Sea-Level on Oblate Spheroid

[Vanamonde]

Watch the tempers, guys. We're all friends here. (Or at least we all should be.) 

[YNM]

Alright, you were (and are) witnessing one of the worst forum members in ever. My apologies for that.

 

But back to seriousness :

I can see what's @arkie87 was referring to : During the mass reconfiguration, is there any "energy" expended ?

This might sounds odd, but the answer is yes. You need some work for it. That work is however many was the forces required to do it multiplied by the radial distance it was done.

A similar thing happens during orbital plane changes : Despite the overall potential and kinetic energy (orbital energy) remains the same through the entire process, changing the velocity vector does mean some work needs to be done.

Though I'd defer to some of the vastly more knowledgeable forum members than I for details of explanation.

[arkie87]

9 hours ago, YNM said:

Alright, you were (and are) witnessing one of the worst forum members in ever. My apologies for that.

Are you referring to me?

[YNM]

1 hour ago, arkie87 said:

Are you referring to me?

Myself. I have like a dozen of warning points or so.

[arkie87]

So to summarize so far:

(1) Ignoring gravitational potential, the energy required to walk from the equator to the pole (equivalent to pulling oneself to the center of a spinning object) is exactly equal to the kinetic energy due to rotation, and that kinetic energy is half of the potential energy difference due to the difference in height.

(2) The other half is supposedly due to different gravitational potential of a perfect ellipsoid vs. a perfect sphere.

Does anyone know the formula for the gravitational force of an ellipsoid. A linearization when a~=b would probably be sufficient to predict the difference of potential for 20km/6000km. 

[arkie87]

Well, i think i have figured it out. Thanks to all for your help.

The main mistake we were making is the assumption that gravitational force scales with the inverse square law-- this is only true outside the body; but inside the body, it increases linearly.

My solution is as follows:

The gravitational acceleration inside the earth is:

g=g_0*r/R

where g_0 is the gravitational acceleration on the surface at R.

The specific energy required to ascend from the center of the earth up to the north pole is:

E_p/m = integral(g_0*r/R_p*dr,0,R_p)= 1/2 g_0 *r^2/R_p (@r=0,R_p) = 1/2*g_0*R_p

Similarly, the specific energy required to ascend from the center of the earth up to the equator is:

E_e/m = integral((g_0*r/R_e-w^2*r)*dr,0,R_e)= 1/2*(g_0/R_e-w^2)*r^2 (@r=0,R_e) = 1/2*g_0*R_e - 1/2*w^2*R_e^2

Plugging in values for

g_0 = 9.81 m/s2

R_p = 6356 km

R_e = 6378 km

w = 2*pi/(60*60*24) = 7.272e-5 rad/s

We get

E_p/m = 31.17618 MJ/kg

E_e/m = 31.17653 MJ/kg

Alternatively, we can write:

1/2 (w*R_e)^2 = 1/2*g_0*(R_e-R_p)

1/2 (w*R_e)^2 = 107,559 J/kg

1/2*g_0*(R_e-R_p)  = 107,910 J/kg

So we have energy balance within 351 J, the error of which is probably due our assumption of g_0.

Thus, the integral of the potential energy from the center of the earth contributes the needed factor of 1/2 to make centripetal energy and gravitational potential energy balance ( @m4v  @sevenperforce). Accordingly, it appears the gravitational potential energy of an oblate spheroid has nothing to do with it. Sorry again,  @Snark

Edited June 17, 2018 by arkie87

[sevenperforce]

1 hour ago, arkie87 said:

Well, i think i have figured it out. Thanks to all for your help.

........

1/2 (w*R_e)^2 = 107,559 J/kg

1/2*g_0*(R_e-R_p)  = 107,910 J/kg

So we have energy balance within 351 J, the error of which is probably due our assumption of g_0.

