Extended Features:

  1. To be a bit more elaborate, you can consider a general astro-body as defining celestial frame since the center of mass is in free fall under the external gravitational field.

    The general formula for the tidal force per unit mass (i.e., the tidal force field) at a point is the difference between the external gravitational field at that point and the mass-weighted average external gravitational field of the astro-body as defining celestial frame.

    In the limit that the external gravitational field has only linear variation over the astro-body, the mass-weighted average external gravitational field is equal to the external gravitational field at the center of mass of the astro-body. This limit is often approached to good approximation for dense astro-bodies: i.e., asteroids, moons, planets, stars, etc..

  2. If the size scale of the celestial frame Δr is much smaller than the distance size scale between the celestial frame and an external gravitational field source R (i.e., Δr/R << 1), then the tidal force falls off as 1/R**3 (i.e., an inverse-cube law) rather than as 1/R**2 (i.e., an inverse-square law) as for the gravitational force between point masses or spherically symmetric objects.

    This relatively rapid fall off of the tidal force plus cancelation between whole surrounding array of external gravitational field sources, means that in many cases the tidal force will be negligible. For example, for most planetary systems (e.g., the Solar System), the tidal force due external gravitational fields is negligible.

    However, in many cases, the tidal force is NOT negligible at least on some scale. For example, the tidal forces of the Moon and, secondarily, the Sun cause the tides, the Earth tide, and atmospheric tide on Earth. Of course, these tides are very small compared to the Earth, but NOT compared to humans.

    For more on the tides, see tide_ideal.html.

  3. Note also that the tidal force varies as Δr/R, and so the smaller Δr/R, the smaller the tidal force, and so small astro-bodies are less affected than large astro-bodies by the tidal force.

    For example, artificial satellites and spacecraft in general have absolutely negligible tidal forces on them.

  4. There is one important special case of the tidal force that it is interesting to elaborate on. But it is beyond the scope of an intro astro course, and so you need NOT read further if you are in such a course.

    1. Consider a two-body system where one astro-body is sufficiently small that the center of mass of the two-body system is effectively at the center of the larger astro-body (i.e., the primary). So the smaller astro-body (i.e., the secondary) just orbits the center of the primary relative to the observable universe. We also assume a circular orbit. The Earth-Moon system is approximately like this case.

      We assume both astro-bodies are spherically symmetric.

    2. We now assume the secondary is tidally locked to the primary. This is a realistic case in that it occurs frequently all over the observable universe for moon systems and for planets very close to their host stars. The Earth-Moon system is of this case as are all (or almost all) moon systems in Solar System at least for larger moons.

    3. The tidal force per unit mass g on the secondary along the radial direction between the centers of the astro-bodies is

          g = -(GM/r**2) - [-(GM/R**2)] = GM(1/R**2-1/r**2) ,

      where r is the radial distance measured from center of the primary, R is the center-to-center distance between the primary and the secondary, gravitational constant G = 6.67430(15)*10**(-11) (MKS units), and M is the mass of the primary.

    4. We Taylor expand g to 1st order about R in small Δr/R (with Δr=r-R) and obtain

          g_1st = GM[1/R**2-1/(R+Δr)**2] = (GM/R**2)[1-(1-2*Δr/R)] = (GM/R**2)*2*(Δr/R) ,

      where we note that g_1st falls off as 1/R**3 and as Δr/R as expected.

      Note g_1st is positive/negative for Δr positive/negative, and so the tidal force gives a stretching force as expected.

    5. There is actually another significant effect on the shape of the secondary: the centrifugal force due its axial rotation.

      For simplicity and also because it is usually approximately true in such cases, we assume the rotation axis is perpendicular to the line between the centers of the astro-bodies.

      Let ρ be the radius of the secondary.

      The centrifugal force per unit mass is (ω**2)*ρ, where ω is the angular frequency of the axial rotation (i.e., its frequency times 2π). The centrifugal force per unit mass points radially outward from the center of the secondary.

      However, because of the tidal locking ω is also the angular frequency of the circular orbit (i.e., its frequency times 2π).

      This means that (GM/R**2) = (ω**2)*R so that the gravitational force of the primary is canceled by the centrifugal force of the secondary orbital motion (i.e., the orbital centrifugal force) in the rotating frame defined by the secondary.

      We can now write rotational centrifugal force per unit mass as

          (ω**2)*ρ = (GM/R**2)*(ρ/R) .

    6. The sum of the tidal force per unit mass and the rotational centrifugal force per unit mass on the far/near side of the secondary is to 1st order in small ρ/R for the radially outward direction from the primary along the center-to-center line between the two astro-bodies is

          (GM/R**2)*2*(±ρ/R) ± (GM/R**2)*(ρ/R) = ±(GM/R**2)*3*(ρ/R) ,

      and so the rotational centrifugal force effectively increases the stretching force along the radially outward direction from the center of the primary.

      However, the rotational centrifugal force is probably best thought of as causing the equatorial bulge of the secondary and NOT as contributing effectively to the tidal force stretching. However, the above analysis shows that tidal force and the rotational centrifugal force are comparable for the simple case of tidal locking we have analyzed.

File: Mechanics file: tidal_force_1bb.html.