Newton's theorem of revolving orbits

The addition of such an inverse-cube force also changes the path followed by the particle. The path of the particle ignores the time dependencies of the radial and angular motions, such as r(t) and θ₁(t); rather, it relates the radius and angle variables to one another. For this purpose, the angle variable is unrestricted and can increase indefinitely as the particle revolves around the central point multiple times. For example, if the particle revolves twice about the central point and returns to its starting position, its final angle is not the same as its initial angle; rather, it has increased by 2×360° = 720°. Formally, the angle variable is defined as the integral of the angular speed

${\displaystyle \theta _{1}\equiv \int \omega _{1}(t)\,dt.}$

A similar definition holds for θ₂, the angle of the second particle.

If the path of the first particle is described in the form r = g(θ₁), the path of the second particle is given by the function r = g(θ₂/k), since θ₂ = k θ₁. For example, let the path of the first particle be an ellipse

${\displaystyle {\frac {1}{r}}=A+B\cos \theta _{1}}$

where A and B are constants; then, the path of the second particle is given by

${\displaystyle {\frac {1}{r}}=A+B\cos \left({\frac {\theta _{2}}{k}}\right).}$

To simplify the equations, Newton writes F(r) in terms of a new function C(r)

${\displaystyle F(r)={\frac {C(r)}{Rr^{3}}}}$

where R is the average radius of the nearly circular orbit. Newton expands C(r) in a series—now known as a Taylor expansion—in powers of the distance r, one of the first appearances of such a series.[27] By equating the resulting inverse-cube force term with the inverse-cube force for revolving orbits, Newton derives an equivalent angular scaling factor k for nearly circular orbits：[24]

${\displaystyle {\frac {1}{k^{2}}}=\left({\frac {R}{C}}\right)\left.{\frac {dC}{dr}}\right|_{r=R}}$

In other words, the application of an arbitrary central force F(r) to a nearly circular elliptical orbit can accelerate the angular motion by the factor k without affecting the radial motion significantly. If an elliptical orbit is stationary, the particle rotates about the center of force by 180° as it moves from one end of the long axis to the other (the two apses). Thus, the corresponding apsidal angle α for a general central force equals k×180°, using the general law θ₂ = k θ₁.

Newton illustrates his formula with three examples. In the first two, the central force is a power law, F(r) = rⁿ⁻³, so C(r) is proportional to rⁿ. The formula above indicates that the angular motion is multiplied by a factor k = 1/√n, so that the apsidal angle α equals 180°/√n.

This angular scaling can be seen in the apsidal precession, i.e., in the gradual rotation of the long axis of the ellipse (Figure 3). As noted above, the orbit as a whole rotates with a mean angular speed Ω=(k−1)ω, where ω equals the mean angular speed of the particle about the stationary ellipse. If the particle requires a time T to move from one apse to the other, this implies that, in the same time, the long axis will rotate by an angle β = ΩT = (k − 1)ωT = (k − 1)×180°. For an inverse-square law such as Newton's law of universal gravitation, where n equals 1, there is no angular scaling (k = 1), the apsidal angle α is 180°, and the elliptical orbit is stationary (Ω = β = 0).

As a final illustration, Newton considers a sum of two power laws

${\displaystyle C(r)\propto ar^{m}+br^{n}}$

which multiplies the angular speed by a factor

${\displaystyle k={\sqrt {\frac {a+b}{am+bn}}}}$

Newton applies both of these formulae (the power law and sum of two power laws) to examine the apsidal precession of the Moon's orbit.

Newton's derivation is found in Section IX of his Principia, specifically Propositions 43–45.[41] His derivations of these Propositions are based largely on geometry.

Proposition 43; Problem 30

Diagram illustrating Newton's derivation. The blue planet follows the dashed elliptical orbit, whereas the green planet follows the solid elliptical orbit; the two ellipses share a common focus at the point C. The angles UCP and VCQ both equal θ₁, whereas the black arc represents the angle UCQ, which equals θ₂ = k θ₁. The solid ellipse has rotated relative to the dashed ellipse by the angle UCV, which equals (k−1) θ₁. All three planets (red, blue and green) are at the same distance r from the center of force C.

