... keeping in mind, of course, that Felix will be burned to a cinder during the course of his journey ...
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Allright...you Geology and Physics Guys
- Thread starter Lonesomedave
- Start date
This whole discussion about the "hole through the planet" idea is an example of a Gedankenexperiment or thought experiment. We've just been talking about the strength of gravity at each point in the journey and conveniently ignoring the temperature and motion of the Earth (in fact, the model assumes a uniformly solid, stationary planet - see post 45). It's very common to use this sort of thought experiment when you want to investigate something and isolate its consideration from other factors.
For JT and Titus - here's another very intuitive way of thinking about the fact that our guy is NOT a point mass. No maths involved in this bit.
If we imagine our guy at some point in his descent, he will of course be subject to the usual tidal forces. His feet will be pulled down a bit harder than his head (because they're closer to the Earth) and so he'll experience some vertical stretching force. He'll also experience a lateral squeezing force because his left and right sides are being pulled inwards a bit.
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Not to scale! In fact, if it was, the tidal forces would be pretty darn strong. It's our small size relative to the planet they keeps tidal forces small enough to be unnoticeable in most respects (ocean tides caused by the moon's gravity - and to some extent the sun's - being the exception).
Now, can the tidal forces acting on Felix become large enough to do him some damage at any point in his journey? Clearly, the size of these tidal forces depends on the overall gravitational field strength. If this was zero, there'd be no tidal forces at all, and if it was very strong, the tidal forces would be correspondingly strong. So the tidal forces must be at a maximum when the overall gravitational field strength is at a maximum. And where does that occur? At the planet's surface. In other words, at no point in his journey will he feel any stretching or squeezing forces stronger than the (unnoticeable) tidal forces we feel on the Earth's surface.
As DerHoggz said, Felix is safe - at least, he's safe gravitationally speaking.
The force acting on you due to gravity would vary according to your description but your weight would always be zero, since you'd be in free fall at every point of your travel.
Your weight IS the force acting on you due to gravity.
I think you' might be confusing the sensation of "weightlessness" with actual weight. You feel "weightless" in freefall, but you will not have a weight of zero in any gravitational field, due to its interaction with your mass. The only place your weight is zero is in the absence of gravity (at the centre of the planet in this case).
Yeah, at the instant your CG and the planet's CG meet, there would be no acceleration due to gravity, so no weight.
I read somewhere that astronauts in space have their mass measured in an inertial chair that kind of shakes them around.
I used to have a fun way of teaching about that - it involved doing a thought experiment in zero gravity. I used to get kids to imagine a ping pong ball floating in front of them and ask them how much it weighed if there was no gravity pulling it down (zero). Then I'd ask if they'd feel anything if I bounced it gently off their forehead - of course, they would (because to change the ball's direction you'd need to exert a force on it - 1st Law - and simultaneously it would exert a force on your head - 3rd Law).
Then I'd get them to do exactly the same "thought experiment", only this time using a house brick.
This was good way of getting them to understand that even though the ball and the brick both had zero weight, they both still had mass - and the brick has a lot more than the ball!
That said, I don't think hbarcat was thinking about mass when he posted - he'd just been bamboozled a little by the misleading term "weightlessness", which is all about the sensation of feeling weightless, rather than actually having a weight of zero newtons. It's confusing because you feel "weightless" when your weight is the only force acting on you (i.e. zero contact force / zero stress and strain).
Weight is the product of the force of gravity and your mass, not just simply 'the force of gravity'.
Let's also clear up this whole 'freefall' thing.
In reality, 2 things happen...
Firstly, initially the downward force of your weight is greater than the upward force of air resistance, hence you accelerate downwards because that is the direction of the resultant force.
Secondly, though your weight does not change, the upwards force of air resistance increases until it balances the downward force of your weight. At this point, both upwards and downwards forces are equal, and there is no resultant force, and therefore no more acceleration. This is what is known as terminal velocity.
Of course our guy falling through the earth in this thought experiment experiences no viscous drag and so continues to be accelerated towards the centre of the earth. Once he passes through that point, according to rules of shm, he will then experience a restoring force directed towards the centre of the earth which decelerates him until he just emerges at the other end of the tunnel.
I was in fact thinking about the difference between mass and weight and was specifically invoking the operational use of the term weightlessness.
This is, as you said, distinct from the purely scientific definition, which is of course W=mg.
Your usage is certainly the more applicable to this context, but I was just mentioning the distinction to add to the discussion. Unfortunately, I think that because I implied (incorectly) that your scientific usage was inappropriate (it isn't) all I added was confusion.
