More on Gravitational potential energy
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 from Chapter 7. We first rewrite Eq. (13.8) as
If the body stays close to the earth, then in the denominator we may replace r1 and r2 by RE, the earth’s radius, so,
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According to Eq. (13.4), g = GmE/RE 2 , so,
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If we replace the r’s by y’s, this is just Eq. (7.1) for the work done by a constant gravitational force. In Section 7.1 we used this equation to derive Eq. (7.2), U = mgy, so we may consider Eq. (7.2) for gravitational potential energy to be a special case of the more general Eq. (13.9).
Frequently Asked Questions
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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: 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: 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: 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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Ans: In 1916 Albert Einstein presented his general theory of relativity, which included
a new concept of the nature of gravitation. In his theory, a massive object actually
changes the geometry of the space around it view more..
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Ans: Think first about the properties of our own sun. Its mass M = 1.99 * 1030 kg
and radius R = 6.96 * 108
m are much larger than those of any planet, but compared
to other stars, our sun is not exceptionally massive view more..
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