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Here's a puzzle to tickle your neurons...
Assume for the moment that the composition of the earth is perfectly uniform and you can tunnel into it completely to the center.
As you descend into this tunnel, does your weight
A. Increase
B. Decrease
C. Remain the same
What is your weight when you reach the center of the earth?
Explain your reasoning.
This is just an educated guess. I would say your weight would be lower as you move closer to the center. Your weight would be approximately zero at the center. Your mass is always the same. Gravity is the phenomenom where mass attracts mass. On the surface the earths mass pulls on your mass. In the center you have equal mass all around you. If you went down a perfectly straight tunnel, there would be slightly more gravity below you than above you. probably not enough to measure. I would expect the gravity pulling from all sides would blow you up like a balloon.
Well Marv if you are influenced by Newton and a few before him the answer is easy. Then you must believe that the force between 2 masses is G(m1 x m2)/r^2
As we approach the centre the product of the masses remains unchanged but r tends to zero, so our weight would tend to infinity. Of course as r gets very close to zero then weird quantum stuff will mess with the answer. I think that you are dismissing quantum stuff in the question so the simple answer is that our weight increases as we dig deeper.
Another way to look at it is to regard the earth as lots of small particles, then you can sum the forces between the digger and each particle. Then you just need to see how the sum of each 1/r^2 varies as you move through the sphere of particles, a simple integration shows that the maximum value of that sum occurs at the centre.
I do not know the proper equation, but I believe tonyfoale is using gravity equations of spherical objects with a separation between the objects. I still hold my theory.
Less mass as you get closer to the center. The Earth's mass at it's central point counteracts with it's self because the Earth is almost spherical. I would guess your weight would be very close to zero.
Ronbo is correct in the result, except that the mass would not have changed.
To put the experiment into perspective, let’s continue tunneling until we break through on the other side. We now have a hole from two antipodal points, going right through the centre of the earth. We know that if we stand above one end of this tunnel (it's a hole in the ground) and drop a rock, it will fall into the hole, and accelerate towards the center of the earth, initially, at some 10m/s each second. Of course, the same would happen if an antipodal resident were to drop a rock from the other end - since it's just a deep hole in the ground over there as well. So what happens if one drops a rock from one end of this tunnel? It will accelerate towards the centre of the earth, but at a decreasing rate, and eventually fly through the tunnel's middle of the earth position, and keep going, and going, but slower and slower until...until it reaches the other end of the tunnel/hole in the ground over there (we're neglecting drag through air in the tunnel, as well as interference with the tunnel). If no one grabs it, the rock will fall again into the hole/tunnel but in the opposite direction. And so on, back and forth, with ever diminishing amplitude, until the oscillations become so small that it can be considered to have stopped...at the centre of the earth, and weightless. Same with the original conditions that Mark put forth, with the difference that the rock would have smashed into the bottom of the tunnel with enough energy that would have enabled it to reach the surface at the other end.
A good Physics 11 question. I would also ask for time to hit bottom, and rock's velocity at that time.
I must learn not to answer things like this quickly at 02:00.
I woke this morning thinking to myself "you fool". I previously quoted the standard expression that the force between 2 bodies is G(m1 x m2)/r^2 which is quite correct but I ignored that "r" also defines the sphere of mass to use. If m1 is the tunneler's mass and m2 is the earth's mass then as m2 is much greater than m1 we can say that m2 = m1+m2 to keep it simple, then we see that the force is proportional to m2/r^2 but the correct m2 to use is the mass within the sphere of radius "r", so m2 is proportional to r^3, resulting in the force being proportional to r^3/r^2 = r. In otherwords the force between the two bodies decreases linearly as we tunnel down.
The net gravitational attraction inside a thin spherical shell is zero. It's easy to see that this is true at the center of the sphere; proving that it's true at any other point in the interior takes a bit of math but is not impossible.
As we tunnel into the earth, the material above us can be considered to be a collection of many thin spherical shells. None of these shells exerts a net attraction on us so their effect can be dismissed. So the only net force we experience is from the mass below us.
A bit of math manipulation with the volume of this mass, proportional to r^3, and our distance from its center, r, will show that net gravity decreases linearly with r as we descend, reaching a value of zero when r = 0 at the center.
Now that we know that gravity varies linearly inside the earth we can address another interesting phenomenon, the gravity subway.
If we bore a hole completely through the earth, a test mass dropped into the hole will experience a spring force, i.e., a force proportional to distance. That means that it will oscillate up and down in that hole from one side of earth to the other. A bit of math...
Hole Through the Earth Example
will show that the period of this oscillation is about 84 minutes. Expressed differently, with such a hole you could reach your antipode on the opposite side of the planet in 42 minutes.
Now, what is truly astounding is that the period of that oscillation is the same for any hole connecting two points on earth - a diametrical hole is not necessary. By building a subway tunnel, you could travel from Los Angeles to Boston in only 42 minutes.
If you would like to read about this concept in more detail, the Wikipedia article is interesting...
https://en.wikipedia.org/wiki/Gravity_train
Ok, there are a few minor problems associated with building a network of gravity subways but this theoretical exercise is helpful in understanding how elementary physics works.
Marv would it not still be 84 minutes for the period of oscillation from LA to Boston since the to points would still share the integer of a center point that being the center of the earth. draw any line place ABC along said line B being in the exact center A&C at either end bend this line in any direction using B as the pivotal then A to B and B to C are equal therefore if said figure were 84 minutes for full oscillation half oscillation at 42 minutes the time factor would never change no matter which direction from center the second half of the oscillation occurred.
Or did I miss something in my assumption which I usually do
Direct integration of the acceleration, being proportional to the distance from the centre of the earth, from the surface to the centre yields time to centre 1266 secs. (Which agrees with Marv's numbers from the spring analogy calcs.) and the velocity as it passes the centre of close to 8000 m/s. For comparison the speed of sound in air at sea level is 332 m/s. Of course the time and velocity calculations assume an evacuated tunnel and so the speed of sound has no meaning there. it is just an indication that it is very fast, 24 times the speed of sound that we experience but only 0.000027 times the speed of light. Click on the pix. for full size versions of the plots of velocity and acceleration.Quote:
Originally Posted by volodar
Attachment 19261
This plot shows the variation in velocity with distance from the centre.
Attachment 19262
This shows the variation in velocity with time.
Attachment 19263
This shows the variation of acceleration with time.