Operational definition:
In the operational definition, the weight of an object is the force measured by the operation of weighing it, which is the force it exerts on its support.[7] This can make a considerable difference, depending on the details; for example, an object in free fall exerts little if any force on its support, a situation that is commonly referred to as weightlessness. However, being in free fall does not affect the weight according to the gravitational definition. Therefore, the operational definition is sometimes refined by requiring that the object be at rest.[citation needed] However, this raises the issue of defining "at rest" (usually being at rest with respect to the Earth is implied by using standard gravity[citation needed]).
http://en.wikipedia.org/wiki/Weight:
Professionals in engineering and scientific disciplines involving accelerations and kinetic energies rigorously maintain the distinctions between mass, force, and weight, as well as their respective units of measure.
http://en.wikipedia.org/wiki/Mass_versus_weight
Sounds like the distinction that used to be made years back between "actual weight" and "apparent weight", which was always a good way to really confuse people - including me.
I was thinking you'd feel half your weight pulling on your head and half pulling on your feet in the opposite direction. Not enough to tear you apart, but still pretty uncomfortable. Not sure about this, though. I'm a chemist really.
I believe the small force (a form of tidal force) would compress you, not stretch you.
If your center of mass and the earth's center of mass coincided, a larger portion of
the earth's mass would lie in the direction of your feet. with respect to your head.
Likewise, with respect to your feet, a larger portion of the earths's mass would lie
in the direction of your head.
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Neutron Star -- A short story by Larry Niven
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Approaching a neutron star within an indestructible starship, tidal forces threaten to tear apart the pilot.
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Approaching a neutron star within an indestructible starship, tidal forces threaten to tear apart the pilot.
Larry Niven is one of my favorite Sci FI writers, Tidal forces, as in that story would stretch you, but I believe he got that one wrong.
The tidal force is due to a gradient in the gravitational force, i.e. a difference in force from one point
in space to another. The most common effect of this is in the height of the tides on the opposite sides
of the earth. (and hence the name)
Centering yourself about the zero gee point doesn't help at all. It is the difference in gee between
your head and feet that matter. And in this case, that difference would be minimum at a point furthest
from the star. That is where the gradient would be less steep.
Finding an orbit and position where your feet are at 2G and your head is at -2G is no better than one
where your feet are at 4G and your head is at 0G. The difference is the same. What you need is a smaller
difference between the head and feet forces. You find that at a point most distant from the star.
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The tidal force is due to a gradient in the gravitational force, i.e. a difference in force from one point
in space to another. The most common effect of this is in the height of the tides on the opposite sides
of the earth. (and hence the name)
Centering yourself about the zero gee point doesn't help at all. It is the difference in gee between
your head and feet that matter. And in this case, that difference would be minimum at a point furthest
from the star. That is where the gradient would be less steep.
Finding an orbit and position where your feet are at 2G and your head is at -2G is no better than one
where your feet are at 4G and your head is at 0G. The difference is the same. What you need is a smaller
difference between the head and feet forces. You find that at a point most distant from the star.
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I just worked out the head-to-feet g force difference for our 2 metre tall jumper Felix at several points during his twelve million metre descent.
At six million metres above the surface (one planet radius, and therefore two planet radii from the centre), g is one quarter what it is at the surface, in accordance with the inverse square law. The head to feet g difference (vertical component of the tidal force) is exactly one-eighth what it will be when he reaches the surface.
Just above the surface, an instant before he enters the hole, he is experiencing normal surface gravity. Here both g and the tidal force due to the head-to-feet g difference are at their maxima.
Once inside the planet, g is no longer governed by the inverse square law but is in simple direct proportion to distance from the centre at any point. Interestingly, this causes the tidal force to drop to half its value at the planet's surface as soon as he enters the hole, and then it stays constant at this value for the remainder of his descent towards the centre and subsequent travel to the exit point on the opposite side of the planet.
It's cool to notice how intuitive ideas about this can be way out and how doing the numbers is the best way to explore the variables.
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At six million metres above the surface (one planet radius, and therefore two planet radii from the centre), g is one quarter what it is at the surface, in accordance with the inverse square law. The head to feet g difference (vertical component of the tidal force) is exactly one-eighth what it will be when he reaches the surface.
Just above the surface, an instant before he enters the hole, he is experiencing normal surface gravity. Here both g and the tidal force due to the head-to-feet g difference are at their maxima.
Once inside the planet, g is no longer governed by the inverse square law but is in simple direct proportion to distance from the centre at any point. Interestingly, this causes the tidal force to drop to half its value at the planet's surface as soon as he enters the hole, and then it stays constant at this value for the remainder of his descent towards the centre and subsequent travel to the exit point on the opposite side of the planet.
It's cool to notice how intuitive ideas about this can be way out and how doing the numbers is the best way to explore the variables.
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Hmmmm ... As part of my job as an engineering geologist, it was necessary for me to be lowered down 30-inch-diameter exploratory borings to depths sometimes exceeding 100 feet. The effect of the tidal force must be very faint, because I never noticed anything as I entered the hole.
For those scratching their heads, I was lowered down the holes inside an aluminum cage with intercom, air supply and light. It was perfectly safe. I only had one hole cave in on me, but it didn't plug the hole, so all was cool. We do this to investigate the subsurface conditions mainly in ancient landslides. I've gone down over 200 holes.
