[hurley_108]

Blue_Nose wrote:

As a sidenote, is that your own graphic, hurley? Good stuff!

Yup. Thanks! No fancy programs either, just powerpoint (to draw it) and paint (to shrink it and save as a gif).

[Blue_Nose]

New one:

The equation for the gravitational force between two objects is

F = G*m1*m2/d^2

F is the force, G is a constant, m1 and m2 are the masses of the two objects, and d^2 is the distance between them squared.

Now, according to this equation, the closer you get to the earth, 'd' becomes smaller and smaller, and the force gets larger.

Now, if you were able to go right to the centre of the earth, the distance would be near 0, and that would cause F to be huge, approaching infinity.

If you were to go to the centre of the earth, would you be crushed to infinity? How can the infinitely large gravity not suck the entire planet into a black hole?

[travior]

Because at the center of the earth you are at an equilibrium point in the earth's mass. Meaning that there is equal amounts (roughly) of the earth's mass in a sphere around you. Therefore, gravity would be negligable at best.

[Blue_Nose]

But that's not what the equation says....

Anyway, I've got another one that's more math-involved.

Let's say you have a super bouncy ball that's 99% efficient - that is, if you drop it from 100 meters it'll bounce back up to 99 meters.

Let's say you drop the ball from 100 meters onto a solid surface and let it bounce an infinite number of times. Assuming this is a mathematical question, so no bounce is too small, how long will it bounce?

[C.M. Burns]

Blue_Nose wrote:

But that's not what the equation says....

Anyway, I've got another one that's more math-involved.

Let's say you have a super bouncy ball that's 99% efficient - that is, if you drop it from 100 meters it'll bounce back up to 99 meters.

Let's say you drop the ball from 100 meters onto a solid surface and let it bounce an infinite number of times. Assuming this is a mathematical question, so no bounce is too small, how long will it bounce?

The equation isn't applicable to the situation you outlined.

[C.M. Burns]

Blue_Nose wrote:

But that's not what the equation says....

Anyway, I've got another one that's more math-involved.

Let's say you have a super bouncy ball that's 99% efficient - that is, if you drop it from 100 meters it'll bounce back up to 99 meters.

Let's say you drop the ball from 100 meters onto a solid surface and let it bounce an infinite number of times. Assuming this is a mathematical question, so no bounce is too small, how long will it bounce?

Infinity.

[hurley_108]

The gravity thing:

The equation given is for point masses, and gives the magnitude only of the force. It can be shown that an evenly distributed spherical mass can be treated as a point. It can also be shown that a mass within a spherically distributed mass feels no net force from that spherical distribution (the Dyson Sphere question I posed earlier).

This all comes out of much calculus and the vector form of the gravity equation. But what it means is that if you are at some point between the center and surface of the earth, the only masses you need to consider are your mass, and the mass of that spherical part of the earth that is below you. At the center, m1, you, is the same, but m2, the mass of the earth below you, is zero. So what you have is basically F=0/0. This doesn't make sense because zero over anything is zero, but anything over zero is infinity.

So, we need to express it differentliy:

F = G*m1*rho*(4/3)*pi*d^3/d^2

What we're looking for is the limit of force as d approaches zero. So we use L'Hopital's rule, and differentiate top and bottom WRT d:

F = G*m1*rho*4*pi*d^2/2d

Still 0/0 so use L'Hopital's rule again and:

lim F = G*m1*rho*8*pi*d/2 = 0
d->0

So the force at the center of the earth is zero.

Now, let's say you bored out a hole directly though the center of the earth, sucked all the air out, and jumped in. How long would it take you to get to the other side?

What if your bore only got as deep as halfway to the center, being drilled diagonally?

[Blue_Nose]

C.M. Burns wrote:

Infinity.

Nope, but that's the obvious guess

It's weird to try to imagine an infinite number of "events" in a finite amount of time, but that is the case here, and there's a finite answer.

[Blue_Nose]

hurley_108 wrote:

So the force at the center of the earth is zero.

Good stuff - I didn't expect the mathematical explanation, just that the equations didn't apply within the objects' radii.

hurley_108 wrote:

Now, let's say you bored out a hole directly though the center of the earth, sucked all the air out, and jumped in. How long would it take you to get to the other side?

