Experiment 12: Planetary Distances

Objective :

To Measure the distances of planets in our Solar System.

Introduction :

In this experiment we shall measure the distance of various planets from the sun. Two different techniques shall be employed here. One for the internal planets, and the other for the external planets. Both will rely on the measurement of a certain parameter known as the elongation of the planet. This shall be measured by noting the coordinates of the planets on the ecliptic co-ordinate system.

Theory:

Far the purpose of this experiment, we shall make a few assumptions. Firstly, we shall take the orbits of all planets to be circular. This assumption is fairly accurate for most planets. The most eccentric orbit is that of Mercury $(e=0.2)$, But even there we shall see that the calculated distance is close to the actual value. Secondly, we assume that the solar system is entirely planar. This is also fairly accurate, with the exception of that of mercury (~${\small 7^\circ}$).

Inferior Planets:

Let S denote the Sun, P denote the planet, and E denote the Earth. Let us consider a frame of reference rotating with an angular velocity equal to that of the earths revolutionary motion. In this frame the line SE will be fixed. Now, the planet P, could be anywhere on the inner circle as shown in Fig 1. We define $\angle SEP$ to be the Elongation of planet P. Note that in the case of an inner planet the elongation cannot attain all values. It can attain all values ranging from ${0^\circ}$ until a certain maximum. Note that this maximum will be attained when the line EP will be tangent to the the inner circle as shown in Fig 1. Let us denote this maximum elongation by ${\varepsilon}$

Now we see that \begin{eqnarray} \sin \varepsilon &=& \frac{SP}{SE}\\ \Rightarrow SP &=& \sin \varepsilon \times SE \end{eqnarray} Since SE is nothing but 1 Astronomical Unit(A.U.), thus we get the distance of the planet from the sun in Astronomical Units.

To measure the angle ${\varepsilon}$ we shall make use of a plug-in called "Angle Measure". This can be enabled in the configuration window.

Procedure (for Mercury) :

• Step 1: Remove the Atmosphere and Ground. (Press A and G)
  
• Step 2: Locate and center on the Sun. (Use Ctrl+F to open the search window)
  
• Step 3: Zoom out until Mercury is also visible. (Ctrl + Down arrow)
  
• Step 4: Switch to Equatorial mount. (Ctrl + M)
  
• Step 5: Increase the time 1 day at a time and go to the day when it appears that Mercury is furthest from the sun. (Use =)
  
• Step 6: Enable the "Angle Measure" Plug-in. (Ctrl+A)
  
• Step 7: Measure the angle from the Sun to Mercury.
  
• Step 8: Measure the angle for a few hours around that time, so as to get the maximum elongation, correct to the time of an hour
  
• Step 9: Use Equation 1 to calculate the Sun Mercury distance
  

For example, doing the above steps with mercury gives ${\varepsilon}$ to be around $25^\circ$ $20$'. This puts the Sun Mercury distance to 0.42 A.U. This has roughly a 10% error. Note that since this is the most eccentric orbit, hence all of our errors for the remaining planets must be less than 10%. Also remember that ${\varepsilon}$ is not constant and varies slightly in different revolutions.

Further Work :

• Repeat the above procedure for Venus

Superior Planets:

In order to understand how to measure the distance to a superior planet, we must first understand the concept of a Synodic period. This is defined as the time between two successive similar configuration of the Sun, Planet, Earth System. For example, 2 successive similar configurations of a general superior planet with the earth is shown in Fig 2. Let us try to understand this better. Let the synodic period of the planet P be Sy. Let the time period of revolution of the Earth be $T_E$, and that of the Planet P be $T_P$. Since the Earth is interior, the earth will revolve faster than planet P. By the Time the Earth takes one complete revolution the planet P will have moved further ahead. The earth would have to move a little further ahead up to position $E_2$ to recover the same configuration.

Let the angular velocity of the Earth be $n_E = 2\pi / T_E$, and similarly that for the planet P be $n_P = 2 \pi/T_P$. Now, as the $n_E > n_P$ the radius vector $SE$ will gain on the radius vector $SP$ at a rate of $n_E-n_P$. In one synodic period the radius vector SE shall gain on the vector $SP$ by $2 \pi$ radians.

Therefore, \begin{eqnarray} &&Sy \times (n_E - n_P)&=& 2\pi\\ & \Rightarrow & n_E - n_P &=& 2 \pi / Sy\\ & \Rightarrow & 2\pi /T_E - 2\pi / T_P &=& 2\pi /Sy\\ & \Rightarrow & 1/Sy &=& 1/T_E - 1/T_P \end{eqnarray} Note that Sy is a quantity that can be measured easily by an observer on earth.

Now, to measure the distance of the planet P to the Sun, we shall start our observations at opposition, i.e. when the elongation is $180^\circ$. This corresponds to situation 1 in Fig 2. Now we take our next observation after some time $t$, at situation 2. In this time the elongation has increased from $0^\circ$ to $\theta$ at a angular velocity of $n_E - n_P$.

Therefore, \begin{eqnarray} \theta &=& (n_E-n_P) \times t\\ &=& t \times (2 \pi/T_E - 2\pi/T_P)\\ &=& 2\pi \times t/S_y\\ \end{eqnarray}

Thus the angle theta can be calculated. Now, $\angle PES$ can be measured from the sky. Thus we can now also calculate $\angle EPS$. Thus, using sine rule,

\begin{eqnarray} \frac{\sin \angle EPS}{SE} = \frac{\sin \angle PES}{SP} \end{eqnarray}

Thus we can calculate the distance to superior planets.