通常没有旋转,重力自然租赁将地球在一个球体的形状。然而事实上地球在赤道凸起,在赤道平面直径是42.72公里超过直径从南极到北极。这是由于地球的旋转。![在这里输入图像描述][1][1]:http://i.stack.imgur.com/707k9.gif在上面的图片中,我们可以看出旋转磁盘似乎凸出在磁盘上的点最远的旋转轴。这是因为为了让磁盘的颗粒保持在轨道上,一定有一个内在的力量,被称为向心力,由:$ F = \压裂{mv ^ 2} {r} $, F是力量,质量m是旋转的身体,v是速度和半径r是旋转轴的粒子。如果磁盘旋转在给定的角速度,说w,切向速度v, v = wr给出。因此,F = m \ω^ 2 r美元因此粒子的半径越大,需要更多的力量来维护这样一个轨道。因此粒子在地球赤道附近,这是最远的轴旋转,将buldge外,因为他们需要一个更大的内在力量来维持他们的轨道。- - -更多细节更多数学文化现在启用mathjax:合力在赤道周围物体旋转半径r美元左右一颗行星的引力美元\压裂{Gm_1m_2} {r ^ 2}的向心力,美元$ f{净}= \压裂{Gm_1m_2} {r ^ 2} - N = m \ω^ 2 r美元,其中N是支持力。重新排列上面的方程为:$ N = \压裂{Gm_1m_2} {r ^ 2} - m \ω^ 2 r美元这里的法向力是向下的力,旋转的身体观察员。 The equation shows that the perceived downward force is lessened due to the centripetal motion. The typical example to illustrate this is there is an appearance of 0 gravity in a satellite orbiting the Earth, because in this situation the centripetal force is exactly balanced by the gravitational force. On Earth however, the centripetal force is much less than the gravitational force, so we perceive almost the whole contribution of mg. Now we will examine how the perceived gravitational force differs at different angles of latitude. Let $\theta$ represent the angle of latitude. Let $F_G$ be the force of gravity. In vector notation we will take the j-direction to be parallel with the axis of rotation and the i-direction to be perpendicular with the axis of rotation. In the absence of the Earth's rotation, $F_G = N = (-\frac{Gm_1m_2}{r^2}cos\theta)\tilde{i} + (-\frac{Gm_1m_2}{r^2}sin\theta)\tilde{j}$ It is easily seen that the above equation represents the perceived force of gravity in the absence of rotation. Now the centripetal force acts only in the i-direction, since it acts perpendicular to the axis of rotation. If we let $R_{rot}$ be the radius of rotation, then the centripetal force is $m_1\omega^2R_{rot}$, which for an angle of latitude of $\theta$ corresponds to $m_1\omega^2r\cos{\theta}$ $N = (-\frac{Gm_1m_2}{r^2} + m_1\omega^2r)\cos{\theta}\tilde{i} + (-\frac{Gm_1m_2}{r^2})\sin{\theta}\tilde{j}$ By comparing this equation to the case shown earlier in the absence of rotation, it is apparent that as $\theta$ is increased (angle of latitude), the effect of rotation on perceived gravity becomes negligible, since the only difference lies in the x-component and $\cos\theta$ approaches 0 as $\theta$ approaches 90 degrees latitude. However it can also be seen that as theta approaches 0, near the equator, the x-component of gravity is reduced as a result of the Earth's rotation. **Therefore, we can see that the magnitude of $N$ is slightly less at the equator than at the poles.** The reduced apparent gravitational pull here is what gives rise to the slight bulging of the Earth at the equator, given that the Earth was not originally as rigid as it is today (see other answer).