Freefall chord through the Earth -- calculus proof

[achap01]

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

 

[haruspex]

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?

 

[achap01]

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

 

[haruspex]

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).

 

[achap01]

Please post your work so far (as required by the forum rules).

 

[haruspex]

Your second equation is a SUVAT equation. They assume constant acceleration.

 

[achap01]

Your second equation is a SUVAT equation. They assume constant acceleration.

Is there a proof of this problem somewhere online that I can reference?

 

[haruspex]

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?

 

[achap01]

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

 

[haruspex]

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?

 

[achap01]

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.

 

[Orodruin]

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.

 

[haruspex]

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.

 

[achap01]

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?

 

[haruspex]

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?

 

[achap01]

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

 

[haruspex]

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.)

 

[achap01]

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

 

[achap01]

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

 

[haruspex]

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?

 

[achap01]

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.

 

[haruspex]

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.

 

[achap01]

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

 

[haruspex]

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?

 

[achap01]

arccos(x/r)=theta
cos(theta)=x/r

 

[haruspex]

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.

 

[achap01]

cos(arccos(x/r))
= x/r

If this is right im gonna feel so braindead

 

[achap01]

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

 

[achap01]

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

 

[haruspex]

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.

 

[Orodruin]

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.

 

[achap01]

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

 

[haruspex]

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.

 

[achap01]

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?

 

[haruspex]

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".

 

[achap01]

so the ODE would be ẍ=a=-gx/R? Im a little confused

 

[haruspex]

so the ODE would be ẍ=a=-gx/R? Im a little confused

Yes, or to put it in its more standard form, $\ddot x+\frac gRx=0$.
Do you recognise that form?

 

[achap01]

Yes, or to put it in its more standard form, $\ddot x+\frac gRx=0$.
Do you recognise that form?

harmonic motion, so angular frequency (rate of oscillation) would be g/R, which have a derivative of 0 with respect to theta?

 

[haruspex]

harmonic motion, so angular frequency (rate of oscillation) would be g/R, which have a derivative of 0 with respect to theta?

Bingo.

 

[kuruman]

Not quite. The angular frequency is $\omega =\sqrt{g/R}.$

 

[achap01]

Bingo.

is there an online or a video proof of this that i can get

 

[kuruman]

What exactly do you want to prove?