Problem 3—
An Elliptical Orbit with the Center of Force at a Focus of the Ellipse

The challenge set forth in Problem 3 is a distinguished one: find the solution to the "Problem of the Planets." Kepler had demonstrated in 1609 that Mars moved in an elliptical orbit with the sun at a focus. The question that remained to be answered, however, was the mathematical nature of the force required to maintain that motion. In On Motion , the solution follows the two preliminary examples without any fanfare. In the Principia , however, Newton called attention to "the dignity of the problem and its use in what follows" by separating Problem 3 from the preceding direct problems and giving it a place of honor at the beginning of a new section. It is the keystone of both works in terms of the dignity of the problem and in the degree of mathematical difficulty.

As in all the previous problems, Newton's diagram identifies the dynamic elements to be employed in the demonstration (see fig. 5.8 in the following statement of the problem): the projected tangential path PR , the elliptical arc PQ , and the deviation QR from the point R on the tangential path to the point Q on the elliptical arc. Note in this diagram that

the deviation QR is constructed parallel to the line of force SP , which is directed toward the focus S , and that the line segment QT is constructed perpendicular to SP . The challenge for this problem is, as it was for the two previous problems, to express the discriminate ratio of the force QR / QT² in terms of the radius SP and / or constants of the orbit. The final expression of the solution appears deceptively simple: the discriminate ratio QR / QT ² is found to be proportional to the reciprocal of the constant latus rectum L of the ellipse (where L = 2BC / AC ), and therefore the force is proportional to the inverse square of the distance, 1/SP² .

where L is a constant of the ellipse. Thus, once it is demonstrated that the discriminate ratio QR / QT² is a constant for the elliptical/focal motion, then the problem is solved. That initial demonstration, however, requires a number of steps, and the reader must remember that the goal is to demonstrate that the discriminate ratio QR / QT² is a constant. Once that goal has been achieved, it is a simple matter to determine from Theorem 3 that the force is inversely proportional to the inverse square of the distance SP .

I have divided Step 2, the analysis, into some sixteen sub-steps between line [2-B] and line [2-Q]. The journey is not for the faint of heart; but then if it had not been challenging there would have been no reward for Newton. In honor of the dignity of the problem, I give the statement of the entire proposition and follow it with a line-by-line analysis. I suggest reading the full statement to get an overview of the problem without attempting to justify each point and then following the details of the proof in the line-by-line analysis.

Problem 3. A body orbits on an ellipse; there is required the law of centripetal force directed to a focus of the ellipse .

Let S be a focus of the ellipse above. Draw SP cutting the diameter of the ellipse DK at E. It is clear that EP is equal to the semi-major axis AC, seeing that, when from the other focus H of the ellipse the line HI is drawn parallel to CE, because CS and CH are equal , ES and EI are equal, and hence EP becomes half the sum of PS and PI, that is, of PS and PH which are conjointly equal to the total axis 2AC. [See fig. 5.8.]

Let drop the perpendicular QT to SP, and, after calling the principal latus rectum (or 2BC² / AC) of the ellipse L, there will be

L × QR to L × PV as QR to PV, that is, as PE (or AC) to PC;
and L × PV to GV × VP as L to GV;
and GV × VP to QV²as CP²to CD² ;
and QV²to QX²as, say , M to N;
and QX²is to QT²as EP²to PF² , that is, as CA²to PF² , or as CD²to CB² .

And when all these ratios are combined ,

L × QR / QT²will be equal to (AC / PC) × (L / GV) × (CP² / CD² ) × (CD² / CB² ),

Figure 5.8
Based on Newton's diagram for Problem 3. A body  P  moves in an
elliptical orbit APQB  about a center of force S  located at a focus of the
ellipse.

that is, as AC × L (or 2BC² ) / (PC × GV) × (CP² / CB² ) × (M / N) or as (2PC / GV) × (M / N).

But with the points P and Q coalescing, the ratios 2PC / GV and M / N become equal [and approach unity ], and therefore the combined ratio of these L × QR / QT² [equals unity (i.e., L = QT² / QR)]. Multiply each part by SP² / QR and there will result L × SP² = SP² × QT² / QR.

