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Apses and apsidal distances

Some results deduced in previous notes :

We have the following equation

(1)

as the energy conservation equation.

Here V(r) is the potential energy per unit mass such that

(1a)

Since ,

(1b)

We also have the angular momentum conservation equation

(2)

These conservation equations are a convenient starting point for investigating the general nature of orbital

motion.

Eliminating from these two equations we obtain the following ordinary differential equation for the radial

distance :

(3)

We call this the radial motion equation for the particle P. Equation (3) (together with the initial

conditions) is sufficient to determine the variation of with , and the angular momentum equation (2) then

determines the variation of with . Unfortunately, for most laws of force, this procedure cannot be carried

through analytically. However, it is still possible to make important deductions about the general nature of the

motion.

For finding the equation of the path we use the transformation,

. (4)

From (2) we also have,

. (5)

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The Path Equation (Polar Form) :

(6)

Its solutions are the polar equations of the paths that the body can take when it moves under the force field

per unit mass.

Applying (4) and (5) in (3) we have an equivalent form of the path equation as

(7)

Despite the appearance of the left side of equation (6), the path equation is not linear in general. This is

because the right side is a function of , the dependent variable. Only for the inverse square and inverse cube

laws does the path equation become linear. It is a remarkable piece of good luck that the inverse square law (the

most important case by far) is one of only two cases that can be solved easily.

Initial conditions for the path equation Suitable initial conditions for the path equation are provided by

specifying the values of and when , say. Since , the initial value of is given

directly by the initial data. The value of is not given directly but can be deduced from equation (5) in the

form

(8)

The Path Equation (Pedal Form):

Let be the length of the perpendicular from the fixed centre of force upon the tangent to the path at P whose

polar coordinates are (see second figure below).

Then we have

, where

, (9)

so that

and

(10)

Differentiating (9) with respect to , we have

, from (6)

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Therefore,

(11)

This is the pedal form of the differential equation of the central orbit.

Apse, apsidal distance, apsidal angle

The points at which an orbit touches its bounding circles are important and are given a special name:

A point of an orbit at which the distance OP achieves its maximum or minimum value is called an apse of the

orbit. These maximum and minimum distances are called the apsidal distances and the angular displacement

between successive apses (the angle α in Figure below (left)) is called the apsidal angle.

Figure: Typical bounded and unbounded orbits

In the special case of orbits around the Sun, the point of closest approach is called the perihelion and the point

of maximum distance the aphelion. The corresponding terms for orbits around the Earth are perigee and apogee.

An apse is also defined as a point in a central orbit at which the normal to the curve passes through the

centre of force.

The apsidal distances, the maximum and minimum distances of P from O, are easily found from the

radial motion equation (3).

Since at an apse the tangent to the curve must be at right angles to the radius vector, i.e.

.

But,

; therefore

.

Also

changes sign as increases through the value that indicates the position of an apse.

At an apse, or equivalently,

and so must satisfy

4

. or

(12)

This is an equation to determine the values of at the apses, supposing the constants and known.

The positive roots of this equation are the apsidal distances.

The following facts may be noted :

(i) Any central orbit has only two apsidal distances. The orbit is a curve touching two concentric circles,

the radii of which are the two apsidal distances.

(ii) Once the orbit between two consecutive apse is known, the whole of the orbit may be constructed by

operations of folding over apsidal radii, as a central orbit is symmetric with respect to the line drawn

from the force centre to an apse.

(iii) The angle subtended at the centre by the arc joining consecutive apses is a constant. It is called the

apsidal angle.

In some cases (i) may be violated. The radius of the inner circle may be zero, or that of the outer circle may

be infinite. But these are to be regarded as exceptional cases.

Let us now consider how the apsidal angle is to be found.

When is increasing, (7) gives

, (13)

where,

. (14)

Thus,

;

and so, if and are the reciprocals of the apsidal distances ( with ), the apsidal angle is

; (15)

Some more results :-

Areal Velocity

We have already obtained that –

areal velocity of a particle moving along a plane curve

(16)

Now, let be the length of the perpendicular from on the straight line and let be the velocity of the

particle at .

Then the areal velocity is

5

(17)

Hence, twice the areal velocity (18)

From (18) we have

Using (9) we have

(19)

Thus the linear velocity of the particle varies inversely as the perpendicular drawn from the centre of force

upon the tangent to the path.

Note: The relation

is often found very useful.

Determination of time

To determine the time, we use (2) or (18) to obtain the formula

. (20)

Velocity from Infinity

The equation of energy for a central force is

or,

.

Therefore the velocity from infinity is given by

.

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If

.

Problems on Central Orbits

The three figures refer to the polar, pedal and Cartesian terms respectively, of which the first two are preferable.

The problems of central orbits are treated most concisely by means of the principle of angular momentum and

the equation of energy and using pedal form of the differential equation.

There are two fundamental problems. We shall use equation, i.e. pedal form.

I. Given the law of acceleration be ; determine the orbit

The equation of energy is then

7

Also ; therefore

(21)

where C is a constant.

This determines the shape and size of the orbit, when C is given, but not its orientation about the centre

of force.

II. Given the orbit, determine the law of acceleration

The equation (21) is given. Differentiating with respect to ,

(22)

This gives the law of acceleration.

