SH particle goes through obliquely drilled hole in Earth

  • #1

Homework Statement



(a) A particle is dropped into a hole drilled straight through the center of Earth.
Neglecting rotational eects, show that the particle's motion is simple harmonic if
you assume Earth has uniform density. Find the period of the oscillation in terms
of Earth mass M and radius R (and G). Find its numerical value in minutes.
SOLVED BY MYSELF
This is needed to describe part (b), which is the problem I am stuck on.

Problem of Concern
(b) Consider the same problem for the case when the hole is drilled obliquely
through Earth (not going through the center). Show that the particle's motion is
simple harmonic with the same period as in (a). Ignore eects of rotation, friction,
etc. Express period through g and REarth.


Homework Equations




[tex] F = m_{p} \ddot{r} = -G \frac{m_{p} M}{r^2}[/tex]

[tex] T = 2 \pi \sqrt{\frac{R_{E}^{3}}{G M_E}} = 84 min[/tex]

The Attempt at a Solution



To solve the first problem was not that difficult. All I did was use the first of the relevant equations where [tex] m_{p} [/tex] will cancel out. I then say that the Mass from 0 to r will be [tex] M = \frac{4}{3} \pi r^3 \rho [/tex]. I ignore the R-r mass assuming that it can be neglected. This results in the wonderful ODE:

[tex] \ddot{r} + (\frac{4}{3} G \pi \rho) r = 0 [/tex]

with

[tex] \omega^2 = \frac{4}{3} G \pi \rho [/tex]

and since [tex] \omega = \frac{2 \pi}{T} [/tex]

Plugging everything in and saying that [tex] \rho = \frac{M_E}{\frac{4}{3} \pi R_{E}^3} [/tex]

we get the second relevant equation.

Now, the second question, part (b), says that this particle is not going to go through the center, that it will have an oblique path.

So, I though that there will be then a minimal r which will be the closest it can be to the center and a maximum r of [tex] R_E [/tex].

But I really don't know where to go with this. Any help would be greatly appreciated especially before 1:00 p.m. Tuesday, November 9th PDT GMT -0800.
 

Answers and Replies

  • #2
fzero
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The gravitational force is no longer parallel to the particle's path, so you will need to find the correct component that acts parallel to the displacement. You can then use the geometrical relationship to write r in terms of the displacement.
 
  • #3
I don't understand what you mean fzero. I mean I understand what you mean by the particle's motion is not parallel to the gravitational force. But would this work as an approach:

Consider the particle moves in a straight line not going through the center. It will have a minimal distance to the center when it's path is perpendicular at a certain point. This distance is r_{min}. Considering this certain point to be the origin of the particle's path, at any time, t, it will be at a point x(t). Therefore we can use Pythagorean Theorem:

r^2 = x^2 + r_{min}^2

Taking the second derivative is easy, but the problem is that if we input this into the first of the relevant equations, it will not simplify to the SH equation:

\ddot{r} + \omega^2 r = 0

Is this the correct approach??? Hints?

I would have used the LaTeX tools on this forum, but for some reason it is acting weird today.
 
  • #4
fzero
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Consider the particle moves in a straight line not going through the center. It will have a minimal distance to the center when it's path is perpendicular at a certain point. This distance is r_{min}. Considering this certain point to be the origin of the particle's path, at any time, t, it will be at a point x(t). Therefore we can use Pythagorean Theorem:

r^2 = x^2 + r_{min}^2

Taking the second derivative is easy, but the problem is that if we input this into the first of the relevant equations, it will not simplify to the SH equation:

\ddot{r} + \omega^2 r = 0

Is this the correct approach??? Hints?

What you want is the equation of motion for the coordinate x, not the equation for r. However you should think of x as the component of the position vector for the particle.
The gravitational force vector points in the direction r, but only the component in the x-direction appears in the equation of motion for x. It's going to be useful to use some trig, not just the Pythagorean equation above.
 
  • #5
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alimerzairan:

I would recommend asking this question tomorrow in discussion so that the rest of the class can go through this problem together. I imagine they are having as much difficulty as you.

