1

Consider an object falling under gravity. To make this an "equilibrium problem", let's say the object has attained terminal velocity (due to air drag). In this situation, how do we apply the first law of thermodynamics to earth. Clearly, earth does positive work on the falling object. According to ∆U = Q - W, total-energy of earth should be decreasing. How is this accounted for?

Putting it another way, is the gravitational potential energy "owned" by the object or earth, in the definition of total-energy U in the above equation?

Qmechanic
  • 220,844
Varidhi Shayana
  • 45
  • 6

4 Answers4

0

Reaching a terminal velocity in air does not make it an equilibrium process, just as dragging a heavy bag on a surface with friction is not a thermodynamical equilibrium process.

Assuming that the size of the mass, $m$, of the object falling in air is small enough so that the Earth's gravitational potential does not change by the objects motion, the gravitational work that the object loses as it drops from height $h_1$ to $h_2$ is exactly $w_{12}=m(gh_1-gh_2)$, and it does not matter whether it stops at $h_2$ or continues on at a constant terminal velocity. If between those heights this $w_{12}$ work lost from the object is used to move something else by said motion, say, an electric motor, or lift another object in a gravitational field, etc., and thereby gaining $w*$ work, then the difference $w_{12}-w^*$ is dissipated into heat.

If the works $w_{12}$ and $w*$ are small so that the heat dissipation is essentially occurring at a constant temperature, then $w_{12}-w^*=q=T\sigma$ where $\sigma=\sigma(h)$ is the locally generated entropy at temperature $T=T(h)$ and it is a positive quantity characteristic of the process. The total dissipated heat is the integral along the process $\int_{h_1}^{h_2} T\sigma.$ In the case of the falling object both the air and the object itself may get heated up by the air resistance.

hyportnex
  • 21,193
0

The potential energy belongs to the Earth-object system. The system's potential energy decreases while its kinetic energy (and then thermal energy from air drag and then ground impact) increases. Since the value of this potential energy $mGM(1/r_1-1/r_2)$ is insignificant compared to the Earth's overall energy, one can say that approximately, Earth's energy is unchanged and the falling object's energy is what changes, and is transferred to the air and ground.

RC_23
  • 12,766
  • 2
  • 17
  • 53
0

First of all, all forms of potential energy are system properties. Potential energy is not “owned” by an object. See https://www.britannica.com/science/potential-energy. So the gravitational potential energy belongs to the object-earth system.

For the falling object the internal energy of the object-Earth system, which when we include the air, is an isolated system and is therefore constant. The loss of gravitational potential energy of the system equals the increase in the kinetic energy of the object part of the system, plus an increase in the internal energy of the object and air (due to friction).

Hope this helps.

Bob D
  • 81,786
0

The more general form of the first law of thermodynamics is $$\Delta U+\Delta(PE)+\Delta(KE)=Q-W$$ The potential energy and kinetic energy terms of typically neglected. But, if $W=W_{ND}+W_D$, where $W_D$ is the work done against air drag and $W_{ND}$ is the remainder of the work that the system does on the surroundings, and, if the system has reached terminal velocity, then $$\Delta (KE)=0$$ and $$\Delta PE=-W_D$$So, we are left with $$\Delta U=Q-W_{ND}$$

Chet Miller
  • 35,124