Applied Science and Innovative Research ISSN 2474-4972 (Print) ISSN 2474-4980 (Online) Vol. 4, No. 2, 2020 www.scholink.org/ojs/index.php/asir 41 Original Paper Excel Files for Newton’s Proposition V Pavlos Mihas1* 1 Retired professor of didactics of physics, Department of Elementary Education, Democritus University, Nea Chili, Alexandroupolis, Gr- 68100, Greece Received: May 3, 2020 Accepted: May 11, 2020 Online Published: May 19, 2020 doi:10.22158/asir.v4n2p41 URL: http://dx.doi.org/10.22158/asir.v4n2p41 Abstract Newton in Principia gives us a mathematical method of finding the center of force for a body moving on an ellipse in Proposition V, Problem I. The same thinking can be applied also to the case of a hyperbola and also a parabola, only that in the last case the center of force is at infinite distance. For the first two cases there are 3 cases of possible forces: a) An force proportional to the distance from the center. For ellipse an attractive force for hyperbola a repulsive, b) a force proportional to the inverse of the square of the distance from the left focus, for ellipse an attractive and for hyperbola a repulsive force, c) an attractive force inversely proportional to the square of the distance, inversely proportional to the square of the distance for both the ellipse and the hyperbola. This method when applied to the case of circular orbits for which we can find the center of force with the same method: Newton studied a semicircular orbit with center of force at infinite distance, and the case of a central force whose center is located on the circular orbit or inside the circle studied the case of a spiral orbit. In each the law of the force was derived by using the law of areas. Keywords conservation of angular momentum, law of areas, central forces, law of force 1. Introduction In Newton’s words: There being given, in any places, the velocity with which a body describes a given figure, by means of forces directed to some common center: to find that center (Newton, n.d.). His proof is simple but also needs some explanation: Newton uses the constancy of angular momentum which he explained in by PROPOSITION I, Theorem. The areas, which bodies made to move describe by radii drawn to an immovable center of force lie unmoving planes, and are proportional to the times in which they are described (Newton, 1999). www.scholink.org/ojs/index.php/asir Applied Science and Innovative Research Vol. 4, No. 2, 2020 42 Published by SCHOLINK INC. Figure 1. Law of Areas Actually this proposition is very useful for teaching purposes. The length of the movement in the unit of time is equal to the velocity. If from the center K we draw the radii KA, KB, KC, then the areas KAB, KBC, are equal, since the triangles KAB, KMN have equal bases and equal heights, but also the triangles KBC and KBM have the same base (KB) and heights DC=NM As we can see in Figure 1, the constancy of the areas is valid for central force which for Newton’s model is given by impulses along the lines which join the end of the velocity vector to the center of force. As is seen in the Figure 1 the law is valid in the case of a central force. 