Does an elliptical orbit of a satellite maintain its orientation as the Earth revolves? Imagine a satellite in an elliptical orbit around Earth. As the earth travels around the sun, does the elliptical orbit of the satellite swing around Earth, as pictured in A, or does it maintain a consistent orientation as pictured in B? Are both of these options wrong?
Option A

Option B

 A: Actually, both are quite possible. In the general case, for an arbitrary elliptical orbit, what you'll tend to find is that B is true (granted there will be some precession, but not usually in line with the Sun). However, it is possible to set up an orbit (such as a Sun-synchronous one) in which A is true. But orbits such as this require planning and precise positioning.
To understand why A does not have to be true (and in fact, why A is not the general case), you should bear in mind that when orbiting a body, your orientation is irrelevant.  The Sun ensures both Earth and the satellite will orbit around it, but it has very little influence over the specific way they are pointing.
Imagine that Earth did not rotate on its axis; you'd see the same stars in the same positions every night. However, you'd still experience the Sun rise and set (once per year). The sun doesn't drag the orientation of orbiters (at least, not in a significantly mention-able way). All that is important from the satellite's perspective is Earth, which has the dominant sphere of influence.
As I mentioned earlier, however, it is possible to establish orbits that follow the Sun. A sun-synchronous orbit can be very useful as it allows a satellite to always be in view of the Sun and thus continuously collect solar power. This is possible due to a combination of several effects including the small amount of sway the Sun's gravity has over the satellite. However, to establish this type of orbit requires the use of a specific angle of incidence (which varies based on altitude). In Low Earth Orbit, the inclination is around $98^\circ$.
An important thing to note is that while most orbits do not generally follow the Sun, they technically do precess (meaning they usually do not look exactly like B but it's a good approximation/generalization). This precession happens at different rates depending mostly on altitude, inclination, and eccentricity of the orbit. But that's just extra info in case you were curious.
P.S.
<3 rocket science
A: The orbit of a satellite is determined by 6 parameters, the orbital elements:


*

*The eccentricity $e$

*The semimajor axis ($a$)

*The inclination ($i$)

*The longitude of the ascending node ($\Omega$)

*The argument of periapsis ($\omega$)

*The mean anomaly at a specific time ($M_0$)



The plane of reference is the equator, and the reference direction is the Vernal Point.
If the reference frame of the Earth were an inertial frame, then these parameters would be constants, and your Figure B would be valid. However, the orbits of satellites are perturbed due to several effects:


*

*The non-sphericity and non-homogeneous mass distribution of the Earth

*The gravitational field of sun and moon

*Solar radiation pressure

*Atmospheric drag (for low-orbit satellites)

*General-Relativistic corrections


The first effect is the largest. The non-sphericity of the Earth can be approximated by a quantity $J_2\approx 1.083\times 10^{-3}$, and it has two consequences:


*

*a rotation of the line of  apsides: the position of the perigee varies as
$$
\dot{\omega} = \frac{3\pi}{T}\frac{J_2\,R_\text{e}^2}{a^2(1-e^2)^2}\left(2 - \frac{5}{2}\sin^2 i\right),
$$
where $T$ is the orbital period of the satellite, and $R_\text{e}$ is the equatorial radius of the Earth. Interestingly, one can choose the inclination $i$ in such a way that $\dot{\omega}=0$, namely by setting $i = \sin^{-1}\sqrt{4/5}$, so that $i=63.43^\circ$ or $i=116.57^\circ$. Such orbits are called critically inclined orbits or Molniya orbits.

*a nodal precession:
$$
\dot{\Omega} = -\frac{3\pi}{T}\frac{J_2\,R_\text{e}^2}{a^2(1-e^2)^2}\cos i.
$$
One can choose the orbital parameters in such a way that the period of this nodal precession is equal to the orbital period of the Earth around the Sun, which implies $\dot{\Omega} =0.9855^\circ/\text{day}$. The result is a Sun-synchronous orbit, which means that the orbital plane of the satellite rotates as in your Figure A.


For more details, see also this link and this link.
