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

[Crush1986]

Homework Statement

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

Homework Equations

[tex] F=\frac{GmM}{r^2} [/tex]

The Attempt at a Solution

I wrote out what I think the differential equation is from the Newtons second law diagram.

[tex] \frac{dr^2}{dt^2}=\frac{GM}{r^2} [/tex]

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

[tex] v\frac{dv}{dr}=\frac{GM}{r^2} [/tex]

I integrate this to get

[tex] \frac{v^2}{2}=\frac{-GM}{r} [/tex]

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!

 

[ehild]

Homework Statement

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

Homework Equations

[tex] F=\frac{GmM}{r^2} [/tex]

The Attempt at a Solution

I wrote out what I think the differential equation is from the Newtons second law diagram.

[tex] \frac{dr^2}{dt^2}=\frac{GM}{r^2} [/tex]

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

[tex] v\frac{dv}{dr}=\frac{GM}{r^2} [/tex]

I integrate this to get

[tex] \frac{v^2}{2}=\frac{-GM}{r} [/tex]

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!

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.

 

[Crush1986]

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. If you get the function v(r) you need one more integral instead of one more derivative.

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.

 

[Crush1986]

So after the integration with limits I get

[tex] \sqrt{\frac{2GM}{R}}*\sqrt{\frac{(R-r)}{r}} = v(r) [/tex]

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.

 

[ehild]

So after the integration with limits I get

[tex] \sqrt{\frac{2GM}{R}}*\sqrt{\frac{(R-r)}{r}} = v(r) [/tex]

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.

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?

 

[Crush1986]

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?

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

[tex] \sqrt{\frac{2GM}{R}}t = \int \sqrt{\frac{R}{R-r}} dr [/tex]

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

[tex] r= R-e^{\frac{-t}{R}\sqrt{\frac{2GM}{R}}} [/tex]

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 :(

 

[ehild]

so can I just treat v(r) as v and just made it dr/dt? /QUOTE]
V(r) is positive, as it is speed but r decreases as the asteroid goes towards he Sun. You need a minus sign.

I just wasn't sure if that was mathematically correct.

I suppose that would yield

[tex] \sqrt{\frac{2GM}{R}}t = \int \sqrt{\frac{R}{R-r}} dr [/tex]

You should multiply with dt instead of t. And use the minus sign.

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

[tex] r= R-e^{\frac{-t}{R}\sqrt{\frac{2GM}{R}}} [/tex]

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 :(

It is totally wrong. Write out the integral correctly. The one with respect r is quite tricky. You can solve it with substitution, but you can cheat a bit, and try wolframalpha, for example.

 

[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.
[tex] \sqrt{\frac{2GM}{R}}*\sqrt{\frac{(R-r)}{r}} = v(r) [/tex]

turn into

[tex] \sqrt{\frac{2GM}{R}}*\sqrt{\frac{(R-r)}{r}} dr = dv [/tex] ?

This might be my problem.

 

[ehild]

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.
[tex] \sqrt{\frac{2GM}{R}}*\sqrt{\frac{(R-r)}{r}} = v(r) [/tex]

turn into

[tex] \sqrt{\frac{2GM}{R}}*\sqrt{\frac{(R-r)}{r}} dr = dv [/tex] ?

This might be my problem.

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

 

[Crush1986]

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

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

[tex] - \sqrt{\frac{2GM}{R}}dt = \sqrt{\frac{r}{R-r}}dr [/tex]

 

[ehild]

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

 

[Crush1986]

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

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.

 

[ehild]

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.

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

 

[Crush1986]

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

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.

 

[Crush1986]

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

When x=R isn't there a problem though? I'm trying to reconcile that. You get an infinity essentially from the square root in the numerator.

 

[ehild]

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

 

[Crush1986]

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

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!

 

[ehild]

You are welcome :)

 

[ehild]

Homework Statement

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.)

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.

 

[Crush1986]

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

[tex] \sqrt{\frac{2GM}{R}} t [/tex]

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.

 

[ehild]

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

 

[Crush1986]

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

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...

 

[ehild]

It should go from R to r_final. Is it equal to the radius of the Earth orbit?
If you want to cheat

Spoiler

http://www.wolframalpha.com/input/?i=int_{30}^1(-sqrt(x/(30-x))dx)

You are not alone https://www.physicsforums.com/threa...roblem-asteroid-on-a-collision-course.836761/

 

[Crush1986]

It should go from R to r_final. Is it equal to the radius of the Earth orbit?
If you want to cheat

Spoiler

http://www.wolframalpha.com/input/?i=int_{30}^1(-sqrt(x/(30-x))dx)

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.

 

[ehild]

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.

That was the original text, was it not?

An asteroid from the Kuiper belt (approximately 30 AU from the Sun)

Do not forget that the result will be in AU.

 

[Crush1986]

That was the original text, was it not?

Do not forget that the result will be in AU.

That was the original text, was it not?

Do not forget that the result will be in AU.

Yah it wants to know time to Earth but I was trying to get a general solution then plug in numbers later.

I've been converting the R to 30*1.5*10^8km.

I'm getting a fall time of about 29 days. Doesn't that seem a tad short?

 

[ehild]

Yah it wants to know time to Earth but I was trying to get a general solution then plug in numbers later.

I've been converting the R to 30*1.5*10^8km.

I'm getting a fall time of about 29 days. Doesn't that seem a tad short?

Convert R to meters.

 

[Crush1986]

Convert R to meters.

Oh yah I did. In the end I was using 4.5*10^12 meters for R. My final equation came to look like this.

[tex] t = \sqrt{\frac{R}{2GM}}*\frac{R}{30}*(\sqrt{29}+15*pi-30*ArcCot(\sqrt{29})) [/tex]

that was basically the equation mathematica gave me when I integrated from R to R/30.

oh jeez... no I'm getting 29 years** I mean.

Doesn't that seem long? I know it's a long distance. But it just seems long. I guess it is starting from rest... with very weak gravitational attraction.

 

[Crush1986]

Oh yah I did. In the end I was using 4.5*10^12 meters for R. My final equation came to look like this.

[tex] t = \sqrt{\frac{R}{2GM}}*\frac{R}{30}*(\sqrt{29}+15*pi-30*ArcCot(\sqrt{29})) [/tex]

that was basically the equation mathematica gave me when I integrated from R to R/30.

oh jeez... no I'm getting 29 years** I mean.

Doesn't that seem long? I know it's a long distance. But it just seems long. I guess it is starting from rest... with very weak gravitational attraction.

yah 29 years* forgot I also divided by 365.25.

 

[ehild]

I think it is realistic. Did not check yet.

 

[Crush1986]

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.

 

[ehild]

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)?

 

[Crush1986]

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.

 

[ehild]

You get v(t) = -dr/dt from conservation of energy. Need integral, but it is easy.

 

[Crush1986]

Hrm, doing it with energy (if I did it right) I get 12.3 years.

I used

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

turned that into

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

I integrate from R to r and 0 to t

I get

[tex] t = 2/3 * \frac{R^{1.5}-r^{1.5}}{\sqrt{2GM}} [/tex]