The equation (22) can also be directly obtained as follows :

Resolving along the normal, we have

,

But

,

and

; therefore

.

In the above two problems we have used pedal form.

They can also be discussed using polar form, for which we use the path equation in polar form given by (6) and

the results (9) and (10).

I. Given the law of force, determine the orbit.

Example 1. Path equation for law of direct distance

Let us consider a particle of mass attracted towards a fixed point by a force . Let us consider the law of

direct distance, meaning thereby that is proportional to . Let us confine our attention to an attractive force,

putting

, (23)

where is a constant.

For rectangular Cartesian coordinates Oxy, the equations of motion are

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, (24)

where .

Since ,

(24) reduces to

. (25)

The general solutions of (25) are

(26)

where the coefficients are constants, to be fixed by the initial conditions.

If we solve (26) for and and eliminate by the identity

we get,

. (27)

This is a central conic and necessarily an ellipse since remain finite, as we see from (26).

Thus the orbit described under a central attractive force varying directly as the distance is an ellipse having its

centre at the centre of force. This motion is called elliptic harmonic .

To illustrate the significance of the constants in (26), let us suppose that at time the particle is at

, moving with velocity in the direction of the y-axis, so that . Putting this

information into the equations (26), first as they stand and then in the form obtained by differentiation , we get

And so the motion is given by

. (28)

Now let us find the apse and apsidal angle of this orbit.

Since ,

from 1(b) we have

(29)

and so (12) may be written as

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. (30)

This quadratic equation in yields roots

, which will be squares of the reciprocals of the semiaxes of

the elliptical orbit.

To evaluate the apsidal angle, we note that in this case

(14) reduces to

which is of the form

,

where and are constants. But at an apse; hence are roots of , and so

.

Thus, by (15)

, (31)

as we already know from the fact that the orbit is a central ellipse.

Example 2. Path equation for the inverse square law

Let us consider the law of inverse square, meaning thereby that is inversely proportional to . Let us confine

our attention to an attractive force, putting

, (32)

where is a constant.

From (21) we have,

(33)

Comparing with the equations of the ellipse, parabola and hyperbola,

, (34)

We see that the orbit is an ellipse, parabola or hyperbola according as is negative, zero or positive.

If a particle is projected with velocity from a point at a distance c from the centre of force, then from (33) we

have

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and therefore is negative, zero or positive, i.e. the orbit is an ellipse, parabola or hyperbola according as

.

Example 3. Path equation for the inverse cube law

The engines of a spaceship have failed and the ship is moving in a straight line with speed V. The crew

calculate that their present course will miss the planet B by a distance p. However, B is known to exert the force

(32)

on any mass in its vicinity. A measurement of the constant reveals that

. (33)

Show that the crew of the spaceship will get a free tour around B before continuing along their original path.

What is the distance of closest approach and what is the speed of the spaceship at that instant?

Solution

For the given law of force, so that .

Also, from the initial conditions, .

The path equation is therefore

(34)

which simplifies to

,

(35)

on using the stated value of . The general solution of this equation is

(36)

The constants and can now be determined from the initial conditions. Take the initial line as shown

in figure below:

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Figure: The path of the spaceship around the planet B

Then:

(i) The initial condition when implies that when . It follows that .

(ii) The initial condition on is given by (8) to be

(37)

when . It follows that . The required solution is therefore

(38)

That is

. (39)

This is the polar equation of the path of the spaceship, as shown in Figure above. The spaceship recedes to

infinity when again, that is when . Thus the spaceship makes one circuit of B before

continuing on as before. The distance of closest approach is and is achieved when . By angular

momentum conservation, the speed of the spaceship at that instant is .

II. Given the orbit, determine the law of force.

1. Force directed to the centre.

Example 4. Circular orbit

Evidently

where is the radius of the circle.

Example 5. Elliptic orbit

The (r,p) equation of the orbit is

.

12

Differentiating, we get

.

Comparing with

, we find that

where

or .

Also,

, and is parallel to where is the semi-diameter conjugate to .

.

2. Force directed to the centre.

Example 6. Elliptic orbit

The (r,p) equation of the orbit is

.

Differentiating, we get

.

Comparing with

, we find that

, where

or , if denotes the semi-latus rectum

.

Also

.

.

Example 7. Parabolic orbit

The (r,p) equation of the orbit is

or

Differentiating, we get

.

13

Comparing with

, we find that

, where

or , if denotes the semi-latus rectum .

Also

.

Example 8. Hyperbolic orbit

The (r,p) equation of the orbit is

.

Differentiating, we get

.

Comparing with

, we find that

, where

or , if denotes the semi-latus rectum

.

Also

.

Note: Thus we see that if the orbit be a conic, the law of force is inverse square of the distance.

Again comparing with Example 2 above, we see that

according as the orbit is

ellipse, parabola or hyperbola.

Example 9. A particle moves with a central acceleration

, being projected from an apse at distance

with a velocity ; show that it describes the curve .

Solution

.

Also when , so that

.

.

Again,

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or,

.

Put

,

Then

, since when

and ,

or, .

Example 10. A particle of mass moves under a central force

.

It is projected from an apse at a distance with a velocity

. Show that the path is .

Solution

.

Also,

when , so that

;

i.e. and .

Again,

or,

or,

,

.

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Also, when ,

so that , or,

.

or,

i.e. .