Weighing the Sun with five photographs

TL;DR

This paper demonstrates that the Sun's mass can be accurately calculated using only five photographs taken over three months, combined with basic astronomical laws and analytical optimization.

Contribution

It introduces a simple, practical method for students to determine the Sun's mass using minimal observational data and straightforward analytical techniques.

Findings

The Sun's mass calculated matches the accepted value closely.

The Earth's orbit parameters are accurately derived from five data points.

The method is accessible for educational purposes and requires minimal equipment.

Abstract

With only five photographs of the Sun at different dates we show that the mass of Sun can be calculated by using a telescope, a camera, and the Kepler's third law. With these photographs we are able to calculate the distance between Sun and Earth at different dates in a period of time of about three months. These distances allow us to obtain the correct elliptical orbit of Earth, proving the Kepler's first law. The analysis of the data extracted from photographs is performed by using an analytical optimization approach that allow us to find the parameters of the elliptical orbit. Also, it is shown that the five data points fit an ellipse using an geometrical scheme. The obtained parameters are in very good agreement with the ones for Earth's orbit, allowing us to foresee the future positions of Earth along its trajectory. The parameters for the orbit are used to calculate the Sun's mass by applying the Kepler's third law and Newton's law for gravitation. This method gives a result wich is in excellent agreement with the correct value for the Sun's mass. Thus, in a span of time of about three months, any student is capable to calculate the mass of the sun with only five photographs, a telescope and a camera.

Tables

Table 1. Table 1: Estimation of distances r 𝑟 r to Sun, given in Eq. ( 9 ), at different dates by using telescope and images measurements.

photo	date	$r_o\text{(in cm)}$	$r_{aS}\text{(in cm)}$	${\varphi }_{aS}$	$r\text{(in Km)}$
1	March 26th, 2019	4.4404	3.7106	0.009331	149118984.1571
2	May 7th, 2019	4.4608	3.6851	0.009224	150840671.5767
3	May 20th, 2019	5.8659	4.8291	0.009192	151364293.7829
4	June 8th, 2019	4.3963	3.6079	0.009164	151840565.2067
5	June 21th, 2019	4.5258	3.7115	0.009157	151950060.0148

Table 2. Table 2: Angular position θ 𝜃 \theta of Earth at different dates, given by Eq. ( 10 ).

photo	$t$	$\theta$
1	83	1.428
2	125	2.150
3	138	2.374
4	157	2.701
5	170	2.924

Equations

$$\frac{a^3}{T^2}=\frac{GM}{4{\pi }^2},$$

$$M_{\odot }$$

$$D_{\odot }$$

$${ϵ}_E$$

$$a_E$$

$$T=365.25[\text{mbox}days]=3.1558\times 10^7[\text{mbox}s].$$

$${\varphi }_T=\left(\frac{\pi }{180}\right)\left(\frac{m{d}_o}{f}\right),$$

$${\varphi }_{aS}=\left(\frac{r_{aS}}{r_o}\right){\varphi }_T.$$

$$r=\frac{D_{\odot }}{{\varphi }_{aS}}=D_{\odot }\left(\frac{r_o}{r_{aS}}\right)\left(\frac{180f}{\pi m{d}_o}\right).$$

$$\theta =\frac{2\pi t}{T},$$

$$r_1$$

$$r_2$$

$$r_3$$

$$r_4$$

$$r_5$$

$$r(\theta )=\frac{a\left(1-{ϵ}^2\right)}{1+ϵ\cos \theta },$$

$$ϵ=\sqrt{1-\frac{b^2}{a^2}},$$

$$E(a,ϵ)=\sum _{i=1}^5{\left(r_i-\frac{a\left(1-{ϵ}^2\right)}{1+ϵ\cos {\theta }_i}\right)}^2,$$

$$\sum _{i=1}^5\frac{r_i}{1+ϵ\cos {\theta }_i}=\sum _{i=1}^5\frac{a\left(1-{ϵ}^2\right)}{{\left(1+ϵ\cos {\theta }_i\right)}^2},$$

