Iridium and gravity

[dichotomy]

I've been just thinking, as I do, and thought....

What volume of iridium would produce the same gravity as earth?

How much smaller would a sphere of this be than the earth?

 

Hope someone can help here.

[big314mp]

I calculated that a sphere of Iridium, with the same mass as the Earth, would have a radius 62.6% the size of Earth's radius. It works out to a radius of roughly 4000km.

 

Basically:

 

Set mass of iridium sphere equal to mass of earth. Use density of iridium to calculate the volume of this sphere. Use the sphere volume equation to calculate the radius of this sphere. Compare to earth's.

[dichotomy]

Wow! that's quite a bit larger than the radius of Mars - 3397 km.

I thought something as heavy as Iridium might be quite a bit smaller and still match the earths gravity.

[D H]

Set mass of iridium sphere equal to mass of earth. Use density of iridium to calculate the volume of this sphere. Use the sphere volume equation to calculate the radius of this sphere. Compare to earth's.

That's too simple a calculation. You are implicitly assuming that the iridium earth will be of constant density. Iridium, like any other material, compresses under pressure. The center of the iridium earth will have a density significantly greater than 22,560 kg/m³ because of the gravitational pressure resulting from all of the material above it.

[big314mp]

Well, if we are going to take that into account, we should also take into account that the gravity at the surface of said Iridium sphere will actually be significantly higher than 1g, due to being closer to the center of the sphere.

 

I figured I'd make a few assumptions in the interest of simplicity (read: the math to take all of those factors into account is beyond me).

 

Hold up...wouldn't a different density not actually change the gravity (assuming a point source), as it is mass only in the gravitational equation?

[insane_alien]

if its the same mass then for the same distance from the center the gravitational field will be the same strength.

 

to calculate surface gravity of the iridium sphere just plug in the new radius and kep the mass term the same as earths.

[big314mp]

Well, I think the OP wanted to go in the other direction. For a given surface gravity, calculate the mass.

[dichotomy]

Basically I thought you could have a sphere of pure iridium floating in space and it would be quite a bit smaller than the earth and still have the same gravity. Why? Because it is a heavier material than what the earth is composed of. I wasn’t sure how much smaller it would need to be to match the earth’s gravity. Hope this clarifies things?

 

Iridium, like any other material, compresses under pressure. .

 

So, an iridium sphere could be quite a bit smaller than Mars, and still achieve the same gravity as earth?

[D H]

At the center of the iridium earth, iridium would be two to three times as dense as the surface density, so assuming a constant density will overestimate the radius.

 

What is it you want -- an iridium sphere that has 1g surface gravity, or an iridium sphere that would have the moon orbiting at one orbit per sidereal month?

[big314mp]

2-3 times as dense seems a bit much...how exactly would you calculate such a number?

[D H]

Iridium's bulk modulus is 320 gigapascal at STP. The pressure at the center of the Earth is about 360 gigapascal. Ignoring temperature changes in density, ignoring that the bulk modulus varies with pressure, and ignoring that the pressure at the center of the iridium earth would not be the same as the pressure at the center of the real Earth, that bulk modulus and that pressure leads to a density three times that at STP.

[big314mp]

I understand (and am quite surprised) that the iridium will significantly compress, but I don't follow the math:

 

Change in volume = pressure applied / bulk modulus

360GPa/320GPa = 1.125

 

This doesn't make logical sense, as the iridium can't decrease in volume by more than 100%...so what's missing?

[D H]

You are applying the relationship incorrectly; it is not linear. The bulk modulus is defined by

 

[math]K = -V\,\frac{\partial P}{\partial V}[/math]

 

Assuming a constant bulk modulus, this leads to

 

[math]V(P) = V_0\,e^{-P/K}[/math]

 

or

 

[math]\rho(P) = \rho_0\,e^{P/K}[/math]

[dichotomy]

What is it you want -- an iridium sphere that has 1g surface gravity, or an iridium sphere that would have the moon orbiting at one orbit per sidereal month?

 

An iridium sphere that has 1g surface gravity.

[D H]

If you assume a constant density, you should be able to solve this yourself. All it takes is some simple high school algebra. I'll give you a start.

