Ridiculously difficult Newtonian physics problem

[Mason Pulinthanathu]

Homework Statement

Derive n expression for the time for an object to move in space from point a to b as a result of the gravitational force of another object

Relevant Equations

MmG/r^2=F | -MmG/r=PE| PE1+KE1=PE2+KE2

I have been trying out with this one for a long time with no luck getting the exact expression. I am pretty sure this is somehow related to Kepler's third law but unable to establish the connection and the integrals I get from energy conservation yield different results. Any ideas?

 

[PeroK]

Homework Statement:: Derive n expression for the time for an object to move in space from point a to b as a result of the gravitational force of another object
Relevant Equations:: MmG/r^2=F | -MmG/r=PE| PE1+KE1=PE2+KE2

View attachment 257518I have been trying out with this one for a long time with no luck getting the exact expression. I am pretty sure this is somehow related to Kepler's third law but unable to establish the connection and the integrals I get from energy conservation yield different results. Any ideas?

This is quite a well known problem. What do you know about differential equations?

Conservation of energy should give you a differential equation to get you started.

 

[rude man]

You're right, this is a ridiculously difficult problem.
It's a non-linear ODE whether you approach it from force, or from energy as post 2 suggests.
Wolfram Alpha gives you an answer for the force way. Ugly!
Then you have r(t) but you want t(r).
That's the best I can think of.
My sympathies!

 

[haruspex]

Since you are given the solution and are only asked to show it is correct, "all" you need to do is to demonstrate that it satisfies the differential equation and boundary conditions.

 

[Mason Pulinthanathu]

Thanks all for the responses. I think this the right path to derive it, but I am unable to solve this integral.

 

[Mason Pulinthanathu]

This is quite a well known problem. What do you know about differential equations?

Conservation of energy should give you a differential equation to get you started.

Thank you for the response. I have a solid understanding of Calc 1 and can solve basic differential equations. Can you help me please with this integral?

 

[Mason Pulinthanathu]

You're right, this is a ridiculously difficult problem.
It's a non-linear ODE whether you approach it from force, or from energy as post 2 suggests.
Wolfram Alpha gives you an answer for the force way. Ugly!
Then you have r(t) but you want t(r).
That's the best I can think of.
My sympathies!

I would be really thankful if you could send me the Wolfram Alpha solution. Thanks in advance.

 

[haruspex]

I would be really thankful if you could send me the Wolfram Alpha solution. Thanks in advance.

You 'loved' my post #4 but did not act on it.
There's a fairly obvious trig substitution that makes the RHS of the expression for t much simpler. Differentiating that leads in a few lines to the energy equation. You can then reverse the process to see how the ODE can be solved.

 

[PeroK]

Thank you for the response. I have a solid understanding of Calc 1 and can solve basic differential equations. Can you help me please with this integral? View attachment 257578

You have:
$$\frac{dr}{dt} = -\sqrt{2GM(\frac 1 r - \frac 1 {r_1}) }= -\sqrt{2GM(\frac{r_1 - r}{rr_1})}$$
$$-\int_{r_1}^{r_2} \sqrt{\frac{rr_1}{r_1 - r}} dr =\int_0^{t_2}\sqrt{2GM}dt = t_2\sqrt{2GM}$$
Note that I've used $r_2$ as the final radius to avoid confusion with the variable $r$.

The key to such integrals is to look for a suitable trig substitution. You might look first at $r = r_1\cos \theta$, which works for an inverse cube law problem. That doesn't quite work here. Any inspiration?

 

[kuruman]

You have:
$$\frac{dr}{dt} = \sqrt{2GM(\frac 1 r - \frac 1 {r_1}) }= \sqrt{2GM(\frac{r_1 - r}{rr_1})}$$

Shouldn't there be a negative sign in front of the radical? The radial distance $r$ is positive and $\dfrac{dr}{dt}$ must be negative since $r$ decreases with time. This means choosing the negative sign when taking the square root.

 

[PeroK]

Shouldn't there be a negative sign in front of the radical? The radial distance $r$ is positive and $\dfrac{dr}{dt}$ must be negative since $r$ decreases with time. This means choosing the negative sign when taking the square root.

Very true. I was trying to adjust my solution to the OP's notation and forgot about that.

 

[Mason Pulinthanathu]

Very true. I was trying to adjust my solution to the OP's notation and forgot about that.

Thanks for the reply. Here what I've done. (Have no clue how to un-substitute or what to do after this!)

 

[PeroK]

That doesn't look quite right. For example, $1 - \frac{r_1}{r}$ is negative, but you have it in a square root.

You are making things much more difficult than they need be.

I showed you how to simplify that expression with the square roots on the denominator. Also, you have used:
$$\frac 1 r = \frac 1 {r_1} \sec^2 \phi$$
Which is more simply written as:
$$r = r_1 \cos^2 \phi$$

I would start with the simplified expression in the integral I gave you:

$$-\int_{r_1}^{r_2} \sqrt{\frac{rr_1}{r_1 - r}} dr =\int_0^{t_2}\sqrt{2GM}dt = t_2\sqrt{2GM}$$

And try the substitution $r = r_1 \cos^2 \theta$. Or use $\phi$ is you prefer.