# Efficiency of tidal friction

Now we will calculate the efficiency of the tidal forces. Let us find out which part of the spent energy of rotation is transferred to another body (for example, from Earth to the Moon), and which part is lost in the form of heat. As far as I know, no one has ever decided this task.

Therefore, it is placed in the heading “Unknown problems of celestial mechanics”. The task is not difficult and the answer is very beautiful, in the sense of very simple. Why nobody solved it, it is not clear. I think that specialists in celestial mechanics do not like tidal forces, as they are accustomed to working with mathematically exact expressions. The tidal forces cannot be accurately calculated. But we will get an EXACT answer.

In the previous publication, we found out that in the satellite-satellite system, the body in the role of the “sponsor” of energy is always a body with a higher angular velocity of rotation. We will calculate how much of the sponsor energy is transferred to another body, and which part goes into heat spending on tidal heating.

We will use the formula (4) from the previous publication (Synchronous orbit):

I remind that *dv* is the change in the satellite’s orbital velocity, *I* is the dimensionless moment of inertia of the planet, *M* is the planet’s mass, *V* is the linear speed of rotation of the planet at the equator, *dV* is the change of this velocity, *m* is the mass of the satellite, *r* is the radius of the satellite’s orbit. Formula (4) follows directly from the law of conservation of the angular momentum: the change in the orbital angular momentum of the satellite is exactly equal to the change with the inverse sign of the angular momentum of the planet.

The kinetic energy *K*_{2} of the satellite is: *K*_{2} = ^{1}/_{2 }*mv*^{2}

Therefore (*K*_{1} is the kinetic energy of the Earth’s rotation, ω_{1} is the frequency of its rotation in radians, ω_{2 }is the angular velocity of the satellite’s orbit): *dK*_{2} = *mvdv*, taking into account formula (1), we obtain:

Since *dK*_{1} = *IMVdV* and *IMdV* = *dK*_{1}/*V*, then:

Since *v*/*r* = ω1 and *V*/*R* = ω_{2}, we obtain:

We again see that energy can only be transferred from a rapidly rotating body to a slowly rotating body (I mean the angular velocity of rotation).

For example, if *ω*_{ 1} > *ω*_{ 2} (the planet rotates faster than the satellite), then |*dK*_{2}| < |*dK*_{1}|. This means that energy can pass only from the planet to the satellite. After all, in the transfer of energy, some of it is irreversibly converted into heat. For *dK*_{2} < 0, the necessary condition is fulfilled:

*dK*_{1} + *dK*_{2} < 0 (5)

The transfer of energy from a slowly rotating satellite to a rapidly rotating planet is impossible. In this case inequality (5) is violated. Therefore, additional energy is needed to implement such a process.

If the satellite rotates faster than the planet (as, for example, Phobos relative to Mars), then in this case ω_{1} > ω_{2}, and taking into account (4) and (5) we obtain: *dK*_{1} > 0. Accordingly, the kinetic energy of the planet grows, and the orbital energy of the satellite decreases. The satellite slowly “falls” on the planet.

So, if the energy is released in the system as heat, the state of the system will change. The system will tend to a stable state, at which energy is not released. A similar stable state is achieved, for example, in the Pluto-Charon system. The same hemisphere of Charon satellite always faces Pluto (like the Moon to the Earth), but also rotates around Pluto with an angular velocity equal to the speed of rotation of Pluto. That is, Pluto will also turn to Charon with the same side.

It is clear that such a state of the system is not accidental. Apparently, earlier Charon was in a different orbit, but due to the action of tidal forces gradually moved to the modern one. It is interesting to know what was the orbit of Charon, when it was formed: lower or higher than the modern one?

It can be seen from equation (4) that the efficiency of tidal forces depends only on the ratio of the frequencies of rotation of the satellite and the planet and, as a rule, it is sufficiently small. For example, in the Earth-Moon system, *ω*_{1}/*ω*_{2} = month/day ≈ 27, so only less than four percent of the Earth’s rotation energy is used to increase the orbital energy of the Moon, and the rest is released as heat dQ inside the Earth and on its surface. Let’s do the calculations. According to the law of conservation of energy:

*dK*_{1} + *dK*_{2} + *dQ* = 0 (6)

We introduce the following definition of the efficiency of the tidal forces. This value is equal to the ratio of the received energy to the transmitted energy. If |*dK*_{2}| < |*dK*_{1}| (energy is transferred from the planet to the satellite), then:

if |*dK*_{2}| > |*dK*_{1}| (energy is transferred from the satellite to the planet), then:

So, in all cases, the efficiency of the energy transfer process is equal to the ratio of frequencies. The efficiency tends to one hundred percent if *ω*_{1} = *ω*_{2}, but in this case no transfer of energy occurs. If the frequencies are very different from each other, the bulk of the transmitted energy is expended on tidal heating, and only a small part is transferred to another body. That is, the process of transferring the energy of rotation is always accompanied by a dissipation of energy, and, consequently, an increase in entropy. This is an irreversible process.

The efficiency of tidal forces can also be expressed through the ratio of periods. If the period of rotation of planet *Т*_{1} (day) is less than the period of revolution of satellite *Т*_{2} (month), then the satellite is removed from the planet, and the efficiency of transmission of energy to it is equal to:

If the period of rotation of planet *Т*_{1} (day) is greater than the period of rotation of satellite *Т*_{2} (month), then the satellite approaches the planet and the efficiency of transmission of energy to it is equal to:

We studied the tidal forces acting from the satellite to the planet. But what about the tidal forces acting on the satellite from the planet? Which of these forces is greater? Even without any calculations it is clear that the latter. After all, all the planets have completely stopped the rotation of their satellites. That is, for all satellites, their hemispheres always face their planets as the lunar hemisphere always faces Earth. The only exceptions are small satellites, rotating at enormous distances from their planets. We will talk about the origin of these strange satellites later.