How Long Does It Take for a Comet With Zero Angular Momentum to Reach the Sun?

  • Thread starter Thread starter esradw
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on calculating the time it takes for a comet with zero angular momentum to reach the Sun, starting from a distance rmax. The participants derive that the motion is parabolic due to the energy being zero, leading to the equation t = (π/2√(2GMs))(rmax)^(3/2) for the time taken to reach the Sun. Additionally, the second part of the discussion explores the comet's motion if it were to pass freely through the Sun, suggesting a pendulum-like oscillation around the Sun's center.

PREREQUISITES
  • Understanding of classical mechanics and orbital dynamics.
  • Familiarity with Kepler's laws of planetary motion.
  • Knowledge of gravitational potential energy and kinetic energy equations.
  • Ability to solve ordinary differential equations (ODEs).
NEXT STEPS
  • Study the derivation of Kepler's Third Law and its applications to parabolic orbits.
  • Learn about the Schwarzschild radius and its implications for black holes.
  • Explore the mathematical techniques for solving ordinary differential equations in physics.
  • Investigate the concept of forced motion in one-dimensional systems.
USEFUL FOR

Students of physics, astrophysics enthusiasts, and anyone interested in celestial mechanics and orbital dynamics.

esradw
Messages
26
Reaction score
0
a) The comet released from rest at a distance rmax from the sun ( angular momentum =0 ). How long comet takes to reach the sun ? (take sun`s radius to be zero )
b) Assuming the comet can somehow pass freely through the sun describe its overal motion and find its period ?


I can't really think how to solve this problem , what I know it that since angular mom is zero, c=0 ( C= l^2/gama*reduced mass ,Gama=GMsMc)
so so if c=o c=rmax*(1-e) e=1 therefore Energy E=0 ( accually E=0 doesn't make sense eather since it has a PE = -Gmsmc/rmax ) so that is not a bonded orbit but parabolic orbit, But How can I use these to find the time ? the hint is use the technique t= integral dx/xdot where we find xdot from the Energy ( KE=mxdot/2=E-U(x) ) but I still don't understand how to use this since my understanding ( e=0 so E=0 )

please help, I would appreciate if anyone help. I still don't feel as if I completeley understood this subject but I have to turn the homework tomorrow :((
 
Last edited:
Physics news on Phys.org
Well, since the angular momentum is zero, this is just a one-dimensional forced motion problem. You know the force as a function of distance, so set this equal to m d2r/dt2 and solve the resulting ODE.
 
the answer is different than what I find

thanks for your reply but when I set the equation, the 't' I find is
t= r max^2/2GMs but the answer is t= (pi/2*root(2GMs))*(r max)^3/2 more like Kepler`s IIIth law. Do you have any suggestion for this one and the the b) ? ?

thanks again
 
How did you arrive at your expression for the time?

Also, don't you expect the answer to resemble Kepler - since Kepler's Law relate to elliptical orbits and the particular orbit here is also an ellipse in the limit that the semiminor radius goes to zero (degenerate)?
 
I actually expect that but I am confused a little. I understand that since angular mom=0 c=0 so c=rmax(1-e) e=1 a (semimajor)=rmax/2 and c=rmin(1+e) for c to be equal to zero rmin ( semiminor must go to zero. Right ? but But I still can not find the right answer :(( ( t= (pi/2*root(2GMs))*(r max)^3/2 )
 
The formula 1.1\sqrt{\frac{a^3}{GM}} provides an impressively accurate estimation.

b. The comet would be become trapped in the Sun's Schwartchild radius, as the Sun would be a black hole if it's radius were 0. But since it said "pass freely through", I'm just guessing, but wouldn't it distance itself from the Sun by the height it was dropped, except 180 degrees away, then fall back to the Sun, and continue this indefinately, like a pendulum?
 

Similar threads

Replies
9
Views
3K
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
13
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
1
Views
6K
  • · Replies 19 ·
Replies
19
Views
3K