Therefore the centripetal force is reciprocally as L × SP² , that is , [reciprocally ] in the doubled ratio of the distance. Which was to be proven .

The concluding statement of Problem 3 sets forth the mathematical nature of the gravitational force; that is, it is reciprocally as the square of the distance SP between the sun and the planet. It is this result that provided the answer to the problem of the planets. I now give a detailed analysis of each portion of Newton's demonstration.

Step 1—
The Diagram

[1-A] Let S be a focus of the ellipse above. Draw SP cutting the diameter of the ellipse DK at E.

The body is located at a general point P , and the center of force is located at a specific point S , the focus of the ellipse (see fig. 5.8). The diameter DK is one of the conjugate diameters to the point P , and it is drawn parallel to the tangent PR and through point C , the center of the ellipse. The

intersection of the diameter DK with the line of force SP defines the point E . Note that the diameter PG is not perpendicular to the diameter DK but that the line segment PF is constructed perpendicular to the diameter DK .^[3]

Step 2—
The Analysis ([2-B] to [2-Q])

In Problem 3, as in the solutions to the two previous direct problems, Newton expressed the discriminate ratio QR / QT² in terms of the radius SP and / or constants of the figure. When the discriminate ratio was determined in terms of SP , then the linear dynamics ratio QR / QT² × SP² was also expressed in terms of SP . So, the proportional dependence of the force F , given by the linear dynamics ratio, was also known, and the direct Kepler problem was solved.

As an overview for the reader, I summarize Newton's determination of the discriminate ratio QR / QT² here in an effort to provide a general guide to the multiple details in the analysis.

Find QR : A relationship for the deviation QR is obtained from the similarity of triangles PXV and PEC , where QR = PX by construction (i.e., PE / PC = PX / PV = QR / PV or QR = PV (PE / PC )). Newton also demonstrated that the line PE (defined in fig. 5.8) is equal to the semi-major axis AC (a useful relationship for an ellipse that was not found in the standard works on conics). Thus, QR can be written as PV (AC / PC ). From the same proposition of Apollonius's used in the solution of Problem 2, Newton obtained an expression for PV (i.e., PV × VG / QV² = PC ² / DC² or PV = (QV² / GV) (PC² / DC² )), and he used it to eliminate PV from the expression for QR (i.e., QR = (QV² / GV ) (PC² / DC² ) (AC / PC )).

Find QT ² : From the similarity of triangles EPF and XQT , he obtained an expression for the second element of the discriminate ratio QT² (i.e., QT / QX = PF / PE or QT² = QX ² (PF² / AC² ), where as above PE = AC ). From Euclid's relationship for circumscribed areas (also used in the solution of Problem 2) one has PF / AC = BC / DC , and thus QT² = QX² (BC ² / DC² ).

Find QR / QT² : The discriminate ratio QR / QT² can then be written from QR = (QV² / GV ) (PC² / DC² ) (AC / PC ) and QT² = QX² (BC ² / DC² ) or QR / QT² = (QV² / QX ² ) (PC / GV ) (AC / BC² ). Given the definition of the constant latus rectum L = 2BC² / AC , the discriminate ratio becomes (QV² / QX² ) (PC / GV ) (2 / L ).

In the limit as the point Q approaches the point P , the line QV approaches the line QX , and the line GV approaches a value of 2PC . Thus, the discriminate ratio QR / QT² approaches 1/L , and the linear dynamics ratio QR /

Figure 5.9
The triangle SIH  below is abstracted from the shaded area of the
ellipse above. Line EC  is parallel to line IH  and thus triangles SEC
and SIH  are similar.

QT² × SP² approaches 1 / (L × SP² ). The force, therefore, is inversely proportional to the square of the distance SP , as was to be demonstrated. What follows is a detailed analysis of Newton's demonstration of the solution.

[2-B] It is clear that EP is equal to the semi-major axis AC,

At least it will be "clear" that EP = AC after Newton's demonstration in the next three steps ([2-C] to [2-E]). This particular relationship for an ellipse, EP = AC , is not found in the standard works on conic sections and appears to be original with Newton.^[4]

[2-C] seeing that, when from the other focus H of the ellipse the line HI is drawn parallel to CE, because CS and CH are equal , ES and EI are equal .

