Photon - gas molecule collision and conservation of energy? I thought up a scenario and I'm having trouble working it out.
Imagine a single greenhouse gas molecule with the direction of its velocity towards the ground.
And a single photon leaving the ground hitting the molecule, exciting an electron briefly, then when the electron returns, leaving the molecule, in this case, lets say it returns in a random direction so we can ignore the exiting part.
The photon has very little momentum, but it has some and hitting the gas molecule that's moving downwards, slows it down somewhat.    This slowing down decreases the temperature of the atmosphere as a whole a tiny little bit.
So, the photon, leaving the ground cools the atmosphere somewhat.   I understand that a sea of photons the impacts on greenhouse gas molecules would average out and warm the atmosphere overall, but I can't get my head around how a single photon can cool the atmosphere, effectively reducing the energy overall and not obeying conservation of energy.
Am I missing something?   Can individual particle collisions break the conservation of energy law?     
 A: 
Imagine a single greenhouse gas molecule with it's direction towards the ground. And a single photon leaving the ground hitting the molecule, exciting an electron briefly, then when the electron returns, leaving the molecule, in this case, lets say it returns in a random direction so we can ignore the exiting part.

We cannot “ignore the the exiting part”. When you are working with conservation laws you need to consider the entire system, and the exiting photon is an essential part of the system and must be considered. Here we have three states of the system: 
1) downward momentum relaxed molecule + upward momentum photon
2) downward moving excited molecule
3) relaxed molecule + random momentum photon

The photon has very little momentum, but it has some and hitting the gas molecule that's moving downwards, slows it down somewhat. This slowing down decreases the temperature of the atmosphere as a whole a tiny little bit.

While it is possible for photons to be used in cooling, this is not the result here. When the photon is absorbed the molecule is excited and that energy in the excited state is still very much part of the thermal energy of the gas. When the random photon leaves the energy is still part of the thermal energy of the atmosphere unless it escapes the atmosphere. If it is absorbed by another atmospheric gas molecule then the photon is just another degree of internal freedom for the atmosphere as a whole. 

I can't get my head around how a single photon can cool the atmosphere, effectively reducing the energy overall and not obeying conservation of energy

The only way for a single photon to cool the atmosphere is for it to leave the atmosphere and carry the thermal energy away as black body radiation. 
Colliding with a molecule does not remove energy from the system. It just changes the energy from the molecular KE and photon energy of state 1 then into the molecular KE and internal energy of state 2 and finally back to molecular KE and photon energy of state 3. Energy is conserved at each stage and the temperature of the atmosphere is constant throughout. 
A: Paradoxes arise in physics when theoretical frameworks are mixed .

This slowing down decreases the temperature of the atmosphere as a whole a tiny little bit.

Here is the confusion between three frameworks, the framework of quantum mechanics, the framework of statistical mechanics, and the framework of thermodynamics. Temperature is a thermodynamic variable, it can be shown mathematically that  the average kinetic energy in an ideal gas is connected with the temperature, but note, average. 

So, the photon, leaving the ground cools the atmosphere somewhat. 

As cooling is a thermodynamic observation of the change in temperature, i.e. an average,it cannot be used in this example. Every motion of the zillions of atoms/molecules moving in the atmosphere instantaneously can be thought as cooling and heating in this case.
Energy is always conserved and the answer by Dale deals with the specific energy conservation case in detail.
