Orbital Radii of planets and moons of jupiter

[L²Cc]

Homework Statement

Finally, develop models for only the four innermost planets and the four Galilean moons by reducing all the orbital radii by a factor that makes the orbital radius of the first planet or moon equal to one. Give possible reasons for the similarities and/or differences in two models.

Homework Equations

x = position of planets relative to sun
y = orbital radii

Mercury Venus Earth Mars Asteriod Jupiter Saturn Uranus Neptune Pluto
x 1 2 3 4 5 6 7 8 9 10
y 57.9 108.2 149.6 227.9 T 778.3 1249 2871 4504 5914

x = position of galilean moons relative to jupiter
y = orbital radii

Lo Europa Ganymede Callisto
X 5 6 7 8
Y 422 670.9 1070 1883

The Attempt at a Solution

This math portfolio asks us to find an equation relating x and y. I find this by plotting the points, and the best fit line. The equation is exponential. I solve for T (the orbital raidus of asteriod), T = 469 millions of kilometers.

For this last question, I don't understand what they mean by the four "innermost" planets...And finding a factor (a common factor?)...

The model for case A is: 1.19x^3.71
The model for case B is: 1.18^3.54

Also, is there another way, other than the graphical method, to find the equation relating x and y??

 

[HallsofIvy]

Homework Statement

Finally, develop models for only the four innermost planets and the four Galilean moons by reducing all the orbital radii by a factor that makes the orbital radius of the first planet or moon equal to one. Give possible reasons for the similarities and/or differences in two models.

Homework Equations

x = position of planets relative to sun
y = orbital radii

Mercury Venus Earth Mars Asteriod Jupiter Saturn Uranus Neptune Pluto
x 1 2 3 4 5 6 7 8 9 10
y 57.9 108.2 149.6 227.9 T 778.3 1249 2871 4504 5914

x = position of galilean moons relative to jupiter
y = orbital radii

Lo Europa Ganymede Callisto
X 5 6 7 8
Y 422 670.9 1070 1883

The Attempt at a Solution

This math portfolio asks us to find an equation relating x and y. I find this by plotting the points, and the best fit line. The equation is exponential. I solve for T (the orbital raidus of asteriod), T = 469 millions of kilometers.

WHY solve for "the orbital radius of asteroid" (and which asteroid?)? There was no mention of asteroids here. (Nor is there a "T".)

For this last question, I don't understand what they mean by the four "innermost" planets...

The four "innermost" planets are Mercury, Venus, Earth, and Mars- that is what is meant by "position of planets relative to sun".

And finding a factor (a common factor?)...

You are asked to "reduce all the orbital radii by a factor that makes the orbital radius of the first planet or moon equal to one". "Reduce by a factor" means to divide them all by the same thing. You are told that the distance from the sun to Mercury (the first planet) is 57.9 (milliion meters?). What do you need to divide by to make that equal to 1? Now divide the next 3 by that same thing.

The model for case A is: 1.19x^3.71
The model for case B is: 1.18^3.54

Also, is there another way, other than the graphical method, to find the equation relating x and y??

Do you mean y= 1.19x^3.71? If so, note that, in y= Ax^b, log y= b log x+ log A, a linear equation. You could use a "least squares method" to find b and log A so that gives as good a line as possible through the (log x, log y) positions and then find your equation. I'm not sure what you mean by "the graphical method".

 

[L²Cc]

Thank you hallsofivy. I have one question, though. (Refering to your method)Do I use values from the given data to find the best fit line? For example,
log422 = b logx + logA. Then ?

I do mention T (asteroid) in the table above. When I copy pasted the table off word, the grids disappeared, thus making it hard to follow. It's in sequential order...you just have to correspond the first x value to the y value in the second row...sorry for the misunderstanding.

 

[HallsofIvy]

Thank you hallsofivy. I have one question, though. (Refering to your method)Do I use values from the given data to find the best fit line? For example,
log422 = b logx + logA. Then ?

Yes, that would be the best way to do this.

I do mention T (asteroid) in the table above. When I copy pasted the table off word, the grids disappeared, thus making it hard to follow. It's in sequential order...you just have to correspond the first x value to the y value in the second row...sorry for the misunderstanding.

You are right- I didn't see the third set of data.

 

[L²Cc]

I have been researching your 'least squares methods" but I'm still struggling to apply it to my set of data! Help highly appreciated! Thank you!

 

[coding_dude]

You should consider doing your IB Type II Assessment on your own.

 

[L²Cc]

I wish to clarify a few things: I did do my Type II assessment on my own. Additionally, HallsOfIvy simply verified my work. S/he didn't, in any way, give away answers. And just for the record, I did not determine the equation algebraically until the day before yesterday when I came upon a method to do so as I was researching for a chemistry-related topic.

Thank you, however, for your concern.

 

[coding_dude]

The least squares method is basically finding the line of best fit (the square of the errors form the smaller number, therefore least squares). If you have a TI graphing calculator, you do linear regression.

You can do what hallsofivy did and enter in the data using logs and still use linear regression. What you do to find y at the end is raise the other side to a power of the base.

However, that's the long way to get the equation. The simpler way is simply find exponential regression using the calculator.

By the way, I thought this assessment was completely pointless. Aren't we trying to find a relationship that doesn't exist?

 

[Dick]

You should consider doing your IB Type II Assessment on your own.

What is this 'Assessment'? I think the historical background for this problem is testing Bode's law. For which I don't think there is any theoretical basis, but was at one time taken very seriously.