Modeling an Asteroid's Trajectory Towards the Sun Using Differential Equations

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SUMMARY

The forum discussion centers on modeling an asteroid's trajectory towards the Sun using differential equations, specifically starting from the Kuiper belt. The participants derive the differential equation using Newton's second law, resulting in the equation dr²/dt² = GM/r². They explore integration techniques to find the velocity function v(r) and subsequently the time function t(r). The discussion highlights the importance of correctly applying limits during integration and emphasizes the need for careful consideration of signs in gravitational equations.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with differential equations
  • Knowledge of gravitational force equations, specifically F = GMm/r²
  • Basic integration techniques and limits in calculus
NEXT STEPS
  • Study the derivation of gravitational potential energy and its application in orbital mechanics
  • Learn about the conservation of energy in gravitational fields
  • Explore advanced integration techniques for solving differential equations
  • Investigate numerical methods for simulating celestial mechanics
USEFUL FOR

Students and educators in physics, particularly those focusing on celestial mechanics, astrophysics, and differential equations. This discussion is also beneficial for anyone interested in the mathematical modeling of astronomical phenomena.

  • #31
OK thanks! Well if all else fails I'm pretty sure I have the other 4 parts of the question right.

The more I think about it the more reasonable I guess it could be since it is starting from a stop from such a long ways away. It's initial acceleration is soooo tiny. I guess it would hang around the Kuiper belt for a very long time before it gained any appreciable speed.
 
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  • #32
I got even more, but now I am sleepy. Try the approximation of E=0, and getting v from conservation of energy. It is much easier.
By the way, did you set your calculator to RAD when calculating the arctan(1/29)?
 
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  • #33
I got using both methods that the speed of the asteroid as it passes Earth is about 1300 km/s. One way I got 1311 and another 1333 km/s.

I don't know how you get time from there though. Since the acceleration isn't constant I obviously can't use a kinematic equation or anything.
 
  • #34
You get v(t) = -dr/dt from conservation of energy. Need integral, but it is easy.
 
  • #35
Hrm, doing it with energy (if I did it right) I get 12.3 years.

I used

1/2 mv^2 = \frac{GMm}{r}

turned that into

-dr/dt = \sqrt{\frac{2GM}{r}}

I integrate from R to r and 0 to t

I get

t = 2/3 * \frac{R^{1.5}-r^{1.5}}{\sqrt{2GM}}
 
  • #36
It looks good. What did you get for t?
 
  • #37
I get 12.3 years from that. I had 29 years the other way. But the other way was much more mistake prone.
 
  • #38
Crush1986 said:
I get 12.3 years from that. I had 29 years the other way. But the other way was much more mistake prone.
Yes, it was easy to make mistakes, but I got the same results as you at the end. The initial conditions were different. The first method assumed zero initial speed, negative total energy. The second one assumed zero total energy, but that meant nonzero initial speed. As you noticed, the asteroid gains speed very slowly at the beginning. That can cause the difference between the times.
 
  • #39
ehild said:
Yes, it was easy to make mistakes, but I got the same results as you at the end. The initial conditions were different. The first method assumed zero initial speed, negative total energy. The second one assumed zero total energy, but that meant nonzero initial speed. As you noticed, the asteroid gains speed very slowly at the beginning. That can cause the difference between the times.
I see. If we made crude approximations that the P.E. at this distance is zero, and assumed the asteroid had zero speed at it's beginning. Method 2 would be ok, right?
 
  • #40
Crush1986 said:
I see. If we made crude approximations that the P.E. at this distance is zero, and assumed the asteroid had zero speed at it's beginning. Method 2 would be ok, right?
Both methods have sense. The complicated one assumed that the asteroid had zero speed, so its energy was negative. The simple method assumed zero total energy, which meant it got some initial speed towards the Sun.
The spaceship New Horizons arrived to Pluto in about 9 years, in a backward track. http://pluto.jhuapl.edu/Mission/The-Path-to-Pluto/Mission-Timeline.php. So times of a few decades have sense.
 
  • #41
I see thanks!

This problem definitely showed me a lot of things I have to learn. I was completely taken by surprise by the negative differentials. I guess it makes sense though, if the force is attractive it's negative, so the acceleration has to be negative. As a consequence dr/dt is also negative with how the problem is describing what the asteroid is doing.
 
  • #42
The velocity and acceleration are first and second time derivatives of the position vector. Putting the origin into the sun, ##\vec v = \frac{d \vec r }{dt}##. But we worked with the scalar r, distance between Sun and asteroid. It decreased with time. That is why we used the negative sign: v, the speed was v=-dr/dt.
 
  • #43
Hrm, ok. I think that is how I was thinking about it. I just said it horribly. I think I understand all that went on in here, haha. I'm definitely going to be mulling it over at work tonight. Thanks! Hopefully all this time will greatly increase my understanding.
 

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