Particle Motion (Astrophysics)

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A particle p is moving at the speed of light (v1 = c) towards another particle q, which moves slower (v2) and has a velocity dependent on the distance r between them. The equation for v2 is given as v2(r) = v0hr, where h is the Hubble constant and v0 is the initial velocity. The goal is to determine when particle p catches up with q, which occurs when the distance r becomes zero. The discussion highlights the importance of reference frames in measuring time and distance, as the observer and q will perceive these differently due to relativistic effects. The problem was further addressed using the Lambert W function to find a solution.
Gregorski
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Homework Statement


This is new for me, so forgive me my clumsiness. I am working on the following problem:
A particle p is moving with a velocity v1 = c (speed of light) towards an object q, which is moving in the same direction with the speed v2, where v1>v2. Now, v2 is a function of the distance r between p and q. I need to find the time when p "catches up" with q.

Homework Equations


v2(r)= v0hr, where h - the Hubble constant, v0 - initial velocity, and r - distance between p and q

The Attempt at a Solution


Particle p will "catch up" with q when the distance between them is 0, so we have:
r - s =0 and r = s (1)
v2(r)= v0hr
Differential equation:
ds/dt = v0hr

Separating variables:
ds/r= v0h dt
Integrating:
log r = v0ht + c
r = ev0ht+c
r = ecev0ht
ec=R
r=Rev0ht

Plugging into (1)
Rev0ht = v1t

I am not sure how to proceed from here or if any of it makes sense.
 
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Gregorski said:

Homework Statement


This is new for me, so forgive me my clumsiness. I am working on the following problem:
A particle p is moving with a velocity v1 = c (speed of light) towards an object q, which is moving in the same direction with the speed v2, where v1>v2. Now, v2 is a function of the distance r between p and q. I need to find the time when p "catches up" with q.

Homework Equations


v2(r)= v0hr, where h - the Hubble constant, v0 - initial velocity, and r - distance between p and q

The Attempt at a Solution


Particle p will "catch up" with q when the distance between them is 0, so we have:
r - s =0 and r = s (1)
v2(r)= v0hr
Differential equation:
ds/dt = v0hr

Separating variables:
ds/r= v0h dt
Integrating:
log r = v0ht + c
r = ev0ht+c
r = ecev0ht
ec=R
r=Rev0ht

Plugging into (1)
Rev0ht = v1t

I am not sure how to proceed from here or if any of it makes sense.
Welcome to PF Gregorski!

There are two reference frames. The observer frame, relative to which q is moving at speed v2, and q's rest frame. It is not clear from the question as to the reference frame in which r or time is being measured.

According to Special Relativity, p must have 0 rest mass and must be moving at speed c relative to all inertial reference frames. The observer and q would measure the time for p to reach q differently.

AM
 
Andrew,
Thank you for your input; you're absolutely right there are two frames. I managed to do the last step by applying Lambert W function.

Greg
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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