Or, ES = EI . The lines CS and CH are equal because they both locate a focus (S or H ) relative to the center C . The equality of ES and EI follows from the similarity of triangles SEC and SIH and the equality of line segments CS and CH (see fig. 5.9).

Figure 5.10
The sum (SP  + PH ) is a constant for any given ellipse,
and when the point P  goes to A  that sum is equal to
2AC .

Similarity: HI was constructed parallel to CE , and these two parallel lines are cut by the two transversals SCH and SEI , as demonstrated in figure 5.9 by the small diagram abstracted from the full diagram. Hence, all the angles a, b , and c in triangles SEC and SIH are equal and thus the triangles are similar.

Because the triangles are similar, C bisects SH , then E bisects SI , and thus ES = EI , as was to be demonstrated.

[2-D] and hence EP becomes half the sum of PS and PI,

Or, EP = 1/2(PS + PI ). From figure 5.9, PS = EP + ES = EP + EI (because ES = EI from [2-C]). Moreover, PS + PI = (EP + EI ) + PI = EP + (EI + PI ) = 2EP . Thus, EP = 1/2(PS + PI ).

[2-E] that is, of PS and PH which are conjointly equal to the total axis 2AC.

Or, 2EP = (PS + PI ) = (PS + PH ) = 2AC . From the properties of conics (Apollonius, Proposition 48, Book 3)^[5] the angles made by the tangent and focal lines (i.e., angles RPS and APH ) are equal, and since the line IH is constructed parallel to the tangent RPZ , the line PI = PH . An ellipse is a curve such that the sum of the distances from two fixed points (the foci) to a general point is given (i.e., the sum (PS + PH ) is a constant). When the general point P is on the major axis A (see fig. 5.10) then the sum (PS + PH ) equals the sum (AS + AH ) equals (2AC ) because AH = A' S . Thus, (PS + PH ) = 2AC . Therefore, from [2-D] 2EP = (PS + PI ) = (PS + PH ) = 2AC , or "clearly" EP = AC , as was to be demonstrated in [2-B].

[2-F] Let drop the perpendicular QT to SP, and, after calling the principal latus rectum (or 2BC² / AC) of the ellipse L,

In [2-F], Newton calls upon the Apollonian definition of the latus rectum L , which states that the latus rectum L is in the same ratio to the minor diameter 2BC as the minor diameter 2BC is to the major diameter 2AC (i.e., L : 2BC :: 2BC : 2AC or L / 2BC = 2BC / 2AC ) or L = 2BC² / AC as given above.

[2-G] there will be L × QR to L × PV as QR to PV,

The deviation QR is introduced in the format (L × QR ) / (L × PV ) = (QR ) / (PV ), which will be extended next. Given (QR ) / (PV ), then simply multiply by (L / L ), which can be written as (L / L ) (QR / PV ) = (L × QR ) / (L × PV ) = (QR ) / (PV ).

[2-H] that is, as PE (or AC) to PC;

Or, L (QR ) / L (PV ) = AC / PC . The expression for the deviation QR is extended. Triangles PEC and PXV are similar because XV and EC are parallel lines (see fig. 5.11). Thus, from the ratio of sides of the similar triangles, PX /PV = PE / PC . Moreover, PX = QR because QRPX is a parallelogram, and PE = AC , as demonstrated in [2-E]. Therefore, the ratio of sides can be written as QR / PV = AC / PC . Thus, from this result and [2-H], one obtains L (QR ) / L (PV ) = AC / PC , as required.

[2-I] and L × PV to GV × VP as L to GV

The element PV is introduced in a similar format to that employed to introduce QR in [2-G]. Given L / GV , simply multiply by PV / PV , which can be written as (PV / PV ) (L / GV ) = (L × PV ) / (GV × VP ) = L / GV .

