The forces on the mass will be the gravitational force, mgo, and the reaction force, R, which is needed te keep the mass at rest relative to the Earth's surface (if the mass m is placed on a scale, R would be the force that the scale exerts on the mass). Thus, F=R+mgo. Since the mass m is assumed to be at rest =0,and,O≡B, we have, R+mgo-mnX(SXTA/B)=0, or, R=-m[ X(SX TA/Bl Thus, an observer at rest on the surface of the Earth will observe a gravitational acceleration given by g=go-QX( XTA/B). The term -nx(QXTA/B) has a magnitude S-d=S-Re cos L, and is directe normal and away from the axis of rotation An alternative choice of reference frames which is sometimes more convenient when working with the Earth s a rotating reference frame is illustrated in the figure below a y The fixed ayz axes are the same as before, but now the rotating observer B is situated on the surface of the Earth. A convenient set of rotating axes is that given by Nort-West-South directions ry'z'. If we assume that the mass m is located at B, then we have a= B, and the above expression(2)reduces to R+mgo -maB=0 mIgo -aB=-mg It is straightforward to verify that aB=Sx(Sx rB), which gives the same expression for R, as expected If we call g the gravity acceleration vector, which combines the fact that the Earth is not spherical and that it is rotating. the g is gl g≈9780327(1+0.005279sin2L+0.00046in4L) where L is the latitude of the point considered and g is given in m/s. The coefficient 0.005279 has two components: 0.00344, due to Earth's rotation, and the rest is due to Earth,s oblateness(or lack of sphericity)The forces on the mass will be the gravitational force, mg0, and the reaction force, R, which is needed to keep the mass at rest relative to the Earth’s surface (if the mass m is placed on a scale, R would be the force that the scale exerts on the mass). Thus, F = R + mg0. Since the mass m is assumed to be at rest, Ω˙ = 0, and, O ≡ B, we have, R + mg0 − m Ω × (Ω × rA/B) = 0 , or, R = −m[g0 − Ω × (Ω × rA/B] = −mg Thus, an observer at rest on the surface of the Earth will observe a gravitational acceleration given by g = g0 − Ω × (Ω × rA/B). The term −Ω × (Ω × rA/B) has a magnitude Ω2d = Ω2Re cosL, and is directed normal and away from the axis of rotation. An alternative choice of reference frames which is sometimes more convenient when working with the Earth as a rotating reference frame is illustrated in the figure below. The fixed xyz axes are the same as before, but now the rotating observer B is situated on the surface of the Earth. A convenient set of rotating axes is that given by Nort-West-South directions x ′y ′ z ′ . If we assume that the mass m is located at B, then we have A ≡ B, and the above expression (2) reduces to R + mg0 − m aB = 0 , or, R = −m[g0 − aB] = −mg . It is straightforward to verify that aB = Ω × (Ω × rB), which gives the same expression for R, as expected. If we call g the gravity acceleration vector, which combines the fact that the Earth is not spherical and that it is rotating, the magnitude of g is given by g ≈ 9.780327(1 + 0.005279 sin2 L + 0.000024 sin4 L), where L is the latitude of the point considered and g is given in m/s2 . The coefficient 0.005279 has two components: 0.00344, due to Earth’s rotation, and the rest is due to Earth’s oblateness (or lack of sphericity). 4