WEIGHT

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 RE,
the weight of a small body at the earth’s surface (a distance RE from its center) is

WEIGHT

 

 

 

 

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

 

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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 mE, so this relationship allows us to compute the mass of the earth. Solving Eq. (13.4) for mE and using RE = 6370 km = 6.37 * 106 m and g = 9.80 m/s 2 , we find

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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 - RE above the surface), the weight of a body is given by Eq. (13.3) with RE replaced by r:

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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.

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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

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

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(Compare to the density of water, 1000 kg/m3 = 1.00 g/cm3.) 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/m3 for sedimentary rocks to about 3300 kg/m3 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/m3 . According to geophysical models of the earth’s interior, the maximum density at the center is about 13,000 kg/m3.
Figure 13.9 is a graph of density as a function of distance from the center.

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CALCULATION OF GRAVITY ON MARS:

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Frequently Asked Questions

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Ans: gravitational forces are negligible between ordinary household-sized objects but very substantial between objects that are the size of stars. Indeed, gravitation is the most important force on the scale of planets, stars, and galaxies view more..
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Ans: HERE ARE SOME SOLVED EXAMPLES TO CLEAR YOUR CONCEPTS view more..
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Ans: To determine the value of the gravitational constant G, we have to measure the gravitational force between two bodies of known masses m1 and m2 at a known distance r. The force is extremely small for bodies that are small enough to be brought into the laboratory, but it can be measured with an instrument called a torsion balance, which Sir Henry Cavendish used in 1798 to determine G. view more..
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Ans: 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 view more..
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Ans: When we first introduced gravitational potential energy in Section 7.1, we assumed that the earth’s gravitational force on a body of mass m doesn’t depend on the body’s height. This led to the expression U = mgy view more..
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Ans: As a final note, let’s show that when we are close to the earth’s surface, Eq. (13.9) reduces to the familiar U = mgy view more..
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Ans: Artificial satellites orbiting the earth are a familiar part of technology But how do they stay in orbit, and what determines the properties of their orbits? We can use Newton’s laws and the law of gravitation to provide the answers. In the next section we’ll analyze the motion of planets in the same way. view more..
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Ans: A circular orbit, like trajectory 4 in Fig. 13.14, is the simplest case. It is also an important case, since many artificial satellites have nearly circular orbits and the orbits of the planets around the sun are also fairly circular view more..
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Ans: The name planet comes from a Greek word meaning “wanderer,” and indeed the planets continuously change their positions in the sky relative to the background of stars. One of the great intellectual accomplishments of the 16th and 17th centuries was the threefold realization that the earth is also a planet, that all planets orbit the sun, and that the apparent motions of the planets as seen from the earth can be used to determine their orbits precisely view more..
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Ans: First consider the elliptical orbits described in Kepler’s first law. Figure 13.18 shows the geometry of an ellipse. The longest dimension is the major axis, with half-length a; this half-length is called the semi-major axis. view more..
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Ans: In a small time interval dt, the line from the sun S to the planet P turns through an angle du. The area swept out is the colored triangle with height r, base length r du, and area dA = 1 2 r2 du in . The rate at which area is swept out, view more..
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Ans: We have already derived Kepler’s third law for the particular case of circular orbits. Equation (13.12) shows that the period of a satellite or planet in a circular orbit is proportional to the 3 2 power of the orbit radius. view more..
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Ans: We have assumed that as a planet or comet orbits the sun, the sun remains absolutely stationary. This can’t be correct; because the sun exerts a gravitational force on the planet, the planet exerts a gravitational force on the sun of the same magnitude but opposite direction. In fact, both the sun and the planet orbit around their common center of mass view more..
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Ans: We have stated without proof that the gravitational interaction between two spherically symmetric mass distributions is the same as though all the mass of each were concentrated at its center. Now we’re ready to prove this statement. Newton searched for a proof for several years, and he delayed publication of the law of gravitation until he found one view more..
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Ans: We start by considering a ring on the surface of a shell , centered on the line from the center of the shell to m. We do this because all of the particles that make up the ring are the same distance s from the point mass m. view more..
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Ans: Any spherically symmetric mass distribution can be thought of as a combination of concentric spherical shells. Because of the principle of superposition of forces, what is true of one shell is also true of the combination. So we have proved half of what we set out to prove: that the gravitational interaction between any spherically symmetric mass distribution and a point mass is the same as though all the mass of the spherically symmetric distribution were concentrated at its center. view more..
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Ans: We assumed at the beginning that the point mass m was outside the spherical shell, so our proof is valid only when m is outside a spherically symmetric mass distribution. view more..
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Ans: Because the earth rotates on its axis, it is not precisely an inertial frame of reference. For this reason the apparent weight of a body on earth is not precisely equal to the earth’s gravitational attraction, which we will call the true weight w 0 of the body. Figure 13.26 is a cutaway view of the earth, showing three observers. Each one holds a spring scale with a body of mass m hanging from it. view more..
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