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Calculation of the distance between the earth and the Sun from measurements taken in occasion of a transit of Venus
By Dr. Carme Jordi i Nebot
1. IntroductionThe main goal of this document is to describe the grounds on which the calculation of the distance EarthSun from the observation of a transit of Venus is founded. To describe the rigorous calculation is not intended, because it involves a level of mathematical knowledge that the most students and general public have not still acquired. Who is interested in this rigorous treatment can check the bibliography at the end of the document. The determination of the distance EarthSun is based on the perspective effect by which, from two different locations, Venus is projected over two distinct points of the solar disc. Therefore, observations from distant places on Earth must be combined: the perspective effect is the more important the further this two localions are, and so a more accurate measurement of the astronomical unit can be derived. Observations must be complemented by the laws of Kepler which describe the orbits of planets around the Sun and were deduced by Johannes Kepler from very numerous observations of the planetary motion. The law of Universal Gravitation, formulated by sir Isaac Newton, applied to the case of two bodies moving around a common centre of mass explains the three empirical laws of Kepler. In this document, a simplified method to calculate the distance between the Earth and the Sun is described. It needs simultaneous observations of the transit from two distant locations on the surface of Earth, in order to measure the distance between the centre of Venus over the solar disc as seen from each point. Another method to carry out this calculation is founded on the comparison of the time Venus spends on crossing the solar disc from side to side (the duration of the transit) when observed from different places. The mathematical approach of both methods is similar, but the last one is quite more demanding from an observational and mathematical point of view. 2. Using simultaneous observations from different placesThis method is founded on knowing the distances between the centre of Venus, projected over the solar disc, in a given instant, as seen from two different locations. Therefore, to carry out simultaneous observations from two different places on Earth (the more distant the better) is imperatively needed to use this simplified method, and the easiest way to achieve it is to take a photograph of the Sun at the same moment from both locations. From two different places M_{1} and M_{2} (see Figure 1) and at the same time t, Venus is projected on two different positions V_{1} and V_{2} over the solar disc due to the perspective effect. Here is happening the same than when we put a finger very near our eyes and look at it closing alternatively one eye or the other. Then, we see the finger projected over the background in a different position with each eye. In the case of the transit of Venus, the two observers are equivalent to the eyes, Venus to the finger and the Sun to the background. Depending on the place from where we observe, Venus is projected over a point or another of the solar disc. The measurement of the distance between V_{1} and V_{2} in terms of the solar radius makes possible to determine the average distance EarthSun, but not in an easy way, in spite of the simplifications we adopt here.
Let's present the mathematical formulation. 2.1 Determination of the distance in the moment of the observationLet us suppose that O is the centre of the Earth, C the centre of the Sun and V_{1} and V_{2} the observed centres of the Venus' projections as seen from points M_{1} and M_{2},respectively, the angles D_{1} and D_{2}, the angular separations between each centre of Venus and the centre of the Sun as seen again from M_{1} and M_{2}, respectively; that is to say, the parallax angles CM_{1}V_{1} and CM_{2}V_{2}. Analogously, we can define the angles π_{s} and π_{v} as the angular separations between M_{1} and M_{2} as seen from the Sun and from Venus, respectively; that is, the angles M_{1}CM_{2} and M_{1}VM_{2}. Given that the four points M_{1}, M_{2}, C and V are not in the same plane (the most usual case is that M_{1} and M_{2} are not on the same meridian and the Earth, Venus and the Sun not perfectly aligned), the geometry of the problem becomes a bit complicated, in such a way that, as can be seen in figure 2, the distance Δπ between the two centres of Venus is the only observable magnitude. It corresponds to Δπ = π_{v} – π_{s} and makes possible to calculate the distance to the Sun. The carrying out of the measurement of Δπ from the two photographs can be done by finding the position of the centre of Venus with respect to a common point of reference on the solar disc (a spot, for instance), and then comparing the difference with the total size of the disc. Measurements on the photographs are taken in length units (mm for instance) and we must transform them to angles. So, we need to know the angular size of the Sun. Let's call (x_{1},y_{1}) and (x_{2},y_{2}) to the spaces, in mm, between the centre of Venus and the reference spot in the horizontal and vertical directions for each photograph. Then, the spaces in arcseconds are obtained by multiplying each magnitude x_{1} and y_{1} by the factor Ã'(arcseconds)/Ã'_{1}(mm) and each magnitude x_{2} and y_{2} by Ã'(arcseconds)/Ã'_{2}(mm) where Ã'(arcseconds) and Ã'(mm) are the diameter of the Sun expressed in arcseconds and mm, respectively. Ã'_{1}(mm) and Ã'_{2}(mm) have the same value if both photographs share the scale. The distance between the centres of Venus in the two photographs is: Δπ(segons d'arc) = [ (x_{2} – x_{1})^{2}+ (y_{2} – y_{1})^{2} ]^{1/2} [1] If the photographs are taken with telescopes that give the same scale and the solar disc is situated exactly in the same place of the photographs, then a corner of the photograph can be taken as a reference point. Besides, in this case Ã'_{1}(mm) = Ã'_{2}(mm) and the previous equation can be written as: Δπ(segons d'arc) = [ (x_{2} – x_{1})^{2}+ (y_{2} – y_{1})^{2} ]^{1/2}Â·Ã'(segons d'arc)/Ã'(mm) [1bis] This is the supposition adopted from now on.
