The Beginning of Modern Astronomy

1. The Beginning of Modern Astronomy

2. Isaac Newton (1642-1727) • did research on optics (the properties of light) • invented calculus • discovered the three laws of Motion • discovered the law of universal gravitation Motion Concepts • inertia: resistance to a change in motion. • mass: numerical measure of inertia, amount of material in the object. • speed: how fast something moves. • velocity: speed and direction. • acceleration: rate of change of velocity.Acceleration in the direction of motion speeds up the object. Acceleration perpendicular to the object’s path changes the direction of its motion. • momentum: mass × velocity.

3. Newton’s Laws of Motion • An object moves with constant velocity unless acted on by an unbalanced external force. • When an object is acted on by an unbalanced force, it accelerates in the direction of the unbalanced force. The magnitude of the acceleration is related to the magnitude of the net (unbalanced) force by the equation Fnet = ma, where m is the mass of the object and a is its acceleration. • When two objects interact, they exert equal and opposite forces on each other. FA on B = - FB on A.

4. The centripetal acceleration is calculated using the formula where a is the centripetal acceleration, v is the speed, and r is the radius of the circle. Circular Motion, Centripetal Acceleration, and Centripetal Force • An object in uniform circular motion is accelerating even if its speed is constant. • In this case, there is no acceleration along the path of the object; the acceleration, called centripetal acceleration ,is perpendicular to the path (toward the center of the circle). • The force that causes a centripetal acceleration is called a centripetal force. Example 1The average distance from Earth to the Moon is 3.84×108 m, and the moon’s average orbital speed is 1022 m/s. Calculate its centripetal acceleration.

5. M r m Newton’s Law of Gravity Any two particles in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the distance between them. The minus sign reminds us that the force is attractive. G = 6.673×10-11 Nm2/kg2 M and m are the masses of the two particles. r is the distance between their centers. A spherically symmetric object is one whose mass is distributed equally in all directions. The force on a particle outside an object with spherical symmetry is the same as if all of of the object’s mass were concentrated at its center. This allows Newton’s law of gravity to be used for things like planets, which are almost spherically symmetric.

6. Consider the falling object shown at the right. If air resistance is negligible, the only force acting on it is gravity. Fnet = FG M = mass of Earth m = mass of object R = radius of earth Relation Between Weight and Mass weight= the gravitational force on an object. Fnet = ma The acceleration due to gravity is directly proportional to the mass of the planet and inversely proportional to the square of the distance from the planet’s center. It is usually denoted by the symbol g. W = weight = mg Near earth’s surface, a = g = 9.8 m/s2. When air resistance is negligible, the acceleration of a falling body does not depend on its weight.

7. Newton’s second law of motion and his law of gravity enable us to determine the masses of planets and stars. Example 2 The acceleration due to gravity at the surface of the Earth is 9.8 m/s2, and the radius of the Earth is 6380 km. What is the mass of Earth?

8. Example 4 From the moon’s orbital speed and its distance from the center of Earth, Newton knew that the centripetal acceleration of the Moon is 0.0027 m/s/s; this is the experimental value of the moon’s acceleration. Since the (average) distance from Earth to the Moon is about 60 times the radius of Earth, Newton’s law of gravity predicts that the acceleration should be Example 3 An astronaut whose weight is 150 lb on Earth is launched to an altitude of twice earth’s radius. What is his weight at that altitude? The agreement between this theoretical value and the experimental value was an important confirmation of Newton’s law of gravity.

9. Conservation of Angular Momentum P O When the net force on a particle is always directed toward a fixed point, its angular momentum relative to that point does not change with time; i.e., its angular momentum is conserved. The gravitational force of the Sun on a planet is always directed toward the Sun, so the angular momentum of the planet relative to the sun is conserved. It can be shown that the conservation of the angular momentum of a planet is equivalent to the statement that the line from the Sun to the planet sweeps out equal areas in equal times, so Kepler’s second law of planetary motion is equivalent to the law of conservation of angular momentum applied to a planet in orbit around the Sun.

10. Kinetic Energy, Radiative Energy, and Potential Energy Energy is the ability to move an object while exerting a force on it. The energy of an object due to its motion is called kinetic energy. It is defined by the equation Potential Energyis the energy that a group of objects has because of their relative positions. There is no single formula for potential energy. When you exert a force on an object and cause it to move, you put energy into it. The process of putting mechanical energy into an object is called doing work on the object. Radiative Energyis the energy of the electric and magnetic fields in electromagnetic radiation.

11. P = the sidereal period of the planet (or satellite). m = the mass of the planet (or satellite). M = the mass of the Sun (or planet). G = 6.673×10-11 Nm2/kg2. Solving for M The law of conservation of energy can be used to prove Kepler’s third law of planetary motion and add some detail to it. Example 5 The sidereal period of the Moon is 27.32 days, and its average distance from Earth is 384,000 km. Calculate the mass of Earth. Assume that the mass (m) of the Moon is negligible compared to that of Earth (M). P = 27.32 days = 2.732×101×8.64×104 s = 2.360×106 s a = 3.84×105 km = 3.84×105×103 m= 3.84×108 m G = 6.673×10-11Nm2/kg2 M = 6.02×1024 kg

12. The orbital speed of a satellite in a circular orbit of radius r around a planet of mass M is where G = 6.67310-11Nm2/kg2. r M Circular Velocity Example 6 Calculate the orbital speed of a satellite in a circular orbit 150 km above the surface of Earth. Assume that the radius of earth is 6380 km and its mass is 5.98×1024 kg. r = 150 km + 6380 km = 6530 km r = 6.53×103 x 103 m r = 6.53×106 m M = 5.98×1024 kg Vc = 28,500 km/hr = 17,500 mi./hr

13. If an object is at a distance r from the center of a planet of mass M, it can escape from the planet if its speed is at least equal to Ve where Ve is called the escape velocity. r = 6.38×106 m M = 5.98×1024 kg Escape Velocity Example 7 Calculate the escape velocity from the surface of Earth.

14. Tides The length of a day increases by about 0.0023 seconds per century, and the Moon moves farther from Earth by about 3.8 cm per year. Why? 900 million years ago, earth’s day was 18 hours long.