$$\sum _{i=1}^5\frac{r_i\left(2ϵ+\left(1+{ϵ}^2\right)\cos {\theta }_i\right)}{{\left(1+ϵ\cos {\theta }_i\right)}^2}$$

$$=\sum _{i=1}^5\frac{a\left(1-{ϵ}^2\right)\left(2ϵ+\left(1+{ϵ}^2\right)\cos {\theta }_i\right)}{{\left(1+ϵ\cos {\theta }_i\right)}^3}.$$

$$ϵ$$

$$a$$

$$R^2=\left(\sum _{i=1}^5{\left(r({\theta }_i)-\bar{r}\right)}^2\right){\left(\sum _{i=1}^5{\left(r_i-\bar{r}\right)}^2\right)}^{-1}=0.9923,$$

$$\text{mbox}Diff(\theta )=\left|\frac{a_E\left(1-{ϵ}_E^2\right)}{1+{ϵ}_E\cos \theta }-\frac{a\left(1-{ϵ}^2\right)}{1+ϵ\cos \theta }\right|.$$

$$M=\frac{4{\pi }^2{a}^3}{G{T}^2}=1.9859\times 10^{30}\text{mbox}[Kg],$$

$$\frac{d{\theta }_E}{dt}=\frac{2\pi }{T}\frac{{\left(1+{ϵ}_E\cos \theta \right)}^2}{{\left(1-{ϵ}_E^2\right)}^{3/2}}.$$

$$\frac{d\theta }{dt}=\frac{2\pi }{T}.$$

Full text

I Introduction

Our solar system is completly dominated by the Sun. Its huge mass bounds the eight planets in elliptical orbits around it. The proper explanation of the motion of each planet can be found from the Newton’s law of universal gravitation. One of the greatest triumph of Newton’s law is to provide the physics foundation to the empirical observations encompassed in the Kepler’s three laws.

Understanding Kepler’s laws allow us to calculate the relation between the shape of the orbit of each planet around Sun and the amount of time (the period) that takes one revolution around it. This relation depends only on the mass of the Sun in our solar system, and usually is used to calculate the distances to it or the period of the celestial object orbiting it. However, it can be used in the opposite way. If the perihelion and aphelion distances, and the period are known, then the mass of the central star can be calculated.

The purpose of this work is to draw attention that the mass of the Sun can be accurately calculated with only five photographs of the Sun taken from Earth. Our aim is to show that any student can perform this project in an very easy way in an span of time of about three months. Only five photographs are needed as they provide five data points for the Earth’s position around the Sun. With those five points, always a conic curve can be found to fit them. As we will show below, that conic curve will correspond to an ellipse. Thus, knowing the ellipse properties, we are able to calculate the Sun’s mass.

All this can be done by using Kepler’s third law for planetary motion. This law relates the parameters of the elliptical orbit of a planet with the mass of the central star in which it revolves. When the star is at one of the focus of the elliptical orbit, the third law tell us that the cube of the semimajor axis $a$ of the orbit is proportional to the square of the period $T$ of one revolution around that star, i.e., the ratio $a^{3}/T^{2}$ is a constant. Finding the value of that constant was one of the greatest achievements of Newton’s law of gravitation. Mathematically, the Kepler’s third law can be derived from first principles from Newton’s law of gravitation thomasc , to simply be written as

[TABLE]

where $G=6.6726\times 10^{-11}$[$\mbox{m}^{3}\mbox{s}^{-2}/\mbox{Kg}$] is the universal gravitational constant nrl , and $M$ is the mass of the central star.

It is the calculation of parameters $a$ and $T$ (and their combination) that allow us to estimate the mass of the Sun. Some of the parameters are need to be measured (the purpose of this work), while others are assumed to be known. For our proposal, we consider known quantities as the universal gravitational constant, the period of time of Earth (the time that takes to perform one revolution), and the diameter of Sun. With these known physical quantities, we are able to estimate the semimajor distance of Earth’s orbit, its eccentricity and the Sun’s mass, both from five photographs.