Best
 
  • #6
@ dr_sasha: Welcome to Physics Forums professor. I will raise this question in discussion. It seems to be more of a trig problem but looking at the homework assignment, it should be fairly simple since it is only 5 pts. Perhaps I am missing something very crucial. It was not that hard to solve the first part of the problem.
 
  • #7
What you want is the equation of motion for the coordinate x, not the equation for r. However you should think of x as the component of the position vector for the particle.
The gravitational force vector points in the direction r, but only the component in the x-direction appears in the equation of motion for x. It's going to be useful to use some trig, not just the Pythagorean equation above.

fzero: Just thinking about this a bit more. Since we know that the gravitational force is acting radially, then we can break it down into the components:

(sorry for the LaTeX syntax. The (tex) functions in HTML are not working for me for some reason)

F_r^2 = F_x^2 + F_y^2 where I will consider F_y^2 to be the F_rmin^2 with r_min being the closest approach to the center of the Earth along the path of the particle.

And I know from trig that r^2 = x^2 + r_min^2.

So I can say that:

m \ddot{x} = F_x = \sqrt{F_r^2 - F_rmin^2}

This will result to:

\ddot{x} = G \sqrt{M_r^2 / r^4 - M_rmin^2/rmin^4}

This equation includes two new variables, M_r & M_rmin. This would be the masses of the Earth at positions r and rmin.

Saying that M_r = 4/3 \pi r^3 \rho and M_rmin = 4/3 \pi rmin^3 \rho with \rho being the spherical density of the Earth, we get this lovely result:

\ddot{x} = 4/3 \pi \rho G \sqrt{r^2 - rmin^2}

But we know that r^2 = x^2 + rmin^2, so:

\ddot{x} = 4/3 \pi \rho G x

This will result in that wonderful SH equation I have been looking for.

Is my logic correct???

The main assumption I am making is taking into account that at any position along the path of the particle, there will be a certain value for the mass of the earth. Basically, it is like part (a).
 
  • #8
fzero
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Sure that works. I would have noted that [tex]x=r\sin\theta[/tex] and [tex]F_x = F\sin\theta[/tex], where [tex]\theta[/tex] is the inclination of the particle, but your computation is equivalent.
 
  • #9
I was thinking of that route as well. But would you not have to also know how \theta will change as the particle follows its path? In your equation, it seems that both F will change due to r and \theta will change as well.

Also, should it not be cosine instead of sine? I mean in part (a) we had \theta = 0 because the particle went right through the center.
 
  • #10
I mean the way I did it just assumed r and x would change which will also is related to the angle you mentioned. I guess they are equivalent, just different way of writing it.
 
  • #11
fzero
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I was thinking of that route as well. But would you not have to also know how \theta will change as the particle follows its path? In your equation, it seems that both F will change due to r and \theta will change as well.

You actually don't since you compute [tex]\ddot{x}[/tex] and you only use [tex]\sin\theta[/tex] to convert [tex]r\sin\theta =x[/tex].

Also, should it not be cosine instead of sine? I mean in part (a) we had \theta = 0 because the particle went right through the center.

If you choose the angle opposite to my [tex]\theta[/tex], you will have cosine instead. If it were easier to post a figure, I would have been clearer. Either way the computation works the same way.
 
  • #12
True. But we should end up to the same result. I will try doing it your way as well. But if we know that F_x = F sin \theta and x = r sin \theta

Then

m \ddot{x} = F_x = F sin \theta.

with F = -G M m_p /r^2

and M = 4/3 \pi r^3 \rho

we get:

m \ddot{x} = - G 4/3 \pi \rho r sin \theta
and
r = x / sin\theta

with this plugged in, we get the same result, except with an extra negative sign???
 
  • #13
Oh, sorry, figured it out. The F in F_x = F sin\theta is the magnitude. Meaning:

||F|| = G M m/r^2.

Got it.
 
  • #14
Thank you fzero for the help & dr_sasha I will still present this question in discussion since you requested, even though this problem is now officially solved for me.

It is not that difficult. I think the main leap in this problem is conceptually figuring out what is going on and translating that concept into the necessary mathematics.

Thank you again fzero for the help.
 
  • #15
2
0
Good job
 
  • #16
Thanks professor. See you tomorrow.
 

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