2. Proposition V, Problem I for Ellipse In the case of the ellipse we consider as Newton (1981) points P, Q, R where we draw the velocity vectors VP, VQ, VR. One question that can be asked is what kind of force caused this movement (Figure 2)? The students can check the figure and guess the force. They can think from the size of the 3 velocities. The teacher can move a little bit the points and the students can try to think of where is the center of force. This problem was used in attempts to teach Principia (Alfred, 1964). Figure 2. Where is the Center of Force that Produces the Velocities? www.scholink.org/ojs/index.php/asir Applied Science and Innovative Research Vol. 4, No. 2, 2020 43 Published by SCHOLINK INC. It is known from history that Newton was asked by Halley, what he thought the Curve would be that would be described by the Planets supposing the force of attraction towards the Sun to be reciprocal to the square of their distance from it (Newton and Kepler Ex Libris, n.d.). Newton answered that this curve is an ellipse. In the present problem we have an ellipse but the answer is not always a force reciprocal to the square. Newton gave in Principia the proposition which can be answered by either gravitational forces directed towards either one of the foci or by a force proportional to the distance from the center and directed to it (which was proposed by Hooke (Michael, 1994)). The motion around the center will have a constant angular momentum, or the areas described by the radius drawn from the center of the force to the particle will be equal in equal times. To find the center we precede either by calculating with the use of analytic geometry or with Newton’s method (the equations from which we derive the center of force, are presented in the end of the paper). We draw the three lines PT, TQZ, ZR which touch the curve at the points, P, Q, R, and meet in T and Z (see Figure 3) On the tangents erect the perpendiculars PA, QB, RC, reciprocally proportional to the velocities of the body in the points P, Q, R, from which the perpendiculars were raised. So 𝐴 = π‘˜ 𝑉𝑃 , 𝑄𝐡 = π‘˜ 𝑉𝑄 , 𝑅𝐢 = π‘˜ 𝑉𝑅 . From A, B, we draw perpendiculars to PA and QB correspondingly. These perpendiculars meet at D. We draw from C a perpendicular to RC which meets BD at E. We draw from D the perpendicular DP” to VP, from D the perpendicular DT” to VQ, and form E the perpendicular ER” to VR. So we have V P Β· P A = V P Β· DP” = V Q Β· BQ = V Q Β· DT” So for point D we have DP”· V P = DT” Β·V Q. Also for point E we have V Q Β· EZ” = V R Β· ER”. These relations resemble the law of areas but actually they are valid only on pairs of velocities. We can repeat the procedure for different values of k. For k β†’ 0 the limiting lines are the tangents at P, Q, R which meet