Let's suppose that r_{V} and r_{T} are the distances between the centre of the Sun and the centres of Venus and the Earth, respectively, at the moment t of the observation. As the projection d of the distance between M_{1} and M_{2} over the plane perpendicular to OC is small if compared with the distances EarthSun and EarthVenus, we can approximate:
π_{s} = d/r_{T} and from this, the following is deduced:
π_{v} = π_{s} r_{T}/(r_{T} r_{V}) and so, π_{s} = d/r_{T} = Δπ (r_{T}/r_{V}  1) This last formula clearly express that, if we know the angular distance Δπ between the centres V_{1} and V_{2}, and the ratio r_{T}/r_{V} between the distances EarthSun and VenusSun, the parallax π_{s} can be deduced; and that if we know the projected distance d between the two locations where observations are carried out, the distance r_{T} can be calculated. (In these expressions, the values of π_{v}, π_{s} and Δπ are given in radians. To transform them to arcseconds to make them compatible with equation [1], multiplying by number 64800/π is enough). Δπ is the observable magnitude, d can be determined as is explained beneath and then, the only magnitude we have to know to solve the problem is the ratio r_{T}/r_{V} between the distances EarthSun and VenusSun. The orbits of the Earth and Venus around the Sun are slightly elliptical and so, the ratio r_{T}/r_{V} is not constant in time. To know this ratio at the exact time t, we need to use the first law of Kepler, which states that the Sun is situated on one of the ellipse's focuses. Then, the distance to the planet is given by: r_{p}(t)=R_{p} (1  e_{p} cos E_{p}(t)) where R_{p} is the semimajor axis of the orbit, e_{p} the eccentricity and E_{p}(t) the eccentric anomaly at the instant t. According to this: r_{T}/r_{v}=[R_{T} (1  e_{T} cos E_{T})] / [R_{V} (1  e_{V} cos E_{V})] The third law of Kepler gives a relation between the semimajor axes of the orbits and its correspounding periods P_{p}: (R_{T} / R_{V})^{3} = (P_{T} / P_{V})^{2} in such a way that r_{T}/r_{v}=(P_{T} / P_{V})^{2/3} (1  e_{T} cos E_{T}) / (1  e_{V} cos E_{V}) [2] Until this point, we have been able to determine π_{s} and r_{T}; the parallax and the distance EarthSun at the instant t of the observation. 2.2 Determination of the average distanceTo determine the average distance EarthSun (R_{T}) and the correspounding average parallax π_{o}, which are related by means of the equatorial radius R according to: π_{o} ≈≈ R/R_{T} some other considerations must be done. If the projection d of the distance between M_{1} and M_{2} on the plane normal to the direction EarthSun is expressed in terms of the equatorial terrestrial radius, and the distance EarthSun in terms of the average distance, we have: π_{s} = [(d/R) / (r_{T} /R_{T})] (R/R_{T}) ≈ [(d/R) / (r_{T} /R_{T})] π_{o} The ratio r_{T}/R_{T} can be deduced from the first law of Kepler as: r_{T}/R_{T} = 1  e_{T} cos E_{T}(t) and so, we only have to calculate d/R (see Figure 3).
