Re: why people pretend that the moon landing was for real

Koobee Wublee wrote:
"Bilge" <dubious@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message news:slrnd8v6mr.ii.dubious@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Now, another mystery deals with the temperature of an enclosed bubble of
air
in vacuum. According to the 1st law of thermodynamics, energy must be
conserved. However, due to black body radiation, the temperature will
slowly decrease. So, what is the mechanism to allow this bubble of air
to
lose 20 degrees C in a few hours?

Radiation.

That type?

Do you mean ``What type?'' Electromagnetic.


Yes, I meant "what type". Now, how do you figure the ambient temperature of Apollo 13 would lose 20 degrees in a few hours due to electromagnetic radiation? What spectrum of the electromagnetic radiation do you have in mind? In the meantime, the spacecraft is still receiving a lot of energy from the sun. Have you tried to get into your car in a hot day with all the windows closed? 

So you are referring to Apollo 13.
I suspected that much.
I have seen this argument before:
"In Ron Howard's 1995 science fiction movie, Apollo 13,
 the astronauts lose electrical power and begin worrying
 about freezing to death.  In reality, of course, the relentless
 bombardment of the Sun's rays would rapidly have overheated
 the vehicle to lethal temperatures with no atmosphere into
 which to dump the heat build up."

So what will happen - in reality?

The power radiated from a body with temperature T is:
Pe = E*sigma*A*T^4
where:
 E is the emissivity
 sigma = 5.67*10^-8 W/(m^2*K^4)
 A is the surface area of the body.

The power absorbed by a body 1AU from the Sun is:
Pa = E*So*Ac
where
 E is the emissivity
 So is the solar constant = 1.37*10^3 W/m^2
 Ac is the area of the cross section of the body
    perpendicular to the direction to the Sun

The body will be in thermal equilibrium when Pe = Pa
E*sigma*A*T^4 = E*So*Ac
T^4 = (So/sigma)*Ac/A = 2.42*10^10*Ac/A  K^4

The ratio Ac/A depend on the shape and orientation of the body.

For a sphere Ac/A = 0.25
T = 279K or 6 deg C. Chilly, but not freezing.

For a cube Ac/A = 1/6 = 0.167
T = 252K or -21 C. Freezing!

For a cylinder with length 4 radii:
Ac/A = ca. 0.25 if its length axis is perpendicular to the Sun.
T = 279K or 6 C.
Ac/A = 0.1 if its length axis points to the Sun.
T = 222K, or -51 C.  Freezing indeed!!

So what temperature Apollo 13 would approach depend
on its exact shape and orientation. But it would
most probably be below zero degree C.

BTW, why didn't it occur to you to make this simple
calculations before gulping up undigested claims
you have read somewhere?
Are your too ignorant of elementary physics to do so?

Paul



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