Newtonian Gravity- Body 'falling' into the sun integration problems

In summary, the problem of finding the time for a body to fall into the sun can be solved using Kepler's laws of planetary motion, without the need for integration.
  • #1
Romeo
13
0
The problem is this:

Given sun of mass M and a body of mass m (M>>m) a distance r from the sun, find the time for the body to 'fall' into the sun (initially ignoring the radius of the sun).


Our first equation is therefore [tex] \ddot{r} = \frac {GM}{r^2} [/tex].

I am able to integrate this, giving:
[tex] \dot{r} = - {\sqrt{2GM}}{\sqrt{1/r - 1/R}} [/tex],

where R is the inital distance of the body from the sun. However, I am unable to integrate this again. I have shoved it into wolfram's integrator for an indicator of what to aim for, but cannot come close.

Any thoughts would be greatly appreciated.

Regards

Romeo
 
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  • #2
With G and M constant:

[tex] \int {\frac{GM}{r^2}dr = GM\int{\frac{1}{r^2}}dr[/tex]

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

Use the power rule for integration to find this integral. There are no square roots involved.
 
  • #3
Appreciated Woozum, but you mis-read my post. Maybe i should clarify further:

[tex] \dot{r} = \frac {dr}{dt}, [/tex]

and that i solved the first integral of this second order differential equation, but cannot, in particular, solve this:

[tex] \int \frac {1}{\sqrt{1/r - 1/R}} dr = -\sqrt{2GM}\int dt, [/tex]

reminding that 'R' is the initial distance of the body from the sun and that 'r' is variable distance that you integrate to.

So, any ideas anyone? :smile: :confused:

Regards

Romeo
 
  • #4
Romeo said:
The problem is this:

Given sun of mass M and a body of mass m (M>>m) a distance r from the sun, find the time for the body to 'fall' into the sun (initially ignoring the radius of the sun).


Our first equation is therefore [tex] \ddot{r} = \frac {GM}{r^2} [/tex].

I am able to integrate this, giving:
[tex] \dot{r} = - {\sqrt{2GM}}{\sqrt{1/r - 1/R}} [/tex],

where R is the inital distance of the body from the sun. However, I am unable to integrate this again. I have shoved it into wolfram's integrator for an indicator of what to aim for, but cannot come close.

Any thoughts would be greatly appreciated.

Regards

Romeo
whozum's right, but, more importantly, how is that going to get your time?

Your first equation is your acceleration. To get your velocity, your integration should be:

[tex]\dot{r}=\int{a}\, dt[/tex]

rdot is your velocity, which you can integrate wrt time to get your position.

That get's you an initial value that you can solve for time (just in case you didn't know the equation, already).
 
  • #5
BobG said:
whozum's right, but, more importantly, how is that going to get your time?

Your first equation is your acceleration. To get your velocity, your integration should be:

[tex]\dot{r}=\int{a}\, dt[/tex]

rdot is your velocity, which you can integrate wrt time to get your position.

That get's you an initial value that you can solve for time (just in case you didn't know the equation, already).

BobG, I understand what you're intending to say, but haven't quite got what I'm trying to say:

I have the second order differential equation [tex] \ddot{r} = -GM /r^2 [/tex] to solve, aiming to find an expression for 'r', which will then simply be be used to find an expression for the time it takes for a body to 'fall' into the sun from a general distance.

I neglected to show the work involved to solve the latter, but I am pretty certain that the expression for [tex] \dot{r} [/tex] is correct.

With that in mind, any ideas folks? :smile: :confused:

Regards

Romeo
 
  • #6
Honestly don't know what your trying to do, but

How does [itex] \dot{r} = - {\sqrt{2GM}}{\sqrt{1/r - 1/R}} [/tex] stem from [itex] \ddot{r} = \frac {GM}{r^2} [/tex]

For constant acceleration,
[tex] x(t) = x_0 + v_0t + \frac{gt^2}{2} \mbox{ where } g = \frac{GM}{r^2} \mbox{ so then } [/tex]

[tex] x(t) = \frac{GMt^2}{2r^2}[/tex] with [tex] x_0 = v_0 t = 0 [/tex]

and [tex] t = \sqrt{\frac{2xr^2}{GM}} [/tex]
 
  • #7
Dog Maddit ! :mad:Do not double post !You got the same problem solved by Arildno in the Diff.Eq.forum.

