0.1 Gravity  (Page 3/6)

Shape of the earth

Earth is not a sphere. It is an ellipsoid. Its equatorial radius is greater than polar radius by 21 km. A point at pole is closer to the center of Earth. Consequently, gravitational acceleration is greater there than at the equator.

Besides, some part of Earth is protruded and some part is depressed below average level. Once again, factors of mass and distance come into picture. Again, it is the relative impact of two factors that determine the net effect. Consider a point right at the top of Mt. Everest, which is about 8.8 km from the mean sea level. Imagine incrementing radius of Earth’s sphere by 8.8 km. Most of the volume so created is not filled. The proportionate increase in mass (mass of Everest mountain range) is less than that in the squared distance from the center of Earth. As such, gravitational acceleration is less than its average value on the surface. It is actually 9.80 $m/s^2$ as against the average of 9.81 $m/s^2$ , which is considered to be the accepted value for the Earth’s surface.

Rotation of earth

Earth rotates once about its axis of rotation in 1 day and moves around Sun in 365 days. Since Earth and a particle on Earth both move together with a constant speed around Sun, there is no effect in the measured acceleration due to gravity on the account of Earth’s translational motion. The curved path around Sun can be approximated to be linear for distances under consideration. Hence, Earth can serve as inertial frame of reference for the application of Newton’s law of motion, irrespective of its translational motion.

However, consideration of rotation of Earth about its axis changes the nature of Earth’s reference. It is no more an inertial frame. A particle at a point, “P”, is rotating about the axis of rotation. Clearly, a provision for the centripetal force should exist to meet the requirement of circular motion. We should emphasize here that centripetal force is not an additional force by itself, but is a requirement of circular motion, which should be met with by the forces operating on the particle. Here, gravitational force meets this requirement and, therefore, gets modified to that extent.

Here, we shall restrict our consideration specifically to the effect of rotation. We will ignore other factors that affect gravitational acceleration. This means that we consider Earth is a solid uniform sphere. If it is so then, measured value of acceleration is equal to reference gravitational acceleration ( $g_0$ ) as modified by rotation.

As we have studied earlier, we can apply Newton’s law in a non-inertial reference by providing for pseudo force. We should recall that pseudo force is applied in the direction opposite to the direction of acceleration of the frame of reference, which is centripetal acceleration in this case. The magnitude of pseudo force is equal to the product of mass of the particle and centripetal acceleration. Thus,

$$F_P=m{\omega }^{2}r$$

After considering pseudo force, we can enumerate forces on the particle at “P” at an latitude “φ” as shown in the figure :