Solve Inverse Cubed Force: Find A & B in r(θ) for E & L

In summary, the body of mass moves in a repulsive inverse cubed force given byF = K/r^3. The values of constants A and β in terms of E, L are found to be K/2r^2 and ω^2, respectively.
  • #1
ghostops
14
0
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 don't really know how to move forward. any help would be useful.

Thank you in advance
 
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  • #2
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
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
The inverse square term should not be there.
 
  • #5
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
 

FAQ: Solve Inverse Cubed Force: Find A & B in r(θ) for E & L

What is the formula for finding A and B in r(θ)?

The formula for finding A and B in r(θ) is:
A = E/(L^3 * cos^3θ)
B = 1/L

What do E and L represent in the formula?

E represents the electric field strength and L represents the distance between two charged particles.

How does the inverse cubed force differ from the inverse square force?

The inverse cubed force is an attractive force that decreases faster with distance compared to the inverse square force. This means that the force between two charged particles decreases at a much faster rate as the distance between them increases.

How is the inverse cubed force used in scientific applications?

The inverse cubed force is often used in studying the interactions between charged particles, such as in the fields of electromagnetism and astrophysics. It is also used in the development of technologies such as particle accelerators and nuclear fusion reactors.

What are the limitations of using the inverse cubed force formula?

The inverse cubed force formula assumes that the charged particles are point charges and that the medium between them is uniform. In reality, this may not always be the case and can lead to inaccuracies in calculations. Additionally, the formula only applies to the inverse cubed force and cannot be used for other types of forces.

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