It is required to make a body move in a curve that revolves about the center of force in the same manner as another body in the same curve at rest.[42]

Newton's derivation of Proposition 43 depends on his Proposition 2, derived earlier in the Principia.[43] Proposition 2 provides a geometrical test for whether the net force acting on a point mass (a particle) is a central force. Newton showed that a force is central if and only if the particle sweeps out equal areas in equal times as measured from the center.

Newton's derivation begins with a particle moving under an arbitrary central force F₁(r); the motion of this particle under this force is described by its radius r(t) from the center as a function of time, and also its angle θ₁(t). In an infinitesimal time dt, the particle sweeps out an approximate right triangle whose area is

${\displaystyle dA_{1}={\frac {1}{2}}r^{2}d\theta _{1}}$

Since the force acting on the particle is assumed to be a central force, the particle sweeps out equal angles in equal times, by Newton's Proposition 2. Expressed another way, the rate of sweeping out area is constant

${\displaystyle {\frac {dA_{1}}{dt}}={\frac {1}{2}}r^{2}{\frac {d\theta _{1}}{dt}}=\mathrm {constant} }$

This constant areal velocity can be calculated as follows. At the apapsis and periapsis, the positions of closest and furthest distance from the attracting center, the velocity and radius vectors are perpendicular; therefore, the angular momentum L₁ per mass m of the particle (written as h₁) can be related to the rate of sweeping out areas

${\displaystyle h_{1}={\frac {L_{1}}{m}}=rv_{1}=r^{2}{\frac {d\theta _{1}}{dt}}=2{\frac {dA_{1}}{dt}}}$

Now consider a second particle whose orbit is identical in its radius, but whose angular variation is multiplied by a constant factor k

${\displaystyle \theta _{2}(t)=k\theta _{1}(t)\,\!}$

The areal velocity of the second particle equals that of the first particle multiplied by the same factor k

${\displaystyle h_{2}=2{\frac {dA_{2}}{dt}}=r^{2}{\frac {d\theta _{2}}{dt}}=kr^{2}{\frac {d\theta _{1}}{dt}}=2k{\frac {dA_{1}}{dt}}=kh_{1}}$

Since k is a constant, the second particle also sweeps out equal areas in equal times. Therefore, by Proposition 2, the second particle is also acted upon by a central force F₂(r). This is the conclusion of Proposition 43.

Proposition 44

The difference of the forces, by which two bodies may be made to move equally, one in a fixed, the other in the same orbit revolving, varies inversely as the cube of their common altitudes.[44]

To find the magnitude of F₂(r) from the original central force F₁(r), Newton calculated their difference F₂(r) − F₁(r) using geometry and the definition of centripetal acceleration. In Proposition 44 of his Principia, he showed that the difference is proportional to the inverse cube of the radius, specifically by the formula given above, which Newtons writes in terms of the two constant areal velocities, h₁ and h₂

${\displaystyle F_{2}(r)-F_{1}(r)=m{\frac {h_{1}^{2}-h_{2}^{2}}{r^{3}}}}$

Proposition 45; Problem 31

To find the motion of the apsides in orbits approaching very near to circles.[24]

In this Proposition, Newton derives the consequences of his theorem of revolving orbits in the limit of nearly circular orbits. This approximation is generally valid for planetary orbits and the orbit of the Moon about the Earth. This approximation also allows Newton to consider a great variety of central force laws, not merely inverse-square and inverse-cube force laws.

