# How much energy is required to hold the earth’s atmosphere up against the forces of gravity?

My understanding is that the earth’s atmosphere was originally formed by the molten earth itself, and the sun was 70% weaker that present. Then as the earth cooled the energy required to keep the atmosphere up against the force of gravity was provided by incoming solar radiation.

Presumably this energy is in excess of that incoming from the sun and outgoing to space. Can anyone tell me how much solar energy is required to kind of keep the sky up.

So say we started with the present sun and current atmospheric mass with the earth at its blackbody temperature with the current solar energy hitting its surface - 255K. How much additional solar energy is required to lift those gasses not already off the ground at this temperature up into the sky?

I realise that the current energy reaching the earth’s surface is due to the atmosphere absorbing some of the energy but I’m interested in the excess energy in the earth/atmosphere system required to keep the atmosphere up.

• it is not the energy from the sun that keeps the atmosphere up it is the lower density of the air,the atmosphere had still been there without the sun only in liquid and frozen form. – trond hansen Jan 5 at 14:06
• To get the current atmosphere in gaseous form does take initial energy so that it's not solid/liquid... but after that the only energy needed is to counteract any net energy that the atmosphere itself radiates away to space, very tiny. Certainly in a cooler atmosphere the molecules move less, thus it's less tall... so the height changes based on temperature... but as long as the molecules remain gases, the atmosphere has height due to the amount of gas... much like when you fill a bucket, the top of the column isn't held up by some special energy, but just the rest of the matter below – JeopardyTempest Jan 5 at 17:07

I can not comment on whether your understanding about the formation of the atmosphere (your first paragraph) is correct. However, given that the complete atmosphere lied at the surface at some point, it is possible to calculate the amount of energy to lift this towards its current state.

The calculations will be done for a small surface area of one meter squared and choosing values corresponding roughly to the global averages.

First we calculate the gravitational potential energy of the atmospheric column as follows

$$E_{atmos} = m_{atmos} \cdot g \cdot h_{atmos}$$

with $$h_{atmos}$$ the height of the center of mass of the atmospheric column. Surface atmospheric pressure ($$P_{surface}$$) is the force resulting from the mass of the atmospheric column under gravitational acceleration. Therefore, the above can be rewritten as

$$E_{atmos} = P_{surface} \cdot h_{atmos}$$

The centre of mass at the atmosphere is approximately at $$h_{atmos}$$ = 5 km (one can look at maps of the 500 hPa geopotential heights for example). For surface pressure we can take the standard atmosphere (101325 Pa). Plugging in the numbers:

$$E_{atmos} = 101325\,Pa \cdot 5000\,m = 5.1e8\,N m^{-1} = 5.1e8\,J\,m{-2}$$

To get to the total energy, this number needs to be multiplied by the surface area of the earth. However, it is more insightful to get a certain timescale out of this value. We can calculate how long it would take for the sun to deliver this energy to the earth's surface.

$$\tau_{sun} = \frac{E_{atmos}}{SI_{avg}}$$

With $$SI_{avg}$$ the global average incoming solar radiation (not taking into account atmospheric absorption). Plugging in the values

$$\tau_{sun} = \frac{5.1e8\,J\,m{-2}}{340\,W\,m{-2}} = 1.5e6 s = 17.5 days$$

So it would take about 17-18 days for the sun in order to 'lift up' the atmosphere from the surface to its current height.

• Please support your claim that the entire atmosphere once lay on the surface (presumably in non-gaseous form). – Spencer Jan 12 at 19:27