Does the gravitational rate of acceleration increase within a planet?

[sjbauer1215]

Thank you. I appreciate the work you put in to come up with this answer.

Agreed, but the reduction of field strength inside the inner sphere will always be greater than the reduction of field strength outside of the innermost sphere [i.e. from planet surface to innermost sphere]. So the gravitational rate of acceleration will always be greater inside the innermost sphere even as distance decreases. So over all, in a nonlinear model of stratified layers of increasing density, gravity increases from the planet surface to innermost sphere towards the center of volume as long as there is any distance whatsoever.

Therefore any object near the center of planet will have to deal with increased gravity and pressure, much more so than was encountered on the planet's surface.

 

[Ibix]

So the gravitational rate of acceleration will always be greater inside the innermost sphere even as distance decreases.

Talking of reading comprehension, greater than what? I have been reading you as meaning "greater than the gravity at the surface of the inner sphere", in which case this is wrong - it falls linearly to zero. But perhaps you mean to compare to something else.

 

[PeroK]

Thank you. I appreciate the work you put in to come up with this answer.

Agreed, but ...

There's always a "but".

 

[sjbauer1215]

Talking of reading comprehension, greater than what? I have been reading you as meaning "greater than the gravity at the surface of the inner sphere", in which case this is wrong - it falls linearly to zero. But perhaps you mean to compare to something else.

 

[sjbauer1215]

Yes again, reading comprehension:
"... the reduction of field strength inside the inner sphere will always be greater than the reduction of field strength outside of the innermost sphere [i.e. from planet surface to innermost sphere]." The use of 'than' to indicate comparison.

Per you statement, 'it falls linearly to zero'. I already agreed to this overriding concern for each stratified layer of increasing density towards the center. However over all, in a nonlinear model of stratified layers of increasing density, gravity increases from the planet surface to innermost sphere towards the center of volume as long as there is any non-zero distance whatsoever.

Therefore any object near the center of planet will have to deal with increased gravity and pressure, much more so than was encountered on the planet's surface.

 

[sjbauer1215]

There's always a "but".

Can't avoid getting the problem right. so you can't just answer the question you want to answer.

 

[Mark44]

Thread moved from the Introduction section to Classical Physics. @sjbauer1215, the Introduction section is intended to be solely for introducing oneself.

 

[Ibix]

Yes again, reading comprehension:
"... the reduction of field strength inside the inner sphere will always be greater than the reduction of field strength outside of the innermost sphere [i.e. from planet surface to innermost sphere]." The use of 'than' to indicate comparison.

...but I thought we agreed that the gravitational field strength increases with depth outside the inner sphere. And you have quoted yourself talking about reductions in field strength which is not the same as the "the gravitational rate of acceleration" you were talking about in the post I quoted.

However over all, in a nonlinear model of stratified layers of increasing density, gravity increases from the planet surface to innermost sphere towards the center of volume as long as there is any distance whatsoever.

Only if your density approaches infinity faster than $1/r$ for all radii, which isn't plausible.

 

[Bandersnatch]

Talking of reading comprehension, greater than what? I have been reading you as meaning "greater than the gravity at the surface of the inner sphere", in which case this is wrong - it falls linearly to zero. But perhaps you mean to compare to something else.

I think the idea is that you take the 4/5 mass, 1/5 radius inner sphere, and divide itself into the outer 1/5 mass shell and 4/5 mass sphere. And so on, ad infinitum, in steps. Which, yeah. If so constructed would result in acceleration increasing without bound as the radius decreases. But it's of course unphysical, as density of real world objects doesn't increase to infinity towards the centre.
And in particular, all it takes for the density $\rho (r)$ to increase with decreasing radius at a slower rate than $\rho (r) \propto 1/r$ for the gravity to keep decreasing with radius.

edit: we seem to have posted at the same time

 

[Ibix]

I think the idea is that you take the 4/5 mass, 1/5 radius inner sphere, and divide itself into the outer 1/5 mass shell and 4/5 mass sphere.

Ok. I didn't get that from the OP's writing at all, but it at least explains why he thinks his density is increasing without bound.

 

[DaveC426913]

Man, this thread would be only one page long if it were easy to draw and post sketches like these, rather than all this writing out of paragraphs wrought with ambiguity...

What f we just posted one generic diagram and label its parts, segments and curves and agree to refer to that?

 

[sjbauer1215]

I think the idea is that you take the 4/5 mass, 1/5 radius inner sphere, and divide itself into the outer 1/5 mass shell and 4/5 mass sphere. And so on, ad infinitum, in steps. Which, yeah. If so constructed would result in acceleration increasing without bound as the radius decreases. But it's of course unphysical, as density of real world objects doesn't increase to infinity towards the centre.
And in particular, all it takes for the density $\rho (r)$ to increase with decreasing radius at a slower rate than $\rho (r) \propto 1/r$ for the gravity to keep decreasing with radius.

edit: we seem to have posted at the same time

I think the idea is that you take the 4/5 mass, 1/5 radius inner sphere, and divide itself into the outer 1/5 mass shell and 4/5 mass sphere. And so on, ad infinitum, in steps. Which, yeah. If so constructed would result in acceleration increasing without bound as the radius decreases. But it's of course unphysical, as density of real world objects doesn't increase to infinity towards the centre.
And in particular, all it takes for the density $\rho (r)$ to increase with decreasing radius at a slower rate than $\rho (r) \propto 1/r$ for the gravity to keep decreasing with radius.

edit: we seem to have posted at the same time

Infinity really? At least lets keep the numbers real, and the 'ad infinitum' argument is unnecessary. I don't believe we will ever reach the extreme density of a black hole.

