# WEIGHT

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We defined the weight of a body in Section 4.4 as the attractive gravitational force exerted on it by the earth. We can now broaden our definition and say that the weight of a body is the total gravitational force exerted on the body by all other bodies in the universe. When the body is near the surface of the earth, we can ignore all other gravitational forces and consider the weight as just the earth’s gravitational attraction. At the surface of the moon we consider a body’s weight to be the gravitational attraction of the moon, and so on.

If we again model the earth as a spherically symmetric body with radius R_{E},

the weight of a small body at the earth’s surface (a distance R_{E} from its center) is

But we also know from Section 4.4 that the weight w of a body is the force that causes the acceleration g of free fall, so by Newton’s second law, w = mg. Equating this with Eq. (13.3) and dividing by m, we find

The acceleration due to gravity g is independent of the mass m of the body because m doesn’t appear in this equation. We already knew that, but we can now see how it follows from the law of gravitation.

We can measure all the quantities in Eq. (13.4) except for m_{E}, so this relationship allows us to compute the mass of the earth. Solving Eq. (13.4) for m_{E} and using R_{E }= 6370 km = 6.37 * 10^{6} m and g = 9.80 m/s ^{2} , we find

This is very close to the currently accepted value of 5.972 * 1024 kg. Once Cavendish had measured G, he computed the mass of the earth in just this way. At a point above the earth’s surface a distance r from the center of the earth (a distance r - R_{E} above the surface), the weight of a body is given by Eq. (13.3) with R_{E} replaced by r:

The weight of a body decreases inversely with the square of its distance from the earth’s center (Fig. 13.7). Figure 13.8 shows how the weight varies with height above the earth for an astronaut who weighs 700 N at the earth’s surface.

The apparent weight of a body on earth differs slightly from the earth’s gravitational force because the earth rotates and is therefore not precisely an inertial frame of reference. We’ve ignored this relatively small effect in our discussion but will consider it carefully in Section 13.7. While the earth is an approximately spherically symmetric distribution of mass, it is not uniform throughout its volume. To demonstrate this, let’s first calculate the average density, or mass per unit volume, of the earth. If we assume a spherical earth, the volume is

The average density p (the Greek letter rho) of the earth is the total mass divided by the total volume:

(Compare to the density of water, 1000 kg/m^{3} = 1.00 g/cm^{3}.) If the earth were uniform, rocks near the earth’s surface would have this same density. In fact, the density of surface rocks is substantially lower, ranging from about 2000 kg/m^{3 }for sedimentary rocks to about 3300 kg/m^{3} for basalt. So the earth cannot be uniform, and its interior must be much more dense than its surface in order that the average density be 5500 kg/m^{3 }. According to geophysical models of the earth’s interior, the maximum density at the center is about 13,000 kg/m^{3}.

Figure 13.9 is a graph of density as a function of distance from the center.

CALCULATION OF GRAVITY ON MARS:

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