Does the gravitational rate of acceleration increase within a planet?

[sjbauer1215]

Example: The radius of the Earth is 6371 km. It has an average density of 5.5 g/cm3. Earth's inner core has the highest density at 12.9 g/cm3 [more than double the average]. Its surface gravity is measured in units of acceleration, which, in the SI system, are meters per second squared. It may also be expressed as a multiple of the Earth's standard surface gravity, which is equal to g = 9.80665 m/s2. Near Earth's surface, the gravity acceleration is approximately 9.81 m/s2 (32.2 ft/s2), which means that, ignoring the effects of air resistance, the speed of an object falling freely will increase by about 9.81 meters (32.2 ft) per second every second.

If 80% of a planet's density exists within 20% of the planet's innermost sphere, what is the standard surface gravity of this innermost sphere? Additionally what is the gravitational rate of acceleration near this innermost sphere?

 

[PeroK]

You can find the profile of the Earth's gravity by depth here:

https://en.m.wikipedia.org/wiki/Gravity_of_Earth

 

[sjbauer1215]

Yes, I have seen this before. And thank you, but it doesn't answer my question.

 

[PeroK]

If 80% of a planet's density exists within 20% of the planet's innermost sphere, what is the standard surface gravity of this innermost sphere? Additionally what is the gravitational rate of acceleration near this innermost sphere?

That depends on the mass and radius of the planet and what you mean by 20% innermost.

Have you heard of Newton's Shell Theorem?

 

[sjbauer1215]

Take a planet with the mass of 2.0 x10^30 with a radius of 8.0 x 200m. Figuring the surface gravity of the planet and its gravitational rate of acceleration, you get the baseline for the planet.

Then take 80% of this existing mass and calculate it within a radius which is 20% of the original planet radius [its innermost radius].

So while the baseline has not changed, the innermost sphere within the planet would now demonstrate an increase in the gravitational rate of acceleration. Therefore the gravitational force of the inner sphere is greater than the gravitational force of the planet surface.

 

[PeroK]

Take a planet with the mass of 2.0 x10^30 with a radius of 8.0 x 200m. Figuring the surface gravity of the planet and its gravitational rate of acceleration, you get the baseline for the planet.

Then take 80% of this existing mass and calculate it within a radius which is 20% of the original planet radius [its innermost radius].

So while the baseline has not changed, the innermost sphere within the planet would now demonstrate an increase in the gravitational rate of acceleration. Therefore the gravitational force of the inner sphere is greater than the gravitational force of the planet surface.

That's entirely possible.

The inner surface gravity would be 20 times greater than the surface gravity in that case.

 

[Ibix]

The gravitational field strength for Earth as a function of depth (according to our model of its density) is shown in the Wiki link @PeroK provided. Direct link: https://en.wikipedia.org/wiki/Gravity_of_Earth#/media/File:EarthGravityPREM.svg. It tops out somewhere near $11\mathrm{ms^{-2}}$ at the outside of the core.

Given your simple 80:20 model, you'd expect a peak gravity of $0.8/0.2^2=20$ times higher than surface gravity. That's not consistent with Earth (you are overestimating the relative density of the core and underestimating its radius), but you are correct that the "interior" gravity would be a maximum at the interface between the regions.

 

[sjbauer1215]

Thanks - So, in this example, where the impact of colliding rates of gravitational acceleration increases towards the center, which would cause great pressure, and hence the gravity at the center would be greater than zero.

 

[Ibix]

hence the gravity at the center would be greater than zero.

No, it's still zero, per the shell theorem.

 

[sjbauer1215]

The gravitational field strength for Earth as a function of depth (according to our model of its density) is shown in the Wiki link @PeroK provided. Direct link: https://en.wikipedia.org/wiki/Gravity_of_Earth#/media/File:EarthGravityPREM.svg. It tops out somewhere near $11\mathrm{ms^{-2}}$ at the outside of the core.

Given your simple 80:20 model, you'd expect a peak gravity of $0.8/0.2^2=20$ times higher than surface gravity. That's not consistent with Earth (you are overestimating the relative density of the core and underestimating its radius), but you are correct that the "interior" gravity would be a maximum at the interface between the regions.

I didn't say I wanted and answer for Earth. I was just use Earth as a model from which to extrapolate my example of 80-20. But thank you for looking into that as well.

 

[sjbauer1215]

No, it's still zero, per the shell theorem.

The shell theorem has to do with two mass bodies attracting. This is a single body problem which pressents a contray assumption from the liner model where the distance from the center decreases, the acceleration due to gravity also decreases. In the linear model, at the center where the distance is zero, gravity would be zero.

But in the nonlinear model, where the distance from the center decreases, the acceleration due to gravity is increasing. Therefore the gravity near the center would be greater than zero.

 

[DaveC426913]

...the gravity at the center would be greater than zero.

...the gravity near the center would be greater than zero.

You moved the goalpost.

Ibix was correct for where the goalpost was in the original post #9.

 

[Ibix]

The shell theorem has to do with two mass bodies attracting.

No. It lets you calculate the gravitational field strength due to a spherically symmetric mass distribution as a function of distance from its center. The answer is that the acceleration at radius $r$ is proportional to the mass enclosed within a sphere of radius $r$ centered at the origin. Since the mass enclosed in a sphere of radius zero is zero the acceleration at the center is zero.

Apart from anything else, if it were non-zero which way would the acceleration point? All directions are the same - how do you pick one?