[2-J] and GV × VP to QV²as CP²to CD² ;

Or, GV × VP / QV² = PC ² / DC² . This property of an ellipse will be used to eliminate PV from the expression for QR . The particular reference is to Proposition 15 of Book 1 of Apollonius's Conics , in which it is demonstrated that any chord parallel to a diameter of an ellipse is bisected by the conjugate diameter and, moreover, its square is equal to the product of the portions of the conjugate diameter. In figure 5.12, if the chord is given by QVQ ' and the conjugate diameter by PVG , then (PV × VG ) = (QV × VQ '), where QV = VQ '. The same relationship holds for the chord DK, that is (DC × CK ) = (BC × CG ), where DC = CK and BC = CG . Thus, the ratio PV × VG / QV² about point V is equal to the ratio PC² / CD² about point C , as required above.

[2-K] and QV²to QX²as, say , M to N;

The ratio M / N is simply a definition of the ratio of QV / QX , a step that Newton eliminates in the 1687 edition of the Principia .

[2-L] and QX²is to QT²as EP²to PF² ,

Figure 5.11
The shaded area CEP  from the ellipse above is abstracted as the similar triangles
PXV  and BEC  below. The parallelogram QRPX  from the ellipse is shown enlarged
below.

The second element of the discriminate ratio QT² is now introduced. Because triangle EPF is similar to triangle XQT (see fig. 5.13), the ratio of similar sides gives the ratio QX / QT = EP / PF .

Similarity: Lines QX and EF are parallel lines cut by the transversal EP , as demonstrated in the small drawing abstracted from the full drawing in figure 5.13. Thus, angles PEF and QXT are equal, as are the right angles EFP and XQT . Thus, all the angles of triangles EPF and XQT are equal and the triangles are similar.

Thus, QX / QT = EP / PF and as given above, QX² / QT² = EP² / PF ² .

Figure 5.12
The diameter PG  bisects the chords QQ ' and
DK . From Proposition 15 of Book 1 of
Apollonius's Conics , the ratio of PV  ×  VG  / QV²
is equal to the ratio PC²  / DC² .

Figure 5.13
The shaded area in the ellipse is abstracted as the triangles  EPF  and XQT . Since the lines
QX  and EF  are parallel, the triangles are similar.

[2-M] that is, as CA²to PF² , or as CD²to CB² . [Finally, that is , QX²to QT²as CD²to CB² .]

From [2-B] or [2-E], EP = CA , thus EP / PF = CA / PF . Also, from Lemma 1, area 4(CA ) (CB ) = area 4(CD ) (PF ) (see fig. 5.14), or CA / PF = CD / CB . Thus, from [2-L] QX / QT = EP / PF = CA / PF = CD /CB , which can be written in the square as in line [2-L] (i.e., QX² / QT ² = CD² / CB² ).

Figure 5.14
The area of parallelogram A (= 4BC  × CA ) is equal to the area of parallelogram B
(= 4CD  × PF ).

[2-N] And when all these ratios are combined , L × QR / QT²

Multiply together the five preceding relationships (i.e., lines [2-H] to [2-L]) and consider the products of the left- and right-hand sides separately. The product of the left-hand side of [H] [I] [J] [K] [L] is equal to the product (L × QR ) (L × PV ) (GV × PV ) (QV² ) (QX² ) divided by the product (L × PV ) (GV × PV ) (QV² ) (QX² ) (QT² ). By simple cancellation of equals, this ratio of products reduces to (L × QR ) / QT² .

[2-O] will be equal to (AC / PC) × (L / GV) × (CP² / CD² ) × (M / N) × (CD² / CB² ),

The product of the right-hand side of [H] [I] [J] [K] [L] equals the product (AC ) (L ) (PC² ) (M ) (CD² ) divided by the product (PC ) (GV ) (CD² ) (N ) (CB² ), which is equal to (AC /PC ) × (L /GV ) × (CP² / CD² ) × (M /N ) × (CD² / CB² ).

[2-P] that is, as AC × L (or 2BC² ) / (PC × GV) × (CP² / CB² ) × (M / N)

That is, by simple cancellation of CD² in the ratio given in [2-O], one has (AC ) (L ) (PC ² ) (M ) divided by (PC ) (GV ) (N ) (CB² ). Also, from the definition of the latus rectum L (see [2-F]), AC × L = 2BC ² . Substituting 2BC ² for AC × L , the ratio can be further reduced to (2BC² ) (CP² ) (M ) divided by (PC ) (GV ) (N ) (CB² ). This ratio can be rearranged into the ratio given above (i.e., [(2BC² ) / (PC × GV )] × [(CP² / CB² ) × (M /N )]).

