Annual and diurnal aberration.

CHAPTER IX.

ABERRATION.

104. The apparent position of the stars is affected by two sources of optical deception, so that they are not in the direction in which they appear to be.

The first of these sources is the motion of the earth, and the corresponding correction is called aberration.

Aberration, like the earth's motion, is either annual or diurnal.

105. Problem. To find the aberration of a star.

Solution. The apparent direction of a star is obviously that of the telescope, through which the star is seen. Let S (fig. 47) be the star, and O the place of the observer at the instant of the observation; SO is the true direction of the star, or the path of the particle of light which proceeded from the star to the observer, and it would be the direction of the telescope if he were stationary. But if he is moving in the direction OP, the direction of the telescope OT must be such, that the end T was at the point R, in the line OS, at the same instant in which the particle of light was at this point. The length RT is, therefore, the distance gone by the observer while the light is describing the line OR.

$I=— ROT the aberration from the true place,

=

106. Problem. To find the annual aberration in latitude and longitude.

Solution. The earth is moving in the plane of the ecliptic at nearly right angles to the direction of the sun. Hence if TP (fig. 48) is the ecliptic, T the point towards which the earth is moving, S the true star, S the apparent star,

O the sun's longitude,

A

= the star's longitude. § 4 — the aberration in long. L= the star's latitude, Lthe aberration in lat. we have

STI, SPL,

=

long. of T-90°, PT 90°-44

0

PPTP-TP, L- S'P'- SP

cos. T=cotan. I tang. 4, cotan. (I+I) tang. (4,— 8 4),

whence

1

and

sin. L sin. (II) sin. I sin. (L+ L), (655)

=

=—m sin. (O — 1) sin. L. (656)

107. Problem. To find the annual aberration in distance and direction from the vernal equinox.

Aberration from vernal equinox.

Solution. Let A (fig. 48) be the vernal equinox, and let

Again; the triangles ASS" and ATS' give by (302),

108. Problem. To find the annual aberration in right ascension and declination.

Solution. If AT (fig. 48) were the equator, we should

have

D =

= SP, R =

= AP,

?

Aberration in right ascension and declination.

and if we put

N1 = SAP, ∞ = obliquity of ecliptic,

we have

N1= N + ",

and the triangles ASP, AS'P' give

sin. D = sin. M sin. N1

(663)

sin. (DD) = sin. (M+8 M) sin. (N,+8 N) (664)

1

1

1

cos. D3 D = sin. Mcos. N1 & N+cos. M sin. N1 & M (665) = B sin. M cos. M sin. N1

C (sin. N cos. N1 cos.2 M sin. N1 cos. N),

1

(666)

(667)

a'-(sin.Ncos. N1-cos.2 M sin N, cos.N) sec. D sec., (668)

= sin. D sin. R cotan. D sin. R cos. D

=

b' sin D cos. R cos. D sec. D = sin. D cos. R (672)

a'=—[sin. (N—N1) + sin.2 M sin. N1cos. N] sec. D sec. ↔ =[sin. -sin.2 Msin.2 N, sin. ] sec. D sec. w

-sin.2 M sin. N1 cos. N, cos. w sec. D sec. w