How the apparent weight varies due to the rotational motion of Earth?

Question

I learned that as the earth rotates about its axis, the bodies on the earth also follow a circular path. In most books I read, they give the example of a person standing on a weight balance at the equator... and I did understand that. However, by doing the following calculation, I am seeing that the apparent weight at other points on the earth (apart from the poles) is the same

This is the picture on my mind:

At B,

$$W-N=m{\omega}^2 R$$

At A, a component of weight will provides the centripetal force to rotate around the circle with the radius $r$, $$W\cos{\theta} - N\cos\theta =m{\omega}^2r $$ as $r=R\cos\theta$, $$W\cos{\theta} - N\cos\theta =m{\omega}^2R\cos\theta$$

The equation eventually ends up as... $$W - N =m{\omega}^2R $$

So from this, I think that the normal reaction force which is the apparent weight remains the same as to the apparent weight at the equator.

However the book states that the apparent weight varies along $A$ and $B$.

Also, we assume for simplicity that the Earth is spherical. (I know this is not a realistic assumption, cf. e.g. this, this and this Phys.SE posts, but that's another story.)

I am really sorry for making the question so long. Could someone please tell me which part of my concept is wrong?

Answer 1

If we look at the $x$- and $y$-components of the forces, we see that $$ \begin{align} W_x - N_x &= m\omega^2 R\cos\theta,\\ W_y - N_y &= 0. \end{align} $$ The normal force $\vec{N}$ is a radial force, so if we assume that the Earth is spherically symmetric, we indeed have $N_x = N\cos\theta$ and $N_y = N\sin\theta$, so that $N=N_R$. But the weight force $\vec{W}$ is not purely radial. It also has a tangential component, precisely because of the rotation of the Earth: $$ \begin{align} W_R &= W_x\cos\theta + W_y\sin\theta,\\ W_\theta &= -W_x\sin\theta + W_y\cos\theta. \end{align} $$ So in general $W_x\neq W\cos\theta$. For the normal force, we have $$ \begin{align} N_R &= N_x\cos\theta + N_y\sin\theta = N\cos^2\theta + N\sin^2\theta = N,\\ N_\theta &= -N_x\sin\theta + N_y\cos\theta = 0. \end{align} $$ From $$ \begin{align} W_x\cos\theta - N_x\cos\theta &= m\omega^2 R\cos^2\theta,\\ W_y\sin\theta - N_y\sin\theta &= 0, \end{align} $$ we find the centripetal force $$ W_R - N = m\omega^2 R\cos^2\theta, $$ but note that there's also a tangential force $$ W_\theta = -m\omega^2 R\sin\theta\cos\theta. $$

Answer 2

Shape of earth is not exactly spherical. It is somewhat like this (oblate spheroid- closest approximation than sphere):

Spherical shape is the close approximation for the true figure of the earth and satisfactory for many purposes (except your case and few other). little closest approximation for true figure of earth is oblate spheroid (3-D image formed by the rotation of ellipse about its minor axis).

So, now you would have understood that radius of earth is not constant, distance of any point on the poles from the center is less than the distance of any point on the pole to the equator. Thus, force felt by the object at the pole is greater than the force felt by the same object at the equator. As a consequence, weight (i.e nothing but force) of the same object is more at pole than at the equator. You can also prove this by your calculations now.

Answer 3

Your acceleration anywhere on Earth would be $g_{net} = -g\hat{n} + \omega^2r\hat{r}$

Where $\hat{n}$ is the unit vector perpendicular on the surface (outward). This is also why rocket launch sites are usually located close to the equator :-)

Answer 4

If the object is not moving Earth drags the object with it and they have the same angular velocity related to constant radius $r$ that varies as a vector because it must depend on angle of its spinning plane and the apparent weight $N=mg-mr\omega^2\cos^2(\theta)$. By the other hand, if the object also moves things are complete different because the magnitude of $R$ is constant but its vector is not anymore and we have to consider the angular velocities due to $r$ and $R$ so $\omega = \omega_1 + \omega_2$ where $\omega_1 = v_1/r\cos(\theta))$ in which $\omega_1$ is due to $R$ radius and $\omega_2$ due to $r$ radius as $\omega$ is the resulting angular velocity that now we have $N= mg - mv^2/r = mg - mr(v^2/r^2\cos^2(\theta) + 2v\omega/rcos(\theta) + \omega^2)$. Note: the 1 and 2 were omitted for simplicity.

$$N = mg - mr(v^2/r^2\cos^2(\theta) + 2v\omega/r\cos(\theta) + \omega^2)$$

Answer 5

The thing you are doing wrong is that there is a centrifugal force bcs we humans are on rotating earth. And a body on rotating platform experience a centrifugal force. And the component of centrifugal force along normal = mrw²cos(THETA) and equate the forces in equilibrium to get the correct relationship.

Answer 6

It is the normal reaction that has tangential component.Gravitational force is always directed towards centre of earth. In inertial I e nonrot frame centrifugal force does not exist.Balace these forces in plane of rotation to give centripetal force. Use another eqn to balance forces in drn perpendicular to rot plane. Solve. Some answers have mixed another phenomena of oblate spheroid which has nothing to do with this problem. In fact observed variation in g on pole and that on equator is a sum of oblate shape and rotation worked out separately.