And as you have already agreed that the innermost sphere would have a greater gravitational rate of acceleration that the surface rate, in this nonlinear model, you should be able to agree that "any object near the center of planet will have to deal with increased gravity and pressure, much more so than was encountered on the planet's surface."

 

[Ibix]

Man, this thread would be only one page long if it were easy to draw and post sketches like these, rather than all this writing out of paragraphs wrought with ambiguity...

View attachment 320230

Yeah, that's in the Wiki page PeroK linked in #2, although left-right reversed.

 

[DaveC426913]

Yeah, that's in the Wiki page PeroK linked in #2, although left-right reversed.

Yes, I just think we could use one that's tailored more to the OP's scenario.

 

[Ibix]

Infinity really?

If you want the gravity to increase with depth for all depths, it's a requirement that your density increase without bound.

 

[Ibix]

Yes, I just think we could use one that's tailored more to the OP's scenario.

Well, if we use what I understand it looks like this, if you'll forgive my lousy finger sketching (that's meant to be a $\rho$ on the vertical axis:

If @Bandersnatch is correct it's a power-law decay with steps.

 

[sjbauer1215]

I'm not sure why you are trying to make this out to reflect a reverse of the linear law. I already agreed that upon each stratified layer that gravity decreases with distance. However, over all [and not linearly] the gravity near the center of the planet would be greater than that near the surface of the planet, in this non-linear model.

 

[DaveC426913]

Well, if we use what I understand it looks like this, if you'll forgive my lousy finger sketching (that's meant to be a $\rho$ on the vertical axis:

I'd sketch it up if but I think we couldn't communicate succinctly enough to not pollute the thread with a sidebar...

OP?

 

[sjbauer1215]

I am not sure of whom I am discussing this with, but you all are familiar with the measure wherein the distance of 3470 km from Earth's center is where gravity is at its strongest (or equivalently, a depth of about 2900 km as measured from the Earth's surface), which is exactly what the graph from the PREM model data tells us as well (see the graph from earlier)?

Now, the notable thing about the above function is that it is a polynomial with respect to r. In other words, it is not just a simple linear function, but it actually has a maximum value at about 3470 km (see the graph below).

This corresponds to a gravitational acceleration of about 10.7 m/s2 (which is comparable to the gravity at the surface of Saturn!). For comparison, the gravitational acceleration at Earth’s surface is 9.81 m/s2.

 

[jbriggs444]

I always have been talking about an innermost radius of 20% the size of the original volume having a mass density that is 80% of the total volume's density. And I have been stating near the center and not at the center, all this time.

While you can take the extreme where zero distance provides for zero gravity, anything other than zero demonstrates that gravity increases toward the center of volume for this problem.

Do I understand that you are assuming that the 80/20 ratio holds good at all radii, all the way down to the center?

So the idea is that 80% of the entire earth mass is within 20% of the earth's volume. This would put it within a sphere of 58% of the earth's radius -- $\sqrt[^3]{0.2} = 0.58$. So the density of this fraction is $\frac{0.8}{0.2} =$ four times the average density of the whole thing.

[You cannot have meant that 80% of the earth's volume fits in a sphere with $0.20^3 = \frac{1}{125}$ of the earth's volume. That would make for a inner density 100 times the average density. Not even compressed nickel/iron is 550 g/cc. The OP says 12.9 g/cc for the core and 5.5 g/cc for the whole thing]

But you would make this recursive. So 80% of that 80% is within 20% of 20% of the earth's volume. So the density of this fraction of a fraction would be 16 times the average density of the whole thing.

And so on.

In this model, the gravitational acceleration at the center of the earth is still zero by symmetry but increases without bound when viewed as a limit. If the recursive model were correct, there would be a discontinuity in gravitational acceleration at the center of the earth.

 

[A.T.]

I don't believe we will ever reach the extreme density of a black hole.

Why don't you simply write down the formula for density as function of radius that you have in mind?

 

[DaveC426913]

This corresponds to a gravitational acceleration of about 10.7 m/s2 (which is comparable to the gravity at the surface of Saturn!). For comparison, the gravitational acceleration at Earth’s surface is 9.81 m/s2.

Saturn may be physically large but it is not very dense. Given a sufficiently large bathtub, it would float.
I weigh 85kg. On Saturn, I would only weigh 8kg more.

 

[jbriggs444]

I don't believe we will ever reach the extreme density of a black hole.

The idea that black holes have a density at all is problematic. You could divide a black hole mass by $\frac{4}{3}\pi r^3$ where r is its Schwarzschild radius. But this turns out not to be a constant. The larger the hole, the lower the "density" in this sense.

You could consider the local density in the interior of a black hole. But black holes are "vacuum solutions". They are vacuum everywhere. Zero density everywhere. There is no mass anywhere. Just space-time curvature.

You could consider what happens as one gets closer and closer to the singularity. But the geometry of a black hole is strange. Its interior volume is infinite. That $\frac{4}{3}\pi r^3$ formula works for Euclidean geometry, not for the curved pseudo-Riemannian manifold of the Schwarzschild solution.

 

[berkeman]

Thread is closed for Moderation...

 

[PeterDonis]

To the extent I can discern a coherent question that the OP is asking, it appears to be answered.

Thread will remain closed.