 

[sjbauer1215]

The intent was not to move any goal post. The intent was to define a model, that when taken in a nonlinear matter would demonstrate a differing result from the linear model. The problem has always been clear about how the density would be distributed in the nonlinear model.

 

[Ibix]

But in the nonlinear model, where the distance from the center decreases, the acceleration due to gravity is increasing. Therefore the gravity near the center would be greater than zero

In your model the peak gravitational field strength is at the outside of your inner sphere. It falls to zero at the center from there.

 

[Ibix]

The intent was not to move any goal post. The intent was to define a model, that when taken in a nonlinear matter would demonstrate a differing result from the linear model. The problem has always been clear about how the density would be distributed in the nonlinear model.

...and this isn't a non-linear model. It's simply one with a variable density.

 

[sjbauer1215]

No. It lets you calculate the gravitational field strength due to a spherically symmetric mass distribution as a function of distance from its center. The answer is that the acceleration at radius $r$ is proportional to the mass enclosed within a sphere of radius $r$ centered at the origin. Since the mass enclosed in a sphere of radius zero is zero the acceleration at the center is zero.

Apart from anything else, if it were non-zero which way would the acceleration point? All directions are the same - how do you pick one?

The acceleration point would be the innermost sphere with a radius which was 20% smaller and containing a mass density of appx 80% of the total volume.

 

[sjbauer1215]

And where you presuppose a value of zero for the radius at the center, you ignored a radius of 1 wherein the gravitational rate of acceleration of this innermost sphere is much greater than the rate for the outermost surface sphere. Sure you can always pretend to reduce anything by introducing a zero in the equation.

 

[PeroK]

The shell theorem has to do with two mass bodies attracting. This is a single body problem which pressents a contray assumption from the liner model where the distance from the center decreases, the acceleration due to gravity also decreases. In the linear model, at the center where the distance is zero, gravity would be zero.

But in the nonlinear model, where the distance from the center decreases, the acceleration due to gravity is increasing. Therefore the gravity near the center would be greater than zero.

In which direction would be the gravitational force at centre?

 

[Ibix]

The acceleration point would be the innermost sphere with a radius which was 20% smaller and containing a mass density of appx 80% of the total volume.

So you are saying the acceleration points towards the center. Which way points to the center when you are at the center? The acceleration has to be zero at the center. It can certainly be higher at some depths inside a sphere than at its surface, but it must be zero at the exact center.

 

[Ibix]

And where you presuppose a value of zero for the radius at the center,

I'm not "presupposing" anything about radius. Radius is defined as distance from the center of a sphere. How can distance from the center be anything other than zero at the center?

 

[sjbauer1215]

So you are saying the acceleration points towards the center. Which way points to the center when you are at the center? The acceleration has to be zero at the center. It can certainly be higher at some depths inside a sphere than at its surface, but it must be zero at the exact center.

So where a nonexistent point at the center has a value of zero, there is no gravity. And yet an object whose size is greater than zero, and at the center of the volume, is suddenly feeling the pressure of the colliding effects of a gravitational acceleration rate of greater than the rate on the surface.

 

[DaveC426913]

And where you presuppose a value of zero for the radius at the center, you ignored a radius of 1 wherein the gravitational rate of acceleration of this innermost sphere is much greater than the rate for the outermost surface sphere. Sure you can always pretend to reduce anything by introducing a zero in the equation.

Are we all talking past each other?

When you say "radius at the center", do you mean the "some smaller, inner but non-zero-radius sphere?"

Because I think the rest of us are thinking that "radius at the centre" is, by definition, zero radius.

Can you confirm for our sakes?

 

[sjbauer1215]

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.

 

[PeroK]

Talking about gravity, this thread has gone rapidly downhill!

 

[sjbauer1215]

It is about reading comprehension. What is the actual problem, as opposed to the one that you want to answer.

 

[PeroK]

It is about reading comprehension. What is the actual problem, as opposed to the one that you want to answer.

It's about physics comprehension!

 

[sjbauer1215]

It's about physics comprehension!

Yes, that as well. At least you came up with the answer at the very beginning of this thread.
So do the others have an answer now that they understand the question?

 

[PeroK]

Yes, that as well. So do you have an answer now that you understand the question?

The gravitational force cannot increase indefinitely as the distance from centre tends to zero. The formula for the magnitude of the force on a test unit mass is $$F = \frac{GM}{r^2}$$Where $M$ is the mass enclosed by radius $r$.

The point is that mass is proportional to radius cubed. I.e. $$M = \frac 4 3 \rho \pi r^3$$ Where $\rho$ is the average density. This gives us
$$ F = \frac{4\rho \pi r}{3}$$ Which tends to zero as $r$ tends to zero for any finite density.

 

[Ibix]

So where a nonexistent point at the center has a value of zero, there is no gravity. And yet an object whose size is greater than zero, and at the center of the volume, is suddenly feeling the pressure of the colliding effects of a gravitational acceleration rate of greater than the rate on the surface.

I'm sorry, I cannot make any sense of that. The maths is quite straightforward. Say your planet has radius $R$ and surface gravity is $g$. There are three regimes:

Inside your inner sphere the acceleration at a distance $r$ from the center is $100gr/R$.

Outside your inner sphere but still inside your planet the acceleration is $\left(\frac{99R^2}{124r^2}+\frac{25r}{124R}\right)g$.

Outside the planet it is $R^2g/r^2$.

Note that the field strength is a maximum at the outer surface of the inner sphere and decreases both inside and outside it.