According to NASA nasasunshhet , the values for the Sun’s mass and diameter are

[TABLE]

Also, Earth’s orbit is elliptical, with an orbit eccentricity $\epsilon_{E}$ and a semimajor axis $a_{E}$ given by nasasunshhet2

[TABLE]

Below, we show how with five photgraphs, the values of parameters (4) and (5) can be accuratelly determined. In order to perform the estimation for the Sun’s mass, only the following few assumptions are required:

Earth revolves around Sun. 2. 2.

The time that Earth takes to perform one revolution around the Sun (a year) is known, given by

[TABLE] 3. 3.

It is assumed that the Sun’s diameter (3) is known.

The above reasonable assumptions are the key to estimate the Sun’s mass. The photographs taken to the Sun are used to calculated the distance from Earth to it at five different dates. These distances and dates correspond to the position of Earth with respect to Sun, allowing us to calculate the Sun’s mass using Kepler’s laws. In the precceding sections, we will show how from the five photographs we are able to estimate the Sun’s mass (2) with a percentage error of about 0.1%.

The most sensitive part of this project is the photograph analysis, which is thoroughly explained in Sec. II. It is from it that distances from Sun to Earth can be obtained by relating the telescope angles (telescope features) and the physical angles and distances between the two astronomical objects. Thus, by measuring angles in the images, we are able to measure different distances. Then, in Sec. III, the approach to use the data obtained from the photographs is explained, showing that it allows us to find an ellipse with parameters in agreement with (4) and (5). Besides, in Sec. IV we use the results from previous sections to calculate the Sun’s mass, and to discuss how our data adjust to the Kepler’s three laws. Finally, in the last section, we comment the validity of our assumptions.

II Photographs of the Sun and its distance estimation

During 2019, from March to June, five photographs of the Sun were taken from Santiago, Chile. The photographs are shown in Fig. 2. The used telescope for this experience is a Celestron Nexstar 8SE. Its focal length is $2032$[mm], and its magnification is [math] with the ocular focal length of $25$[mm]. It was used a Eclipsmart Solar Filter 8” SCT attached to the telescope. Also, an Iphone 6 was used to take the photographs attached to the telescope with a NexYZ 3-Axis Universal Smartphone adapter. The photographs, taken in several different dates, were used to calculate the distance from Earth to Sun. The only requirement is that the image of the Sun is completely inside of the ocular of the telescope. Once that condition is achieved, the method to calculate the distance is straightforward.

By using the features of the telescope, and compare them with the images obtained, the physical distances to the observed objects can be inferred. The procedure is explained below, and it is depicted in Fig. 1. A given telescope has an ocular focal length $d_{o}$, a focal length $f$, and an apparent field of view $m$. The telescope magnification is determined as $f/d_{o}$, and the effective ocular angle of the telescope $\phi_{T}$ can be obtained as

[TABLE]

measured in radians. For the telescope used in this work, $d_{o}=25$[mm], $f=2032$[mm], and $m=$ [math], giving $\phi_{T}\approx 0.011166$.

Once $\phi_{T}$ is found, it can be used to obtain the aparent ocular angle $\phi_{aS}$ of any object that image is lying completely inside the ocular size of the telescope, as it is shown in Fig. 1(a). Let us assume that for a given photograph (such as those shown in Fig. 2) we calculate the effective radius $r_{o}$ of the ocular and the effective radius $r_{aS}$ of the image (of the Sun in our cases). As it is shown in Fig. 2, those radii can be calculated with GeoGebra software geogebra using its own scale. However, this is not relevant as it is only important their relative magnitude.

