General Physics 1 12 Q2 Module-2- Final PDF

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Senior High School

GENERAL PHYSICS 1 Quarter 2 – Module 2

GRAVITY

What I Know Multiple Choice. Select the letter of the best answer from among the given choices. 1. The gravitational constant has a symbol of __________. A. C C. k B. G D. E 2. Who is the proponent of the Universal Law of Gravitation? A. Newton C. Archimedes B. Einstein D. Pascal 3. The symbol Fg is referred as the magnitude of the ___________. A. Electrical Force C. Gravitational Force B. Rotational Force D. Magnetic Force 4. What is the value of G (in three significant figures)? A. 6.36 x 10-8 N C. 6.67 x 10-11 N m2 / kg -36 B. 9.11 x 10 kg D. 9.8 m / s2 5. In which of the following cases would you feel weightless? A. Traveling through space in an accelerating rocket B. Falling from an airplane with your parachute open C. Walking on the moon D. Taking an exam 6. A useful way to describe forces that act a distance is in terms of __________. A. Newtons C. distance B. Period D. field 7. If your mass is 65 kg on Earth, what would your mass be on the Moon? A. 11 kg B. 11 lb C. 65 kg

D. 65 N

8. How much GPE does a 2-kg block have if it is lifted 12.5 m high? A. 480 J B. 490 J C. 500 J

D. 510 J

9. Which of the following is NOT one of Kepler’s Laws of Planetary Motion? A. The square of a planet’s period is proportional to its distance from the sun cubed. B. The area of a planet’s orbital plane is inversely proportional to its speed. C. A planet sweeps out equal area in an equal time interval. D. Planets move around the sun in elliptical orbits. 10. He was a Danish astronomer who worked with Kepler. His data collected were used by Kepler in his laws. A. Aristotle C. Tycho Brahe B. Archimedes D. None of them

Lesson

1

NEWTON’S LAW OF UNIVERSAL GRAVITATION

Some of the earliest investigations in Physical Science started with questions that people asked about the night sky. Why doesn’t the moon fall to the earth? Why do the planets move across the sky? Why doesn’t the earth fly off into space rather than remaining in orbit around the sun? The study of gravitation provides the answers to these and many related questions. Gravitation is one of the four classes of interactions found in nature, and it was the earliest of the four to be studied extensively. Newton discovered in the 17 th century that the same interaction that makes an apple fall out of a tree also keeps the planets in their orbits around the sun. In this module, you will learn the basic law that governs gravitational interactions. This law is universal: Gravity acts in the same fundamental way between the earth and your body, between the sun and the planet, and between a planet and one of its moons. We’ll apply the law of gravitation to phenomena such as the variation of weight with altitude, the orbits of satellites around the earth, and the orbits of planets around the sun.

What’s In The example of gravitational attraction that’s probably most familiar to you is your weight, the force that attracts you toward the earth. During his study of the motions of the planets and of the moon, Newton discovered the fundamental character of the gravitational attraction between two any bodies. Along with his three laws of motion, Newton published the Law of Gravitation in 1687. It may be stated as follows:

Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. Translating this into an equation, we have

where Fg is the magnitude of the gravitational force on either particle, m1 and m2 are their masses, r is the distance between them, and G is a fundamental physical constant called the gravitational constant. The numerical value of G depends on the system of units used. This equation tells us that the gravitational force between two particles decreases with increasing distance r: If the distance is doubled, the force is only one-fourth as great, and so on. Although many of the stars in the night sky are far more massive than the sun, they are so far away that their gravitational force on the earth is negligibly small.

What’s New Determining the Value of G 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. The figure below shows a modern version of the Cavendish torsion balance. A light, rigid rod shaped like an inverted T is supported by a very thin, vertical quartz fiber. Two small spheres, each of mass m1, are mounted at the ends of the horizontal arms of the T. When we bring two large spheres, each of mass m2, to the positions shown, the attractive gravitational forces twist the T through a small angle. To measure this angle, we shine a beam of light on a mirror fastened to the T. The reflected beam strikes a scale, and as the T twists, the reflected beam moves along the scale. After calibrating the Cavendish balance, we can measure gravitational forces and thus determine G. The presently accepted value is

G = 6.67428 (67) x 10-11 N ∙ m2 / kg2 To three significant figures, G (Because 1 N = 1 kg

= 6.67 x 10-11 N ∙ m2 / kg2.

∙ m/s2, the units of G can also be expressed as m3 / kg ∙ s2.)

Gravitational forces combine vectorially. If each of two masses exerts a force on a third, the total force on the third mass is the vector sum of the individual forces of the first two.

