Determing a polar orbit for a non-inverse square central force

In summary, the conversation discusses the projection of a particle with a unit mass and initial velocity at a distance from the origin of a center of attractive force. The resulting orbit can be described by a polar equation, but the differential equation for the orbit is difficult to solve due to a non-linear term. Various attempts have been made to solve the equation, including using an energy relation, but none have been successful so far. The discrepancy between f(u) and f(u^-1) has also been pointed out.
  • #1
^_^physicist
235
1

Homework Statement


A particle of unit mass is projected with a velocity v-(sub 0), at right angle to the radius vector at a distance 'a' from the origin of a center of attractive force given by:

f(r)= -k*(4/(r^3)+ (a^2)/(r^5)).

If (v-(sub 0))^2 = (9*k)/(2*(a^2)) find the polar equation of the resulting orbit.

Homework Equations


F=ma

Treat 1/r = u

acceleration in polar coordinates= (r*{double dot} - r*((theta) {dot})^2 * e-sub r) + (r*(theta) {double dotted} + 2*r {dotted} + (theta) {dot})e-sub theta. Where e-sub are units vectors.

theta{dot} = l*u^2, where l is angular momentum per mass.

The Attempt at a Solution



Ok this is kind of long so I may skip few steps (sorry):

m*((r{double dot}) - r*((theta){dot})^2 = f(r), since the angular componet of acceleration is zero for this situtation.

m*(r{double dot} - (theta{dot})^2) = f(u^-1)

m*(r{double dot} - (theta{dot})^2) = -k*(4/(u^3)+ (a^2)/(u^5)).

m* [ -l^2*u^2*d^2*u/d(theta)^2-1/u*(l^2*u^3)]=k*(4/u^3 +a^2/u^5)

d^2*u/d(theta)^2 + u= (k*(4/u^3 + a^2/u^5))/(m*l^2*u^2)

and then I get stuck. I have tried multiple avenues for trying to solve this diff. eq, but none of them seem to cut it.

Anyone have any ideas? If the suggestion for problem goes to using an energy relation, I have tried that too and I get stuck in a similar problem.
 
Last edited:
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  • #2
wot... gives me a headache viewing non-latexed equations... anyway,

[tex]\frac{-4k}{r^3}\neq \frac{-4k}{u^3}[/tex]

so basically, you have:
[tex]u''=-u-\frac{m}{l^2u^2}F(u)[/tex]

you'll have an ugly non-linear term of u^3...
 
Last edited:
  • #3
Sorry about the non-latex (I am working on it, I swear...but I use maple in the lab to write up equations so don't really have a need to use latex (except here).

Thanks for help, however, could you clarify with the f(u) statement: are you referring to the f(u^(-1)) or are you actually meaning f(u).

Because when I found the general form of the differential equation of an orbit in my textbook it gave it as f(u^(-1)).

But again thanks.
 
  • #4
What tim_lou is pointing out is that, going from

^_^physicist said:
m*(r{double dot} - (theta{dot})^2) = f(u^-1)

m*(r{double dot} - (theta{dot})^2) = -k*(4/(u^3)+ (a^2)/(u^5)).

is incorrect, because f(u^-1)= -k*(4/((u-1)^3)+ (a^2)/((u-1)^5)), and not as you have written.
 

1. What is a polar orbit?

A polar orbit is a type of orbit in which a satellite or object orbits the Earth from one pole to the other, passing over the north and south poles. This type of orbit is often used for Earth observation and mapping purposes.

2. How is a polar orbit determined for a non-inverse square central force?

A polar orbit for a non-inverse square central force is determined by first analyzing the force acting on the object and then using mathematical equations to calculate the necessary velocity and direction for the object to maintain a stable polar orbit.

3. What factors affect the determination of a polar orbit for a non-inverse square central force?

The determination of a polar orbit for a non-inverse square central force is affected by factors such as the strength and direction of the central force, the mass and velocity of the object, and the distance between the object and the center of the force.

4. How does a polar orbit for a non-inverse square central force differ from a polar orbit for an inverse square central force?

The main difference between a polar orbit for a non-inverse square central force and a polar orbit for an inverse square central force is the strength and direction of the central force. In a non-inverse square central force, the force varies with distance, while in an inverse square central force, the force remains constant regardless of distance.

5. What are some applications of determining a polar orbit for a non-inverse square central force?

Determining a polar orbit for a non-inverse square central force has many applications, including satellite and space exploration, Earth observation and mapping, and studying the effects of non-uniform gravitational forces. It is also used in astrophysics to study the behavior of celestial bodies under different types of central forces.

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