correspondingly at T and Z. If we increase the value of k we get a smaller distance of DE and we can reach the point where the two lines ZE and TD meet. At this point S we have SR’ Β· V R = ST’ Β· V Q = SP’· V P. So we find the point S which is the center of force. So PA/QB = V Q/V P, and QB/RC = VR/VQ. Figure 3. Newton’s Method for Finding the Center of Force www.scholink.org/ojs/index.php/asir Applied Science and Innovative Research Vol. 4, No. 2, 2020 44 Published by SCHOLINK INC. 2.1 Newton’s Proof as Explained by Chandraseckar Chandraseckar (1995) gave a proof based on Principia easier to understand. For the points A, B, C we note that P A : QB : RC = V P(βˆ’1) : V Q(βˆ’1) : V E(βˆ’1) If S should be the (yet unspecified) center of attraction, drop the perpendiculars SP’ and ST’ to the tangents at P and Q. And draw (as illustrated) CE, EBD, and DA parallels to the tangents at R, Q, and P and intersecting at the points E and D. Now drop the perpendiculars DP” and DT” to the tangents at P and Q. Then by the 𝑆𝑃’ 𝑆𝑇’ = 𝑉𝑄 𝑉𝑃 = 𝐴𝑃 𝐡𝑄 = 𝐷𝑃’’ 𝐷𝑇’’ or 𝑆𝑃’ 𝐷𝑃 ’’ = 𝑆𝑇’ 𝐷𝑇 ’’ So S, D, and T lie on the same line. 2.2 Finding the Law of the Force Newton in principia gives us a mathematical method of finding the center of force for a body moving on an ellipse in Proposition V, Problem I. The same thinking can be applied also to the case of a hyperbola and also a parabola, only that in the last case the center of force is at infinite distance. The forces that studied Newton in the next pages of principia, for elliptical motion could be: a) An elastic force with center at the origin, b) a gravitational force with center at the left focus or c) a gravitational force with center at the right focus. In this article Newton’s methods were extended to the case of hyperbola, where the forces could be a) an attractive gravitational force with center one pole of the hyperbola, b) a repulsive force with center the other pole of the hyperbola, c) an elastic force with center at the origin. It is possible to prove using the conservation of the angular momentum to prove that the force is proportional the inverse of the square of distance r from the focus. To find the law of force we use the law of areas in the form π‘Ÿ Β· π‘π‘œπ‘ πœ” Β· 𝑉 = β„Ž where Ο‰ is the angle between the line r=SM (see Figure 3) and the perpendicular on the ellipse. To find the cosΟ‰ we use the parametric equations of the ellipse with semi-axes a and b as x = a Β· cosΟ†,y = b Β· sinΟ† so 𝑑𝑦 𝑑π‘₯ = βˆ’ π‘ŽΒ·π‘ π‘–π‘›πœ‘ π‘Β·π‘π‘œπ‘ πœ‘ , with unit vector 𝑒 = π‘ŽΒ·π‘π‘œπ‘ πœ‘+𝑐 π‘Ÿ , π‘Β·π‘ π‘–π‘›πœ‘ π‘Ÿ for perpendicular οΏ½βƒ—βƒ—οΏ½ = π‘Β·π‘π‘œπ‘ πœ‘ √ 𝛼2·𝑠𝑖𝑛2πœ‘+𝑏2Β·π‘π‘œπ‘ 2πœ‘ , π‘ŽΒ·π‘ π‘–π‘›πœ‘ βˆšπ‘Ž2·𝑠𝑖𝑛2πœ‘+𝑏2Β·π‘π‘œπ‘ 2πœ‘ The Force that acts as centripetal: πΉπ‘˜ = 𝐹 Β· π‘π‘œπ‘ πœ” = π‘š Β· 𝑉2 𝜌 = π‘šΒ·β„Ž2 (π‘ŸΒ·π‘π‘œπ‘ πœ”)2·𝜌 where ρ is the radius of curvature 𝜌 = √(π‘Ž2·𝑠𝑖𝑛2πœ‘+𝑏2Β·π‘π‘œπ‘ 2πœ‘)3 π‘ŽΒ·π‘ so 𝐹 = π‘šΒ·β„Ž2 π‘Ÿ2Β·π‘π‘œπ‘ 3πœ”Β·πœŒ . Physical entity Center