Space methods - Physical Geodesy - page 281

comprises the components of the angular velocity and is used to describe the
orientation of the gradiometer. Since Ω is skewsymmetric, the tensor ΩΩ
is symmetric. Finally, ∆ x in (7-49) is the vector from the intersection of
the three coordinate axes to the respective accelerometer (where the same
length is assumed), and f ng comprises all nongravitational effects (air drag,
solar radiation pressure, etc.).
Now we once add (“common mode”) and once subtract (“differential
mode”) the two accelerations in (7-49) and obtain

( a 1 + a 2 ) / 2= f ng ,

(7-52)

a 2 ) / 2= M +

˙ Ω + ΩΩ ∆ x ,

( a 1 −

where we can extract the nongravitational effects f ng in the common mode.
Introducing the quantity

˙ Ω + ΩΩ

Γ = M +

(7-53)

and assuming a known geometry of the gradiometer, i.e., ∆ x may safely
assumed to be known, then the remaining task is to extract the gravity
gradient tensor M from Γ . This can be achieved by the two relations

Γ T ) / 2= ˙ Ω ,
( Γ + Γ T ) / 2= M + ΩΩ ,

( Γ

−

(7-54)

where the superscript T denotes transposition. To verify these relations, a
little matrix calculus is needed. If, generally, K is a symmetric matrix, then
we have K = K T .If K is a skewsymmetric matrix, then we have K = K T .

Referring now to (7-53), we know that M is symmetric, ˙ Ω is skewsym-
metric, and ΩΩ is symmetric. Therefore, transposing (7-53) yields

−

˙ Ω + ΩΩ .

Γ T = M − (7-55)

Using now (7-53) and (7-55), we get immediately

Γ T =2 ˙ Ω Γ − (7-56)

and, finally,

Γ T ) / 2= ˙ Ω , ( Γ − (7-57)

which completes our proof for the first relation of (7-54). To prove the second
relation of (7-54), we add Eqs. (7-53) and (7-55):

Γ + Γ T =2 M +2 ΩΩ (7-58)