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ActaPhysica
TL;DR Summary
The usual approach for solving for the time it takes to fall to the center of the earth neglects air resistance & uses Hooke's Law. But if you solve it "the hard way"...
The usual approach for solving for the time it takes to fall to the center of the earth neglects air resistance & uses Hooke's Law. But if you solve it "the hard way"...
Gme*m/r^2 = ma
me = 4/3 rho pi r^3
G * 4/3 rho pi r = a
separate variables, integrate twice
r*Ln(r) - r = 2/3 G rho pi t^2
The lower limit of integration (0) has Ln(0) getting large and negative much more slowly than r going to 0, so there's no problem here.
But, this answer gives a very large time. What am I doing wrong?

ActaPhysica said:
G * 4/3 rho pi r = a
This is your ”Hooke’s law”. You are not integrating correctly.

It's not my Hooke's Law it's Hooke's Law :)
Can you point out my mistake?

ActaPhysica said:
It's not my Hooke's Law it's Hooke's Law :)
Can you point out my mistake?
The point is: It is not Hooke’s law. Hooke’s law relates to the elastic force in a spring or material. It is not simply any linear force.

I already pointed out your mistake: You did not integrate correctly. If you want more detail than that you need to provide your actual computation because your current one is only saying ”integration gives” and then a wrong statement.

Well if you integrate 1/r that gives Ln(r)
And if you integrate Ln(r) it gives rLn(r) - r
The other side is pretty straightforward.

ActaPhysica said:
Well if you integrate 1/r that gives Ln(r)
And if you integrate Ln(r) it gives rLn(r) - r
The other side is pretty straightforward.
But that is the wrong thing to integrate. So again please show your actual work step by step. If you stubbornly refuse to provide this you are making it impossible to help you.

G * 4/3 rho pi r = a = d^2r / dt^2
4/3*G*rho*pi dt^2 = 1/r d^2r
4/3*G*rho*pi *t dt = Ln (r) dr
4/3*G*rho*pi * t^2/2 = r*Ln(r) - r
Is this detailed enough?

Last edited:
ActaPhysica said:
TL;DR Summary: The usual approach for solving for the time it takes to fall to the center of the earth neglects air resistance & uses Hooke's Law. But if you solve it "the hard way"...

The usual approach for solving for the time it takes to fall to the center of the earth neglects air resistance & uses Hooke's Law. But if you solve it "the hard way"...
Gme*m/r^2 = ma
me = 4/3 rho pi r^3
G * 4/3 rho pi r = a
separate variables, integrate twice
r*Ln(r) - r = 2/3 G rho pi t^2
The lower limit of integration (0) has Ln(0) getting large and negative much more slowly than r going to 0, so there's no problem here.
But, this answer gives a very large time. What am I doing wrong?
What is your definition for the variable, me ?,

ActaPhysica said:
What am I doing wrong?
1. Try to use LaTeX to make your formulas more readable. For example, your expression for the interior-mass ##m_e## of the earth between its center and the radius ##r(t)## can be written:$$m_{e}(t)=\frac{4\pi}{3}\rho\,r^{3}(t)\tag{1}$$where ##\rho## is the density of the earth (assumed constant).
2. You wrote the following differential-equation based on Newton's Second Law:$$\frac{4\pi}{3}Gmr(t)=m\ddot{r}(t)\tag{2}$$Think about the sign of the force here. Is gravity repulsive or attractive?
3. You can't solve a second-order differential equation like (2) by performing two ordinary integrations over each variable ##r## and ##t## separately. Instead, find its solution the same way you'd solve the simple-harmonic-oscillator equation:$$\ddot{x}(t)+\omega^{2}x(t)=0\tag{3}$$

Vanadium 50, PeroK, Orodruin and 1 other person
ActaPhysica said:
G * 4/3 rho pi r = a = d^2r / dt^2
4/3*G*rho*pi dt^2 = 1/r d^2r
4/3*G*rho*pi *t dt = Ln (r) dr
4/3*G*rho*pi * t^2/2 = r*Ln(r) - r
Is this detailed enough?
You can't seperate variables this way (with second derivatives).

ActaPhysica said:
G * 4/3 rho pi r = a = d^2r / dt^2
4/3*G*rho*pi dt^2 = 1/r d^2r
4/3*G*rho*pi *t dt = Ln (r) dr
4/3*G*rho*pi * t^2/2 = r*Ln(r) - r
Is this detailed enough?
That’s simply not how separation of variables work with second derivatives. Your separation of variables is incorrect, leading to an incorrect result.

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