at left focus Center at the center of ellipse Center at the right focus r π‘Ž + 𝑐 Β· π‘π‘œπ‘ πœ‘ √ 𝛼2 Β· 𝑠𝑖𝑛2πœ‘ + 𝑏2 Β· π‘π‘œπ‘ 2πœ‘ π‘Ž βˆ’ 𝑐 Β· π‘π‘œπ‘ πœ‘ CosΟ‰ 𝑏 Β· π‘π‘œπ‘ πœ‘ √ 𝛼2 Β· 𝑠𝑖𝑛2πœ‘ + 𝑏2 Β· π‘π‘œπ‘ 2πœ‘ π‘Ž + 𝑏 Β· π‘π‘œπ‘ πœ‘ π‘Ÿβˆš 𝛼2 Β· 𝑠𝑖𝑛2πœ‘ + 𝑏2 Β· π‘π‘œπ‘ 2πœ‘ 𝑏 Β· π‘π‘œπ‘ πœ‘ √ 𝛼2 Β· 𝑠𝑖𝑛2πœ‘ + 𝑏2 Β· π‘π‘œπ‘ 2πœ‘ F π‘š Β· π‘Ž βˆ™ β„Ž2 π‘Ÿ2 Β· 𝑏2 π‘š Β· β„Ž2 βˆ™ π‘Ž βˆ™ 𝑏 βˆ™ π‘Ÿ (π‘Ž + 𝑏)3 π‘š Β· π‘Ž βˆ™ β„Ž2 π‘Ÿ2 Β· 𝑏2 By using the Excel file, the user can see that this method gives correctly the center of the force for www.scholink.org/ojs/index.php/asir Applied Science and Innovative Research Vol. 4, No. 2, 2020 45 Published by SCHOLINK INC. either forces directed to the foci of the ellipse (gravitational attractive proportional to the reciprocal of the distance from the focus) or to the center of the ellipse (proportional to the distance of the particle from the center). 3. Finding the Center in the Case of Hyperbola In the case of hyperbola there are three cases of force (We consider the movement on the right branch of the hyperbola). A) An attractive force with center of force the focus on the right side of the hyperbola. This force is reciprocal to the square of the distance from the focus. B) A repulsive force with center of force the left focus, which also is reciprocal to the square of the distance from the focus. C) A repulsive force with center of force the origin of the coordinates. In the figure the particle is moving to the left. It is clear that for attractive force the velocity is increasing as the particle moves to the left. For the case of elastic force the velocity increases fast as the body moves away to the right, while for repulsive force the velocity increases but there is a limit to the velocity. We proceed as in the case of the ellipse. We find the points T and Z where the lines of the velocities 𝑉𝑃⃗⃗⃗⃗⃗⃗ , 𝑉𝑄⃗⃗⃗⃗ βƒ—βƒ— and 𝑉𝑄⃗⃗⃗⃗ βƒ—βƒ— , 𝑉𝑅⃗⃗⃗⃗ βƒ—βƒ— βƒ— cut. We find as in ellipse the points, E and D and draw the lines TD and ZE and find the point S where the two lines cut each other. At the point S the law of areas holds for all points, so this is the center of force. . Figure 4. Application of Newton’s Ideas for Hyperbola 3.1 Finding the Law of Force To find the law of the force we need to calculate the Velocity and the radius of curvature ρ. The radius of curvature is ρ = (π‘Ž2βˆ™π‘ π‘–π‘›(πœ‘)2+𝑏2) 3 2 π‘Žβˆ™π‘βˆ™π‘π‘œπ‘ (πœ‘)3 where Ο† is the angle from which we calculate the point of the curve: (π‘₯, 𝑦) = ( π‘Ž π‘π‘œπ‘ (πœ‘) , 𝑏 βˆ— tan(πœ‘)) . The tangent is: 𝑦 𝑑π‘₯ = 𝑏 π‘Žβˆ—π‘ π‘–π‘›πœ‘ , and the unit vector on π‘Ÿ is www.scholink.org/ojs/index.php/asir Applied Science and Innovative Research Vol. 4, No. 2, 2020 46 Published by SCHOLINK INC. 