<p>haha…and the above crazy math posting is one of the many reasons I WANT to go to MIT</p>
<p>though…if we were discussing science instead of math…I would actually be able to participate…maybe</p>
<p>All righty, then. </p>
<p>My friend’s friend was asked on her interview to explain what would happen if she was to fall through a hole bored through the center of the Earth all the way to the opposide side of the world.</p>
<p>Assuming that there is no air resistance and that she would not eventually be burned to a crisp, of course!</p>
<p>(By the way, interviewers arn’t going to be displeased or anything if you politely reply that you haven’t the slightest inkling of a correct (albeit theoretical) answer.)</p>
<p>She would fall at a decreasing rate of gravitational acceleration proportional to the distance between her and the center of the earth. She would be going really, really fast by the time she reaches the center, at which point in time she would have no gravity acting on her for an instant. Then she would continue at a high speed, through the earth’s center and toward the other side, losing speed at an increasing rate proportional to the distance between her and the center of the earth. By the time she reaches the other side, she will come to a rest, and hopefully she will hang on or someone will grab her, or else she will fall back and forth through the center of the earth forever. And oh yeah, she’ll be upside down when she comes out the other end.</p>
<p>MIT interview question of the day: Show that the union of two proper subgroups of a group cannot be the entire group. </p>
<p>One of my friends was asked this question in his interview. He actually solved it on his first try and the interviewer was impressed…but he still didn’t get in.</p>
<p>My MIT interview is going to be at a library. I guess well just be whispering to each other X-D</p>
<p>tripNip - Her acceleration due to gravity would increase as she approached the center of the earth. The force she experiences from gravity is G(m-earth)(m-her)/r^2. Thus, her acceleration due to gravity would be G(m-earth)/r^2.</p>
<p>sonofsam,</p>
<p>There is no net gravitational force inside a uniform shell. (I forgot to mention, I was assuming that the Earth is a sphere of uniform density, which it isn’t in actuality).</p>
<p>So as she comes closer to the center of the earth, the shell around her (portion of the earth that does not affect her gravitationally) thickens, so the “M-earth” used in your equation decreases. Since mass is proportional to volume (through density) and volume is proportional to the cube of the radius, the acceleration is directly proportional to the radius, when inside the Earth.</p>
<p>a = GM/r^2
a = (G(density)(4/3 pi)(r^3)) / r^2
a = (G)(density)(4/3)(pi)(r)
a = kr, where k is some constant</p>
<p>The cumulative force would increase as she approaches the center of the earth, but the net force would decrease since the directional correlation would weaken as the earth’s mass becomes spread out around you.</p>
<p>Assume either that she “phased” through the earth, or that the hole she fell through was just barely large enough for her (either condition establishes we can consider the earth a sphere, and we can with relative saftety assume it is of uniform radial mass distribution). Then she would neither accelerate nor decelerate once within the earth. </p>
<p>Hints as to why this is so:
- Gravity ~1/r^2, area ~r^2
(Major spoiler: 2. remember the proof for a point charge inside a charged sphere not experiencing any electrical force?)</p>
<p>sran,
- gravity works in 3 dimensions, not just 2.
volume ~r^3 - read my post #47 above</p>
<p>I did. We did this exact problem in 8.012 the other day.</p>
<p>You are correct, gravity is a force in 3-space; that is not germane to this argument. We are concerned with the mass “generator” of gravity and the distance which governs the magnitude of the interaction, which do end up going as r^2. Let me know if you want a sketch of the full solution as opposed to hints.</p>
<p>I think she will reach her highest speed when she passes the center, at the same time the net force on her will equal to 0. As she went pass the center of the earth, the acceleration on her will become negative resulting a decrease of speed. please correct me if i am wrong.</p>
<p>I agree with youknowme.</p>
<p>sran,
I still don’t know what you’re getting at. Let’s see your sketch of the whole solution then.</p>
<p>Yay, <em>shake hands</em></p>
<p>OK, this is a sort of strange picture to draw. It’s possible that Kleppner & Kolenkow’s textbook has a solid picture. Otherwise, my description will have to do.</p>
<p>LEMMA: The net gravitational force on a particle inside a solid spherical shell with uniform density is 0
PROOF: The argument centers around “canceling out” radially opposed gravitational forces. So, consider such a particle within the spherical shell. Consider the “cone” which has apex at the particle and has its base at the shell itself (the shape is not quite a cone because of the curvature of the sphere, but it gives an idea of what I’m trying to show and does not affect the math). Consider the “cone” radially opposite to this cone projected through the point; this cone is “similar” to the other cone. Divide the surface of the sphere marked off by each cone into an equal number of equally sized pieces (differential analysis style). Consider pieces radially opposite to each other. The area of piece 1 is (r1/r2)^2<em>the radially opposite piece 2 (r1, r2 distances of each of the pieces to the point), and thus the mass of piece 1 is (r1/r2)^2</em>the m2, the mass of piece 2 (uniform mass distribution). Now, consider:
Fpiece 2, point = G m2 mpoint/r2^2
Fpiece 1, point = G (r1/r2)^2 m2 mpoint/(r1^2) = G m2 mpoint/r2^2</p>
<p>So the two forces are equal, and they cancel out; so do all the other pieces cancel out with their counterparts. So the net force on the particle is zero. This concludes our lemma.</p>
<p>The main proof follows readily from this. Recall, we assume uniform radial mass distribution of the earth. So the earth can be considered composed of nested shells with uniform mass distribution. Each one contributes zero net gravitational force, so taken together they will also not contribute net gravitational force.</p>
<p>Very nice. So, from your excellent proof, we can conclude that there is no net gravitational force in the exact center of the Earth.</p>
<p>But what about a point halfway between the Earth’s center and it’s surface, at say, d = .5r-earth? Wouldn’t this point be inside half of the nested uniform shells of the earth, but outside those between itself and the earth’s center? In other words, if I am halfway to the Earth’s center, wouldn’t a sphere of a radius equal to the distance remaining between me and the center still exert a gravitational force, since I would not be inside these nested shells? It would be as if I were standing on the surface of an Earth with half the radius.</p>
<p>I think sran’s proof also claims to show there is no net gravitational force epsilon below the surface of the earth, where epsilon > 0 is an arbitrarily small number, which is clearly wrong.</p>
<p>“LEMMA: The net gravitational force on a particle inside a solid spherical shell with uniform density is 0” From that i can conclude everything under the earth surface will experience a gravitational force of 0. So when i go down to my basement today, i ll be weightless?</p>
<p>Hahaha.</p>
<p>Actually, a solid spherical shell is different from a solid sphere. I agree that the net gravitational force on a particle inside a solid spherical shell amounts to 0, but the net gravitational force on a particle inside a solid sphere does not necessarily equal 0.</p>
<p>If the entirety of the Earth’s mass were on its surface, i.e. if the Earth were hollow, then sran’s assertion would be true.</p>
<p>
So, they teach incorrect physics at MIT? Maybe I shouldn’t apply then…</p>
<p>Wait, sorry. My lemma remains correct, but my statement regarding the cumulative effect is incorrect; tripNpip’s argument holds. </p>
<p>And yes, apparently they do goof in teaching physics occasionally, at least when taking diversions; the ‘proof’ I gave was based on what my TA told me outside of class when talking through different implications. I didn’t realize the flaw till I thought it through here.</p>