# Homework Help: Radius of Mercury's Core

1. Jan 21, 2010

### Katte

1. The problem statement, all variables and given/known data
After determining the average density of Mercury as 5.43 x 10^3(kg/m3), you may calculate the size of Mercury's core. Assume Mercury consists of a core and a mantle+crust, with core composition being similar to Earth (7,800(kg/m3)). From space missions and studying meteorites we know that the density of the mantle+crust is about 2,600 (kg/m3). Calculate the radius of Mercury, using the total radius of Mercury from NASA.

2. Relevant equations
My problem is that I don't know what equation to use. I am attempting a geophysics course in university after having finished high school physics three years ago, which I did by correspondence, teaching myself out of a book! This is all a little new to me; I still have to look up the equations for volume on Wikipedia.

3. The attempt at a solution
I spoke to my professor about the problem. We determined that the known variables are:

Density of Mercury: 5.43 x 10^3 (kg/m3)
Density of Mercury's Core: 7,800 (kg/m3)
Density of Mercury's Mantle: 2,600 (kg/m3)

The unknown is the radius. He told me that the equation I will need to solve the problem will involve my known variables. (I wrongly assumed I could look something up very simply.) I have thought about using v=4/3(pi)r^3, but I don't know the volume of the core. I thought about calculating the volume of the mantle and subtracting it from the total volume of Mercury to find the volume of the core, but I don't think I have enough information to do that. I've considered using a ratio, but any ratio I can think of, I don't have enough information for. Can anyone point me in the right direction?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 21, 2010

### mgb_phys

You are on the right lines
You can write three equation for the volume of the core, the mantle and the total planet in terms of the total radius and the core radius.
Then you can find the mass of each part in terms of their density and volume.
Obviosuly the mass of the planet must be the sum of the mass of the core and mass of the mantle

Then you should only have one unknown - the core radius

3. Jan 21, 2010

### Katte

So I have calculated that the mass of Mercury is 3.303 x 10^23 kg, and the volume of Mercury is 6.08 x 10^19 m3, But I'm unsure of how to calculate the volume or mass of the mantle, because wouldn't I need the volume or mass of the core to do so?

4. Jan 21, 2010

### Katte

Would I possibly use the total radius to determine Mercury's moment of inertia to calculate core size? I've just read about it and unsure of the process.

5. Jan 21, 2010

### mgb_phys

No,
Simply write an equation for the mass of the core in terms of 'r' the core radius
Then one for the mantle in terms of 'R' the total radius and 'r'
And one for the total mass of the planet in terms of 'R'

post them here.

6. Jan 21, 2010

### Katte

Total Mass
= v x d
= (6.081 x 10^19 m3)(5.43 x 10^3 kg/m3)

Mantle Mas
= v x d
= (4/3(pi)R-r)3 x d
= (4/3 (pi) 2.439x10^6m - r)3 x (2,600 kg/m3)

Core Mass
=v x d
=(4/3(pi)r)3 x d
= (4/3 (pi) r)3 x (7,800 kg/m3)

7. Jan 21, 2010

### mgb_phys

That's where you are going wrong.
The volume of a shell isn't (4/3(pi)(R-r)3 it's 4/3piR^3 - 4/3pir^3 = 4/3pi(R^3-r^3)

So total mass
Mt = 4/3pi R^3 * density_ave
Mc = 4/3pi r^3 * density_core
Mm = 4/3pi (R^3 - r^3 ) * density_mantle

Total mass Mt = Mc + Mm
4/3pi R^3 * density_ave = 4/3pi r^3 * density_core + 4/3pi (R^3 - r^3 ) * density_mantle

You can lose the 4/3pi and so:
R^3 * density_ave = r^3 * density_core + (R^3 - r^3 ) * density_mantle

Last edited: Jan 21, 2010
8. Jan 21, 2010

### Katte

Okay, let's see if I got this.

(R^3)(density_ave)=(r^3)(density_core)+(R^3-r^3)(density_mantle)

((2.4397x10^6m)^3)(5.43x10^3kg/m^3)=(r^3)(7,800kg/m^3)+((2.4397x10^6m)^3-r^3)(2,600kg/m^3)

Which calculates to (rounded for reading's sake):

(7.89x10^22)=(7,800r^3)+(3.78x10^22)-(2,600r^3)

4.1096x10^22=5,200r^3

7.903x10^18=r^3 => Cube root

r=1,991,884.369

Does that even make sense? That would mean the core radius is more than 80% of the planet!

9. Jan 21, 2010

### mgb_phys

Rather an optimistic number of significant figures!
But yes - see ( http://en.wikipedia.org/wiki/Mercury_(planet [Broken]) )

Last edited by a moderator: May 4, 2017
10. Jan 21, 2010

### Katte

Oh. Wow! Well, thank you for all the help! I realize now the answer was pretty simple, all things considered. Thanks for steering me in the right direction! =)