𝑒 = ( π‘Ÿπ‘₯ π‘Ÿ , π‘Ÿπ‘¦ π‘Ÿ ). The perpendicular is found from οΏ½βƒ—βƒ—οΏ½ = ( 𝑏 βˆšπ‘2+Ξ±2βˆ™sin2 πœ‘ , βˆ’π‘Žβˆ™π‘ π‘–π‘›πœ‘ βˆšπ‘2+Ξ±2βˆ™sin2 πœ‘ ) and the angle Ο‰ between 𝑒 and οΏ½βƒ—βƒ—οΏ½ is found from π‘π‘œπ‘ πœ” = οΏ½βƒ—βƒ—οΏ½ βˆ™ 𝑒 = π‘Ÿπ‘₯βˆ™π‘βˆ’π‘Ÿπ‘¦βˆ™π‘Žβˆ™π‘ π‘–π‘›πœ‘ π‘Ÿβˆ™βˆšπ‘2+Ξ±2βˆ™sin2 πœ‘ . The velocity V is found from the law of areas: If β„Ž = π‘Ÿ Γ— οΏ½βƒ—βƒ—οΏ½ then 𝑉 = β„Ž π‘ŸΒ·π‘π‘œπ‘ πœ” and we find the law of the force from the centripetal force πΉπ‘˜ = 𝐹 βˆ™ π‘π‘œπ‘ πœ” = π‘š βˆ™ 𝑉2 𝜌 = π‘š βˆ™ β„Ž2 πœŒβˆ™(π‘Ÿβˆ™π‘π‘œπ‘ πœ”)2 so 𝐹 = π‘š βˆ™ β„Ž2 π‘Ÿ2βˆ™cos3 πœ”βˆ™ (π‘Ž2βˆ™π‘ π‘–π‘›(πœ‘)2+𝑏2) 3 2 π‘Žβˆ™π‘βˆ™π‘π‘œπ‘ (πœ‘)3 . Center at left focus Center at the center of ellipse Center at the right focus R π‘Ž Β· π‘π‘œπ‘ πœ‘ + 𝑐 π‘π‘œπ‘ πœ‘ βˆšπ›Ό2 βˆ™ cos2 πœ‘ + 𝑏2 βˆ™ sin2 πœ‘ π‘Ž Β· π‘π‘œπ‘ πœ‘ βˆ’ 𝑐 π‘π‘œπ‘ πœ‘ cosΟ‰ 𝑏 βˆ™ π‘π‘œπ‘ πœ‘ βˆšπ‘2 + Ξ±2 βˆ™ sin2 πœ‘ π‘Žπ‘ βˆ™ π‘π‘œπ‘ πœ‘ π‘Ÿ βˆ™ βˆšπ‘2 + Ξ±2 βˆ™ sin2 πœ‘ 𝑏 Β· π‘π‘œπ‘ πœ‘ βˆšπ‘2 + Ξ±2 βˆ™ sin2 πœ‘ F π‘š βˆ™ β„Ž2π‘Ž π‘Ÿ2 βˆ™ 𝑏2 π‘š Β· β„Ž2 βˆ™ π‘Ÿ (π‘Žπ‘)2 π‘š Β· β„Ž2 Β· π‘Ž 𝑏2 Β· π‘Ÿ2 4. Parabola 4.1 Constant Force By applying the same method we get two parallel lines which show that the center of force is at an infinite distance. The user can change the initial speed and the angle of the initial velocity with the horizontal. As in the case of ellipse he can change the number of steps and see the construction of the figure. The lines TD and ZE are now parallels and so the center of force is at infinite distance. Figure 5. Application of Newton’s Ideas for a Parabola for a Force Parallel to y Axis www.scholink.org/ojs/index.php/asir Applied Science and Innovative Research Vol. 4, No. 2, 2020 47 Published by SCHOLINK INC. We can prove that T and D have the same x_ component. The tangent at P is 𝑖𝑝 = 𝑑𝑦 𝑑π‘₯ = tan(πœ‘) βˆ’ 𝑗 𝑣0π‘₯ 2 (π‘₯𝑝 βˆ’ π‘₯0) where 𝑗 = 𝑔 𝑣0π‘₯ 2 and Ο† the angle of the initial velocity v0 with the horizontal. We find that the point T of the intersections of the lines of the velocities is given by: π‘₯𝑇 = π‘†π‘π‘ž π‘–π‘βˆ’π‘–π‘ž where π‘†π‘π‘ž βˆ’ π‘¦π‘ž βˆ’ 𝑦𝑝 + 𝑖𝑝π‘₯𝑝 βˆ’ π‘–π‘žπ‘₯π‘ž. The points A, and B are having π‘₯𝐴 = π‘₯𝑝 + 𝑑𝑝 𝑠𝑖𝑛(atan(𝑖𝑝)) and π‘₯𝐡 = π‘₯π‘ž + π‘‘π‘ž 𝑠𝑖𝑛(atan(π‘–π‘ž)) where 𝑑𝑝 = π‘˜ 𝑣𝑝 , π‘‘π‘ž = π‘˜ π‘£π‘ž We find for the point D that it has as x – component π‘₯𝐷 = (π‘†π‘π‘ž βˆ’ π‘‘π‘ž(1 + π‘–π‘ž 2) 0,5 + 𝑑𝑝(1 + 𝑑𝑝 2) 0,5 )/(𝑖𝑝 βˆ’ π‘–π‘ž) but the last two terms on the numerator have a sum of zero since βˆ’π‘‘π‘ž(1 + π‘–π‘ž 2) 0,5 + 𝑑𝑝(1 + 𝑑𝑝 2) 0,5 = βˆ’ π‘˜ 𝑣0π‘₯ + π‘˜ π‘£π‘œπ‘₯ = 0 since in the parabola the horizontal component is constant. For Ellipse and hyperbola the expressions become very complicated. In the case of a force having a constant direction we can prove that this force is constant. If the parabola is described by the equation y = aΒ·xΒ² then the radius of curvature is 𝜌 = |√1+( 𝑑𝑦 𝑑π‘₯ ) 2 | 3 |𝑑2𝑦/𝑑π‘₯2 | = |√1+tan2 πœ”| 3 2π‘Ž = 1 2π‘ŽΒ·cos3 πœ” where Ο‰ is the angle of the tangent with the horizontal. Here the horizontal component of the velocity remains constant. So the velocity V is found from Ο‰ by V= V0/cosΟ‰, where V0 is the horizontal component of the velocity. The law for the force that acts on the body is found by: 𝐹 Β· π‘π‘œπ‘ πœ” = π‘š Β· 𝑉2 𝜌 = 2π‘Žπ‘š Β· 𝑉0 2π‘π‘œπ‘ πœ” so F is a constant. 