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Could someone tell me what the final velocity of an object will be if it fell from the Earth's surface to its center (assuming no air resistance). Since it involves a constant changing acceleration due to the object's distance to the Earth's center decreasing, I'm sure it involves calculus (which I don't know how to do).

I wrote a program that recalculates the acceleration at every 0.3m, and it gave me a final velocity of 9.6 x 10^7 m/s. Is this close to the exact answer?

Qmechanic
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Inertial Ignorance
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  • As you'll see from my answer, I think your answer is wrong. If you want help diagnosing the problem, we'll need to know how you were calculating the acceleration. It is proportional to the distance from true centre of the Earth, assuming the Earth to be a homogeneous sphere. – Philip Wood Apr 18 '17 at 17:45
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    This page might help. – StephenG - Help Ukraine Apr 18 '17 at 17:46
  • +Philip Wood I made a computer program to do my calculations. Essentially, at every metre of the object's fall, I recalculated the acceleration value using: g = GM / r^2. Then for that 1m distance interval, I calculated the object's final velocity at the end of the 1m distance using vf^2 = vi^2 + 2ad. The vf from this was then the vi I used in the next 1m distance interval. I kept doing this until the object's distance from the center of Earth was zero. – Inertial Ignorance Apr 18 '17 at 17:55
  • Check @StephenG's link. Your formula for acceleration isn't right, it should be $a = g \frac{r}{R}$ – nnovich-OK Apr 18 '17 at 18:12
  • I assume the mine shaft going down is frictionless, so you can ignore coriolis force. You know what $g$ is at the surface radius $R$. When $r = R/2$, $1/r^2$ is 4 times as strong, but the mass pulling is $M/8$ (because all the outer mass cancels out). So the force of gravity is linear in $r$, just like a spring, so that says it is simple harmonic motion. No? – Mike Dunlavey Apr 18 '17 at 19:03

2 Answers2

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Your procedure works, just be aware of the systematic error (the smaller the steps, the more accurate the result, of course). This is a version of numerical modelling, a well-recognized prediction method.

But a mistake shows in your method from the comments.

  • The formula $$g=G\frac{M}{r^2}$$ only holds true outside Earth. The Earth mass is constant, but it's gravitational pull "spreads out" over a quadratically larger area (a spherical surface area is $4\pi r^2$, so e.g. doubling distance quadruples the area, which the same mass must pull in). In more correct terms, the gravitional field, which is what $g$ represents, is "spread more out" and becomes quadratically weaker with distance.

  • Inside Earth on the other hand some mass cancels out. It is not the same mass pulling in you all the time, which the above formula assumed. In the centre, gravity is zero, since there is the same amount of mass in all directions pulling equally. If you are on the way down towards the centre, it turns out that all mass "above" you (the spherical shell above you) cancels out on itself and leaves all mass "below" you (the sphere below you) as if it was alone - as if it was a smaller planet attracting you.

The same formula can therefore be used inside Earth, if the entire Earth mass is replaced by the mass "below":

$$g=G\frac{M_{below}}{r^2}=G\frac{V_{below}\varrho}{r^2}=G\frac{\frac43\pi r^3\varrho}{r^2}=G\frac43 \varrho \pi r$$

Here mass is split into density and volume $M=V\varrho$ and sphere-volume is $V=\frac43 \pi r^3$. This does though assume constant density throughout the depth of Earth, which isn't entirely the case. A more realistic gravity variation is shown in the graph in an answer to this question. But for your project, the assumption might be enough. (The average density can be looked up on Google.)

Steeven
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    Great, thank you for the very detailed answer... it makes much more sense now. You're right, I had incorrectly assumed the g = GM / r^2 formula applied inside the Earth as well. – Inertial Ignorance Apr 18 '17 at 19:49
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I make it $7900\ \text{m}\ \text{s}^{-1}.$ This is using Simple Harmonic Motion theory.

Philip Wood
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  • Guess I'm off by a huge amount then... Is there a reason basic kinematic calculations wouldn't work for this type of problem? – Inertial Ignorance Apr 18 '17 at 17:44
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    The motion would not be simple harmonic. Please let us know what you did. I'm going to guess that this answer is wrong. – garyp Apr 18 '17 at 18:27
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    @garyp Value itself is correct, I got the same one by finding potential energy of body on the surface (integrate gravity work on small paths $dr$) and getting kinetic energy of body in the center out of it. However, answer without any explanation confuses a lot with no way easy to check calculations. – nnovich-OK Apr 18 '17 at 18:51
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    @garyp Actually for a uniform-density sphere, the $\sim 1/r^2$ gravitational force conspires with the $\sim r^3$ enclosed mass to make the force vary like $F\propto r$, so the motion in that simple model is simple harmonic. The other answer does this correctly. – rob Apr 18 '17 at 20:57
  • I'm a bit puzzled by the negative votes for my short answer (above). Inertial Ignorance asked whether or not his answer was right; I gave a very different answer and a hint as to how I'd reached it. What did I do wrong? Should I have given the complete solution then and there for the Earth as a homogeneous sphere? – Philip Wood Apr 19 '17 at 12:02