Gravity fields and $J_2$

Now it is time to talk about gravity fields. The equatorial bulge of a planet is due to its rotation. Faster rotation implies larger bulge. One consequence of this is, that the gravity field of the planet is not that of a point mass. In particular, the field does not have the spherical symmetry, which we assumed earlier. This will not have a significant impact on a satellite which moves in the equatorial plane. However, if the satellite moves in a plane which is at an anle to the equatorial plane, one can imagine that at a point when the satellite is above or below the plane there will be a pull toward the equator due to the excess mass on the equator. This will result in a changing of the plane of motion of the satellite, in particular in a change of the mode.

We start by giving the general expression for a gravity field for a rotationally flattened (oblate) planet:

$$U({\bf r}) = -\frac{\mu}{r}\left[1-J_2 \left(\frac{R}{r}\ri... ... \left(\frac{R}{r}\right)^3 P_3(\sin \delta) \ldots \right].$$

This is an approximation. Here $\delta$ is the latitude, or declination, or angle of elevation above the equator, $R$ is the radius of the planet and $P_i$ are Legendre polynomials, i.e.,

$$P_0(x) = 1$$

$$P_1(x) = x$$

$$P_2(x) = \frac{1}{2}(3x^2-1)$$

and so on. Since $J_2$ is usually much larger than $J_3$ and the rest, we will include only that term.

First, we have

$$\sin \delta = \frac{z}{r}$$

and we have

$$U({\bf r}) = -\frac{\mu}{r} - \frac{\mu J_2 R^2}{2r^3} + \frac{3 \mu J_2 R^2}{2r^5} z^2$$

The first term is the usual Newtonian attraction. We can observe that in the equatorial plane, $z=0$ we have

$$U({\bf r}) = -\frac{\mu}{r} - \frac{\mu J_2 R^2}{2r^3}$$

which corresponds to a central force field of the type of perturbation we have discussed above. Hence, we know how the pericenter precesses for an orbit in the equatorial plane.