# Inverse Cubed Force

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1. Nov 14, 2015

### ghostops

Question: A body of mass m is moving in a repulsive inverse cubed force given by
F = K/r^3 where K > 0​
show the path r(θ) of the body given by
1/r = A cos[β(θ-θο)]
Find the values of constants A and β in terms of E, L.

Work done so far:
we did a similar problem with inverse square so I am attempting to modify that for use with cubes

so far I have
F = K/r^2+K/r^3
V(r) = ∫ F(r)dr
V(r) = K/r + K/2r^2
dr/dθ = r^2/L√[2u(E-L^2/2ur^2-K/r - K/2r^2)]

distributing and moving L
dθ=dr/r^2 1/√[2uE/L^2 - 1/r^2 - uK/L^2r^2 - 2uK/L^2r]^-½
substituting 1/r for ω and dr/r^2 for dω
dθ=dω1/√[2uE/L^2 - ω^2 - uKω^2/L^2 - 2uKω/L^2]^-½
from here our professor wants us to reduce this to dθ=dω/√(a^2+x^2) so we can use trig sub we are running into issues and dont really know how to move forward. any help would be useful.

2. Nov 14, 2015

### Chandra Prayaga

Your question says that the force is inverse cube. In your work, you have two forces, an inverse square, + an inverse cube. Why did you do that?

3. Nov 14, 2015

### ghostops

I was trying to apply Newton's theorem of revolving orbits and I believe I just went completely off from where I should have been heading

4. Nov 14, 2015

### Staff: Mentor

The inverse square term should not be there.

5. Nov 14, 2015

### ghostops

ok so without that it becomes

F = K/r^3
V(r) = ∫ F(r)dr
V(r) = K/2r^2
dr/dθ = r^2/L√[2u(E-L^2/2ur^2- K/2r^2)]

distributing and moving L​
dθ=dr/r^2 1/√[2uE/L^2 - 1/r^2 - uK/L^2r^2]^-½

substituting 1/r for ω and dr/r^2 for dω
dθ=dω1/√[2uE/L^2 - ω^2 - uKω^2/L^2]^-½
consolidating
dθ=dω1/√[2uE/L^2 - (1+ uK/L^2)ω^2]^-½
I still do not where to go from here however

again thank you for any help provided