This geometry free-source software allows to select points in the image, which are used to calculate the different radii. These radii can then obatined using geometry, and the general method is presented in Fig. 3. In general, only three points are need to determine the radius of the circumscribed circumference cox . Consider a circumference with undetermined radius. Let us choose three points in its perimeter, as $S_{1}$, $S_{2}$ and $S_{3}$. The segments $\overline{S_{1}S_{2}}$ and $\overline{S_{2}S_{3}}$ connect these three points. Now, GeoGebra alow us to calculate the mediatrix to each segment at points $M$ and $N$. The key step is to notice that mediatrixes intersect at $C_{S}$. In this form, we can construct two isosceles triangles, that share one side. Therefore, the intersection point is the center of the circle, the sides represent the radius of the circumference circumscribed to the three initial points, and Geogebra measures the segments in units of the software. We do this for the image of Sun and the ocular.

Once the two radii are known, then the apparent ocular angle of the image (of the Sun) is given simply by

[TABLE]

The apparent ocular angle $\phi_{aS}$ is the most important parameter to be determined in an image, as it allow us to relate the real physical size of the image with its physical distance $r$ to the observer that took the photograph. As it can been seen in Fig. 1(b), if the physical size of the image (in this case the diameter $D_{\odot}$ of the Sun) is known, then it can be modelled as the arc sustended by the ocular angle $\phi_{aS}$ at distance $r$, such that $D_{\odot}=r\,\phi_{aS}$. This is possible as the scales of the Sun are smaller than its distance from Earth. Therefore, we find that the distance from Earth to the Sun is

[TABLE]

In our case, when the diameter of the Sun is assumed with a value given in Eq. (3), we can readily calculate distances $r$ by using photographs and Eq. (9). In Table 1, we tabulate the different photographs (ordered by date) with their corresponding values of $r_{o}$ and $r_{aS}$ measured directly from the photographs (see Fig. 2), together with their calculated values of $\phi_{aS}$ and $r$ obtained from Eqs. (8) and (9). The radii $r_{o}$ and $r_{aS}$ depicted in Table 1 are measured in centimeters units using the GeoGebra software. The measurement units of these quantities is not important, as we only require the information of its ratio. The radii $r_{o}$ and $r_{aS}$ have four significant figures due to the camera of Iphone 6 has photographs with 3264$\times$2448 pixels. Thus, the distance $r$ in Eq. (9) have only four significant figures.

At different dates, the Sun is at different distances $r$ from Earth. As we know that Earth is in orbit around the Sun, then the data shows that the orbit is not circular. In order to calculate the shape of the Earth’s orbit we need to know the position of Earth during the whole timespan of the data acquisition. As the orbital trajectory is a two-dimensional curve, we need two coordinates to describe the position of Earth. As it is shown in Fig. 4, the coordinates can be defined using the distance from Sun to Earth and its angular position (called polar coordinates), or using the Earth position with respect to the Sun in a cartesian plane (called cartesian coordinates). Both systems are explored in the following sections.

From Table 1, we already have the magnitude of the distance $r$ for different dates, but we lack the information on the angular position $\theta$ of Earth along the trajectory (see Fig. 4). This angle can be defined from the perihelion (the closest distance from Earth to Sun) by choosing it when $\theta=0$. We can estimate the relation between date and the angular position of Earth in a very simple manner. First, as we known that Earth is revolving in a closed orbit, then let us assume that $\theta=2\pi$ in one year. Second, let us assume that Earth revolves around Sun at a constant rate. With these considerations, the angular position $\theta$ can be related with the date of each photograph in the simplest possible way as

[TABLE]

measured in radians. Here, $T$ is the time that last one terrestial year given in days by Eq. (6), and $t$ is the time (in the same units than $T$) that have passed from the perihelion of the orbit. For Earth, during 2019, the perihelion occurred approximatelly in the night of January, 2nd. The relation (10) between the angular position $\theta$ on the Earth’s orbit and the date along the year is not correct in principle, as we will show in Sec. IV. Earth does not revolve at constant rate during the complete orbit (that is the Kepler’s second law). However, we will show that $\theta$ given by Eq. (10) is a very good approximation for Earth’s orbit.

By using the dates of each photograph, in Table 2 we show that angular position of Earth along the trajectory for different dates in which the photographs were taken.

