User:Mark Widmer/sandbox

Sandbox. Mark Widmer (talk) 01:17, 5 August 2021 (UTC)

Benchmark quantities

Heat equation

Heat_equation

Define variables when equation is used for temperature: u=temperature, k = k_therm / (c*rho)

Define variable when equation refers to diffusion: u = density???

Draft for additions to Hill_sphere New sections:

Hill sphere and L1 Lagrange point

-- added note in Formulas section

Hill sphere of the Sun

-- added to article

Hill sphere of objects that orbit Earth

The Moon -- added to article

Artificial satellites in low-Earth orbit -- added to article

L1 Lagrange point for comparable-mass objects

Usually, derivations of the L1 point assume a planetary mass that is much less than the star's mass. This no longer applies if the orbiting objects have comparable masses. This is the case for many binary star systems. For example, in the Alpha Centauri system, the stars Alpha Centauri A and B have masses that are 1.1 and 0.9 times that of the Sun, respectively, or a mass ratio of about 0.8.

For two equal-mass objects, let R be the distance between the objects. Each object is then in a circular orbit of radius R/2 about the center of mass, which is halfway between them.

Outline:

We follow the derivation for small planet/star mass ratio given at http://www.phy6.org/stargaze/Slagrang.htm, without making the small-ratio approximations that are incorporated there.

Planet/star mass ratio $\mu =m/M$ , with $0<\mu <=1$

Equate the gravitational force (which acts at a distance R) with the centripetal force (for a circle of radius $R/(1+\mu )$ ):

${\frac {GmM}{R^{2}}}={\frac {mv^{2}}{\frac {R}{1+\mu }}}={\frac {mv^{2}(1+\mu )}{R}}$

Mult by R/m:

${\frac {GM}{R}}=(1+\mu )v^{2}$

Substitute for $v={\frac {2\pi {\frac {R}{1+\mu }}}{T}}$

${\frac {GM}{R}}={\frac {(1+\mu )4\pi ^{2}R^{2}}{T^{2}(1+\mu )^{2}}}$

${\frac {GM}{R^{3}}}={\frac {4\pi ^{2}}{T^{2}(1+\mu )}}$

${\frac {GM}{R^{3}}}={\frac {4\pi ^{2}}{(1+\mu )T^{2}}}$

An small-mass object at the L1 point, a distance r from object m, will have an orbit with radius ${\frac {R}{1+\mu }}-r$ and the same period T:

${\frac {GM}{(R-r)^{2}}}-{\frac {Gm}{r^{2}}}={\frac {v^{2}}{{\frac {R}{1+\mu }}-r}}$

$v={\frac {2\pi ({\frac {R}{1+\mu }}-r)}{T}}$ ,

so

${\frac {GM}{(R-r)^{2}}}-{\frac {Gm}{r^{2}}}={\frac {4\pi ^{2}({\frac {R}{1+\mu }}-r)^{2}}{T^{2}}}{\frac {1}{{\frac {R}{1+\mu }}-r}}={\frac {4\pi ^{2}({\frac {R}{1+\mu }}-r)}{T^{2}}}$

Since T is the same for the planet and an object at the L1 point,

${\frac {4\pi ^{2}}{T^{2}}}={\frac {GM(1+\mu )}{R^{3}}}={\frac {GM}{(R-r)^{2}({\frac {R}{1+\mu }}-r)}}-{\frac {Gm}{r^{2}({\frac {R}{1+\mu }}-r)}}$

Divide through by GM

${\frac {(1+\mu )}{R^{3}}}={\frac {1}{(R-r)^{2}({\frac {R}{1+\mu }}-r)}}-{\frac {\mu }{r^{2}({\frac {R}{1+\mu }}-r)}}$

${\frac {(1+\mu )}{R^{3}}}({\frac {R}{1+\mu }}-r)={\frac {1}{(R-r)^{2}}}-{\frac {\mu }{r^{2}}}$

${\frac {R-r(1+\mu )}{R^{3}}}={\frac {1}{(R-r)^{2}}}-{\frac {\mu }{r^{2})}}$

${\frac {R-r(1+\mu )}{R}}{\frac {(R-r)^{2}}{R^{2}}}=1-{\frac {\mu (R-r)^{2}}{r^{2}}}$

$(1-\rho (1+\mu ))(1-\rho )^{2}=1-\mu ({\frac {1}{\rho }}-1)^{2}$

$\rho ^{2}(1-\rho (1+\mu ))(1-\rho )^{2}=\rho ^{2}-\mu (1-\rho )^{2}$

Didymos

The asteroid Didymos and its smaller, satellite asteroid Dimorphos comprise a binary asteroid system within the solar system.

Pole-in-the-barn Paradox

The lack of simultaneity in special relativity is illustrated by the pole-in-the-barn paradox. The scenario includes a long, horizontal pole and a barn with both a front and a back door. The pole's length is Lp in its rest frame, and the distance between the barn's two doors is Lb < Lp. As such, when at rest the pole cannot fit inside the barn. When the pole is partially inside the barn and at rest, at least one of the two barn doors must be open, as one or possibly both ends of the pole will extend outside of the barn.

Now imagine that the pole is moving horizontally at a constant relativistic speed toward the barn, whose doors are both open. The speed is fast enough so that, in the barn's rest frame, the pole is length-contracted to Lp' < Lb. As the pole moves through the barn, there is a brief amount of time where it is completely inside the barn as seen by an observer at rest inside the barn. When the pole is completely inside the barn, the two barn doors are both briefly shut at the same time, and then reopened before the pole starts exiting the rear of the barn.

Next we look at things as observers moving with the pole. In this reference frame, the length of the barn is contracted to Lb' < Lb, which is still less than Lp. In other words, the pole is seen to be longer than the space inside the barn. But then how is it possible for the two barn doors to close without either hitting / getting hit by the pole?

This paradox is resolved by noting that, in pole's rest frame, the two doors do not get shut at the same instant. Instead, the rear door is shut and reopened before the front of the pole reaches the rear door and the rear of the pole has yet to enter the barn. A short time later, when the rear of the pole is inside the barn, the front door is shut and reopened -- and by this time the front of the pole has passed outside, beyond the rear door.

The fact that the two doors are shut and reopened at the same time in the barn's frame, but at different times in the pole's rest frame, illustrates the idea that [something about simultaneity].