}{d\phi}$ - which it isn't. Starting afresh ...

$ \nabla g = \frac{1}{r} \frac{dg}{d\phi}\hat{\phi} = \frac{g_0}{r} (0.0053)(2sin\phi cos\phi)$

$ \therefore \frac{dg}{g_0}= (0.0053)(2sin\phi cos\phi) d\phi \hat{\phi} $

$ \therefore d\phi=\frac{1}{10^8 (0.0043) (2sin\phi cos\phi) }$ ?

$ ds = rd\phi = \frac{6,371,000}{10^8 (0.0043) (2sin\phi cos\phi) } = 1.4 m$ sure this is wrong :-( I think I am very muddled now, please point out the 'bleedin obvious',

 

[I like Serena]

An $r$ got lost between the first and second line.
The formula for $ds$ should not contain an $r$.

Furthermore, $0.0053$ transformed inexplicably into $0.0043$ in the following lines.

 

[ognik]

Just a quick note because I have a few threads with you helping - yesterday my PC upgraded to WIN 10, got stuck in an endless auot repair loop, the usual lack of support from MS - so I was trying to work (old laptop) while trying to rescue my PC, so pls excuse any carelessness or lack of focus on my part. Stayed up till after midnight, the eventual outcome was I had to rebuild the PC with Win 7, everything lost - so busy restoring etc.