Now we have all the required data to calculate the form of the orbit of Earth. We are able to obtain the five distances $r$ and the angular positions $\theta$ extracted from Tables 1 and 2 for the five photographs of the Sun

[TABLE]

where the subindex of distances and the angular positions is related to the number of each photograph. These data will be useful in the next sections in order to find the correct form of the orbit. In Sec. III, we show that the data for the position obtained from the photographs describe a ellipse in polar coordinates. Also, we show that the same elliptical orbit can be obtained in cartesian coordinates. This orbit is in very good agreement with the ellipse with correct values for eccentricity and semimajor axis of the Earth’s orbit (4) and (5).

III Orbit of the Earth from five photographs

Using polar coordinates, we can develop an analytical process in which the equation for an ellipse is shown to be the equation that fit the previous data (II). This is achieved obtaining the ellipse parameters by an optimization scheme that resut to be in agreement with values (4) and (5).

The equation of an ellipse, written in polar coordinates, is bate

[TABLE]

where $r$ and $\theta$ are depicted and defined in Fig. 4. Here, $a$ is the semimajor axis of the ellipse. The eccentricity is defined as

[TABLE]

in terms of the semiminor axis $b$ (see Fig. 4). This equation implies that the Sun is at one of the focus, and that the distance from Sun to Earth changes during the trajectory (as the angle changes). An eccentricity $\epsilon=0$ implies that the orbit is a circle, as $b=a$.

The purpose of this section is to show that the data points (II) are points in an ellipse (12) for parameters $a$ and $\epsilon$ that are in agreement for those of Earth’s orbit. In order to prove this, we construct an optimization procedure that allow us to calculate the best fit of parameters $a$ and $\epsilon$ to the data (II). Let us define the error function

[TABLE]

where the sum is on $r_{i}$ and $\theta_{i}$ for $i=1,2,3,4,5$, the five data points displayed in (II). This error function measures how far the data points are from fit an ellipse equation (12). To have the best fit parameters, this function must be minimum. This is achieved by applying a minimization process to $E$ with respect to $a$ and $\epsilon$. Analytically, this implies that the partial derivatives of $E$ with respect to $a$ and to $\epsilon$ are both simultaneously null. The condition $\partial E/\partial a=0$ can be written as

[TABLE]

while the condition $\partial E/\partial\epsilon=0$ leads to

[TABLE]

Eqs. (15) and (III) must be solved for $a$ and $\epsilon$. Those values minimize the error function $E(a,\epsilon)$ given in (14). Performing both sumation (15) and (III) on the five data points (II) in order to solve the above equations for $\epsilon$ and $a$ can be done by analytical methods or by using computational programs. Thereby, using data (II), we obtain that the parameters that minimize the error function $E$ are

[TABLE]

These values correspond to eccentricity and the semimajor axis of the ellipse described by Eq. (12) that best fit the data points (II). Note the excellent agreement with the correct values (4) and (5) for Earth’s orbit. We can calculate the error of our prediction with respect to the values (4) and (5). We find that for the eccentricity, the percentage error is about of $\left|({\epsilon_{E}-\epsilon})/{\epsilon_{E}}\right|\approx 1.15\%$. For the semimajor axis, the percentage error is about of $\left|({a_{E}-a})/{a_{E}}\right|\approx 0.05\%$ On the other hand, the error of our fit can be quantitative estimated by calculating the coefficient of determination

[TABLE]

where $\bar{r}=(1/5)\sum_{i=1}^{5}r_{i}$ is the average of our data, and $r(\theta_{i})$ is the function (12) evaluated in the data points (II) with the found parameters (III). The value for coefficient $R^{2}$ tells us that our fit is very good.

Results (III) imply that the five points (II) correspond to points in an ellipse. This is shown in Fig. 5 where the predicted elliptical orbit with parameters (III) (yellow dashed line) is compared with the correct Earth’s orbit (blue solid line). Notice how the data points (II), in red solid circles, fit very well in both curves.