Modern derivations of Newton's theorem have been published by Whittaker (1937)[45] and Chandrasekhar (1995).[42] By assumption, the second angular speed is k times faster than the first

${\displaystyle \omega _{2}={\frac {d\theta _{2}}{dt}}=k{\frac {d\theta _{1}}{dt}}=k\omega _{1}}$

Since the two radii have the same behavior with time, r(t), the conserved angular momenta are related by the same factor k

${\displaystyle L_{2}=mr^{2}\omega _{2}=mr^{2}k\omega _{1}=kL_{1}\,\!}$

The equation of motion for a radius r of a particle of mass m moving in a central potential V(r) is given by Lagrange's equations

${\displaystyle m{\frac {d^{2}r}{dt^{2}}}-mr\omega ^{2}=m{\frac {d^{2}r}{dt^{2}}}-{\frac {L^{2}}{mr^{3}}}=F(r)}$

Applying the general formula to the two orbits yields the equation

${\displaystyle m{\frac {d^{2}r}{dt^{2}}}=F_{1}(r)+{\frac {L_{1}^{2}}{mr^{3}}}=F_{2}(r)+{\frac {L_{2}^{2}}{mr^{3}}}=F_{2}(r)+{\frac {k^{2}L_{1}^{2}}{mr^{3}}}}$

which can be re-arranged to the form

${\displaystyle F_{2}(r)=F_{1}(r)+{\frac {L_{1}^{2}}{mr^{3}}}\left(1-k^{2}\right)}$

This equation relating the two radial forces can be understood qualitatively as follows. The difference in angular speeds (or equivalently, in angular momenta) causes a difference in the centripetal force requirement; to offset this, the radial force must be altered with an inverse-cube force.

Newton's theorem can be expressed equivalently in terms of potential energy, which is defined for central forces

${\displaystyle F(r)=-{\frac {dV}{dr}}}$

The radial force equation can be written in terms of the two potential energies

${\displaystyle -{\frac {dV_{2}}{dr}}=-{\frac {dV_{1}}{dr}}+{\frac {L_{1}^{2}}{mr^{3}}}\left(1-k^{2}\right)}$

Integrating with respect to the distance r, Newtons's theorem states that a k-fold change in angular speed results from adding an inverse-square potential energy to any given potential energy V₁(r)

${\displaystyle V_{2}(r)=V_{1}(r)+{\frac {L_{1}^{2}}{2mr^{2}}}\left(1-k^{2}\right)}$

Although Newton states that the problem was to be solved by Proposition 6, he does not use it explicitly. In the following, simplified proof, Proposition 6 is used to show how the result is derived.

Newton's detailed proof follows that, and finally Proposition 6 is appended, as it is not well-known.

Proposition 44 uses Proposition 6 to prove a result about revolving orbits. In the propositions following Proposition 6 in Section 2 of the Principia, he applies it to specific curves, for example, conic sections. In the case of Proposition 44, it is applied to any orbit, under the action of an arbitrary force directed towards a fixed point, to produce a corresponding revolving orbit.

In Fig. 1, MN is part of that orbit. At point P, the body moves to Q under the action of a force directed towards S, as before. The force, F(SP) is defined at each point P on the curve.

In Fig. 2, the corresponding part of the revolving orbit is mn with s as its centre of force. Assume that initially, the body in the static orbit starts out at right angles to the radius with speed V. The body in the revolving orbit must also start at right angles and assume its speed is v. In the case shown in Fig. 1, ${\displaystyle v>V}$ and the force is directed towards S. The argument applies equally if ${\displaystyle v<V}$. Also, the force may be directed away from the centre.

Let SA be the initial direction of the static orbit, and sa, that of the revolving orbit. If after a certain time the bodies in the respective orbits are at P and p, then the ratios of the angles ${\displaystyle {\frac {asp}{ASP}}={\frac {psq}{PSQ}}={\frac {v}{V}}}$; the ratios of the areas ${\displaystyle {\frac {asp}{ASP}}={\frac {psq}{PSQ}}={\frac {v}{V}}}$; and the radii ${\displaystyle SP=sp}$, ${\displaystyle SQ=sq}$, ${\displaystyle SA=sa}$.

The figure pryx and the arc py in Fig. 2 are the figure PRQT and the arc PQ in Fig. 1, expanded linearly in the horizontal direction in the ratio ${\displaystyle {\frac {v}{V}}}$, so that ${\displaystyle PU=pu}$, ${\displaystyle UT=ux}$, and ${\displaystyle qt={\frac {v}{V}}QT}$. The straight lines qt and QT should really be circular arcs with centres s and S and radii sq and SQ respectively. In the limit, their ratio becomes ${\displaystyle {\frac {v}{V}}}$, whether they are straight lines or arcs.