4.2 Force Directed to a Center In this case the force is directed to the focus. The distance r of the point (x,y) on the parabola from the focus is found to be: π‘Ÿ = π‘Ž Β· π‘₯2 + 1 4Β·π‘Ž (Since 𝑦 = π‘Ž Β· π‘₯2 + 𝑐, 𝑓 = 𝑐 + 1 4Β·π‘Ž ). The angle ΞΈ between the r, and the perpendicular is found from π‘π‘œπ‘ πœƒ = ( π‘₯ π‘Ÿ , π‘¦βˆ’π‘“ π‘Ÿ ) Β· ( 𝑑𝑦 𝑑π‘₯ √1+( 𝑑𝑦 𝑑π‘₯ ) 2 , βˆ’ 1 √1+( 𝑑𝑦 𝑑π‘₯ ) 2 ) = 1 √1+( 𝑑𝑦 𝑑π‘₯ ) 2 , www.scholink.org/ojs/index.php/asir Applied Science and Innovative Research Vol. 4, No. 2, 2020 48 Published by SCHOLINK INC. Figure 6. Parabolic Motion with a Force Directed towards the Focus and Proportional to 1/r 2 Since this case the particle stars with zero velocity at infinity and gets its maximum velocity at the apex. Since 𝜌 = |√1+( 𝑑𝑦 𝑑π‘₯ ) 2 | 3 | 𝑑2𝑦 𝑑π‘₯2| = 1 ⌊cos3 πœƒβˆ™2π‘ŽβŒ‹ we see that πΉπ‘˜ = π‘š Β· 𝑉2 πœŒΒ·π‘π‘œπ‘ πœƒ = π‘šβˆ™π‘‰π‘œ 2 π‘Ÿ2βˆ™πœŒβˆ™cos3 πœƒ = 2π‘Žβˆ™π‘šβˆ™π‘‰0 2 π‘Ÿ2 and so 𝐹_π‘˜ ∝ 1 π‘Ÿ2. In this case if the particle starts with zero velocity at infinity and has maximum velocity at the apex. 5. Circular Movements 5.1 Circle and Forces which are Parallel In this case a body describes a semicircle under the influence of a force which has a direction parallel to the y axis (Figure 7). Figure 7. Application of Newton’s Ideas for a Force Parallel to y-axis www.scholink.org/ojs/index.php/asir Applied Science and Innovative Research Vol. 4, No. 2, 2020 49 Published by SCHOLINK INC. In this case Newton proves that the force is inversely proportional to y3. We can use the constancy of vx to find the Centripetal Force. If v0 is the velocity at the highest point then the horizontal component of the velocity v is vx= V0 = VΒ· cosΟ‰ where Ο‰ is the angle of the radius at the point P with the perpendicular. Then the centripetal force for the circular movement is 𝐹 Β· π‘π‘œπ‘ πœ” = π‘šΒ·π‘‰2 𝑅 = π‘šΒ·π‘‰0 2 π‘π‘œπ‘ 2πœ”Β·π‘… . Thus 𝐹 ∝ 1 π‘π‘œπ‘ 3 πœ” ∝ 1 𝑦3 since y = R Β·cosΟ‰ Newton gives examples of circular movements as an introduction for the way to determine the centripetal attraction from the orbit. His examples can be also used to demonstrate that his method for finding the center of the force is applicable to these circular orbits. A body describes a semicircle under the influence of a force which has a direction parallel to the y axis. In this case Newton proves that the force is inversely proportional to y3. We can use the constancy of vx to find the Centripetal Force. If v0 is the velocity at the highest point then the horizontal component of the velocity v is vx=V0=V*sinΟ‰ where Ο‰ is the angle of the radius at the point P with the horizontal. Then the centripetal force for the circular movement is 𝐹 βˆ™ π‘ π‘–π‘›πœ” = π‘šπ‘‰2 𝑅 = π‘šπ‘‰0 2 sin2 πœ”βˆ™π‘… so F ~1/sinΟ‰3 ~1/y3, since y =RΒ·sinΟ‰. 