1-dimensional problem in Newtonian gravity- HELP

In summary, the problem is finding the time for a body to fall from a certain distance, given the mass of the sun and the body's initial distance from the sun. The equation is integrable, but difficult to solve for the time. However, using a change of variables and Kepler's law, you can get an approximate answer.
  • #1
Romeo
13
0
1-dimensional problem in Newtonian gravity- HELP!

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] \frac {d^2r}{dt^2} = \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
You can think of this as a zero-width elliptical orbit, so you can use Kepler's law relating the orbital period to the length semi-major axis, and then take 1/4 of the orbital period of any orbit with the same semi-major axis.

If you insist on carrying out the integration, my textbook also suggests the substition

r = a(1-e*cos(psi))

in this case I think the eccentricity, e would be 1, so you'd have

r = a(1 - cos(psi))

I haven't tried this out, though.
 
  • #3
Thank you pervect, it's an approach I hadn't considered.

Regards

Romeo
 
  • #4
Aah, the triple post.
This has already been solved in Diff-eq.
 
  • #5
Here is my try at it. Can you check for errors?
A rock is dropped from [tex] r_0 [/tex] and falls straight down to R.
How long does it take. Where R is Earth radius and [tex]r_0 >> R [/tex].

The gravity force equation [tex] F= \frac{GMm}{r^2} [/tex] ,

integrates to a specific energy equation [tex] W= -\frac{u}{r}[/tex].

The velocity at some intermediate r is given by:

[tex] V=\sqrt{V_0^2+\frac{2u}{r} -\frac{2u}{r_0}}[/tex] .

This is a hard differential equation to solve for r(t). I believe I have a form which resembles your integral.

Change of symbols:

[tex] A=2u,\ \ \ \ B = -\frac{2u}{r_0},\ \ \ \ V_0=0[/tex]

[tex] V(r) = \sqrt{\frac{A}{r}+B}= \frac{dr}{dt}[/tex]

Now the part which I have never trusted.

[tex] \frac{dt}{dr}=\frac{1}{\sqrt{\frac{A}{r}+B}}[/tex]

Then [tex] \int_R^r dt= \int_\frac{A}{R}^\frac{A}{r} \frac{1}{\sqrt{\frac{A}{r}+B}}\ \ dr [/tex]

So [tex] T = \int_\frac{A}{R}^\frac{A}{r} \frac{1}{\sqrt{\frac{A}{r}+B}}\ \ dr [/tex]

A change of variables [tex] \frac{A}{r}=x,\ \ r=\frac{A}{x},\ \ dr=\frac{-A}{x^2}dx [/tex]

[tex]
T = \int_\frac{A}{R}^\frac{A}{r} \frac{\frac{-A}{x^2}}{\sqrt{x+B}}\ \ dx = -A \left[ \frac{-\sqrt{B+x}}{Bx} -\frac{1}{2B} \left( \frac{2}{\sqrt{-B}} \arctan{ \sqrt{\frac{B+x}{-B}}} \right) \right]_\frac{A}{R}^\frac{A}{r} [/tex]

[tex] = \left[ \frac{-r_0 \sqrt{x-\frac{2u}{r_0}}}{x} -\frac{2r_0}{\sqrt{\frac{2u}{r_0}}} \arctan{\sqrt{\frac{x-\frac{2u}{r_0}}{\frac{2u}{r_0}}}} \right]_\frac{2u}{R}^\frac{2u}{r_0} =
[/tex]

[tex] = \frac{-r_0 \sqrt{\frac{2u}{r_0}-\frac{2u}{r_0}}}{\frac{2u}{r_0}} -\sqrt{\frac{2}{u}} r_0^\frac{3}{2} \arctan{\sqrt{\frac{\frac{2u}{r_0}-\frac{2u}{r_0}}{\frac{2u}{r_0}}}} \
+\ \frac{r_0 \sqrt{\frac{2u}{R}-\frac{2u}{r_0}}}{\frac{2u}{R}} +\sqrt{\frac{2}{u}} r_0^\frac{3}{2} \arctan{\sqrt{\frac{\frac{2u}{R}-\frac{2u}{r_0}}{\frac{2u}{r_0}}}}
[/tex]

[tex] = \frac{Rr_0}{\sqrt{2u}} \sqrt{\frac{1}{R}-\frac{1}{r_0}}+\sqrt{\frac{2}{u}} r_0^\frac{3}{2} \arctan{\sqrt{\frac{r_0}{R}-1}}= \sqrt{\frac{Rr_0}{2u}} \sqrt{r_0-R} +\sqrt{\frac{2}{u}} r_0^\frac{3}{2} \arctan{\sqrt{\frac{r_0}{R}-1}} [/tex]
This of course means you have to work the problem backwards usually as an iteration, but I would use a binary search which I found converges rapidly for the Kepler problem. It should be ok here too.
 
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Related to 1-dimensional problem in Newtonian gravity- HELP

1. What is a 1-dimensional problem in Newtonian gravity?

A 1-dimensional problem in Newtonian gravity refers to a scenario where objects are only moving along a single axis, such as up or down. This simplifies the equations and calculations involved in understanding the gravitational force acting on the objects.

2. What are the key principles of Newtonian gravity?

The key principles of Newtonian gravity include the law of universal gravitation, which states that all objects with mass are attracted to each other, and the three laws of motion, which describe how objects move under the influence of gravity.

3. How is a 1-dimensional problem in Newtonian gravity different from a 2-dimensional or 3-dimensional problem?

In a 1-dimensional problem, objects are only moving along a single axis, while in a 2-dimensional or 3-dimensional problem, objects can move in multiple directions. This means that the calculations and equations involved in understanding the gravitational force will be more complex for higher dimensions.

4. How do you solve a 1-dimensional problem in Newtonian gravity?

To solve a 1-dimensional problem in Newtonian gravity, you need to use the equations and principles of Newtonian gravity, such as the law of universal gravitation and the three laws of motion. You will also need to apply the appropriate mathematical methods, such as vector addition, to calculate the gravitational force acting on the objects.

5. What are some real-world examples of 1-dimensional problems in Newtonian gravity?

Some real-world examples of 1-dimensional problems in Newtonian gravity include a ball being thrown straight up into the air and a satellite orbiting the Earth in a circular path. In both cases, the objects are only moving along a single axis, making them 1-dimensional problems in Newtonian gravity.

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