Astronomy Principles and Practice Fourth Edition.pdf

Now
But θ is a small angle so we may write
Hence,
Then
sin TOC = cos θ =
cos θ = 1 − θ 2
2 .
Correction for the observer’s altitude 119
R ⊕
r ⊕ + h .
1 − θ 2
2 = R ⊕
R ⊕ + h .
θ 2
2 = R ⊕ + h
R ⊕ + h −
R ⊕
R ⊕ + h =
h
R ⊕ + h
so that
√
θ =
2h
R ⊕ + h . (10.14)
Now h is much less than R ⊕ so we may write
θ =
√
2h
R ⊕
rad.
If θ is now expressed in minutes of arc, and we may take 1 radian to be 3438 minutes of arc,
√
2h
θ = 3438 minutes of arc.
R ⊕
Taking the unit distance to be the metre, we find that R ⊕ = 6·372 × 10 6 m, so that
θ = 1·93 √ h minutes of arc
h being given in metres.
When refraction is taken into account, the path of the ray from the horizon at T ′ is curved as
shown and, therefore, appears to come from a direction OD, so that the distance to the horizon is
greater and the angle of dip is less.
Theangleofdipθ ′ is then found to be given by the expression
θ ′ = 1·78 √ h minutes of arc
where h is in metres.
It is of interest to consider at this stage a quantity related to the angle of dip, namely the distance
to the apparent horizon.
In figure 10.5, this is the distance OT, neglecting refraction. Now
or
since θ is small.
cos TOC = OT
OC
OT = (R ⊕ + h) sin θ = (R ⊕ + h) × θ