Figure 3. Projection of the distance between M_{1} and M_{2} on the plane normal to the direction EarthSun. Calculating the vectorial product between vectors M_{1}M_{2} and OC, the value of sin θ is obtained, due to M_{1}M_{2} Ã— OC = M_{1}M_{2} r_{T} sin θ The next expression is clear in figure 3: d = M_{1}M_{2} cos (90 – θ) = M_{1}M_{2} sin θ and then, d = M_{1}M_{2} Ã— OC / r_{T} Now we just have to find M_{1}M_{2} Ã— OC. Calculation of the vector OCThis vector can be expressed in terms of the equatorial coordinates of the Sun (α,δ) at the moment of the observation as:
x=r_{T} cos δ cos α Calculation of the vector M_{1}M_{2}The position of each observer can be expressed as (see figure 4):
x=R cos φ cos (λ+T_{G}) where φ and λ are the geographical coordinates (latitude and longitude) and T_{G}=T_{G}(0) + 1.00273791 t. The time t of the observation must be expressed in the scale of universal time (UT). For the most countries in Europe UT = official time – 2^{h} in June.
Figure 4. Positions of a celestial object (for instance, the Sun) and of an observer on Earth in equatorial coordinates. The coordinates of the vector M_{1}M_{2} can be easily found as:
X=x_{1} – x_{2} Practical exampleOne of the programs available on https://gaia.am.ub.es/Twiki/bin/view/ServiAstro/AgendaAstronomica2004 is founded on this formulation and can be used as a practical example. The second program is based on the comparison of the duration of the transit as observed from two different locations. 3. For more information

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Calculation of the distance between the earth and the Sun from measurements taken in occasion of a transit of Venus
By Dr. Carme Jordi i Nebot
1. IntroductionThe main goal of this document is to describe the grounds on which the calculation of the distance EarthSun from the observation of a transit of Venus is founded. To describe the rigorous calculation is not intended, because it involves a level of mathematical knowledge that the most students and general public have not still acquired. Who is interested in this rigorous treatment can check the bibliography at the end of the document. The determination of the distance EarthSun is based on the perspective effect by which, from two different locations, Venus is projected over two distinct points of the solar disc. Therefore, observations from distant places on Earth must be combined: the perspective effect is the more important the further this two localions are, and so a more accurate measurement of the astronomical unit can be derived. Observations must be complemented by the laws of Kepler which describe the orbits of planets around the Sun and were deduced by Johannes Kepler from very numerous observations of the planetary motion. The law of Universal Gravitation, formulated by sir Isaac Newton, applied to the case of two bodies moving around a common centre of mass explains the three empirical laws of Kepler. In this document, a simplified method to calculate the distance between the Earth and the Sun is described. It needs simultaneous observations of the transit from two distant locations on the surface of Earth, in order to measure the distance between the centre of Venus over the solar disc as seen from each point. Another method to carry out this calculation is founded on the comparison of the time Venus spends on crossing the solar disc from side to side (the duration of the transit) when observed from different places. The mathematical approach of both methods is similar, but the last one is quite more demanding from an observational and mathematical point of view. 2. Using simultaneous observations from different placesThis method is founded on knowing the distances between the centre of Venus, projected over the solar disc, in a given instant, as seen from two different locations. Therefore, to carry out simultaneous observations from two different places on Earth (the more distant the better) is imperatively needed to use this simplified method, and the easiest way to achieve it is to take a photograph of the Sun at the same moment from both locations. From two different places M_{1} and M_{2} (see Figure 1) and at the same time t, Venus is projected on two different positions V_{1} and V_{2} over the solar disc due to the perspective effect. Here is happening the same than when we put a finger very near our eyes and look at it closing alternatively one eye or the other. Then, we see the finger projected over the background in a different position with each eye. In the case of the transit of Venus, the two observers are equivalent to the eyes, Venus to the finger and the Sun to the background. Depending on the place from where we observe, Venus is projected over a point or another of the solar disc. The measurement of the distance between V_{1} and V_{2} in terms of the solar radius makes possible to determine the average distance EarthSun, but not in an easy way, in spite of the simplifications we adopt here.