Since in the limit the forces are parallel to SP and sp, if the same force acted on the body in Fig. 2 as in Fig. 1, the body would arrive at y, since ry = RQ. The difference in horizontal speed does not affect the vertical distances. Newton refers to Corollary 2 of the Laws of Motion, where the motion of the bodies is resolved into a component in the radial direction acted on by the whole force, and the other component transverse to it, acted on by no force.

However, the distance from y to the centre, s, is now greater than SQ, so an additional force is required to move the body to q such that sq = SQ. The extra force is represented by yq, and f is proportional to ry + yq, just as F is to RQ.

${\displaystyle RQ=SP-PU-ST}$, ${\displaystyle rq=sp-pu-st=RQ+ST-st}$.

The difference, ${\displaystyle ST-st=yq}$, can be found as follows:

${\displaystyle ST^{2}=SQ^{2}-QT^{2}}$, ${\displaystyle st^{2}=sq^{2}-qt^{2}}$, so ${\displaystyle (ST-st)(ST+st)=qt^{2}-QT^{2}}$.

And in the limit, as QT and qt approach zero, ${\displaystyle ST+st}$ becomes equal to ${\displaystyle SQ+sq}$ or 2SP so

${\displaystyle yq*2SP={\frac {QT^{2}(v^{2}-V^{2})}{V^{2}}}}$.

Therefore, ${\displaystyle rq=RQ+ST-st=RQ+{\frac {QT^{2}(v^{2}-V^{2})}{2SP*V^{2}}}}$.

Since from Proposition 6 (Fig.1 and see below), the force is ${\displaystyle {\frac {k*QR}{(SP^{2}*QT^{2})}}}$. Divide by ${\displaystyle {\frac {(SP^{2}*QT^{2})}{k}}}$, where k is constant, to obtain the forces ${\displaystyle f(sp)=F(sp)+{\frac {k(v^{2}-V^{2})}{2sp^{3}*V^{2}}}}$.

In Fig. 3, at the initial point A of the static curve, draw the tangent AR, which is perpendicular to SA, and the circle AQD, which just touches the curve at A. Let ρ be the radius of that circle. Since angle SAR is a right angle, the centre of the circle lies on SA. From the property of a circle: ${\displaystyle QT^{2}=AT(2\rho -AT)=RQ(2\rho -RQ)}$, and in the limit as Q approaches A, this becomes ${\displaystyle \rho =QT^{2}/2RQ}$.

Hence, ${\displaystyle F(SA)={\frac {k}{(2\rho *SA^{2})}}}$.

And since F(SA) is given, this determines the constant k. However, Newton wants the force at A to be of the form ${\displaystyle cV^{2}/SA^{2}}$, where c is a constant, so that ${\displaystyle f(sp)=F(sp)+{\frac {c\rho (v^{2}-V^{2})}{sp^{3}}}}$, where ${\displaystyle c={\frac {F(SA)*SA^{2}}{V^{2}}}}$.

The expression for f(sp) above is the same as Newton's in Corollary 4 of Proposition 44, except that he uses different letters. He writes ${\displaystyle {\frac {G}{F}}={\frac {v}{V}}}$ (where G and F are not necessarily equal to v and V respectively), and uses the letter “V” for the constant corresponding to “c”, and the letter “X” for the function F(sp).

The above geometric proof shows very clearly where the additional force arises from to make the orbit revolve with respect to the static orbit.

Newton's proof is complicated, in view of the simplicity of the above proof. As an example, his proof requires some deciphering, as the following sentence shows:

“And therefore, if with centre C and any radius CP or Cp a circular sector is described equal to the total area VPC which the body P revolving in an immobile orbit has described in any time by a radius drawn to the centre, the difference between the forces by which the body P in an immobile orbit and body p in a mobile orbit revolve will be to the centripetal force by which some body, by a radius drawn to the centre, would have been able to describe that sector uniformly in the same time in which the area VPC was described as G² - F² to F².”