5.2 Attractive Force towards a Point on a Circumference Figure 8. Application of Newton’s Ideas for the Case of Force Directed to a Point on the Circumference A circular (almost) orbit under the influence of a force directed to a point directed to a point on the circumference. In this case the force is proportional to the inverse of the 5th power of the distance SR of the body R from the center of force S. Here we apply the law of constant angular momentum: In the highest point V = V0 so π‘š Β· 𝑉0 βˆ™ 2 Β· 𝜌 = π‘šπ‘‰ Β· π‘Ÿ Β· π‘π‘œπ‘ πœƒ where ΞΈ is the angle of the r = SR=2ρcosΞΈ with the radius ρ at point R. The component of F o R is the centripetal Force so mV2/ρ = FΒ·cosΞΈ or www.scholink.org/ojs/index.php/asir Applied Science and Innovative Research Vol. 4, No. 2, 2020 50 Published by SCHOLINK INC. mV0 2/(2ρ*cos2ΞΈ)= F*cosΞΈ so 𝐹~ 1 π‘π‘œπ‘ πœƒ3 ~ 1 𝑆𝑅3. 5.3 Circular orbit with the Center of the Force inside the Circle A circular orbit on a center which is inside the circle. In this case the force is expressed as 𝐹 = π‘˜ 𝑅𝑉3βˆ—π‘†π‘…2 where SR is the distance of the center of force S from the body R and RV is the distance of the body P from the point V found by extending the line SR and finding the point V on the circle. Again here 𝑣 = 𝑉0 βˆ™ 𝜌 βˆ™ 1+𝛼 π‘†π‘ƒβˆ™π‘ π‘–π‘›πœ€ If Ξ΅ is the angle of the velocity at R with SR then 𝐹 βˆ™ π‘ π‘–π‘›πœ€ = π‘šβˆ™π‘‰0 2(1+𝛼)2 (π‘†π‘…βˆ™π‘ π‘–π‘›πœ€)2 π‘ π‘–π‘›πœ€ = 𝑅𝑉 2𝜌 . By combining these we find that 𝐹~ 1 𝑆𝑅2βˆ—π‘…π‘‰^3 . 6. Newton’s Isogonal Spiral Figure 9. Application of Newton’s Ideas for a Force Directed to a Center not on the Circumference Figure 10. Application of Newton’s Ideas for a Movement under the Influence of a Force Directed towards the Origin www.scholink.org/ojs/index.php/asir Applied Science and Innovative Research Vol. 4, No. 2, 2020 51 Published by SCHOLINK INC. The last application of the ideas is on Newton’s isogonal spiral which is Proposition 9 Problem 3 Let a body revolve in a spiral PQS intersection all its radii SP, SQ, … Of a given angle; it is required to find the law of the centripetal force tending toward the center of the spiral. We can prove that the force is proportional to r-2 where r is the distance of the point to the point S. We apply again the law of areas and if at an initial point the velocity is V0 and the initial radius of curvature is ρ0 then at the point Z ρ·V = ρ0Β·V0 To find the law of the force we note that the centripetal force will be FΒ·sinΞ΅ =mΒ·V2/ρ = mΒ·V0 2 ρ0Β²/ρ 3. Since the radius of curvature is ρ = r/sinΞ΅ and we find that 𝐹~1/π‘Ÿ3. 7. Fixing the Velocities in Excel To move the body around we need to have velocities that really correspond to the ellipse with semi-axes a,b and foci at c,-c: The velocity in the farthest point is given the value voo and then by the constancy of the angular momentum the velocity at the nearest point 𝑣_π‘œπ‘œ = π‘£π‘œπ‘œ Β· π‘Ž+𝑐 π‘Žβˆ’π‘ and then by the law of conservation of energy we can find the constant of the law by 1 2 Β· π‘š Β· π‘£π‘œπ‘œ2 βˆ’ π‘˜Β·π‘š