Let's present the mathematical formulation. 2.1 Determination of the distance in the moment of the observationLet us suppose that O is the centre of the Earth, C the centre of the Sun and V_{1} and V_{2} the observed centres of the Venus' projections as seen from points M_{1} and M_{2},respectively, the angles D_{1} and D_{2}, the angular separations between each centre of Venus and the centre of the Sun as seen again from M_{1} and M_{2}, respectively; that is to say, the parallax angles CM_{1}V_{1} and CM_{2}V_{2}. Analogously, we can define the angles π_{s} and π_{v} as the angular separations between M_{1} and M_{2} as seen from the Sun and from Venus, respectively; that is, the angles M_{1}CM_{2} and M_{1}VM_{2}. Given that the four points M_{1}, M_{2}, C and V are not in the same plane (the most usual case is that M_{1} and M_{2} are not on the same meridian and the Earth, Venus and the Sun not perfectly aligned), the geometry of the problem becomes a bit complicated, in such a way that, as can be seen in figure 2, the distance Δπ between the two centres of Venus is the only observable magnitude. It corresponds to Δπ = π_{v} – π_{s} and makes possible to calculate the distance to the Sun. The carrying out of the measurement of Δπ from the two photographs can be done by finding the position of the centre of Venus with respect to a common point of reference on the solar disc (a spot, for instance), and then comparing the difference with the total size of the disc. Measurements on the photographs are taken in length units (mm for instance) and we must transform them to angles. So, we need to know the angular size of the Sun. Let's call (x_{1},y_{1}) and (x_{2},y_{2}) to the spaces, in mm, between the centre of Venus and the reference spot in the horizontal and vertical directions for each photograph. Then, the spaces in arcseconds are obtained by multiplying each magnitude x_{1} and y_{1} by the factor Ã'(arcseconds)/Ã'_{1}(mm) and each magnitude x_{2} and y_{2} by Ã'(arcseconds)/Ã'_{2}(mm) where Ã'(arcseconds) and Ã'(mm) are the diameter of the Sun expressed in arcseconds and mm, respectively. Ã'_{1}(mm) and Ã'_{2}(mm) have the same value if both photographs share the scale. The distance between the centres of Venus in the two photographs is: Δπ(segons d'arc) = [ (x_{2} – x_{1})^{2}+ (y_{2} – y_{1})^{2} ]^{1/2} [1] If the photographs are taken with telescopes that give the same scale and the solar disc is situated exactly in the same place of the photographs, then a corner of the photograph can be taken as a reference point. Besides, in this case Ã'_{1}(mm) = Ã'_{2}(mm) and the previous equation can be written as: Δπ(segons d'arc) = [ (x_{2} – x_{1})^{2}+ (y_{2} – y_{1})^{2} ]^{1/2}Â·Ã'(segons d'arc)/Ã'(mm) [1bis] This is the supposition adopted from now on.
Let's suppose that r_{V} and r_{T} are the distances between the centre of the Sun and the centres of Venus and the Earth, respectively, at the moment t of the observation. As the projection d of the distance between M_{1} and M_{2} over the plane perpendicular to OC is small if compared with the distances EarthSun and EarthVenus, we can approximate:
π_{s} = d/r_{T} and from this, the following is deduced:
π_{v} = π_{s} r_{T}/(r_{T} r_{V}) and so, π_{s} = d/r_{T} = Δπ (r_{T}/r_{V}  1) This last formula clearly express that, if we know the angular distance Δπ between the centres V_{1} and V_{2}, and the ratio r_{T}/r_{V} between the distances EarthSun and VenusSun, the parallax π_{s} can be deduced; and that if we know the projected distance d between the two locations where observations are carried out, the distance r_{T} can be calculated. (In these expressions, the values of π_{v}, π_{s} and Δπ are given in radians. To transform them to arcseconds to make them compatible with equation [1], multiplying by number 64800/π is enough). Δπ is the observable magnitude, d can be determined as is explained beneath and then, the only magnitude we have to know to solve the problem is the ratio r_{T}/r_{V} between the distances EarthSun and VenusSun. The orbits of the Earth and Venus around the Sun are slightly elliptical and so, the ratio r_{T}/r_{V} is not constant in time. To know this ratio at the exact time t, we need to use the first law of Kepler, which states that the Sun is situated on one of the ellipse's focuses. Then, the distance to the planet is given by: r_{p}(t)=R_{p} (1  e_{p} cos E_{p}(t)) where R_{p} is the semimajor axis of the orbit, e_{p} the eccentricity and E_{p}(t) the eccentric anomaly at the instant t. According to this: r_{T}/r_{v}=[R_{T} (1  e_{T} cos E_{T})] / [R_{V} (1  e_{V} cos E_{V})] The third law of Kepler gives a relation between the semimajor axes of the orbits and its correspounding periods P_{p}: (R_{T} / R_{V})^{3} = (P_{T} / P_{V})^{2} in such a way that r_{T}/r_{v}=(P_{T} / P_{V})^{2/3} (1  e_{T} cos E_{T}) / (1  e_{V} cos E_{V}) [2] Until this point, we have been able to determine π_{s} and r_{T}; the parallax and the distance EarthSun at the instant t of the observation. 2.2 Determination of the average distanceTo determine the average distance EarthSun (R_{T}) and the correspounding average parallax π_{o}, which are related by means of the equatorial radius R according to: π_{o} ≈≈ R/R_{T} some other considerations must be done. If the projection d of the distance between M_{1} and M_{2} on the plane normal to the direction EarthSun is expressed in terms of the equatorial terrestrial radius, and the distance EarthSun in terms of the average distance, we have: π_{s} = [(d/R) / (r_{T} /R_{T})] (R/R_{T}) ≈ [(d/R) / (r_{T} /R_{T})] π_{o} The ratio r_{T}/R_{T} can be deduced from the first law of Kepler as: r_{T}/R_{T} = 1  e_{T} cos E_{T}(t) and so, we only have to calculate d/R (see Figure 3).