He initially regards the infinitesimal ${\displaystyle \tau }$ as fixed, then the areas SPQ and spq are proportional to V and v, respectively; therefore, ${\displaystyle QT*SP\propto V\tau }$ and ${\displaystyle qt*sp\propto v\tau }$ at each of the points P and p, and ${\displaystyle yq={\frac {qt^{2}-QT^{2}}{2sp}}\propto {\frac {1}{sp^{3}}}}$ so the additional force varies inversely as the cube of the radius.

In Fig.1, XQ is a circular arc, with centre S and radius SQ, meeting SP at X. The perpendicular XY meets RQ at Y, and ${\displaystyle YQ={\frac {QT^{2}}{2SP}}}$.

Let ${\displaystyle \phi (SQ)}$ be the force required to make a body move in a circle of radius SQ, if it has the same speed as the transverse speed of the body in the static orbit at Q.

${\displaystyle {\frac {f(sp)-F(sp)}{\phi (sp)}}={\frac {yq}{YQ}}={\frac {(qt^{2}-QT^{2})}{QT^{2}}}={\frac {(v^{2}-V^{2})}{V^{2}}}}$

at every point, P and in particular at the apside, A:

${\displaystyle {\frac {f(SA)-F(SA)}{\phi (SA)}}={\frac {(v^{2}-V^{2})}{V^{2}}}}$.

But at A, in Fig. 3., the ratio of the force that makes the body follow the static curve, AE, to that required to make it follow the circle, AB, with radius SA, is inversely as the ratio of their radii of curvature, since they are both moving at the same speed, V, perpendicular to SA: ${\displaystyle {\frac {F(SA)}{\phi (SA)}}={\frac {SA}{\rho }}}$. From the first part of the proof, ${\displaystyle f(sp)-F(sp)={\frac {(f(sa)-F(sa))sa^{3}}{sp^{3}}}={\frac {\rho F(sa)sa^{2}(v^{2}-V^{2})}{V^{2}sp^{3}}}}$.

Substituting Newton's expression for F(SA), gives the result obtained previously.

“To find the motion of the apsides in orbits approaching circles.”

Proposition 44 was devised expressly to prove this Proposition. Newton wants to investigate the motion of a body in a nearly circular orbit attracted by a force of the form ${\displaystyle f(sp)=\gamma *sp^{n-3}}$.

He approximates the static curve by an ellipse with an inverse square force, F(SP), directed to one of the foci, made to revolve by the addition of an inverse cube force, according to Proposition 44.

For the static ellipse, with the force varying inversely as SP squared, ${\displaystyle F(SP)={\frac {cV^{2}}{SP^{2}}}}$, since c is defined above so that ${\displaystyle F(SA)={\frac {cV^{2}}{SA^{2}}}}$.

With the body in the static orbit starting from the upper apside at A, it will reach the lower apside, the point closest to S, after moving through an angle of 180 degrees. Newton wants a corresponding revolving orbit starting from apside, a, about a point s, with the lower apside shifted by an angle, α, where ${\displaystyle {\frac {(180+\alpha )}{180}}={\frac {v}{V}}}$.

The initial speed, V, at A must be just less than that required to make the body move in a circle. Then ρ can be taken as equal to SA or sa. The problem is to determine v from the value of n, so that α can be found, or given α, to find n.

${\displaystyle f(sp)=F(sp)+{\frac {c\rho (v^{2}-V^{2})}{sp^{3}}}={\frac {c(sp*V^{2}+sa(v^{2}-V^{2}))}{sp^{3}}}}$

Letting ${\displaystyle sp=sa-X}$,

${\displaystyle f(sp)={\frac {c(s*av^{2}-XV^{2})}{sp^{3}}}}$.

Then “by our method of converging series”: ${\displaystyle f(sp)={\frac {\gamma *sp^{n}}{sp^{3}}}={\frac {\gamma (sa-X)^{n}}{sp^{3}}}={\frac {\gamma *sa^{n-1}(sa-nX)}{sp^{3}}}}$ plus terms in X² and above which can be ignored because the orbit is almost circular, so X is small compared to sa.