This one's a good one, I'll have to take a second to do the math... we'll have to assume the rotation of the earth doesn't affect the fall, as it would in reality... I'd imagine you'd actually need some sort of elliptical path between the two point to account for this, but I'm not sure.

[hurley_108]

Blue_Nose wrote:

hurley_108 wrote:

So the force at the center of the earth is zero.

Good stuff - I didn't expect the mathematical explanation, just that the equations didn't apply within the objects' radii.

NOBODY EXPECTS THE MATHEMATICAL EXPLANATION!

Quote:

hurley_108 wrote:

Now, let's say you bored out a hole directly though the center of the earth, sucked all the air out, and jumped in. How long would it take you to get to the other side?

This one's a good one, I'll have to take a second to do the math... we'll have to assume the rotation of the earth doesn't affect the fall, as it would in reality... I'd imagine you'd actually need some sort of elliptical path between the two point to account for this, but I'm not sure.

Yea, assume no rotation. Just a sphere and a bore and you jumping in. For the diagonal one, assume frictionless rails or something.

[Blue_Nose]

Okay, here we go:

Using hurley's equation for the force of gravity, we can remove our own mass and find the acceleration (F=ma),

Attachment:

eqn1.PNG [ 1.98 KiB | Viewed 78 times ]

Note that the acceleration is given as negative because we'll say that 'd' is the distance outward from the centre to our starting point, and gravity is acting inward.

Since acceleration, a, is the second derivative of position, d, we'll call it d" and we have a second order differential equation. I rearranged the terms a bit and substituted all the constants (like G, rho, etc) into one constant 'k', so we get

d" + kd = 0

which we now have to solve to find d in terms of time.

I assumed the path would be cyclic, and since it starts up at the surface, goes down to the middle, then goes out to the other surface, I chose a general cos function:

d = a*cos(b*t)

where we have to find a and b for our situation.

We know that at t=0 we're at the outside of the earth, r, so we can easily find that a = r.

to find b, we need to find d" and substitute that into our equation above and solve that.

d' = -r*b*sin(b*t)
d" = -r*b²*cos(b*t)

substituting we get:

-r*b²*cos(b*t) + k*a*cos(b*t) = 0

bunch of math to solve that for b, we get:

Attachment:

eqn2.PNG [ 1.21 KiB | Viewed 78 times ]

Now all that's left is to find the first time, t, when the distance, d, is zero:

Attachment:

eqn3.PNG [ 7.1 KiB | Viewed 78 times ]

We know that G is 6.67300*10^(-11) and the density of the earth is about 5500kg/m3, and we get a time of 1267 seconds, or a little over 21 minutes.

If we were to fall all the way through and fall back again, the total trip would be four times that, or about 1 hour and 24 minutes.

Not doing the second half of the question, hurley

[hurley_108]

I am impressed. That's exactly it. I was going for the time just to the other side, 42 minutes (See HHGTTG and my sig), but your math is spot on. As it happens, it's the same result no matter how you drill the hole. 42 minutes to the other side. I'm not doing the math because I did it once in University and I'm not doing it again.

[Blue_Nose]

Blue_Nose wrote:

Let's say you have a super bouncy ball that's 99% efficient - that is, if you drop it from 100 meters it'll bounce back up to 99 meters.

Let's say you drop the ball from 100 meters onto a solid surface and let it bounce an infinite number of times. Assuming this is a mathematical question, so no bounce is too small, how long will it bounce?

Anyone want to take a shot at this one?

I admittedly used a spreadsheet to add up all the times, so it's not an exact solution...

[Tricks]

Would it not be forever? It would never reach zero would it?

[hurley_108]

Blue_Nose wrote:

Blue_Nose wrote:

Let's say you have a super bouncy ball that's 99% efficient - that is, if you drop it from 100 meters it'll bounce back up to 99 meters.

Let's say you drop the ball from 100 meters onto a solid surface and let it bounce an infinite number of times. Assuming this is a mathematical question, so no bounce is too small, how long will it bounce?

Anyone want to take a shot at this one?

I admittedly used a spreadsheet to add up all the times, so it's not an exact solution...

Holy smokes it looks like it converges. 1797.053 seconds, or just shy of half an hour. Why does it converge? Another thing I used to know - we did infinite series in second or third year calculus. There are tests you can do to prove it, but I've completely forgotten what they are. If I get REALLY bored on the weekend I might pull out my text, but for now, I'm gonna say half an hour.