Figure 3. Projection of the distance between M_{1} and M_{2} on the plane normal to the direction EarthSun. Calculating the vectorial product between vectors M_{1}M_{2} and OC, the value of sin θ is obtained, due to M_{1}M_{2} Ã— OC = M_{1}M_{2} r_{T} sin θ The next expression is clear in figure 3: d = M_{1}M_{2} cos (90 – θ) = M_{1}M_{2} sin θ and then, d = M_{1}M_{2} Ã— OC / r_{T} Now we just have to find M_{1}M_{2} Ã— OC. Calculation of the vector OCThis vector can be expressed in terms of the equatorial coordinates of the Sun (α,δ) at the moment of the observation as:
x=r_{T} cos δ cos α Calculation of the vector M_{1}M_{2}The position of each observer can be expressed as (see figure 4):
x=R cos φ cos (λ+T_{G}) where φ and λ are the geographical coordinates (latitude and longitude) and T_{G}=T_{G}(0) + 1.00273791 t. The time t of the observation must be expressed in the scale of universal time (UT). For the most countries in Europe UT = official time – 2^{h} in June.
Figure 4. Positions of a celestial object (for instance, the Sun) and of an observer on Earth in equatorial coordinates. The coordinates of the vector M_{1}M_{2} can be easily found as:
X=x_{1} – x_{2} Practical exampleOne of the programs available on https://gaia.am.ub.es/Twiki/bin/view/ServiAstro/AgendaAstronomica2004 is founded on this formulation and can be used as a practical example. The second program is based on the comparison of the duration of the transit as observed from two different locations. 3. For more information
 
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> >  Calculation of the distance between the earth and the Sun from measurements taken in occasion of a transit of Venus
By Dr. Carme Jordi i Nebot
1. IntroductionThe main goal of this document is to describe the grounds on which the calculation of the distance EarthSun from the observation of a transit of Venus is founded. To describe the rigorous calculation is not intended, because it involves a level of mathematical knowledge that the most students and general public have not still acquired. Who is interested in this rigorous treatment can check the bibliography at the end of the document. The determination of the distance EarthSun is based on the perspective effect by which, from two different locations, Venus is projected over two distinct points of the solar disc. Therefore, observations from distant places on Earth must be combined: the perspective effect is the more important the further this two localions are, and so a more accurate measurement of the astronomical unit can be derived. Observations must be complemented by the laws of Kepler which describe the orbits of planets around the Sun and were deduced by Johannes Kepler from very numerous observations of the planetary motion. The law of Universal Gravitation, formulated by sir Isaac Newton, applied to the case of two bodies moving around a common centre of mass explains the three empirical laws of Kepler. In this document, a simplified method to calculate the distance between the Earth and the Sun is described. It needs simultaneous observations of the transit from two distant locations on the surface of Earth, in order to measure the distance between the centre of Venus over the solar disc as seen from each point. Another method to carry out this calculation is founded on the comparison of the time Venus spends on crossing the solar disc from side to side (the duration of the transit) when observed from different places. The mathematical approach of both methods is similar, but the last one is quite more demanding from an observational and mathematical point of view. 2. Using simultaneous observations from different placesThis method is founded on knowing the distances between the centre of Venus, projected over the solar disc, in a given instant, as seen from two different locations. Therefore, to carry out simultaneous observations from two different places on Earth (the more distant the better) is imperatively needed to use this simplified method, and the easiest way to achieve it is to take a photograph of the Sun at the same moment from both locations. From two different places M_{1} and M_{2} (see Figure 1) and at the same time t, Venus is projected on two different positions V_{1} and V_{2} over the solar disc due to the perspective effect. Here is happening the same than when we put a finger very near our eyes and look at it closing alternatively one eye or the other. Then, we see the finger projected over the background in a different position with each eye. In the case of the transit of Venus, the two observers are equivalent to the eyes, Venus to the finger and the Sun to the background. Depending on the place from where we observe, Venus is projected over a point or another of the solar disc. The measurement of the distance between V_{1} and V_{2} in terms of the solar radius makes possible to determine the average distance EarthSun, but not in an easy way, in spite of the simplifications we adopt here.
