Presentation on theme: "Chapter 5 Part 2 Newton’s Law of Universal Gravitation, Satellites, and Weightlessness."— Presentation transcript:
Chapter 5 Part 2 Newton’s Law of Universal Gravitation, Satellites, and Weightlessness
Newton’s ideas about gravity Newton knew that a force exerted on an object causes an acceleration. Most forces occurred because of contact with an object. The idea of a force without contact met resistance. Newton theorized the force that holds all things to the Earth, is the same force that acts to hold the moon in its nearly circular path.
Force at a distance Newton compared the force Earth exerts on objects at its surface to the force Earth exerts on the moon. On Earth’s surface, acceleration due to gravity, g = 9.80m/s 2. The centripetal acceleration of the moon is a R =v 2 /R. In terms of g, that is about 1/3600g. The moon is 60 times farther from Earth’s center than objects are from the surface. Newton concluded F g on any object decreases with distance from Earth’s center by 1/d 2.
What about mass? Newton reasoned, according to his 3 rd law of motion, that the force of gravity on an object must also be directly proportional to the masses of BOTH objects. m B is the mass of a body, m E is the mass of Earth, and r is the distance between their centers.
Force vs distance applied In analyzing gravity, Newton found, by examining the orbits of planets around the sun, the force required to hold different planets in orbit around the Sun seemed to diminish as the inverse square of their distance from the Sun.
Universal Gravitation In examining the planets, Newton stepped further analyzing gravity. Newton theorized the gravitational force that attracts an apple to the Earth and the moon to the Earth, is the same force that acts between the Sun and other planets to keep them in their orbits. If gravity acts here, why not between ALL things?
Finalizing the Universal law of Gravitation Every particle in the Universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. This force acts along the line joining the two particles. Universal Gravitational constant, G, (very small number) was experimentally discovered to calculate the EXACT force of attraction between 2 objects. G = 6.67 x 10 -11 N-m 2 /kg 2 (a constant like pi)
Example 1 A 50 kg person and a 75 kg person are sitting on a bench so that their centers are 50 cm apart. Estimate the magnitude of the gravitational force each exerts on the other.
Example 1 Solution Using the equation for force, F = 1.0 x 10 -2 N Which is unnoticeably small unless delicate instruments are used.
Example 2 What is the force of gravity acting on a 2000kg spacecraft when it orbits two Earth radii from the Earth’s center (a distance r E = 6380 km above Earth’s surface)? The mass of the Earth is m E = 5.98 x 10 24 kg.
Example 2 Solution Instead of plugging in all the numbers for our equation, we could take a simpler approach. The spacecraft is twice as far from the Earth’s center as when at the surface of the Earth. Therefore, since the force of gravity decreases as the square of the distance (and ½ 2 = ¼ ), the force of gravity on it will be only ¼ its weight at the Earth’s surface. F G = ¼ mg = ¼ (2000 kg)(9.80m/s 2 ) = 4900 N
Geophysical applications of Gravity near Earth’s surface Applying the Universal law between Earth and objects at Earth’s surface, the F G becomes the object’s weight. Thus we rewrite the formula: Cancelling mass we get: Acceleration of gravity at Earth’s surface depends on m E and r E. Once G was known, the Earth’s mass was determined.
Applications When dealing with objects at Earth’s surface, we calculate weight by mg. If we want to consider objects far from Earth’s surface, we can calculate the acceleration due to gravity there by including their mass and distance from Earth’s surface. EX: Estimate the effective value of g on the top of Mt. Everest, 8848m above the Earth’s surface. What is the accel due to gravity of objects that freely fall at this altitude?
Example Solution Calling the acceleration due to gravity there, g’, we replace r E with r = 6380 km + 8.8km = 6389 km = 6.389 x 10 6 m: =9.77m/s 2 Which is a reduction of about 0.3%.
Satellites and “Weightlessness” Artificial satellites are put into Earth orbit by high accelerations to give them a high tangential speed, v. If tangential speed is too fast, the satellite escapes Earth orbit. If too slow, it accelerates downward to Earth due to Earth’s gravitational pull. What keeps a satellite ‘up’ is a combination between its tangential speed and Earth’s a R. Satellites moving in an approx circle have an acceleration of a R = v 2 /r. This force is caused by the force of gravity acting on it.
Newton’s laws on Satellites For an orbiting satellite, the only force present is the force due to gravity. Using Newton’s Second law: ΣF R =ma R, we find Note that r is the sum of Earth’s radius and the satellite’s height above Earth: r = r E + h.
Geosynchronous Satellites A geosynchronous satellite is one that stays above the same point on the equator of the Earth. Such are used for purposes like cable tv, weather forecasting, and communication relays. Determine (a) the height above Earth’s surface a satellite must orbit and (b) such a satellite’s speed. See page 130 for worked out solution.
Weightlessness Standing on a scale in a stationary elevator, mg ON the scale is equal to the support force, w, BY the scale. If the elevator accelerates upward, the net force on you = ma. So we have w – mg = ma. Solving for w = mg + ma and you would “weigh” more than normal. In a freely falling elevator, the scale falls at the same rate as you and cannot push up (support) your weight, so it would read zero. (Apparent weightlessness).
Weightlessness in satellites Since satellites orbiting Earth are essentially “falling” around Earth, a passenger experiences the same apparent weightlessness that you would find in a freely falling elevator.
Kepler’s Laws (1571-1630) Before Isaac Newton, a German astronomer named Johannes Kepler spent his lifetime studying planets and their motion. He developed 3 laws of planetary motion: 1 st Law: The path of each planet about the Sun is an ellipse with the Sun at one focus. 2 nd Law: Each planet moves so that an imaginary line drawn from the Sun sweeps out equal areas in equal periods of time.
Kepler’s Third Law 3 rd Law: The ratio of the squares of the periods of any two planets (time for one revolution around the sun) is equal to the ratio of the cubes of their mean distances from the Sun. If T 1 and T 2 are the periods for any two planets and r 1 and r 2 are their average distances from the Sun, then If we rewrite this, then r 3 /T 2 should be the same for each planet.
Your turn to Practice Please do Chapter 5 Review pg 141 #s 25-30, 39, and 42. Bonus # 53.