Location of the magnetopause using Chapman-Ferraro equation

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  • #1
Kovac
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Hello,

Lets say you need to calculate the location of the magnetopause subsolar point on earth and you only have this information:

> Solar wind proton number density: 10 cm−3

> Solar wind speed: 700 km s−1

Chapman_ferraro equations:

What is the difference between the above chapman-ferraro equations? Why does one of them have 2^1/3 in front and one doesnt? What does the "pl" & "E" stand for?

Which one is more applicable to my case scanario?
 

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  • #2
It might help to know what source those two equations came from.
 
  • #3
My suspicion is that the simple CF equation gives the midday (subsolar) radius, and that the added term of; 2^(1/3) = 1.26; gives the dawn or dusk radius.
 
  • #4
Ibix said:
It might help to know what source those two equations came from.
These equations are coming from lecture slides.
 
  • #5
Baluncore said:
My suspicion is that the simple CF equation gives the midday (subsolar) radius, and that the added term of; 2^(1/3) = 1.26; gives the dawn or dusk radius.
So if you want realistic values, is it the second equation to be used? Because both give different results
 
  • #6
Kovac said:
Because both give different results
They are different because they are applied at different times of the day.
The magnetopause is not a spherical surface with one radius.
 
  • #7
Ibix said:
It might help to know what source those two equations came from.
Here are both equations mentioned in the slides.

 

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  • #8
Baluncore said:
They are different because they are applied at different times of the day.
The magnetopause is not a spherical surface with one radius.
Alright, could you help me understand what the signs in the denominator stands for.
What is the B in the numerator?
I assume the last term in the denominator is the solar wind velocity, the middle term pressure? And the u0 term is what?
 
  • #9
https://en.wikipedia.org/wiki/Magnetopause

According to this link the p= density of the solar wind, v = velocity, B=magnetic field strength of the planet.
How do I get the μ0? Can I get it through the proton number density?
 
  • #11
Kovac said:
https://en.wikipedia.org/wiki/Magnetopause

According to this link the p= density of the solar wind, v = velocity, B=magnetic field strength of the planet.
How do I get the μ0? Can I get it through the proton number density?
##B_{E}## and ##\mu_{0}## characterize the magnetic properties of the planet. The magnetic field of the Earth can be modeled as a magnetic dipole (https://en.wikipedia.org/wiki/Dipole_model_of_the_Earth's_magnetic_field), with the value of the field at the Earth's surface along the equator taking the value ##B_{E}=3.12\times10^{-5}\text{ tesla}##. And the magnetic permeability of vacuum, ##\mu_{0}=1.26\times10^{-6}{\rm \ N/A^{2}}##, is a basic constant of electromagnetism.
 
  • #12
renormalize said:
##B_{E}## and ##\mu_{0}## characterize the magnetic properties of the planet. The magnetic field of the Earth can be modeled as a magnetic dipole (https://en.wikipedia.org/wiki/Dipole_model_of_the_Earth's_magnetic_field), with the value of the field at the Earth's surface along the equator taking the value ##B_{E}=3.12\times10^{-5}\text{ tesla}##. And the magnetic permeability of vacuum, ##\mu_{0}=1.26\times10^{-6}{\rm \ N/A^{2}}##, is a basic constant of electromagnetism.
Are you sure about u0? Because it seems it should be 4pi * 10^-7 as it is the magnetic permiability of free space:
http://www.sp.ph.imperial.ac.uk/~mkd/AdvancedOption3solutions.pdf
 

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  • #13
Here are the first 4 pages of chap-8, The Handbook of Geophysics and Space Environments.
It defines the CF equation you should use and the variables.
 

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  • #15
renormalize said:
Try multiplying out ##4\times 3.14159\times 10^{-7}##. What do you get?
Yes correct, my bad!
 
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  • #16
Baluncore said:
Here are the first 4 pages of chap-8, The Handbook of Geophysics and Space Environments.
It defines the CF equation you should use and the variables.
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).
B = M/r^3 where M= magnetic dipole of the planet in question, r= radius of the planet in question.
μ0= 4pi x 10^-7 Vs/Am [magnetic permiability of free space]
u= solar wind velocity.

But I still dont understand why the second equation has 2^1/3 in front? Which one is more correct if you want to calculate for earth?
 
  • #17
Kovac said:
But I still dont understand why the second equation has 2^1/3 in front? Which one is more correct if you want to calculate for earth?
In post #7, the lower RHS of your attachment reads, as best as I can OCR;
"Assuming B=0 In the magnetosheath the induced Bmp, must cancel the geomagnetic dipole field in this region. This yields

Bmp = Bdipole(Rmp)

However, just inside the magnetosphere, B will increase the total B to

B = 2⋅Bdipole⋅(Rmp) = 2^(1/3) * ....
"
Do you want the Rmp, or do you want B ?
 

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