A remote spherical planet the size of Earth is made of pure gold, and has gravity g = 100m/s²?

Question

The atmosphere of this planet is pure helium He4, and its pressure is so low, that each atom of helium almost always unable to collide with another atom of heluim between two bounces off the surface of the planet. In other words, the mean free path significantly exceeds the height of the atmoshpere. The atoms of heluim therefore follow parabolic trajectories between bouncing off the surface.

Temperature of the surface of the planet is To = 128.5 K.
What temperature will be measured by thermometer placed at height H = 1000 above the surface?

(Assume that thermometer 'absolutely reflects' all radiation,
and its temperature is entilery determined by collisions with atmosperic helium atoms)

H = 1,000 meters = 1km
Sorry.

Dr. R:

I would not be certain

Answer 1

The distribution of kinetic energies for the two horizontal degrees of freedom (v_x and v_y) have to be the same as at the planet surface. Let n0(v_z)dv_z and n1(v_z)dv_z be the number densities of atoms with upward vertical speeds in the interval (v_z,v_z+dv_z) at the surface and at height H respectively. An atom at height H has upward vertical velocity u_z > 0 if and only if it had velocity v_z > 0 at the surface, with:

mv_z^2/2 = mu_z^2/2 + mgH

Current continuity requires:

v_z (n0(v_z)dv_z) = u_z (n1(u_z)du_z)
-->
n0(v_z) = n1(u_z)
-->
n1(u_z) = constant*exp(-(mu_z^2/2 + mgH)/kTo) = (other constant)exp(-(mu_z^2/2)/kTo)

In other words a Maxwell Boltzmann distribution with temperature To - the same as at the surface except with the total density reduced by a factor exp(-mgH/kTo)

This seems like a satisfactory result since if the top of the atmosphere had a lower temperature than the planet and there were a spherical metal shell at height H held at temperature To, there would be a flow of heat from the shell through the atoms to the planet even though it's at the same temperature To.

Answer 2

Big bang!! Why couldn't it be a Big Juicy Raspberry!BJR Actually, the shape of the earth is rather flatter at the poles, like a deflating basket ball. The molten core with a high percentage of iron maintains a magnetic field.North & South But the way I understand it is the analogy of a large cement mixer or stones worn smooth by tumbling along the river. It may be the centripital force of spinning and friction which the orbitting body(ies) develop in the style of least resistance that is- rounded or as you say spherical. Star dust as a living thing ? mmmm Again the analogy - you need the right ingredients the right conditions mixed in the right way for the right amount of time with the specific energy requirements maintained for an inordinate amount of eons to raise bubble and form into anything approaching a matrix which invests the ebb and flow of seasonality the pull of tides from a satellite combining the chemical / physical properties of the planet, to even resemble that which we find sensate pulsing reactive reproductive and alive. One heck of a recipe eh! But in a way - why not? God said let there be light! Nothing would exist as we know it without the source.. Pretty fundamental stuff and look at the nuclear reaction of the sun- every thing is made in this cauldren and then consumed but not lost totally but radiates fuelling heat light and cosmic reactions throughout the system, inside a galaxy within our section of the universe which sits on the back of an elephant standing on the back of a turtle!!! BJR I reckon!! Cheers

Answer 3

Vrms = sqrt(3RT/m) = 895 m/s (where M is molar mass = 0.004 kg and R is the UGC = 8.3145).
For vertical bounce, V(H) = sqrt(V0^2 - 2aH) = 775 m/s.
This will register T = MV^2/(3R) = 96.4 K.
The lowest angle that reaches H is 26.5 deg and impact speed is 801 m/s ==> 102.9 K.
Integrating the range of angles between 26.5 and vertical for a mean is something I'd rather not try.
EDIT: Integration is not needed. Doing the math again, the impact speeds are all the same, 775 m/s, down to the lowest angle that reaches H. It takes a minimum vertical velocity of 447.2 m/s to reach H. The lowest angle is 30.0 deg. All these bounces lose the same amount of kinetic energy as answerer 1 realized. His formulation also yields 96.4 K if you use the single-molecule weight, (0.004/Avogadro's number) kg, for mass and the right value (100 not 1000) for g.

Answer 4

Using an energy basis for tempertature and ignoring the skewing of the Maxwellian distribution (your assumptions preclude thermalization),

T=T_0 - 2mgH/(3k)

m = mass of He4
k=Boltzman constant
g=1000 m/s^2

This is based on energy =3kT/2 for a monatomic ideal gas in thermal equilibrium. Numbers left to reader.