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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 F 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 s = m g . This would be the true reading of the weight.

An illustration of the earth, rotating on its north-south axis, with masses on spring scales shown at three locations. The radius of the earth is labeled as R E, its center is labeled as O. One spring scale is above the north pole. An upward force F S N and a downward force m g are shown acting on the mass on this spring scale. A dashed line is shown from the center of the earth to the north pole. Another spring scale is shown to the right of the equator and a dashed line connects the center of the earth to the equator on the right side of the earth. The forces on the mass on this second spring scale are shown as a force F S E to the right and m g to the left. A third spring scale is shown at an angle lambda to the horizontal. A dashed line at this angle is shown from the center to the surface of the earth. The horizontal distance from the surface of the earth at this angle lambda to the vertical dashed line connecting the center to the north pole is labeled as r. The point on the dashed vertical line where r meets it is labeled P. Three forces are shown for the third mass. One force is labeled F S and points radially outward. A second force, labeled m g points radially inward. A third force, labeled F c, points horizontally to the left.
For a person standing at the equator, the centripetal acceleration ( a c ) is in the same direction as the force of gravity. At latitude λ , the angle the between a c and the force of gravity is λ and the magnitude of a c decreases with cos λ .

With rotation, the sum of these forces must provide the centripetal acceleration, a c . Using Newton’s second law, we have

F = F s m g = m a c where a c = v 2 r .

Note that a 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 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 c from Motion in Two and Three Dimensions . Recall that the tangential speed is related to the angular speed ( ω ) by v = r ω . Hence, we have a c = r ω 2 . By rearranging [link] and substituting r = R E , the apparent weight at the equator is

F s = m ( g R E ω 2 ) .

The angular speed of Earth everywhere is

ω = 2 π rad 24 hr × 3600 s/hr = 7.27 × 10 −5 rad/s.

Substituting for the values or R E and ω , we have R E ω 2 = 0.0337 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 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 F = F s m g = m a c , so setting F s = 0 , we get g = a c . Using the expression for a c , substituting for Earth’s radius and the standard value of gravity, we get

a c = v 2 r = g v = g r = ( 9.80 m/s 2 ) ( 6.37 × 10 6 m ) = 7.91 × 10 3 m/s .

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

v = 2 π r T T = 2 π r v = 2 π ( 6.37 × 10 6 m ) 7.91 × 10 3 m/s = 5.06 × 10 3 s .

This is about 84 minutes.

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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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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