Let's present the mathematical formulation. 2.1 Determination of the distance in the moment of the observationLet us suppose that O is the centre of the Earth, C the centre of the Sun and V_{1} and V_{2} the observed centres of the Venus' projections as seen from points M_{1} and M_{2},respectively, the angles D_{1} and D_{2}, the angular separations between each centre of Venus and the centre of the Sun as seen again from M_{1} and M_{2}, respectively; that is to say, the parallax angles CM_{1}V_{1} and CM_{2}V_{2}. Analogously, we can define the angles π_{s} and π_{v} as the angular separations between M_{1} and M_{2} as seen from the Sun and from Venus, respectively; that is, the angles M_{1}CM_{2} and M_{1}VM_{2}. Given that the four points M_{1}, M_{2}, C and V are not in the same plane (the most usual case is that M_{1} and M_{2} are not on the same meridian and the Earth, Venus and the Sun not perfectly aligned), the geometry of the problem becomes a bit complicated, in such a way that, as can be seen in figure 2, the distance Δπ between the two centres of Venus is the only observable magnitude. It corresponds to Δπ = π_{v} – π_{s} and makes possible to calculate the distance to the Sun. The carrying out of the measurement of Δπ from the two photographs can be done by finding the position of the centre of Venus with respect to a common point of reference on the solar disc (a spot, for instance), and then comparing the difference with the total size of the disc. Measurements on the photographs are taken in length units (mm for instance) and we must transform them to angles. So, we need to know the angular size of the Sun. Let's call (x_{1},y_{1}) and (x_{2},y_{2}) to the spaces, in mm, between the centre of Venus and the reference spot in the horizontal and vertical directions for each photograph. Then, the spaces in arcseconds are obtained by multiplying each magnitude x_{1} and y_{1} by the factor Ã'(arcseconds)/Ã'_{1}(mm) and each magnitude x_{2} and y_{2} by Ã'(arcseconds)/Ã'_{2}(mm) where Ã'(arcseconds) and Ã'(mm) are the diameter of the Sun expressed in arcseconds and mm, respectively. Ã'_{1}(mm) and Ã'_{2}(mm) have the same value if both photographs share the scale. The distance between the centres of Venus in the two photographs is: Δπ(segons d'arc) = [ (x_{2} – x_{1})^{2}+ (y_{2} – y_{1})^{2} ]^{1/2} [1] If the photographs are taken with telescopes that give the same scale and the solar disc is situated exactly in the same place of the photographs, then a corner of the photograph can be taken as a reference point. Besides, in this case Ã'_{1}(mm) = Ã'_{2}(mm) and the previous equation can be written as: Δπ(segons d'arc) = [ (x_{2} – x_{1})^{2}+ (y_{2} – y_{1})^{2} ]^{1/2}Â·Ã'(segons d'arc)/Ã'(mm) [1bis] This is the supposition adopted from now on.
