# 13.2 Gravitation near earth's surface  (Page 3/6)

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## Apparent weight: accounting for earth’s rotation

As we saw in Applications of Newton’s Laws , objects moving at constant speed in a circle have a centripetal acceleration directed toward the center of the circle, which means that there must be a net force directed toward the center of that circle. Since all objects on the surface of Earth move through a circle every 24 hours, there must be a net centripetal force on each object directed toward the center of that circle.

Let’s first consider an object of mass m located at the equator, suspended from a scale ( [link] ). The scale exerts an upward force ${\stackrel{\to }{F}}_{\text{s}}$ away from Earth’s center. This is the reading on the scale, and hence it is the apparent weight    of the object. The weight ( mg ) points toward Earth’s center. If Earth were not rotating, the acceleration would be zero and, consequently, the net force would be zero, resulting in ${F}_{\text{s}}=mg$ . This would be the true reading of the weight.

With rotation, the sum of these forces must provide the centripetal acceleration, ${a}_{\text{c}}$ . Using Newton’s second law, we have

$\sum F={F}_{\text{s}}-mg=m{a}_{\text{c}}\phantom{\rule{1em}{0ex}}\text{where}\phantom{\rule{1em}{0ex}}{a}_{\text{c}}=-\frac{{v}^{2}}{r}.$

Note that ${a}_{\text{c}}$ points in the same direction as the weight; hence, it is negative. The tangential speed v is the speed at the equator and r is ${R}_{\text{E}}$ . We can calculate the speed simply by noting that objects on the equator travel the circumference of Earth in 24 hours. Instead, let’s use the alternative expression for ${a}_{\text{c}}$ from Motion in Two and Three Dimensions . Recall that the tangential speed is related to the angular speed $\left(\omega \right)$ by $v=r\omega$ . Hence, we have ${a}_{c}=\text{−}r{\omega }^{2}$ . By rearranging [link] and substituting $r={R}_{\text{E}}$ , the apparent weight at the equator is

${F}_{\text{s}}=m\left(g-{R}_{\text{E}}{\omega }^{2}\right).$

The angular speed of Earth everywhere is

$\omega =\frac{2\pi \phantom{\rule{0.2em}{0ex}}\text{rad}}{24\phantom{\rule{0.2em}{0ex}}\text{hr}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3600\phantom{\rule{0.2em}{0ex}}\text{s/hr}}=7.27\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}\text{rad/s.}$

Substituting for the values or ${R}_{\text{E}}$ and $\omega$ , we have ${R}_{\text{E}}{\omega }^{2}=0.0337\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ . This is only 0.34% of the value of gravity, so it is clearly a small correction.

## Zero apparent weight

How fast would Earth need to spin for those at the equator to have zero apparent weight? How long would the length of the day be?

## Strategy

Using [link] , we can set the apparent weight ( ${F}_{\text{s}}$ ) to zero and determine the centripetal acceleration required. From that, we can find the speed at the equator. The length of day is the time required for one complete rotation.

## Solution

From [link] , we have $\sum F={F}_{\text{s}}-mg=m{a}_{\text{c}}$ , so setting ${F}_{\text{s}}=0$ , we get $g={a}_{\text{c}}$ . Using the expression for ${a}_{\text{c}}$ , substituting for Earth’s radius and the standard value of gravity, we get

$\begin{array}{}\\ \\ {a}_{\text{c}}=\frac{{v}^{2}}{r}=g\hfill \\ v=\sqrt{gr}=\sqrt{\left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)\left(6.37\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{m}\right)}=7.91\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{m/s}.\hfill \end{array}$

The period T is the time for one complete rotation. Therefore, the tangential speed is the circumference divided by T , so we have

$\begin{array}{}\\ v=\frac{2\pi r}{T}\hfill \\ T=\frac{2\pi r}{v}=\frac{2\pi \left(6.37\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{m}\right)}{7.91\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{m/s}}=5.06\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{s}.\hfill \end{array}$

## Significance

We will see later in this chapter that this speed and length of day would also be the orbital speed and period of a satellite in orbit at Earth’s surface. While such an orbit would not be possible near Earth’s surface due to air resistance, it certainly is possible only a few hundred miles above Earth.

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