Why Gravitational forces are important




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 (Fig. 13.6). It is responsible for holding our earth together and for keeping the planets in orbit about the sun. The mutual gravitational attraction between different parts of the sun compresses material at the sun’s core to very high densities and temperatures, making it possible for nuclear reactions to take place there. These reactions generate the sun’s energy output, which makes it possible for life to exist on earth and for you to read these words.

 

 

 

Topics You May Be Interested In
Fluid Mechanics Kepler's Laws (firsts, Second, Third Laws) And The Motion Of Planets
Surface Tension A Point Mass Outside A Spherical Shell
The Continuity Equation The Escape Speed From A Star
Solved Problems A Visit To A Black Hole
Weight Detecting Black Holes

The gravitational force is so important on the cosmic scale because it acts at a distance, without any direct contact between bodies. Electric and magnetic forces have this same remarkable property, but they are less important on astronomical scales because large accumulations of matter are electrically neutral; that is, they contain equal amounts of positive and negative charge. As a result, the electric and magnetic forces between stars or planets are very small or zero. The strong and weak interactions that we discussed in Section 5.5 also act at a distance, but their influence is negligible at distances much greater than the diameter of an atomic nucleus (about 10-14 m)Why Gravitational forces are important

 

 

 

Topics You May Be Interested In
Solving Physics Problems Why Gravitational Forces Are Important
Using And Converting Units More On Gravitational Potential Energy
Solved Examples On Equilibrium The Motion Of Satellites
Pressure In A Fluid Kepler's Laws (firsts, Second, Third Laws) And The Motion Of Planets
Buoyancy Kepler's Second Law

 

 

 

 

Topics You May Be Interested In
Estimates And Order Of Magnitudes Why Gravitational Forces Are Important
Summary Of Equilibrium And Elasticity Gravitational Potential Energy
The Continuity Equation Apparent Weight And The Earth’s Rotation
Deriving Bernoullis Equation Describing Oscillation
Examples On Gravition Amplitude, Period, Frequency, And Angular Frequency

 

A useful way to describe forces that act at a distance is in terms of a field. One body sets up a disturbance or field at all points in space, and the force that acts on a second body at a particular point is its response to the first body’s field at that point. There is a field associated with each force that acts at a distance, and so we refer to gravitational fields, electric fields, magnetic fields, and so on. We won’t need the field concept for our study of gravitation in this chapter, so we won’t discuss it further here. But in later chapters we’ll find that the field concept is an extraordinarily powerful tool for describing electric and magnetic interactions.



Frequently Asked Questions

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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 have stated the law of gravitation in terms of the interaction between two particles. It turns out that the gravitational interaction of any two bodies having spherically symmetric mass distributions 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: 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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