On the other hand, in order to evaluate our error in the determination the Earth’s orbit, we can construct a difference function between the correct orbit [with parameters (4) and (5)] and our calculation due to optimization of the five data points. This function reads

[TABLE]

The absolute value of this difference will tell us information about where our estimation error is larger. The difference function Diff is plotted in Fig. 6, from where we see that the error margin is larger in $\theta=0$ and $2\pi$, i.e., at the perihelion. However, this error is about $\sim 10^{5}$[Km], and thus our estimation has an approximated maximal percentage error of about $0.1$%. On the other hand, near $\theta\sim\pi$, in the lapse of time where we took the photographs, the error decreases.

In order to show that the predicted elliptical orbit of Earth coincide with the real one in a more appealing fashion, we plot the data points (II) in a cartesian plane using GeoGebra software geogebra . This is shown in Fig. 7, where the data points (II) now represent the different position of Earth along the elliptical trajectory. The data points (in cartesian coordinates) are in orange solid circles, while the sun is at one of the focus of the orbit. The black solid line is the elliptical orbit predicted by parameters (III), and as we can see the predicted orbit pass for all the five points. Another interesting feature of Fig. 7 is shown through in blue hollow circles. These points correspond to different future Earth’s positions extracted from Stellarium software estela , and they lie in the orbit fulfilled by parameters (III). In this way, our predicted orbit also foresee the future positions of Earth. Furthermore, in dashed purple line, the elliptical conic curve that fit the five data points (III) is shown. In principle, always a conic curve pass by five points. However, in our case, this conic curve does not fit the elliptical orbit predicted by parameters (III). The main reason is due to the proximity (in time) of our data. In order to use the treatment of finding an elliptical orbit by fitting a conic curve with five points, these data points should be obtained along a more extended span of time (a year), and not only approximatelly three months.

IV Mass of the Sun and Kepler’s laws

In above section, we have proved that the position of Earth (with its distances and angles at several dates) correspond to points in an elliptical orbit where the Sun is in one focus. Our estimations of parameters of the elliptical orbit are in very good agreement with the real values for Earth’s orbit. Therefore, we have proved the Kepler’s first law. This law establishes that elliptical orbits are direct consequence of the Newton central gravitational force (that is inverse to the square of the distance) owe to the central star prentis ; yasou .

Even more interesting is the Kepler’s third law (1), that relates the form of the ellipse with the mass of the central star (in our case, the Sun). By using Eq. (1), we can calculate the Sun’s mass. In Sec. III we already have found $a$ in Eq. (III). Using this, altogheter with the period $T$ given in (6), we can calculate the Sun’s mass. This calculation is simply

[TABLE]

where $T$ have been used in seconds (6). Compare this mass estimation with the real Sun’s mass value given in (2). This is the most important results of this work. With only five photographs, and using Kepler’s laws, we have been able to estimate the mass of Sun, with a percentage error of about $|(M_{\odot}-M)/M_{\odot}|\sim 0.13\%$. This shows that with only five photographs, the mass calculations are strikingly accurately.

Lastly, we are in position to discuss the Kepler’s second law. This law states that a line between the sun and the planet sweeps equal areas in equal times. This is a consequence of the conservation of the angular momentum of the planet. Basically, it establishes that any planet moves faster when is closer to the Sun, and slower when is far, implying that the angular velocity is not constant, i.e., a planet does not revolve around Sun a constant rate. In mathematical terms, the Kepler’s second law for Earth (conservation of its angular momentum) is written as $r^{2}(d\theta/dt)=\sqrt{G\,a_{E}(1-\epsilon_{E}^{2})M_{\odot}}$, where $d\theta/dt$ is the derivative of the angular position, i.e., how it changes in time bate . Using the Kepler’s laws for the elliptical form of the orbit, this equation can be put in the form

[TABLE]

Compare this result with Eq. (10) for an orbital motion at constant rate. Taking the derivative of Eq. (10), we find

[TABLE]