[2-Q] or as (2PC / GV) × (M / N) or [(L × QR) to QT² ]

Upon canceling BC² and PC , [2-P] reduces to (2PC / GV ) (M /N ), as in [2-Q]. Thus, the left-hand side [2-N] equals the right-hand side [2-Q], or (L × QR ) / QT² = (2PC / GV ) (M / N ).

Figure 5.15
As the point Q  approaches the point P , the line QV  approaches the line QX
and the line GV  approaches the line GP , which is equal to 2PC .

Step 3—
The Limit

[3-R] But with the points P and Q coalescing, the ratios 2PC / GV and M / N become equal [and approach unity] ,

In the limit as the point Q approaches the point P , then the line GV approaches the line PG = 2PC (see fig. 5.15), and thus the ratio 2PC / GV approaches unity. Also in the limit, the line QV approaches the line QX and from [2-K], (QV / QX )² = M / N , thus the ratio M / N , also approaches unity.

[3-S] and therefore the combined ratio of these L × QR / QT² [equals unity (i.e ., L = QT² / QR)]

At long last, QR /QT ² = 1/L ! Thus, from [2-Q], one has (L × QR ) / QT² = (2PC / GV ) (M / N ) ® 1, or L = QT² / QR . Recall that Newton concluded Theorem 3 by stating that the force was to be computed from the ratio QR /QT ² × SP² . Everything in this proof has been directed toward showing that the discriminate ratio QR /QT ² is a constant equal to the reciprocal of the constant latus rectum L . The long journey is over.

[3-T] Multiply each part by SP² / QR and there will result L × SP² = SP² × QT² / QR.

To obtain the reciprocal of the linear dynamics ratio (QR / SP² × QT² ), simply multiply the expression for the discriminate ratio in [3-S] by SP² . Thus, SP² (L ) = SP² (QT ² / QR ) = 1 / linear dynamics ratio.

Conclusion

[U] Therefore the centripetal force is reciprocally as L × SP² , that is, [reciprocally ] in the doubled ratio of the distance. Which was to be proved .

From Theorem 3, the force is proportional to the linear dynamics ratio QR / SP² × QT² , or reciprocally as L (SP ² ), where the latus rectum L is a constant of the ellipse. Thus, the force is proportional to the inverse square of the distance SP .

Scholium

[V] Therefore the major planets orbit in ellipses having a focus at the center of the sun, and with their radii having been constructed to the sun describe areas proportional to the times, exactly as Kepler supposed .

By 1609, Kepler had obtained both the area law and the general ellipticity of solar planetary orbits by analyzing Tycho's observations of the motion of the planet Mars. In 1664, Newton demonstrated that both relationships can represent motion in the absence of any resistance. Note, however, that Newton demonstrated only that if the planet moves in an ellipse, and if the force center is at a focus, then the force is as the inverse square of the distance (a solution to the direct problem). He has not here demonstrated that if the force is as the inverse square of the distant, then it necessarily follows that the orbit is an ellipse with the force center at a focus of the ellipse (a solution to the inverse problem). In the solution to Problem 4 (to follow) he demonstrates that if the body is initially moving in an arbitrary ellipse under the action of an inverse square force, and if the speed and angle of inclination are changed to take on all possible values, then the resulting motion will be an ellipse, a hyperbola, or a parabola. Newton did not formally address the question of the uniqueness of the solution of the direct problem in the 1678 edition of the Principia ; however, he did speak to that question in the 1713 edition (to be discussed in chapter 10).

[W] Moreover, the latera recta of these ellipses are QT² / QR, where the distance between the points P and Q is the least possible and, as it were, infinitely small .

For infinitely small displacements the discriminate ratio QT² / QR is equal to the constant latus rectum L . Newton considers this property of elliptical motion (derived in [3-S]) to be so important that it is elevated to the position of a scholium for easy future reference. Given the detailed analysis required to provide the demonstration, it is a result worthy of the elevation.