WJEC Physics for A2: Student Bk

18 We’ll consider the forces on a an object of mass m on the surface of the (spherical) Earth at a latitude, λ , as in Fig. 3.1.20. direction of resultant force mg C R E m Equator Axis of rotation Fig. 3.1.20 Forces on an object at latitude λ . Notice that the ‘normal’ contact force is not directly away from the centre of the Earth. It cannot be because the resultant of C and mg is not zero but points towards the axis of rotation – it is centripetal. We’ll leave the detailed analysis to a Stretch & Challenge question at the end of Exercise 3.1. However we’ll anticipate a result from it. For λ = π 4 , the value of C is almost identical to mg but the angle of C is 0.0017 rad (about 0.1 ° ) away from the ‘vertical’. The effect of this is that the horizontal line at the position of the mass is not at right angles to mg but at right angles to C – with the effect that the Earth bulges outwards towards the equator and is flattened at the poles. Note that this undermines (by a small amount) our initial assumption of a spherical Earth! From Fig. 3.1.21 we see that the magnitude of C (i.e. the object’s apparent weight) is less than mg : this effect gets smaller towards the pole. There is a secondary effect, which we haven’t taken account of: because the Earth now bulges at the equator, actual value of mg is less because the surface is further away from the centre. mg C F 3.1.21 Resultant force 3.1.12 Self-test Given that the rotational period of the Earth is 24 hours and the mean radius is 6731 km. Show that the centripetal acceleration of a point on the equator is approximately 0.03 m s −2 (i.e. about 0.3% of g ). WJEC Physics for A2 Level: Unit 3