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let y = n; then the inclination of the image of the given object to the major axis is had from
b v(b-n). tan. p =
and therefore the inclination to the given axis is known, viz.,
- ( - ).
The portion of the image due to the given object is the arc of the conic section intercepted by the straight lines drawn from the extremities of the given object at right-angles to it.
827. SINCE the day is given, the sun's place in the ecliptic is known, and therefore, by the tables, its declination ; which suppose D. Also let L denote the given latitude of the spectator. Hence the meridian altitude of the sun is
a = 90° - LED + or
- being used according as the given time is in summer or winter. This is the < subtended by the tower; consequently its altitude is
h = a x tan.a = a x cot. (L + D).
828. Once for all let us investigate a general formula for the aberratic curve.
Let p and p denote the radius vector and I upon the tangent of the given orbit ; p and p' the corresponding ones to the aberratic curve. Also let O be the < described by p; ('the Z described by é. Then since the angular velocity of p = the angular velocity of ¢ (Woodhouse's Ašt. 2d edit., p. 304,) we have (see 443)
do': do ::
dp , do
р Again, g' cc velocity of the earth
dp P do' p
w (8-p) § 7 (pe - p)
Nga - po which gives p = p.
Also g' =
р These two equations will give the equation to the aberratic
Ex. (1.) In the question
L på =
4 L being the principal parameter of the parabolic orbit;
cp р ✓Lp
L :: p = ! But if a be the distance of the centre of polar co-ordinates from the centre of a circle, its equation is
que - d
and consequently when that pole is in the circumference
Hence it appears that the aberrative curve is a circle whose radius is
Ex. 2. Let the orbit be an ellipse whose equation is
62 so that the aberratic curve is a circle whose radius is
78 and the distance of whose centre from the centre of co-ordinates is
5 + (z – 8).
Ex. 3. Let the orbit be the logarithmic spiral whose equation is
p = mg. Then
Therefore the aberratic curve is also a logarithmic spiral.
á be the altitudes when the sun is due east and at six o'clock, L the latitude of the place, and D the declination of the sun; then from the two right-angled A whose sides are
90° L, 90° – D, 90 and 90° L, 90°
D, 90 ca we easily get
r sin. D= sin. L sin, a
g sin. á = sin, L. sin. D and eliminating sin. D,
gue sin sin. 'L =
830. Let a, b denote the semi-axis of the planet's orbit ; then its area is
q=(ab) its area will be equal to that of the orbit; and if, with an uniform motion, a body be supposed to go through the whole circumference in the periodic time of the planet, the 2 so described by the radius in an hour, will measure the mean horary motion (s) of the planet in its orbit.
Again if do denote the true described by the radius veetor in the same hour, by Kepler's law of the equable , description of areas, we have
e'de Area described by p =
:: pede = year = aba
aba is gdo =
8 which is the true horary motion of the planet in its orbit.
Let the planet's orbit Np (Fig. 106.) and ecliptic Nm intersect in the node N, then make
P and describe the arcs PM, pm, Pr 1 NM and pn. Now since Pp is very small compared with the whole extent of the orbit, Ppr may be considered rectilinear, and we have
Pr : Pp :: sin. NPM : 1
Mm : Pr:: 1 : cos. PM
:: cos. N:cos. 'PM
cos. 'PM which gives the horary motion required. In like manner the horary motion in latitude is found to be
pr = Pp x tan. PM. cot. NP