This was our assumption of revolutions at constant rate in the calculations of Sec. III. Nevertheless, one can notice that because $\epsilon_{E}\ll 1$, the maximum error of our assumption (occurring in the perihelion) is of the order $\left|({d\theta_{E}/dt-d\theta/dt})/({d\theta_{E}/dt})\right|\sim 2\epsilon_{E}$, i.e., the maximum percentage error of considering Earth moving at constant rate is about of 3.4%. As the Earth moves away from the perihelion, that error can be very small. For example, for photograph 1 at March 26th (day 83 and angle $\theta\approx 1.4278$), the error is $\sim|2\epsilon_{E}\cos\theta|=0.48$%. Therefore, our assumption (10) for revolutions at constant rate is justified, and in agreement with the Kepler’s second law for percentage errors of the order of 1%. This is the reason why our results are so good even when strictly we are violating the Kepler’s second law.

V Conclusions

With this work, we have shown that in a lapse of time of about three months, several features about the Earth’s trajectory and Sun can be obtained from only five photographs. Any student can perform the present analysis under the guidance of a teacher. Students perfoming this experience will realize that observations from Earth allow to calculate distances to other bodies. Most important, this experience combine practical astronomical observations and theoretical physics knowledge with analytical and computational skills. These are the kind of proficiencies that any student should possess.

Several following comments are useful to anyone attempting to repeat this experience. First, with the five point presented here, we were able to calculate the elliptical orbit (12) that assumes that the Sun is at a focus of the ellipse. With more photographs, or the same amount of photographs taken in a large lapse of time, it is even possible to prove that the Sun is at one focus of the ellipse. Secondly, we do not recommend to take data points in a small lapse of time, as that increases the magnitude of the error estimation. Also, the photographs should be taken in a sunny day with the Sun in its highest position, as this decreases the atmospheric aberration. On the other hand, in general, all the photographs in Fig. 2 have errors associated to the measurements of $r_{o}$ and $r_{aS}$. One can improve the accuracy in their calculation, by counting the pixels in the photographs. This allow us to minimize the error associated to the focus of photographs. This is important as the Sun (when is inside the telescope ocular) varies its angular size very little during a year. In Fig. 8 is shown the Sun’s maximum $32^{\prime}32^{\prime\prime}$ and minimum $31^{\prime}27^{\prime\prime}$ angular sizes. Therefore, the difference between Sun’s maximum and minimum angular sizes is $1^{\prime}5^{\prime\prime}$ or [math]. For this experience is imperative a correct calculation of $r_{aS}$ through a careful measuments of the angles.

Lastly, it is important the quality of the telescope solar filter in order to have good photographs. Also, a better focusing of the photographs can be obtained when solar spots are present.

On the other hand, the eccentricity can be measured by only knowing the position of Earth’s perihelion and aphelion. By taking photographs in those points, geometrical method can be used to estimate the eccentricity. However, for this approach, the ellipse orbit is assumed, and those two points do not provide enough information to calculate the orbit’s parameters. On the contrary, following the same procedure of this paper, the orbits of Moon and Jupiter can be calculated, as these bodies are visible from Earth using telescopes. These works are left for the future.

Finally, it is also important to consider that Sun’s diameter $D_{\odot}$ is considered a known variable along this work. Its value can be estimated experimentally, although its precision depends on several solar factors sund , which implies that it needs to be estimated periodically. However, a simple and very accurate manner to estimate Sun’s diameter can be constructed by using an eclipse. The ratio between Sun and Earth radii can be estimated using the tecniques developed in Ref. caerolsasenjomooon . Thus, by knowing Earth’s diameter (for example, through Eratosthenes’ experiment), the Sun’s diameter is readily calculated.

All the above considerations lead to or improve the results shown along this work. We strongly believe that this experience can be developed in a simpler manner by a group of students. The process of taking the photographs, the use and understanding of a telescope, the computational analysis to measure the radii, the analytical methods to evaluate the orbit, and the theoretical use of Kepler’s laws to estimate the Sun’s mass can be used by a teacher to encourage the scientific knowledge pursue in students, and at the same time, to explain to them the different processes that a scientist must do in order to unveil deep truths behind the facts.

Acknowledgements.

F.A.A. was supported by Fondecyt-Chile Grant No. 1180139.