Daniel.
 
  • #8
whozum said:
Honestly don't know what your trying to do, but

How does [itex] \dot{r} = - {\sqrt{2GM}}{\sqrt{1/r - 1/R}} [/tex] stem from [itex] \ddot{r} = \frac {GM}{r^2} [/tex]

For constant acceleration,
[tex] x(t) = x_0 + v_0t + \frac{gt^2}{2} \mbox{ where } g = \frac{GM}{r^2} \mbox{ so then } [/tex]

[tex] x(t) = \frac{GMt^2}{2r^2}[/tex] with [tex] x_0 = v_0 t = 0 [/tex]

and [tex] t = \sqrt{\frac{2xr^2}{GM}} [/tex]

Thank you anyway Whozum. I think maybe I should have stated that there will of course not be constant acceleration, hence your approach is invalid. However, there is now a solution in Diff Equ/ns, so check out what I was trying to do!

Regards

Romeo
 
  • #9
A word of thanks could work wonders, you know.
Not a wholesale devotional of course, but an appreciative nod might be in order..
 
  • #10
Thank you all, apologies for the double post, all is accounted for in the Diff Equ forum. And again, thank you Arildno for the solution.

Regards

Romeo
 
  • #11
The problem can also be solved without performing any integration by applying Kepler's laws of planetary motion.

From Kepler's first law the orbit of any mass around the sun is an ellipse with the sun at one of the two foci. Let's call the semimajor axis of this ellipse 'a' and the semiminor axis of the elipse 'b'.

An object with zero initial velocity falling into the sun is the limiting case of this orbit as the minor axis of the ellipse (b) goes to 0. The distance f from the centre of the ellipse to the foci is f = (a^2 - b^2) ^(1/2). Obviously as b -> 0, f ->a. So the initial distance R from the object to the sun is R = 2a in this limit.

Kepler's third law gives the orbital period T in terms of the semimajor axis length.

T^2 = a^3 (4*Pi^2)/MG

Bearing in mind that on object falling into the Sun only completes half of its 'orbit' and using the relationship between a and R above you can solve the problem very simply.
 

Related to Newtonian Gravity- Body 'falling' into the sun integration problems

1. What is Newtonian Gravity?

Newtonian Gravity is a theory proposed by Sir Isaac Newton in the 17th century to explain the force of gravity between objects. It states that any two objects in the universe attract each other with a force that is directly proportional to their masses and inversely proportional to the square of the distance between them.

2. How does Newtonian Gravity explain a body "falling" into the sun?

According to Newtonian Gravity, the force of gravity between the body and the sun increases as the distance between them decreases. As the body gets closer to the sun, the force of gravity becomes stronger and the body falls towards the sun due to this force.

3. What are integration problems in relation to Newtonian Gravity?

Integration problems in Newtonian Gravity refer to the mathematical process of integrating the equations of motion to predict the trajectory of a body as it falls towards the sun. This involves solving differential equations and can be complex for some situations.

4. Can Newtonian Gravity accurately predict the trajectory of a body falling into the sun?

In most cases, Newtonian Gravity can accurately predict the trajectory of a body falling into the sun. However, for extremely massive or high-speed objects, the theory may not be accurate and Einstein's theory of general relativity may need to be used.

5. Are there any limitations to Newtonian Gravity in regards to bodies falling into the sun?

Yes, Newtonian Gravity has some limitations when it comes to predicting the trajectory of a body falling into the sun. It does not take into account the effects of relativity or the presence of other massive objects, which can affect the trajectory. Additionally, it does not explain the behavior of objects at very small scales, such as at the atomic or subatomic level.

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