Gravity field of the earth - Physical Geodesy - page 114

In a north-south direction, we have

ε = ξ

and ds = ds ϕ = Rdϕ ;

(2-374)

in an east-west direction,

ε = η and ds = ds λ = R cos ϕdλ. (2-375)

In the formulas for ds ϕ and ds λ , we have again used the spherical approxi-
mation; according to (1-30), the element of arc on the sphere r = R is given
by

ds 2 = R 2 dϕ 2 + R 2 cos 2 ϕdλ 2 . (2-376)

By specializing (2-373), we find

dN
ds ϕ

1
R

∂N
∂ϕ ,

ξ = − = −

(2-377)

dN
ds λ

1
R cos ϕ

∂N
∂λ ,

η = − = −

which gives the connection between the geoidal undulation N and the com-
ponents ξ and η of the deflection of the vertical.
As N is given by Stokes' integral, our problem is to differentiate this
formula with respect to ϕ and λ . For this purpose, we use the form (2-317),

2 π π/ 2

R
4 πγ 0

∆ g ( ϕ ,λ ) S ( ψ )cos ϕ dϕ dλ ,

N ( ϕ, λ )=

(2-378)

λ =0 ϕ = −π/ 2

where ψ is defined in (2-318) as a function of ϕ, λ and ϕ ,λ .
The integral on the right-hand side of this formula depends on ϕ and
λ only through ψ in S ( ψ ). Therefore, by differentiating under the integral
sign,

2 π π/ 2

∂N
∂ϕ

R
4 πγ 0

∆ g ( ϕ ,λ ) ∂S ( ψ )
∂ϕ

cos ϕ dϕ dλ = (2-379)

λ =0 ϕ = −π/ 2

is obtained and a similar formula for ∂N/∂λ .Herewehave

∂S ( ψ )
∂ϕ

= dS ( ψ )
dψ

∂ψ
∂ϕ ,

∂S ( ψ )
∂λ

= dS ( ψ )
dψ

∂ψ
∂λ .

(2-380)

Differentiating (2-318) with respect to ϕ and λ ,weobtain

sin ψ ∂ψ

∂ϕ =cos ϕ sin ϕ −

sin ϕ cos ϕ cos( λ −

− λ ) ,

(2-381)

− sin ψ ∂ψ

∂λ =cos ϕ cos ϕ sin( λ − λ ) .