 

The surface gravity of a constant density sphere of radius R is

 

[math]g=\frac {G M}{R^2} [/math]

 

where M is the mass of the sphere and G is the universal gravitational constant and has a numerical value of 6.67300×10^-11 meters³/kilogram/second².

 

The mass of a constant density sphere of radius R is

 

[math]M = \frac 4 3 \pi \rho R^3[/math]

 

Where [math]\rho[/math] is the density of the sphere. The density of iridium is 22.56 grams/cm³, or 22,560 kilograms/meter³.

[big314mp]

We've (my roommate and I) been puzzling over this one for a while, and have run into a rather solid wall:

 

We've been attacking this problem by dividing the sphere into infinitely thin "shells" and trying to come up with various to formulas to integrate over the radius.

 

We've run into a problem on the following (I have no idea how to use LaTeX, so I'll do my best to describe):

 

We are running into problems with writing out a function for pressure, as pressure is a function of radius of the entire sphere, and the depth that is being considered. The radius of the entire sphere, however, is a function of the pressure, as the pressure determines how much the iridium is compressed. So our equations seem to go in an endless loop, with one always turning up inside the integral of the other.

 

Anyone?

/bump

Edited October 30, 2008 by big314mp
multiple post merged

[dichotomy]

Hi big314mp,

Can you work out DH’s simple high school algebra. I’m a virtual mathematical dyslexic. I really only wanted a rough approximate personally.

 

I was originally guestimating something quite a bit smaller than the earth, maybe about the size of the Moon once the iridium sphere's gravity compressed itself.

[D H]

Hi big314mp,

Can you work out DH’s simple high school algebra. I’m a virtual mathematical dyslexic.

The only way you can overcome this problem is by practice. I gave an equation for the gravitational acceleration due to a uniform sphere and an equation for the mass of a uniform sphere:

 

[math]g=\frac {G M}{R^2}[/math]

 

[math]M = \frac 4 3 \pi \rho R^3[/math]

 

All you have to do to solve this simple version of the problem is to substitute the mass in the first equation with the expression on the right hand side of the second equation. Simplify and solve for R. Give it a shot before you give up.

[big314mp]

Also, post back your answer when you get it, and we'll check it for you.

 

D H: Any ideas on how to take compressibility into account, or is that just a mathematical dead end?

[insane_alien]

its not a dead end, it just turns your density into a variable.

[big314mp]

Density becomes a function in terms of pressure.

 

Pressure is a function in terms of force from overlying mass and radius.

 

Radius is a function of total mass and density.

 

Force from overlying mass is a function of underlying mass, radius, and overlying mass.

 

I ended up running in circles, since they are all functions of each other. I think I need an assumption, which will simplify one of the variables. I'm not sure what exactly I can assume though.

[D H]

Density becomes a function in terms of pressure.

Assuming a constant temperature and a constant bulk modulus leads to

 

[math]\rho® = \rho_0 \exp(P®/\kappa) [/math]

 

where [math]\rho_0[/math] is the density at the surface (where P=0).

 

Pressure is a function in terms of force from overlying mass and radius.

 

[math]\frac{dP}{dr} = -\rho® g®[/math]

 

where g® is the acceleration due to gravity at radius r:

 

[math]g®=\frac {GM®}{r^2}[/math]

 

[math]M® = \int_0^r 4\pi \rho(r') r'^2 dr'[/math]

 

Putting it all together,

 

[math]\frac{dP}{dr} =

-\,4 \pi G \rho_0^2

\frac{\exp(P®/\kappa)}{r^2}

\int_0^r \exp(P(r')/\kappa) r'^2 dr'[/math]

 

Radius is a function of total mass and density.

The goal is to find some R such that g® is 9.80665 m/s².

 

I can't find a closed form solution; that doesn't mean that one doesn't exist.

[big314mp]

The goal is to find some R such that g® is 9.80665 m/s².

 

I can't find a closed form solution; that doesn't mean that one doesn't exist.

 

Damn. And it sounded like such a fun problem.

[dichotomy]

I have been confirmed an incurrable mathematical dyslexic by a specialist.

 

Can someone please post the solution the problem:

What size would a sphere of pure iridium need to be, to give it the same gravity as our Earth?

 

You can send me a private message if you don't want to spoil it for others. cheers.