Thus, the integral of the potential energy from the center of the earth contributes the needed factor of 1/2 to make centripetal energy and gravitational potential energy balance ( @m4v  @sevenperforce). Accordingly, it appears the gravitational potential energy of an oblate spheroid has nothing to do with it. Sorry again,  @Snark

The 351 J error is probably attributable to the oblate spheroid.

[arkie87]

5 minutes ago, sevenperforce said:

The 351 J error is probably attributable to the oblate spheroid.

Could be. 351 J out of 107,559 J is about 0.326% , which is comparable to the oblateness of the spheroid 22 km out of  6371 km (~0.345%). Though, on the other hand, if you used the value of g at the equator and poles, though, you might reduce the error as well...?

However, i thought that there was some law that says that the law of gravitation -- GMm/R^2 -- applies to point masses as well as solid bodies, as long as the distance is between centers of mass. Based on that, i'm not sure why an oblate spheroid would have any effect?

 

[YNM]

1 hour ago, arkie87 said:

However, i thought that there was some law that says that the law of gravitation -- GMm/R^2 -- applies to point masses as well as solid bodies, as long as the distance is between centers of mass. Based on that, i'm not sure why an oblate spheroid would have any effect?

I think "tidal force" have something on it. After all, sun-synchronous orbits are only possible because Earth has oblateness... and is rotating ? I don't know precisely.

Edited June 17, 2018 by YNM

[sevenperforce]

3 hours ago, arkie87 said:

Could be. 351 J out of 107,559 J is about 0.326% , which is comparable to the oblateness of the spheroid 22 km out of  6371 km (~0.345%). Though, on the other hand, if you used the value of g at the equator and poles, though, you might reduce the error as well...?

However, i thought that there was some law that says that the law of gravitation -- GMm/R^2 -- applies to point masses as well as solid bodies, as long as the distance is between centers of mass. Based on that, i'm not sure why an oblate spheroid would have any effect?

 

Almost, but you don't have the law quite right. Per the shell theorem, arguably Newton's most vital contribution to classical mechanics, the gravitational field outside a solid body is identical to the gravitational field of a point mass, if the solid body is spherically symmetric. This means that we don't have to calculate different gravitational fields for the core and the mantle and the crust; Earth's gravitational field is equivalent to that of a point mass inasmuch as Earth is spherically symmetric.

However, the Earth is not spherically symmetric; it bulges around the middle. Thus, the mass is distributed in such a way that the resultant gravitational field deviates from that of a point mass. In orbit, this effect is negligible, but at the very surface, it's measurable enough to cause that 350-joule deviation.

[arkie87]

7 hours ago, sevenperforce said:

Almost, but you don't have the law quite right. Per the shell theorem, arguably Newton's most vital contribution to classical mechanics, the gravitational field outside a solid body is identical to the gravitational field of a point mass, if the solid body is spherically symmetric. This means that we don't have to calculate different gravitational fields for the core and the mantle and the crust; Earth's gravitational field is equivalent to that of a point mass inasmuch as Earth is spherically symmetric.

However, the Earth is not spherically symmetric; it bulges around the middle. Thus, the mass is distributed in such a way that the resultant gravitational field deviates from that of a point mass. In orbit, this effect is negligible, but at the very surface, it's measurable enough to cause that 350-joule deviation.

ok, so it only applies to perfect spheres. no other shape?

[sevenperforce]

37 minutes ago, arkie87 said:

ok, so it only applies to perfect spheres. no other shape?

The outer gravitational field of a body is equal to that of a point mass if-and-only-if the body is spherically symmetric. So it could be a hollow shell, or a gigantic water droplet of uniform density, or a matryoshka doll planet of nested shells, or a differentiated sphere like most planets; as long as there is spherical symmetry, it works.

Furthermore, the deviation from a point-mass field is only relative to the deviation from spherical symmetry. Earth is almost a perfect sphere, and so its outer gravitational field is almost exactly that of a point mass. Venus and the Sun deviate even less. Saturn deviates much more. A highly oblate body like Pluto's moon Nix deviates grossly from the point-mass field shape.