Let's suppose that r_{V} and r_{T} are the distances between the centre of the Sun and the centres of Venus and the Earth, respectively, at the moment t of the observation. As the projection d of the distance between M_{1} and M_{2} over the plane perpendicular to OC is small if compared with the distances EarthSun and EarthVenus, we can approximate:
π_{s} = d/r_{T} and from this, the following is deduced:
π_{v} = π_{s} r_{T}/(r_{T} r_{V}) and so, π_{s} = d/r_{T} = Δπ (r_{T}/r_{V}  1) This last formula clearly express that, if we know the angular distance Δπ between the centres V_{1} and V_{2}, and the ratio r_{T}/r_{V} between the distances EarthSun and VenusSun, the parallax π_{s} can be deduced; and that if we know the projected distance d between the two locations where observations are carried out, the distance r_{T} can be calculated. (In these expressions, the values of π_{v}, π_{s} and Δπ are given in radians. To transform them to arcseconds to make them compatible with equation [1], multiplying by number 64800/π is enough). Δπ is the observable magnitude, d can be determined as is explained beneath and then, the only magnitude we have to know to solve the problem is the ratio r_{T}/r_{V} between the distances EarthSun and VenusSun. The orbits of the Earth and Venus around the Sun are slightly elliptical and so, the ratio r_{T}/r_{V} is not constant in time. To know this ratio at the exact time t, we need to use the first law of Kepler, which states that the Sun is situated on one of the ellipse's focuses. Then, the distance to the planet is given by: r_{p}(t)=R_{p} (1  e_{p} cos E_{p}(t)) where R_{p} is the semimajor axis of the orbit, e_{p} the eccentricity and E_{p}(t) the eccentric anomaly at the instant t. According to this: r_{T}/r_{v}=[R_{T} (1  e_{T} cos E_{T})] / [R_{V} (1  e_{V} cos E_{V})] The third law of Kepler gives a relation between the semimajor axes of the orbits and its correspounding periods P_{p}: (R_{T} / R_{V})^{3} = (P_{T} / P_{V})^{2} in such a way that r_{T}/r_{v}=(P_{T} / P_{V})^{2/3} (1  e_{T} cos E_{T}) / (1  e_{V} cos E_{V}) [2] Until this point, we have been able to determine π_{s} and r_{T}; the parallax and the distance EarthSun at the instant t of the observation. 2.2 Determination of the average distanceTo determine the average distance EarthSun (R_{T}) and the correspounding average parallax π_{o}, which are related by means of the equatorial radius R according to: π_{o} ≈≈ R/R_{T} some other considerations must be done. If the projection d of the distance between M_{1} and M_{2} on the plane normal to the direction EarthSun is expressed in terms of the equatorial terrestrial radius, and the distance EarthSun in terms of the average distance, we have: π_{s} = [(d/R) / (r_{T} /R_{T})] (R/R_{T}) ≈ [(d/R) / (r_{T} /R_{T})] π_{o} The ratio r_{T}/R_{T} can be deduced from the first law of Kepler as: r_{T}/R_{T} = 1  e_{T} cos E_{T}(t) and so, we only have to calculate d/R (see Figure 3).
Figure 3. Projection of the distance between M_{1} and M_{2} on the plane normal to the direction EarthSun. Calculating the vectorial product between vectors M_{1}M_{2} and OC, the value of sin θ is obtained, due to M_{1}M_{2} Ã— OC = M_{1}M_{2} r_{T} sin θ The next expression is clear in figure 3: d = M_{1}M_{2} cos (90 – θ) = M_{1}M_{2} sin θ and then, d = M_{1}M_{2} Ã— OC / r_{T} Now we just have to find M_{1}M_{2} Ã— OC. Calculation of the vector OCThis vector can be expressed in terms of the equatorial coordinates of the Sun (α,δ) at the moment of the observation as:
x=r_{T} cos δ cos α Calculation of the vector M_{1}M_{2}The position of each observer can be expressed as (see figure 4):
x=R cos φ cos (λ+T_{G}) where φ and λ are the geographical coordinates (latitude and longitude) and T_{G}=T_{G}(0) + 1.00273791 t. The time t of the observation must be expressed in the scale of universal time (UT). For the most countries in Europe UT = official time – 2^{h} in June.
Figure 4. Positions of a celestial object (for instance, the Sun) and of an observer on Earth in equatorial coordinates. The coordinates of the vector M_{1}M_{2} can be easily found as:
X=x_{1} – x_{2} Practical exampleOne of the programs available on https://gaia.am.ub.es/Twiki/bin/view/ServiAstro/AgendaAstronomica2004 is founded on this formulation and can be used as a practical example. The second program is based on the comparison of the duration of the transit as observed from two different locations. 3. For more information
