Thread: Need help re-arranging an equation!

Remember the rule ? well, if you square both sides of the equation, you'll remove the square root.

Then, you can rearrange to remove the fraction, and solving for should be easier.

Rowe! Your back! I thought you might have gone to bed. Not sure what time zone you are in. Anyway! do I end up with:
or am I completely off track again?

Almost, if my calculations are correct (this is quite a tedious one!) you should have . You want to try and isolate , start by getting rid of that fraction by multiplying both sides by . Once that's "on the flat", simplify / expand where necessary so you can subtract the term on its own side of the equation and work from there.

Originally Posted by rowe

Looks like this is the final/accepted version

Let k = SQRT[(1 - r^2)^2 + (2zr)^2] ; then:

x = MRr^2 / (mk)
k = MRr^2 / (mx) ; square both sides:

k^2 = M^2 R^2 r^4 / (m^2 x^2) ; substitute back in:

(1 - r^2)^2 + (2zr)^2 = M^2 R^2 r^4 / (m^2 x^2) ; rearrange:

4 z^2 r^2 = M^2 R^2 r^4 / (m^2 x^2) - (1 - r^2)^2

z^2 = [M^2 R^2 r^4 / (m^2 x^2) - (1 - r^2)^2] / (4r^2) ; sooooooooooo:

z = SQRT{[M^2 R^2 r^4 / (m^2 x^2) - (1 - r^2)^2] / (4r^2)}

Can be slightly simplified to:
z = SQRT[M^2 R^2 r^4 - (m^2 x^2)(1 - r^2)^2] / (2rmx)