# Homework Help: Asteroid Falling into the Sun

1. Oct 9, 2015

### Crush1986

1. The problem statement, all variables and given/known data
An asteroid is knocked out of the Kuiper belt and starts to fall toward the sun. (Assume it's initial potential energy and kinetic energy is 0.)

a. Write down the differential equation for r(t) giving the position of the asteroid as a function of time.
b. Solve the differential equation for t(r), the time as a function of position

2. Relevant equations
$$F=\frac{GmM}{r^2}$$

3. The attempt at a solution
I wrote out what I think the differential equation is from the Newtons second law diagram.

$$\frac{dr^2}{dt^2}=\frac{GM}{r^2}$$

I think I can rewrite to second derivative of r with respect to time as velocity multiplied by the derivative of velocity with respect to r. Making my equation

$$v\frac{dv}{dr}=\frac{GM}{r^2}$$

I integrate this to get

$$\frac{v^2}{2}=\frac{-GM}{r}$$

I stopped here because I'm not sure if maybe I should do this integration with some limits? Like from R (my starting point) to r (current position). What should I make the limits for velocity? 0 to v?

More importantly, I'm very very new to differential equations. Is all this even remotely correct? I imagine that if I'm correct up to this point I can just solve for v and take one more derivative and I think I'll be very close to finished.

Any help is greatly appreciated!!!

2. Oct 9, 2015

### ehild

Check your signs. What is the direction of the gravitation force? Positive acceleration would increase the distance of the asteroid. Is your highlighted equation true?
The limits are all right, and you need to use them. If you get the function v(r) you need one more integral instead of one more derivative.

3. Oct 9, 2015

### Crush1986

Oh darn. I had it negative to start then thought, "Nooo, I'm not using the vector form." I should have thought that out better. Yes, of course, another integral, oops.

4. Oct 9, 2015

### Crush1986

So after the integration with limits I get

$$\sqrt{\frac{2GM}{R}}*\sqrt{\frac{(R-r)}{r}} = v(r)$$

now I'm a bit confused as to what to do to the right side. will it be dv/dr? Then it can be written 1/dt? That just doesn't make sense. I'm missing something. Just not sure what.

Last edited: Oct 9, 2015
5. Oct 9, 2015

### ehild

The right side is a function v(r). v is the speed, the magnitude of the velocity. How is it related to the time derivative of r, the distance from the Sun?

6. Oct 9, 2015

### Crush1986

so can I just treat v(r) as v and just made it dr/dt? I just wasn't sure if that was mathematically correct.

I suppose that would yield

$$\sqrt{\frac{2GM}{R}}t = \int \sqrt{\frac{R}{R-r}} dr$$

I did the integral and solved for r just to see if it made sense. At this point I had

$$r= R-e^{\frac{-t}{R}\sqrt{\frac{2GM}{R}}}$$

this result doesn't work though. At time = 0 it doesn't work and as time goes to infinity it's even more ridiculous. Again.. I've done something wrong :(

7. Oct 9, 2015

8. Oct 9, 2015

### Crush1986

Oh god I see what I did, it is late I suppose... I completely just.. I don't even know. Ugh. Yah this integral is tough though.
$$\sqrt{\frac{2GM}{R}}*\sqrt{\frac{(R-r)}{r}} = v(r)$$

turn into

$$\sqrt{\frac{2GM}{R}}*\sqrt{\frac{(R-r)}{r}} dr = dv$$ ?

This might be my problem.

9. Oct 9, 2015

### ehild

NO. v=-dr/dt. There should be no dv. Try again.

10. Oct 9, 2015

### Crush1986

omg... I'm so sorry, I am just exhausted I guess. These ridiculous mistakes aren't me usually. Can't go to sleep yet though.

v does equal - dr/dt though? I suppose that is because the asteroid is moving toward the sun?

so it should be it looks like

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

11. Oct 9, 2015

### ehild

It is correct now. I suggest to have some rest :) The integral is not simple.

12. Oct 9, 2015

### Crush1986

Yeah it doesn't look simple. I should take a break for now... I have to stay up late to flip my schedule for my work days anyway, haha. Just, yeah, working on my physics homework just doesn't work that well at this time I guess.

13. Oct 9, 2015

### ehild

If you give up the integral, this may help :) http://www.wolframalpha.com/input/?i=int(-sqrt(x/(R-x))dx)

14. Oct 9, 2015

### Crush1986

Yeah, I had a peep already after having a few ideas that just weren't going to work.

That is the stuff nightmares are made of. Now I definitely can't go to sleep. HAHA. Wow... Amazing someone so innocent turns into that.

15. Oct 9, 2015

### Crush1986

16. Oct 9, 2015

### ehild

Well, even functions which have infinite limit, can have finite integral. You can simplify the result with $\sqrt{R-x}$

17. Oct 9, 2015

### Crush1986

Ok I took the limit and it actually does check out and behaves as it should as x goes to R. That is really cool. Thanks! Hopefully the last parts of this problem won't be a problem tomorrow.

Otherwise I learned a TON from you! Thanks!!

18. Oct 9, 2015

### ehild

You are welcome :)

19. Oct 9, 2015

### ehild

The original problem suggested the approximation that both the initial kinetic energy and potential energy is zero. The speed can be obtained from conservation of energy and with zero energy, the solution becomes quite simple.
It would be interesting to see how close are the times needed to reach the Sun obtained with both methods.

20. Oct 9, 2015

### Crush1986

Right now I'm trying to use the answer to that last integral. Along with the other side of the equation which is

$$\sqrt{\frac{2GM}{R}} t$$

to solve for the time it takes for the asteroid to cross earth's orbit. I'm getting total rubbish though, (negative time). :(

I'm pretty sure I have the right answer for the speed it has when it passes earth orbit, as I did check that against conservation of energy. My equation for velocity from the first integration matches it's answer to within about 1 percent.

21. Oct 9, 2015

### ehild

Did you take the definite integral? Wolframalpha gave the primitive function.

22. Oct 9, 2015

### Crush1986

Ohhhh, wait so the time integral should go from 0 to t and I think the ugly integral should go from R to 0. Or should it go from R to r?

I'm thinking R to r...

23. Oct 9, 2015

### ehild

24. Oct 9, 2015

### Crush1986

Ohhhh I like your idea. A lot. I'm going to try that. R to r was nasty. R to R/30 might be nicer. Going to try it.

25. Oct 9, 2015

### ehild

That was the original text, was it not?
Do not forget that the result will be in AU.