Central Force II

[Nusc]

The orbit of a particle in a central field is known to obey the following relationship:

r = A/(1+sin(theta))

a) determine the form of the central force F(r) that is responsible for this motion.
b) What is the distance of closest approach between the particle and the point that acts as the origin of the force? What is the furthest distance that the particle can be found form the origin of the force?

a) After applying the equation of motion, you get f(r) = -(A^3*l^2*m)/r^2

But for part b, how do I find the r-min?

Also A is not mentioned as a constant so do I assume it is?

[siddharth]

For part b,

You have
r(\theta)=\frac{A}{1+\sin \theta}
So for what value of \theta is r minimum and what is the minimum value?
HINT: You know \sin \theta can only take values between -1 and 1.

[Nusc]

What is the distance of closest approach between the particle and the point that acts as the origin of the force?

R is min at theta = -pie/2, therefore R = 0

What is the furthest distance that the particle can be found form the origin of the force?

R is max at theta = pie/2, therefore R = A/2

Is this correct?

[siddharth]

R is min at theta = -pie/2, therefore R = 0
What is the furthest distance that the particle can be found form the origin of the force?
R is max at theta = pie/2, therefore R = A/2
Is this correct?

No, it is not correct.
When \theta = \frac{-\pi}{2} , \sin \theta is minimum (-1).

Now,
r = \frac{A}{1+\sin \theta} . So 'r' will not be 0 when \theta = \frac{-\pi}{2} .

[Nusc]

Okay so R is minimum when theta is equal to pie/2, thus A/2


But R is max when theta is equal to pie, thus R max = A ?

[siddharth]

Nusc,

What is the value of r when \theta = -\frac{\pi}{6} ?

Compare this value to your "r max".

Can you now figure out what happens to r as \sin \theta approaches -1 ?

[Nusc]

My bad, so when theta = -pie/6 Rmax = 2A
When theta = pie/2 R min=A/2


Can you now figure out what happens to r as sin(theta) approaches -1 ?

sin(theta) = -1 if theta is equal to -pie/2

R = A/(1 + sin(-pie/2)) = A/(1-1) = A/0

How can r not be zero?

[siddharth]

My bad, so when theta = -pie/6 Rmax = 2A
When theta = pie/2 R min=A/2
Can you now figure out what happens to r as sin(theta) approaches -1 ?
sin(theta) = -1 if theta is equal to -pie/2
R = A/(1 + sin(-pie/2)) = A/(1-1) = A/0
How can r not be zero?

No, that's not right at all.

I did not say A/0 (Which is not 0). I meant as the denominator approaches 0 (ie, really close to 0 but not 0).

What is the value of the fraction as the denominator approaches 0? Try using a calculator to find this value for smaller and smaller values of the denominator. Can you see any relation?