Comparing the 2 expressions for f(sp), ${\displaystyle f(sp)={\frac {\gamma *sa^{n-1}(sa-nX)}{sp^{3}}}={\frac {c(sa*v^{2}-XV^{2})}{sp^{3}}}}$ it follows that ${\displaystyle {\frac {v}{V}}={\frac {1}{n^{1/2}}}={\frac {(180+\alpha )}{180}}}$.

Also, ${\displaystyle f(sp)={\frac {cv^{2}*sp^{n-3}}{sa^{n-1}}}}$.

The ratio of the initial forces at a is given by ${\displaystyle {\frac {f(sa)}{F(sa)}}={\frac {1}{n}}}$.

In Fig. 1, a body is moving along a specific curve MN acted on by a (centripetal) force, towards the fixed point S. The force depends only of the distance of the point from S. The aim of this proposition is to determine how the force varies with the radius, SP. The method applies equally to the case where the force is centrifugal.

In a small time, ${\displaystyle \tau }$, the body moves from P to the nearby point Q. Draw QR parallel to SP meeting the tangent at R, and QT perpendicular to SP meeting it at T.

If there was no force present it would have moved along the tangent at P with the speed that it had at P, arriving at the point, R. If the force on the body moving from P to Q was constant in magnitude and parallel to the direction SP, the arc PQ would be parabolic with PR as its tangent and QR would be proportional to that constant force and the square of the time, ${\displaystyle \tau }$.

Conversely, if instead of arriving at R, the body was deflected to Q, then a constant force parallel to SP, with magnitude: ${\displaystyle F\propto {\frac {QR}{\tau ^{2}}}}$ would have caused it to reach Q instead of R.

However, since the direction of the radius from S to points on the arc PQ and also the magnitude of the force towards S will change along PQ, the above relation will not give the exact force at P. If Q is sufficiently close to P, the direction of force will be almost parallel to SP all along PQ and if the force changes little, PQ can be assumed to be approximated by a parabolic arc with the force given as above in terms of QR and ${\displaystyle \tau }$.

The time, ${\displaystyle \tau }$ is proportional to the area of the sector SPQ. This is Kepler's Second Law. A proof is demonstrated in Proposition 1, Book 1, in the Principia. Since the arc PQ can be approximated by a straight line, the area of the sector SPQ and the area of the triangle SPQ can be taken as equal, so

${\displaystyle F\propto {\frac {QR}{\tau ^{2}}}={\frac {k*QR}{(SP^{2}*QT^{2})}}}$, where k is constant.

Again, this is not exact for finite lengths PQ. The force law is obtained if the limit of the above expression exists as a function of SP, as PQ approaches zero.

In fact, in time ${\displaystyle \tau }$, the body with no force would have reached a point, W, further from P than R. However, in the limit QW becomes parallel to SP. The point W is ignored in Newton's proof.

Also, Newton describes QR as the versed sine of the arc with P at its centre and length twice QP. Although this is not strictly the same as the QR that he has in the diagram (Fig.1), in the limit, they become equal.

Notes:

This proposition is based on Galileo's analysis of a body following a parabolic trajectory under the action of a constant acceleration. In Proposition 10, he describes it as Galileo's Theorem, and mentions Galileo several other times in relation to it in the Principia. Combining it with Kepler's Second Law gives the simple and elegant method.

In the historically very important case where MN in Fig. 1 was part of an ellipse and S was one of its foci, Newton showed in Proposition 11 that the limit ${\displaystyle QR/QT^{2}}$ was constant at each point on the curve, so that the force on the body directed towards the fixed point S varied inversely as the square of the distance SP.

Besides the ellipse with the centre at the focus, Newton also applied Proposition 6 to the hyperbola (Proposition 12), the parabola (Proposition 13), the ellipse with the centre of force at the centre of the ellipse (Proposition 10), the equiangular spiral (Proposition 9), and the circle with the centre of force not coinciding with the centre, and even on the circumference (Proposition 7).