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Somewhat silly gravity question for Mechanics

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Homework Statement



This class has given me a number of problems that seem like they'd be ludicrously easy until I start them.

A commuting scheme involves boring a straight tunnel through the crust of the Earth between New York City and San Francisco. If you were to drop a ball down into the tunnel in NYC then how long does it take for the ball to pop up in SF? Assume that there is no friction or air resistance in the tunnel. Assume that the earth is round and has constant density, and that at the surface of the Earth g = 10m/s^2. Finally assume that the triangle connecting NYC, SF and the center of the Earth is an equilateral triangle (60 degrees for each angle). Assume that the radius of the Earth is 6,400 km. Hint: Ignore the Earth's rotation and centrifugal force aects in this problem!

Homework Equations



[tex]\vec{F} = m\vec{a}; \vec{F_g} = -G\frac{mM}{r^2}\hat{r}; \vec{g} = \vec{F}/m [/tex]

The Attempt at a Solution



As best as I can figure, a picture of this would look like so:

http://img180.imageshack.us/img180/648/physprob.png [Broken]

In the picture, R is the radius of the earth, and [tex]\theta[/tex] is the angle formed by the intersection of the radius that touches NYC and the ray from the center of the Earth to the position of the ball. The way I've considered approaching it is to say that the acceleration due to gravity will depend exclusively on the angle [tex]\theta[/tex], which itself depends on time. So

[tex]\ddot{x} = g\cos\theta[/tex]

However, since [tex]\theta[/tex] depends on time in a way that I cannot seem to define, I can't figure out how to integrate this. Am I on the right track? Thanks for your help! :biggrin:
 
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Answers and Replies

  • #2
Pengwuino
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This question makes no sense, you can't drop a ball down such a tunnel because it would be attracted towards the center and simply fall to the ground where you dropped it.
 
  • #3
gabbagabbahey
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This question makes no sense, you can't drop a ball down such a tunnel because it would be attracted towards the center and simply fall to the ground where you dropped it.
The idea is that there is no friction in the tunnel, so as long as there is a non-zero component of the force directed along the length of the tunnel, there will be acceleration along the length.
 
  • #4
gabbagabbahey
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In the picture, R is the radius of the earth, and [tex]\theta[/tex] is the angle formed by the intersection of the radius that touches NYC and the ray from the center of the Earth to the position of the ball. The way I've considered approaching it is to say that the acceleration due to gravity will depend exclusively on the angle [tex]\theta[/tex], which itself depends on time. So

[tex]\ddot{x} = g\cos\theta[/tex]

However, since [tex]\theta[/tex] depends on time in a way that I cannot seem to define, I can't figure out how to integrate this. Am I on the right track? Thanks for your help! :biggrin:
Two things:

(1) The gravitational field inside a uniformly dense sphere (a reasonable model for the earth) is not constant... use Gauss' Law to find what it really is.

(2) You haven't actually labeled what [itex]x[/itex] is, but assuming it is supposed to represent the ball's distance from NYC, along the tunnel, at any given time; it will be related to [itex]\theta[/itex] through some simple trig.
 
  • #5
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Two things:

(1) The gravitational field inside a uniformly dense sphere (a reasonable model for the earth) is not constant... use Gauss' Law to find what it really is.

(2) You haven't actually labeled what [itex]x[/itex] is, but assuming it is supposed to represent the ball's distance from NYC, along the tunnel, at any given time; it will be related to [itex]\theta[/itex] through some simple trig.
1.) Would the force be

[tex]F_g = -\frac{4}{3}\pi G\rho m r[/tex]

? (where r is the distance from the center)

2.) Sorry, yes, that is what it was meant to be. It's related to [tex]\theta[/tex] by [tex]x = R\sin\theta[/tex]. How this relates to time is beyond me.
 
  • #6
gabbagabbahey
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1.) Would the force be

[tex]F_g = -\frac{4}{3}\pi G\rho m r[/tex]

? (where r is the distance from the center)
Yes. You can use some trig to express [itex]r[/itex] in terms of [itex]R[/itex] and [itex]\theta[/itex] at any given time.

2.) Sorry, yes, that is what it was meant to be. It's related to [tex]\theta[/tex] by [tex]x = R\sin\theta[/tex]. How this relates to time is beyond me.
First, that isn't true...for arbitrary [itex]\theta[/itex], the triangle you've drawn in need not be a right-angled triangle...you need to use the law of sines to express [itex]x[/itex] in terms of [itex]\theta[/itex].

Second, you also will have a differential equation in the form [tex]\ddot{x}=\frac{4}{3}\pi G\rho r\cos\theta[/tex]...you can replace [itex]r[/itex] and [itex]\cos\theta[/itex] by an expression involving only [itex]x[/itex], giving you a differential equation involving only [itex]x[/itex]. (Or, you can replace [tex]\ddot{x}[/itex] with an expression involving [itex]\theta[/itex] and its time derivatives)
 
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  • #7
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Oh right. Man, I don't know where my head is tonight.

For the differential equation you gave,

[tex]\ddot{x} = \frac{4}{3}\rho r\cos\theta[/tex]

is it as easy as just using ye olde polar substitution [tex]x = r\cos\theta[/tex] and having the somewhat basic differential equation [tex]\ddot{x} = \frac{4}{3}\rho x[/tex]? While that make sense mathematically, it doesn't make sense trigonometrically (since for the given [tex]\theta[/tex], [tex]r\cos\theta[/tex] corresponds to the line connecting the center of the earth to the line joining NYC and SF.

Thanks again!
 
  • #8
gabbagabbahey
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is it as easy as just using ye olde polar substitution [tex]x = r\cos\theta[/tex]
Not the way that you've defined [itex]r[/itex], [itex]\theta[/itex] and [itex]x[/itex], no.

Also, my DE didn't display correctly, I've edited my previous post to give the correct output. You should be able to easily derive it yourself using the fact that [itex]m\ddot{x}=\textbf{F}\cdot\hat{\mathbf{x}}=||\textbf{F}||\cos\theta[/tex]

Anyways, label [itex]x[/itex] and [itex]r[/itex] on your diagram and get rid of that erroneous right-triangle symbol. Use the Law of sines to express [itex]r[/itex] in terms of [itex]R[/itex] and [itex]\theta[/itex], and do the same thing for [itex]x[/itex]...what do you get?
 
  • #9
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Alright, I fixed the picture. Using the law of sines to get [tex]r[/tex] and [tex]x[/tex] in terms of [tex]R[/tex] and [tex]\theta[/tex], I get [tex]\sqrt{3}/2r = \sin\theta/x = (\sqrt{3}\sin\theta/2 - \cos\theta/2)/R[/tex]. In other words, [tex]r = \frac{\sqrt{3}R/2}{\sqrt{3}\sin\theta/2 - \cos\theta/2}; x = \frac{R\sin\theta}{\sqrt{3}\sin\theta/2 - \cos\theta/2}[/tex]. Now using the chain rule, [tex]\frac{dx}{dt} = \frac{dx}{d\theta}\frac{d\theta}{dt}; \frac{d^2x}{dt^2} = \frac{dx}{d\theta}\frac{d^2\theta}{dt^2} + \frac{d^2x}{d\theta^2}\frac{d\theta}{dt}[/tex], but that gives a very rough looking differential equation (though I suppose at that point, a numerical solution might be in order)
 
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  • #10
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You can do it by computing the velocity on each point of the tunnel using conservation of energy.

First calculate the potential energy at a distance r from the center of the earth, with a potential energy of 0 at the surface. this is:

[tex] \int_r^R {- k m r} [/tex]

with
[tex] k = \frac {4}{3} \pi G \rho [/tex]

This gets you a potential energy E(r)

Now calculate r with pythagoras as a function of the distance travelled in the tunnel x

[tex] r^2 = \frac {3}{4} R^2 + (\frac {R}{2} - x)^2 [/tex]

Substitute in the formula for E(r) to get a formula for E(x)

now use conservation of energy to get the speed v(x)

The time spent on a small distance dx is

[tex] \frac {dx} {v} [/tex]

so the total time spent is

[tex] \int_0^R { \frac {dx} {v} } [/tex]

This will become difficult integral, but it is solvable, if not by hand than by Wolfram,
and you get a very complicated function t = f(x)
Inverting it symbolically seems impossible, but you don't need that to get the travel time for a given x.
 
  • #11
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Thank you for the reply! That seems like a clever solution - I will start working it out right now and let you know where it goes. Where did you get that potential energy function?

Edit: After working it out, I get the final integral to be [tex]\int_0^R \frac{dx}{\sqrt{x^2 - Rx}}[/tex], which will diverge.
 
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  • #12
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Thank you for the reply! That seems like a clever solution - I will start working it out right now and let you know where it goes. Where did you get that potential energy function?

Edit: After working it out, I get the final integral to be [tex]\int_0^R (x^2 - Rx)[/tex], which will diverge.
That integral doesn't diverge, the integrand remains finite. did you mean

[tex]\int_0^R \frac {1}{x^2 - Rx}dx [/tex]

that one does diverge, but you missed a square root somewhere.
 
  • #13
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Oh, haha, I forgot the most important part of that because of the shorthand I was using. The expression for t becomes

[tex]t = \frac{1}{\sqrt{k}}\int_0^R\frac{1}{\sqrt{x^2-Rx}}\ dx[/tex]
 
  • #14
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With regards to gabbagabbahey's method, the differential equation that the problem produces is one that neither I nor MATLAB seem to really like.
 
  • #15
gabbagabbahey
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In other words, [tex]r = \frac{\sqrt{3}R/2}{\sqrt{3}\sin\theta/2 - \cos\theta/2}; x = \frac{R\sin\theta}{\sqrt{3}\sin\theta/2 - \cos\theta/2}[/tex].
I think both of these are a little off; [itex]\sin\left(\frac{2\pi}{3}-\theta\right)=\frac{\sqrt{3}}{2}\cos\theta+\frac{1}{2}\sin\theta[/itex].

After a little algebra, you should (happily) find that [itex]x=2(R-r\cos\theta)[/itex] and so you can easily replace [itex]r\cos\theta[/itex] in the ODE and get something that looks an awful lot like a harmonic oscillator.
 
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  • #16
gabbagabbahey
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Oh, haha, I forgot the most important part of that because of the shorthand I was using. The expression for t becomes

[tex]t = \frac{1}{\sqrt{k}}\int_0^R\frac{1}{\sqrt{x^2-Rx}}\ dx[/tex]
Shouldn't this be

[tex]t = \frac{1}{\sqrt{k}}\int_0^R\frac{1}{\sqrt{Rx-x^2}}\ dx[/tex]

instead?:wink:
 

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