π‘Ž+𝑐 = 1 2 π‘š Β· π‘£π‘œπ‘œ 2 βˆ’ π‘˜Β·π‘š π‘Žβˆ’π‘ . We can proceed on the same manner for the case of hyperbola. In the case of hyperbola we can have for a particle moving on the right branch either an attractive gravitational force directed to the right focus, or a repulsive –1/rΒ² or a repulsive directed to the center. We note that for the repulsive force from the left focus, we can see that the velocity for infinite distance is not zero but equal to π‘£βˆž = 𝑣_00 Β· (𝑐 + π‘Ž)/𝑏 while for the elastic force the velocity is infinite. 8. Analytic Solution The distance of the center of force (sx, sy) from the velocities is 𝑑 = Β±(π‘ π‘–π‘›πœ‘π‘ π‘₯ βˆ’ π‘π‘œπ‘ πœ‘π‘ π‘¦ + π‘π‘œπ‘ πœ‘π‘¦1 βˆ’ π‘ π‘–π‘›πœ‘π‘₯1) where (x1, y1) is the point where the velocity is V1 and Ο† the angle of inclination. For x_1 we take the negative since the numerator is negative (we expect sy to be zero). By demanding that the law of areas is valid we get the equations: 𝑑1𝑉1 = 𝑑2𝑉2 = 𝑑3𝑉3 or: We denote by V_1y=V_1sinΟ†_1, V_1x=V_1cosΟ†_1, etc. (𝑉1𝑦 + 𝑉2𝑦)𝑆π‘₯ βˆ’ (𝑉1π‘₯ + 𝑉2𝑋)π‘†π‘Œ = (π‘Ÿπ‘„βƒ—βƒ— βƒ—βƒ— ⃗𝑋𝑉2 βƒ—βƒ— βƒ—βƒ— + π‘Ÿπ‘ƒβƒ—βƒ—βƒ—βƒ—βƒ— 𝑋 𝑉1 βƒ—βƒ— βƒ—βƒ— ) (𝑉1𝑦 + 𝑉3𝑦)𝑆π‘₯ βˆ’ (𝑉1π‘₯ + 𝑉3π‘₯)π‘†π‘Œ = (π‘Ÿπ‘…βƒ—βƒ—βƒ—βƒ—βƒ—π‘‹ 𝑉3 βƒ—βƒ— βƒ—βƒ— + π‘Ÿπ‘ƒβƒ—βƒ—βƒ—βƒ—βƒ— 𝑋 𝑉1 βƒ—βƒ— βƒ—βƒ— ) By solving these equations we find the center. Excel Files For ellipses and parabolas, the _le is downloadable from: http://www.kyriakosxolio.gr/2_dimensional_motiona/newtonstheoryofellipse2_2.xlsm The corresponding _le for hyperbolas is downloadable from: http://www.kyriakosxolio.gr/2_dimensional_motiona/HYPEROBOLAS_NEWTON_FIND_CENTER_ without_macros.xlsm A zip _e with the excel _les for circular motions and Newton’s spiral is downloadable from: http://www.kyriakosxolio.gr/2_dimensional_motiona/circular_spiral.zip www.scholink.org/ojs/index.php/asir Applied Science and Innovative Research Vol. 4, No. 2, 2020 52 Published by SCHOLINK INC. References Alfred M. B. (1964). Newton in the College Classroom. American Journal of Physics, 32, 959. https://doi.org/10.1119/1.1970038 Bernard, C. (1981). Newton’s Discovery of Gravity. Scientific American, 244(3), 123-133. https://doi.org/10.1038/scientificamerican0381-166 Chandrasekhar, S. (1995). Newton’s Principia for the Common Reader (p. 75). Clarendon Press. Michael, N. (1994). Hooke, Orbital Motion, and Newton’s Principia. American Journal of Physics, 62, 331. https://doi.org/10.1119/1.17576 Murray, R. S. (1967). Theoretical Mechanics (p. 127). SI Edition, Schaum’s Outline Series, McGraw-Hill Book Company. Newton and Kepler Ex Libris. (n.d.). Retrieved from http://www.nonagon.org/ExLibris/newton-keplers-laws Newton, I. (1999). The Principia: The Authoritative Translation and Guide: Mathematical Principles of Natural Philosophy (Bernard Cohen, Anne Whitman, & Julia Budenz, Trans., p. 454). University of California Press. Newton, I. (n.d.). The Mathematical Principles of Natural Philosophy (Andrew Motte, Trans.).