What Is It Calculating Gravitational Force Example: The mass m1 of one of the small spheres of a Cavendish balance is 0.0100 kg, the mass m2 of the nearest large sphere is 0.500 kg, and the center-to-center distance between them is 0.0500 m. Find the gravitational force Fg on each sphere due to the other. Solution: Because the spheres are spherically symmetric, we can calculate Fg by treating them as particles separated by 0.0500 m. Each sphere experiences the same magnitude of force from the other sphere. We use Newton’s law of gravitation to determine Fg:

Acceleration due to Gravitational Attraction Example: Suppose the two spheres in the previous example are placed with their centers 0.0500 m apart at a point in space far removed from all other bodies. What is the magnitude of the acceleration of each, relative to an inertial system? Solution: Each sphere exerts on the other a gravitational force of the same magnitude Fg , which we found in the previous example. We can neglect any other forces. The acceleration magnitudes a1 and a2 are different because the masses are different. To determine these, we’ll use Newton’s second law:

What’s More

PROBLEM SOLVING ACTIVITY 1. The moon has a mass of 7.34  10 22 kg and a radius of 1.74  106 meters. If you have a mass of 66 kg, how strong is the force between you and the moon? 2. A distance of 0.002 m separates two objects of equal mass. If the gravitational force between them is 0.0104 N, find the mass of each object.

Lesson

2

GRAVITATIONAL FIELD

What’s In WHY GRAVITATIONAL FORCES ARE IMPORTANT? 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.

What Is It Gravitation is the most important force on the scale of planets, stars, and galaxies. 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. 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 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). Our solar system is part of a spiral galaxy like the figure below, which contains roughly 1011 stars as well as gas, dust, and other matter. The entire assemblage is held together by the mutual gravitational attraction of all the matter in the galaxy.

What’s More Answer as required. 1. Compare the gravitational attraction between objects on earth and interaction of celestial bodies in space. Which gravitational force is almost negligible? Why? 2. Discuss why the study of a gravitational field is important.

Lesson

3

GRAVITATIONAL POTENTIAL ENERGY What’s In

When we first introduced gravitational potential energy in our previous lessons, we assumed that the gravitational force on a body is constant in magnitude and direction. This led to the expression U = mgh. But the earth’s gravitational force on a body of mass m at any point outside the earth is given more generally by Fg = (GmEm) / r2, where mE is the mass of the earth and r is the distance of the body from the earth’s center.

What Is It Consider a block with mass, and is tied to the end of a rope and goes up over a pulley while the other end is being pulled by a man. If the man lets go of the rope, the rope will be pulled downward with a force equal to the force of gravity of the block. The work performed by the block depends on the weight and height, Δh. The work done will be: W = Fd Since F = mg, then: W = (mg)( Δh) where d = Δh The more work to perform and energy stored in the block, the higher the block is from the ground. Gravitational Potential Energy, Eg , is the energy stored of an object because of its distance above the surface of the Earth. The change in gravitational potential energy of an object is expressed as: Eg = mgΔh where: m – is the mass of the object in kilograms g – is the acceleration due to gravity at 9.8 m/s2 Δh – is the vertical displacement of the object in meters ΔEg - is the object’s change in gravitational potential energy in Joules Sample Problems: 1. How much gravitational potential energy does a 4.0 kg block has if it is lifted 25 m? Eg

= mgΔh = (4.0 kg) (9.80 N/kg) (25 m) = 9.8 x 102 J

2. A 61.2 kg boy fell 0.500 m out of the bed. How much potential energy is lost? Eg

= mgΔh = (61.2 kg) (9.80 N/kg) (-0.500 m) = -299.8 J ≈ 300 J

What’s More PROBLEM SOLVING. Answer as required. 1. How much potential energy does a car gain if a crane lifts the car with a mass of 1,500 kg and 20 m straight up?

2. A basketball of mass 0.0400 kg is dropped from a height of 5.00 m to the ground and bounces back to a height of 3.00 m. a. On its way down, how much potential energy does the ball lose? b. On its way back, how much potential energy does the ball regain?

Lesson

4

ORBITS What’s In

Artificial satellites orbiting the earth are a familiar part of modern 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. We’ll see in the next section that the motion of planets can be analyzed in the same way.

What Is It A circular orbit, 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. The only force acting on a satellite in circular orbit around the earth is the earth’s gravitational attraction, which is directed toward the center of the earth and hence toward the center of the orbit. This means that the satellite is in uniform circular motion and its speed is constant. The satellite isn’t falling toward the earth; rather, it’s constantly falling around the earth. In a circular orbit the speed is just right to keep the distance from the satellite to the center of the earth constant. The radius of the orbit is r, measured from the center of the earth; the acceleration of the satellite has magnitude arad = v2 / r and is always directed toward the center of the circle. By the law of gravitation, the net force (gravitational force) on the satellite of mass m has magnitude Fg = GmEm / r2 and is in the same direction as the acceleration. Newton’s second law (F = ma) then tells us that GmEm = mv2 2 r r Solving this to find the circular orbit, we find

This relationship shows that we can’t choose the orbit radius r and the speed independently; for a given radius r, the speed for a circular orbit is determined. The satellite’s mass m doesn’t appear in the equation above, which shows that the motion of a satellite does not depend on its mass. If we could cut a satellite in half without changing its speed, each half would continue on with the original motion. For example, an astronaut on board a space shuttle is herself a satellite of the earth, held by the earth’s gravitational attraction in the same orbit as the shuttle. The astronaut has the same velocity and acceleration as the shuttle, so nothing is pushing her against the floor or walls of the shuttle. She is in a state of apparent weightlessness, as in a freely falling elevator.

