Calculate magnetic dipole of other planets relative to the Earth

[Kovac]

So I have a few questions regarding the above task.

I will use this equation to get the results for the different planets:

Questions:

1. So in the equation density p= mass of proton x proton density of the solar wind x 1000 000 (conversion between kgcm^-3 to kgm^-3) x 1/r^2 (I assume that I should multiply this equation in here since it says that the solar wind density is decreasing with distance). Question: Is this correctly assumed?
2. B = M/r^3 where M= magnetic dipole of the planet in question, r= radius of the planet in question. For radius I have a table with values, but for M Im having trouble interpreting what to write since it says "relative to earth".
   

Question: What does the M value become for Mercury? Should I take Eart value * Mercury relative value? Or Is it simply 3.8 x 10^-4 or should I take the delta between earth value and the value for Mercury in the table? Its because it says relative to earth what confuses me, if you google the value for mercury its a lot bigger than the table is showing.μ0= 4pi x 10^-7 Vs/Am [magnetic permiability of free space], stays the same (not planet dependent)
u= solar wind velocity (not planet dependent). Will assume a value.

 

[TSny]

1. So in the equation density p= mass of proton x proton density of the solar wind x 1000 000 (conversion between kgcm^-3 to kgm^-3) x 1/r^2 (I assume that I should multiply this equation in here since it says that the solar wind density is decreasing with distance). Question: Is this correctly assumed?

Yes, the factor of $1/r^2$ should be there. [EDIT: Actually, you should NOT have the explicit factor of $1/r^2$if your proton density already includes the $1/r^2$ behavior. Answering the questions below should help clarify this.]

You are given that the proton number density at $r = 1$ AU is $10$ cm^-3. So, how would you express the proton number density as a function of $r$ where $r$ is in AU?

How would you express the mass density $\rho$ of protons as a function of $r$ for $r$ in AU?

Question: What does the M value become for Mercury? Should I take Eart value * Mercury relative value? Or Is it simply 3.8 x 10^-4 or should I take the delta between earth value and the value for Mercury in the table? Its because it says relative to earth what confuses me, if you google the value for mercury its a lot bigger than the table is showing.

For the earth you have the equation

Presumably, $B_E$ is some sort of value of the earth's magnetic field at the earth's surface. Hopefully, you have a value for this. For a different planet, you will need to replace $B_E$ by the planet's surface magnetic field, $B_{planet}$. Try to express $B_{planet}$ in terms of $B_E$ , the radii of the earth and the planet, and the magnetic dipole moments of the earth and the planet. This should allow you to find $B_{planet}$ in terms of $B_E$ and values in your table.

EDIT #2: I just noticed that your table in post #1 gives a value for the earth's magnetic moment at the bottom of the table. So, you can use the table to find the numerical value of the magnetic moment of the planet. Then you can find the magnetic field at the surface of the planet from the magnetic moment and the radius of the planet. So, you will not need to worry about a numerical value for the earth's field $B_E$.

 

[nasu]

Even though B is usually used for magnetic field (in Tesla) here it designates the magnetic moment. And if you have the relative radius on the left hand size it should be relative moment on the right hand size.
https://en.wikipedia.org/wiki/Magnetopause

 

[TSny]

Even though B is usually used for magnetic field (in Tesla) here it designates the magnetic moment. And if you have the relative radius on the left hand size it should be relative moment on the right hand size.
https://en.wikipedia.org/wiki/Magnetopause

The link writes the magnetopause distance as $$r_{mp} \approx \sqrt[6]{\frac {2 B_0^2}{\mu_0 \rho v^2}}$$ where $B_0$ is the magnetic moment. (Not a great notation since the link also used $B$ for magnetic field.)

However, in this problem, the formula given is for the ratio of $r_{mp}$ to the earth's radius $r_E$. So, using the formula from wikipedia, we have $$\frac{r_{mp}}{r_E} \approx \sqrt[6]{\frac {2 (B_0/r_E^3)^2}{\mu_0 \rho v^2}}$$ The quantity $B_0/(r_E)^3$ is the earth's magnetic field at the surface of the earth: $B_E$. So, we get $$\frac{r_{mp}}{r_E} \approx \sqrt[6]{\frac {2 B_E^2}{\mu_0 \rho v^2}}$$ This agrees with the formula given in the first post except for numerical factors.

 

[renormalize]

Question: What does the M value become for Mercury?

This table of planetary moments (relative to Earth's) and magnetopause distances (in terms of planetary radii), from the Wikipedia entry that @TSny cited, should help answer some of your questions and serve as a check on your calculations: