Freefall chord through the Earth -- calculus proof

  • #1
achap01
21
0
Homework Statement
I've gotten homework relating to proving that at any angle, freefalling through the earth will take the same amount of time.
Relevant Equations
L = 2rsin(θ) (chord length)
I haven't really attempted anything worth showing on the calculus side
 
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  • #2
achap01 said:
Homework Statement: I've gotten homework relating to proving that at any angle, freefalling through the earth will take the same amount of time.
Relevant Equations: L = 2rsin(θ) (chord length)

I haven't really attempted anything worth showing on the calculus side
I am sure you can think of more relevant equations. How about something concerning gravity and (uniform) spheres?
 
  • #3
haruspex said:
I am sure you can think of more relevant equations. How about something concerning gravity and (uniform) spheres?
Ive derived the newtonian motion eqs but i can't seem to understand how to get the derivative of acceleration with respect to anything relevant
 
  • #4
achap01 said:
Ive derived the newtonian motion eqs but i can't seem to understand how to get the derivative of acceleration with respect to anything relevant
Please post your work so far (as required by the forum rules).
 
  • #5
haruspex said:
Please post your work so far (as required by the forum rules).
1730953140133.png
 
  • #6
Your second equation is a SUVAT equation. They assume constant acceleration.
 
  • #7
haruspex said:
Your second equation is a SUVAT equation. They assume constant acceleration.
Is there a proof of this problem somewhere online that I can reference?
 
  • #8
achap01 said:
Is there a proof of this problem somewhere online that I can reference?
It's not that hard, and quite interesting.
When it is at distance x from the midpoint of the shaft, what are the forces on it?
What acceleration will that produce?
 
  • #9
haruspex said:
It's not that hard, and quite interesting.
When it is at distance x from the midpoint of the shaft, what are the forces on it?
What acceleration will that produce?
I just think it'd be easier for me to learn with a video or something I can reference online rather than talking about it with you over forums
 
  • #10
achap01 said:
I just think it'd be easier for me to learn with a video or something I can reference online rather than talking about it with you over forums
But that won't teach you how to think about such problems. It will only show you how to do that particular problem.
What are the forces?
 
  • #11
haruspex said:
But that won't teach you how to think about such problems. It will only show you how to do that particular problem.
What are the forces?
I'm saying it'd be faster and easier for me to figure out the problem, and learn from a process someone else used to think about such problems.
 
  • #12
achap01 said:
I'm saying it'd be faster and easier for me to figure out the problem, and learn from a process someone else used to think about such problems.
Then you’ll only be learning how someone else solved a problem. You will not be learning to solve problems yourself. It is a really bad way to learn physics. You might think it is faster and easier for you, but you will not actually be learning. Learning requires effort and working through the problem yourself. Not looking up the solution.

We are happy to help you learn, not to help you get the illusion of learning.
 
  • #13
achap01 said:
I'm saying it'd be faster and easier for me to figure out the problem, and learn from a process someone else used to think about such problems.
I can tell you the first step they made: they figured out which forces apply. But seeing what they got for that won’t show you how they found them.
 
  • #14
haruspex said:
I can tell you the first step they made: they figured out which forces apply. But seeing what they got for that won’t show you how they found them.
the forces would just be gravity(and normal force perpendicular to the chord) gravity would just be linear so (dist from center)/R * weight(at surface). To get an eq relating gravitational force towards the center with force parallel to the chord would require both time and dist which would have to be related seperately right?
 
  • #15
achap01 said:
To get an eq relating gravitational force towards the center with force parallel to the chord would require both time and dist which would have to be related seperately right?
Don't worry about the time.
If at distance r from the Earth's centre and distance x from the midpoint of the shaft, what is the acceleration along the shaft?
 
  • #16
haruspex said:
Don't worry about the time.
If at distance r from the Earth's centre and distance x from the midpoint of the shaft, what is the acceleration along the shaft?
force of gravity towards the center would be (r/R(radius of earth))*weight. Angle of Fg would be θx/2rsin(θ/2), so force parallel to chord would be w(r/R)*sin(θx/2rsin(θ/2)). Then plug in to f=ma
 
  • #17
achap01 said:
Angle of Fg would be θx/2rsin(θ/2),
That's a very strange formula.
Is that ##\theta## still how you defined it in post #1?
If so, ignore that angle and consider the angle between the distances x and r. Call that ##\phi##. What fraction of ##F_g## acts along x?

(No idea why the LaTeX has stopped working.)
 
  • #18
haruspex said:
That's a very strange formula.
Is that ##\theta## still how you defined it in post #1?
I mistypes in the first post. it should be L = 2rsin(θ/2). I got the angle by finding the fraction of the total chord length x is, and then multiplying that with the central angle
 
  • #19
haruspex said:
If so, ignore that angle and consider the angle between the distances x and r. Call that ##\phi##. What fraction of ##F_g## acts along x?
cos(θ) of Fg
 
  • #20
achap01 said:
I mistypes in the first post. it should be L = 2rsin(θ/2). I got the angle by finding the fraction of the total chord length x is, and then multiplying that with the central angle
As I wrote, forget about the angle the entire chord subtends at the centre. Concentrate on the angle between r (not R) and x. If the gravitational force acts along r, what fraction of it acts along x?
 
  • #21
haruspex said:
As I wrote, forget about the angle the entire chord subtends at the centre. Concentrate on the angle between r (not R) and x. If the gravitational force acts along r, what fraction of it acts along x?
cos(theta) of Fg. Is there any way I could see a reference for this proof bc I feel like this convo is going to take a while with the responses. I can ask you about the reference too, but I feel like I'd learn much better with it.
 
  • #22
achap01 said:
cos(theta) of Fg. Is there any way I could see a reference for this proof bc I feel like this convo is going to take a while with the responses. I can ask you about the reference too
Where theta is what, now? If the angle between x and r, yes.
 
  • #23
haruspex said:
Where theta is what, now?
the only way i see to derive that theta(angles between x and r) is with the angle I mentioned previously, but by subtracting it from 90deg. So theta =90-(θx/2rsin(θ/2)). I don't understand how to get it without the central angle
 
  • #24
achap01 said:
the only way i see to derive that theta is with the angle I mentioned previously, but by subtracting it from 90deg. So theta =90-(θx/2rsin(θ/2)). I don't understand how to get it without the central angle
You have a right angled triangle where r is the hypotenuse and x is the "adjacent " and (this) theta is the angle between them. What equation relates them?
 
  • #25
arccos(x/r)=theta
cos(theta)=x/r
 
  • #26
achap01 said:
arccos(x/r)=theta
cos(theta)=x/r
Right, so express the component of the gravitational force along x in terms of x and r.
 
  • #27
cos(arccos(x/r))
= x/r

If this is right im gonna feel so braindead
 
  • #28
haruspex said:
Right, so express the component of the gravitational force along x in terms of x and r.
is there a proof online bc i think im close now
 
  • #29
haruspex said:
Right, so express the component of the gravitational force along x in terms of x and r.
I go to 9.8/R=a but i dont know what to do now
 
  • #30
achap01 said:
I go to 9.8/R=a but i dont know what to do now
That is dimensionally wrong, so cannot be right.
There should be an x in there. Check your working. If you can’t find the error, post it.
 
  • #31
I would suggest you need to work more methodically rather than in short snippets. You have to answer the following questions:
  1. What is the magnitude of the force on a mass ##m## at a distance ##r## from the center?
  2. What are its components in the different directions?
  3. Can you write down Newton's second law for each component of the motion?
You should note that when you write down 3, you obtain equations of motion for each component that is independent of the other components. This is key. The resulting differential equation for each component should be easily solvable if you are familiar with basic ODEs.
 
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  • #32
haruspex said:
That is dimensionally wrong, so cannot be right.
There should be an x in there. Check your working. If you can’t find the error, post it.
sorry i mean 9.8x/R=a
 
  • #33
achap01 said:
sorry i mean 9.8x/R=a
Ok, nearly there!
So write the differential equation relating ##x## to ##\ddot x##.

But it is better to leave it as "g". If you substitute a numeric value you should include units.
 
  • #34
haruspex said:
Ok, nearly there!
So write the differential equation relating ##x## to ##\ddot x##.

But it is better to leave it as "g". If you substitute a numeric value you should include units.
would ẍ just be the second derivative of x with respect to time, so ẍ=a=9.8x/R?
 
  • #35
achap01 said:
would ẍ just be the second derivative of x with respect to time, so ẍ=a=9.8x/R?
Almost, but watch the signs. If we take x as going from ##x=-L## to ##x=+L## then when ##x<0## the acceleration is positive. Correcting that, do you recognise the form of the ODE?

Please use "g", not "9.8".
 

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