Central Force Motion (two-mass orbit problem)

[mat5041]

Homework Statement

1a) Verify that the kinetic energy of two particles can be written as the kinetic energy of a single 'reduced mass' particle with velocity r'equal to their difference.

1b) In the center-o-mass frame, show that theta is the same for r1 as for r, whereas the polar angle for r2 differs by pi radians from that of r

1c)Show that if r(theta) is an ellipse, then r1(theta) is an ellipse with the center of mass at one focus; r2(theta) is an ellipse with one focus at the center of mass; in a (non-inertial) frame in which the Earth is at rest, the sun would move in an ellipse around the sun if they were the only two massive objects.

Homework Equations

1a)
kinetic energy of the system:
T=1/2*m1*|r'1|^2+1/2*m2*|r'2|^2

reduced mass (greek letter mu=u)
u=m1*m2/(m1+m2)

relation of r, r1, and r2
r=r1-r2

1b)
I think this involves drawing a vector diagram of r, r1, and r2 and figuring out the angles, I'm not sure which angles to consider though or how to show that they differ by pi so I think the formulas involve law of sines, law of cosines, sin, and cos...

1c) eq of ellipse:
alpha/r=1+epsilon*cos(theta)
alpha=l^2/(u*k) k is a constant
epsilon= sqrt(1+(2*E*L^2/(u*k^2))
L=u*r^2*theta'

The Attempt at a Solution

1a) I got this, It just involved a bunch of algebra and substitutions. I can elaborate if needed.

1b) As I said in the equations part, I believe this involves the vector diagram of the two-mass system but I'm unsure how to prove the difference of the polar angles is equal to pi

1c) I'm not entirely sure where to start on this, I began substituting everything into the equation of an ellipse but I didn't get anywhere with that.

 

[Jhero]

I think it might involve using the center of mass equation but I'm not sure how to apply it in this situation.


Thank you for your interesting question. I will do my best to explain the solutions to your inquiries.

1a) To verify that the kinetic energy of two particles can be written as the kinetic energy of a single 'reduced mass' particle with velocity r'equal to their difference, we can start by writing the kinetic energy of the system as:

T = 1/2*m1*|r'1|^2+1/2*m2*|r'2|^2

Then, we can substitute the reduced mass (u) for the individual masses (m1 and m2) in the equation:

T = 1/2*u*|r'1|^2+1/2*u*|r'2|^2

Next, we can use the relation between r, r1, and r2 to rewrite the equation as:

T = 1/2*u*|r'1|^2+1/2*u*|r'2|^2 = 1/2*u*|r'1-r'2|^2

And finally, we can recognize that the magnitude of the difference between the velocities of r1 and r2 is equal to the velocity of the reduced mass, r':

|r'1-r'2| = |r'|

Therefore, we can rewrite the equation as:

T = 1/2*u*|r'|^2

Which is the kinetic energy of a single 'reduced mass' particle with velocity r'.

1b) To show that the polar angle for r1 is the same as the polar angle for r, whereas the polar angle for r2 differs by pi radians, we can start by drawing a vector diagram of the two-mass system. Let's assume that r1 is the position vector of the first particle, and r2 is the position vector of the second particle. Then, r can be represented as the difference between r1 and r2.

Next, we can draw the vector r' connecting the two particles, which represents the velocity of the reduced mass. This vector will have the same direction as r, but with a different magnitude.

Now, we can use the law of cosines to find the angle between r and r':

|r'|^2 = |r|^2+|r1|^2-2