True weightlessness would occur only if the astronaut were infinitely far from any other masses, so that the gravitational force on her would be zero. Apparent weightlessness is not just a feature of circular orbits; it occurs whenever gravity is the only force acting on a spacecraft. Hence it occurs for orbits of any shape. We can derive a relationship between the radius r of a circular orbit and the period T, the time for one revolution. The speed is the distance traveled in one revolution, divided by the period:

To get an expression for T, we use the equation above to solve for T and substitute:

We have talked mostly about earth satellites, but we can apply the same analysis to the circular motion of any body under its gravitational attraction to a stationary body. Other examples include the earth’s moon and the moons of other planets. Sample Problem: You wish to put a 1000-kg satellite into a circular orbit 300 km above the earth’s surface. What speed, period, and radial acceleration will it have? Solution: The radius of the satellite’s orbit is r = 6380 km + 300 km = 6680 km = 6.68 x 10 6 m. The orbital speed is

= 7720 m/s To find the orbital period, we have:

Finally, the radial acceleration is

What’s More PROBLEM SOLVING. Answer as required. 1. NASA is expected to send a 2600-kg satellite 450 km above the earth’s surface. (a) What is its radius? (b) What speed will it have? (Hint: Earth’s mass is 5.97 x 1024 kg) (c) What is its orbital period? (d) What is its radial acceleration?

KEPLER’S LAWS OF PLANETARY MOTION

Lesson

5

What’s In 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 precisely determine their orbits. The first and second of these ideas were published by Nicolaus Copernicus in Poland in 1543. The nature of planetary orbits was deduced between 1601 and 1619 by the German astronomer and mathematician Johannes Kepler, using a voluminous set of precise data on apparent planetary motions compiled by his mentor, the Danish astronomer Tycho Brahe.

What Is It The following are the laws developed by Johannes Kepler:

1. LAW OF ORBITS The first law explains that all planets move in elliptical orbits with the sun at one focus. Kepler’s first law means that planets move around the Sun in elliptical orbits. An ellipse is a shape that resembles a flattened circle. How much the circle is flattened is expressed by its eccentricity. The eccentricity is a number between 0 and 1. It is zero for a perfect circle. The eccentricity of an ellipse measures how flattened a circle it is. For a perfect circle, a and b are the same such that the eccentricity is zero. Earth’s orbit has an eccentricity of 0.0167, so it is very nearly a perfect circle.

2.

LAW OF AREAS

The second law describes a line that connects a planet to the sun and sweeps out equal areas in equal times. When a planet is near the sun, it travels faster and sweeps through a longer path in a given time. It follows from Kepler’s second law that Earth moves the fastest when it is closest to the Sun. This happens in early January, when Earth is about 147 million km (91 million miles) from the Sun. When Earth is closest to the Sun, it is traveling at a speed of 30.3 kilometers (18.8 miles) per second. 3. LAW OF PERIODS The third law mathematically expressed as the square of the period of any planet is proportional to the cube of the semi-major axis of its orbit.

Knowledge of these laws, especially the second (the law of areas), proved crucial to Sir Isaac Newton in 1684–1685, when he formulated his famous law of gravitation between Earth and the Moon and between the Sun and the planets, postulated by him to have validity for all objects anywhere in the universe. Newton showed that the motion of bodies subject to central gravitational force need not always follow the elliptical

orbits specified by the first law of Kepler but can take paths defined by other, open conic curves; the motion can be in parabolic or hyperbolic orbits, depending on the total energy of the body. Thus, an object of sufficient energy—e.g., a comet—can enter the solar system and leave again without returning. From Kepler’s second law, it may be observed further that the angular momentum of any planet about an axis through the Sun and perpendicular to the orbital plane is also unchanging.

What’s More MULTIPLE CHOICE. Identify the letter that corresponds to the correct answer. 1. “All planets move in elliptical orbits where the sun s at one focus.” This statement refers to Kepler’s __________ law of Planetary Motion. A. First B. Second C. Third D. None 2. From Kepler’s Third Law, the period (P) squared divided by the cube of semi-major axis (d) is the same for all planets and is expressed as P2 / d3 = k, where k is a constant. One can conclude that: A. In the same amount of time, every planet orbits the sun. B. At the same distance from the sun, every planet orbits the sun. C. It takes more time to orbit the sun if the planets are farthest from the sun. D. It takes less time to orbit the sun if the planets are farthest from the sun.

Summative Test (Post-Test) Multiple Choice. Select the lett...