Re: A newcomer here

On Tue, 13 Sep 2005 02:03:27 GMT, Logan Kearsley
<chrono.surfer@xxxxxxxxxxx> wrote in
<PrqVe.6937$b37.3139@trnddc04">news:PrqVe.6937$b37.3139@trnddc04> in
rec.arts.sf.composition:

> "David Friedman" <ddfr@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
> news:ddfr-DAA931.18240812092005@xxxxxxxxxxxxxxxxxxxxxxxx

>> In article <GepVe.3508$XO6.2586@trnddc03>,
>> "Logan Kearsley" <chrono.surfer@xxxxxxxxxxx> wrote:

>>> With constant crust thickness and density, the surface gravity
>>> increases as the radius increases.

>> No. It stays constant.

>> Area goes as the square of radius, so with constant thickness and
>> density mass goes as the square of radius.

> Except that it doesn't, exactly. It's somewhat off. Assuming constant
> density, the mass is dependent on the volume of the shell, which is the
> volume of the spherical world minus the volume of the empty spherical
> interior. This will only scale with the surface area / square of the radius
> if the ratio between the outer sphere and the inner sphere is constant, but
> if the shell thickness is constant, that ratio won't be.

Let inner radius = r, outer radius = r + t. Then volume =
(4/3)*pi*[(r + t)^3 - r^3] = (4/3)*pi*(3tr^2 + 3rt^2 + t^3),
i.e., mass is proportional to 3tr^2 + 3rt^2 + t^3, and force
is proportional to 3t + 3t^2 / r + t^3 / r^2, which for
constant t decreases as r increases.

Alternatively, let the inner radius be r, the outer radius
kr. The volume is proportional to (k^3 - 1)r^3, the force
to (k^3 - 1)r = (1 + k + k^2)(k - 1)r = (1 + k + k^2)t,
since t = (k - 1)r. Thus, the force is proportional to
1 + k + k^2. But if t is fixed, k = 1 + t/r is decreasing
as r increases, so the force is decreasing.

Brian
.

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