Although geocentrism has long been debunked, we do indeed observe the universe from the perspective of Earth. We must treat the Earth as stationary to satisfy our daily observational needs. In other words, we must use the Earth as a reference frame. In doing so, we effectively reinstate the Earth’s status as the “center of the universe.” The most typical model for this is the so-called celestial coordinate system, which essentially treats the Earth as the center of the universe...
From a physics perspective, the choice of a reference frame is arbitrary, which is equivalent to the arbitrary choice of a coordinate system in mathematics. In some cases, treating the Earth as the “center of the universe” can simplify certain problems. However, it is important to note that I am not promoting geocentrism; I am merely choosing the Earth as a reference frame! Let us see what the orbit of Mars looks like under these conditions.
First, let us discuss the problem of reference frame transformation: In a Cartesian coordinate system, if the trajectories of two different particles m_1 and m_2 are given by [x(t), y(t)] and [X(t), Y(t)] respectively, then if we take m_1 as the origin (i.e., using m_1 as the reference frame), the trajectory of m_2 becomes [X(t)-x(t), Y(t)-y(t)]. The original origin was stationary, but after choosing m_1 as the reference frame, it becomes moving, with a trajectory of [-x(t), -y(t)].
Let us assume the orbits of Mars and Earth are perfect circles centered on the Sun, and use the position of opposition (where the three celestial bodies align in a straight line) as the X-axis. Then, the equations of planetary motion (with the Sun as the reference frame) can be easily derived.
For Mars: \begin{aligned} x &= R_m \cos\left(\frac{v_m t}{R_m}\right) = R_m \cos\left(\frac{2\pi t}{T_m}\right) \\ y &= R_m \sin\left(\frac{v_m t}{R_m}\right) = R_m \sin\left(\frac{2\pi t}{T_m}\right) \end{aligned}
For Earth: \begin{aligned} x &= R_e \cos\left(\frac{v_e t}{R_e}\right) = R_e \cos\left(\frac{2\pi t}{T_e}\right) \\ y &= R_e \sin\left(\frac{v_e t}{R_e}\right) = R_e \sin\left(\frac{2\pi t}{T_e}\right) \end{aligned}
If we choose the Earth as the reference frame, the trajectory is given by the following parametric equations: \begin{aligned} x &= R_m \cos\left(\frac{2\pi t}{T_m}\right) - R_e \cos\left(\frac{2\pi t}{T_e}\right) \\ y &= R_m \sin\left(\frac{2\pi t}{T_m}\right) - R_e \sin\left(\frac{2\pi t}{T_e}\right) \end{aligned}
What kind of curve is this? By inputting the data, we obtain the following curve (where the origin is the Earth):
(From this perspective, geocentrism appears to be a very unwise choice...)
If we square both sides of the parametric equations above and add them together, we get: x^2 + y^2 = R_m^2 + R_e^2 - 2R_m R_e \cos\left(\frac{2\pi t}{T_e} - \frac{2\pi t}{T_m}\right) where x^2 + y^2 is the square of the distance from Mars to Earth, which is a function of time t.
We can find that when t = \frac{1}{1/T_e - 1/T_m}, the left side of the equation reaches its minimum value. In other words, for an outer planet, another “opposition” occurs at a time \frac{1}{1/T_e - 1/T_m} after the previous “opposition.”
When \frac{2\pi t}{T_e} - \frac{2\pi t}{T_m} = \frac{\pi}{2}, i.e., t = \frac{1}{4(1/T_e - 1/T_m)}, the left side of the equation reaches its maximum value. In other words, the planet reaches “conjunction” at a time \frac{1}{4(1/T_e - 1/T_m)} after the “opposition.”
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