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Planet w/ hole drilled through it...(gravity prob)

by ninjagowoowoo
Tags: drilled, hole, itgravity, planet, prob, w or
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ninjagowoowoo
#1
Apr29-05, 11:33 AM
P: 75
Q:
The residents of a small planet have bored a hole straight through its center as part of a communications system. The hole has been filled with a tube and the air has been pumped out of the tube to virtually eliminate friction. Messages are passed back and forth by dropping packets through the tube. The planet has a density of 3640 kg/m3, and it has a radius R=5.39106 m. What is the speed of the message packet as it passes a point a distance 0.460R from the center of the planet?

I can find the mass of the planet, but other than that I'm just stuck.
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OlderDan
#2
Apr29-05, 11:39 AM
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Quote Quote by ninjagowoowoo
Q:
The residents of a small planet have bored a hole straight through its center as part of a communications system. The hole has been filled with a tube and the air has been pumped out of the tube to virtually eliminate friction. Messages are passed back and forth by dropping packets through the tube. The planet has a density of 3640 kg/m3, and it has a radius R=5.39106 m. What is the speed of the message packet as it passes a point a distance 0.460R from the center of the planet?

I can find the mass of the planet, but other than that I'm just stuck.
The usual way to approach this problem is to find the force on an object inside a hollow shell of uniform mass density. Then you can separate the force acting into the force from "outside" the objects radius and "inside" the objects radius.
ninjagowoowoo
#3
Apr29-05, 11:47 AM
P: 75
But to find the force acting on the packets, wouldnt I need to know the mass of the packets? (F=(G)(Mplanet)(Mpackets)/r^2) ???

OlderDan
#4
Apr29-05, 11:53 AM
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Planet w/ hole drilled through it...(gravity prob)

The mass of the packet will not be a factor in the answer, for the same reason that all objects near the earth fall with the same acceleration. Just represent the mass of the packet by an algebraic representation (m) and you sill see it divide out in the end.
squib
#5
Apr29-05, 12:31 PM
P: 40
I solved algebraicly and ended up with a = GM/r^2, however, I tried using this inserted into d=(1/2)at^2, and solving for t to find the speed, but this didn't work... any ideas?
OlderDan
#6
Apr29-05, 01:01 PM
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Quote Quote by squib
I solved algebraicly and ended up with a = GM/r^2, however, I tried using this inserted into d=(1/2)at^2, and solving for t to find the speed, but this didn't work... any ideas?
The acceleration is not constant in this problem, so you cannot use d=(1/2)at^2. After you figure out the contribution to the force from the exterior shell you will realize that the force you are using is not correct unless M, the mass of the planet, is replaced by a factor that also depends on r.
Theelectricchild
#7
Apr29-05, 02:22 PM
P: 258
Yes. You should treat this as a mechanical oscillator problem--- that is why the acceleration is not constant----

You would do well to find the "effective spring constant" for the situation.

I will give you the hint that the effective spring constant involves the acceleration of free fall, the distance d from the center of the earth, and the radius of the earth.
Theelectricchild
#8
Apr29-05, 02:27 PM
P: 258
Task for you: Newtons law of gravity can be written as a "Gauss" law of gravity:

[tex]\int g dA = (4\pi G)\int{\rho_m dV}[/tex]

where [tex]\rho_m[/tex] is the mass density... you should convince yourself that the acceleration inside the earth a distance d away from the center is:

[tex]4 \pi G \rho_m \frac{d}{3}[/tex]

I should point out that G is the gravitational constant you know and love.
squib
#9
Apr29-05, 03:29 PM
P: 40
K, I figured out the period, and thus the answer to the second part... but still unsure how to find the velocity at a specific point.
pixelized
#10
Apr29-05, 03:35 PM
P: 7
ok it's just an energy problem. The difference in potential energy between the surface and the point it asks for is the kinetic energy which is 1/2 mv^2

Now you know that F = GMm/r^2
and Mass = p * 4/3 * pi * r^3

the mass of the object cancels so you don't need it

and potential energy is just the integral of Force with respect to r

Note r at the surface is r and r at the point it passes is constant * r
squib
#11
Apr29-05, 03:57 PM
P: 40
Big thanks there... makes a lot more sense now. I knew mass would cancel out, just couldn't figure out how
Theelectricchild
#12
Apr29-05, 05:06 PM
P: 258
It was interesting because in this problem was on a physics test of mine a while back, and he told use to find the time for the earth, and the moon--- but since the period is independent of the mass the time for something to reach the other side would be the same!
hbomb
#13
May1-05, 03:41 AM
P: 58
what's interesting to me is that this question was asked in the first place. anything "dropped" through the hole wouldn't reach the other side. It would reach the center of the planet. The acceleration that it would achieve traveling down the hole would decrease as it passes the center. the package would oscillate "up" and "down" the hole until it comes to rest at the center.
OlderDan
#14
May1-05, 05:16 PM
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Quote Quote by hbomb
what's interesting to me is that this question was asked in the first place. anything "dropped" through the hole wouldn't reach the other side. It would reach the center of the planet. The acceleration that it would achieve traveling down the hole would decrease as it passes the center. the package would oscillate "up" and "down" the hole until it comes to rest at the center.
In the absence of any forces other than gravity, it would reach the other side. That is where its velocity would be zero, and it would begin its return trip to its starting point. Of course it is a hypothetical problem, as most problems asked of students are.

What would be really interesting is to drill a hole along a chord of the planet, and let a frictionless cart run back and forth through the tunnel. Then compare the period of oscillation for the cart to the period of oscillation of the through-the-center packets.
HallsofIvy
#15
May1-05, 06:21 PM
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Quote Quote by hbomb
what's interesting to me is that this question was asked in the first place. anything "dropped" through the hole wouldn't reach the other side. It would reach the center of the planet. The acceleration that it would achieve traveling down the hole would decrease as it passes the center. the package would oscillate "up" and "down" the hole until it comes to rest at the center.
Assuming any friction at all, yes. Of course, the original problem said "air has been pumped out of the tube to virtually eliminate friction" so I would assume the problem was to be done ignoring friction. Of course, "virtually eliminate friction" is not the same as "eliminate friction" so you are still right- with a tiny amount of residual friction, the packet will not reach the other side of the planet and will eventually stop at the center of the planet.
jdavel
#16
May1-05, 06:33 PM
P: 618
Here's a little riddle someone fooled me with once.

1) Suppose that instead of dropping the packet from the surface of the earth, you go millions of miles out into space and drop the packet, but you aim carefully enough so that it still goes right through the hole. What happens?

2) Now suppose that instead of just dropping the packet from that location, you give it just the slightest transverse force as you're dropping it, just enough initial perpendicular velocity so that instead of going through the hole it barely skims past the earth on one side. Now what happens?
HardestPart
#17
Dec4-09, 05:01 AM
P: 17
I did not understand the solution can anyone explain it agian?


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