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Gravitaton Universal Gravitation Gravity Near the Earth’s Surface Satellites and Weightlessness.

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Presentation on theme: "Gravitaton Universal Gravitation Gravity Near the Earth’s Surface Satellites and Weightlessness."— Presentation transcript:

1 Gravitaton Universal Gravitation Gravity Near the Earth’s Surface Satellites and Weightlessness

2 Any two objects of mass attract each other. This force (gravitational force) acts at a distance; no contact is needed. The gravitational force between objects of mass M and m, separated by a distance R, is given by G is a constant, and has a value of G=6.67x N·m 2 /kg Newton’s Law of Universal Gravitation M m F Mm F mM

3 Solution #1: plug the numbers into the equation on the previous slide. Solution #2: Giancoli’s physicist-type solution—pull the answer seemingly out of nowhere. Solution #3: Use the equation on the previous slide to calculate the force at R E and 2R E, and take the ratio. I’ll do this one on the blackboard. (Note: the space shuttle orbits at a height of a few hundred km.) Example 5-12: What is the force of gravity acting on a 2000 kg spacecraft when it orbits two earth radii from the earth’s center (that is, a distance R E =6380 km above the earth’s surface)? The mass of the earth is M E =5.98x10 24 kg.

4 For the first half of this class we have been assuming g (as in W = mg) is a constant. It is not. We have been treating the earth as an object with all its mass concentrated at a point at its center. Clearly, the value of g depends on how far you are from this center. Weight (on earth) is the force on an object due to the gravitational attraction of the earth. suggests that 5.7 Gravity near the Earth’s Surface

5 An object with a spherically symmetric mass distribution can be treated, for the purposes of calculating gravitational forces, as a point mass with all the mass concentrated at the center. The earth is a flattened spheroid, and its mass distribution is nonuniform. Slightly fatter at equator. Greatly exaggerated in drawing.

6 g is different in the basement of Physics than it is on the 2nd floor, and the difference is measurable. The difference, however, is small enough that we can treat g as a constant for our 3-figure calculations. Near the surface of the earth, g is approximately 9.80 m/s 2. g varies measurably from place to place, depending on how the nearby mass is distributed, and the value of g can be measured quite accurately (say, 6 decimal places) with a pendulum (we’ll see how this works later). See this web page for depictions of the earth’s gravitational field, and a bigger rotating earth image.this web page

7 Of course, the existence of Mount Everest proves that the earth’s mass distribution is not spherically symmetric. Nevertheless, we can use and get g effective =9.77 m/s 2. Gosh, I wonder what you get for g if you plug in R E and M E ? you can’t use the symbol g for this because it is not at the earth’s surface Example 5-14: Estimate the effective value of g at the top of Mount Everest (8848 m above the earth’s surface). Use R E =6380 km, M E =5.98x10 24 kg, and assume (unrealistically) that the earth’s mass distribution is spherically symmetric.

8 A satellite in orbit is acted on by gravity (the centripetal force). You already know how to handle centripetal force problems: Note: mg = F G (done four slides back) is not an OSE; it only works at surface of earth with g = 9.8 m/s 2. Alternatively, if we know G and the radius of the earth, we can calculate the mass of the earth.  F r = F grav,r = ma r = mv 2 /r and the force is directed towards the center of the earth. This is old stuff; nothing new here! 5.8 Satellites and “Weightlessness”

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10 Now, if you have thought much about satellites, you probably have thought you could put a satellite in orbit at any height above the earth you wanted, with any speed you wanted. Not so! I’ll show the simple calculation on the board that proves there is a unique orbital velocity for any orbital radius (or vice versa).

11 Do you know what a geosynchronous satellite is? It’s a satellite that orbits with a speed so that it always stays above the same point above the earth. Such satellites are extremely valuable for communications, weather monitoring, and spying.

12 Huh? How can you run out of room in space? Trouble is, we’ve virtually run out of room in space for geosynchronous satellites. The speed of the satellite is v=2  R/T where T is one day. If you use V=2  a/T for our orbital speed V at height a, and set F G =mV 2 /r, you get another OSE: I’ve used the symbol “a” for the orbital radius, for consistency with Physics 23.

13 You can solve this equation for the orbital radius of a geosynchronous satellite (T = 1 day for this satellite): Everything on the right hand side is a constant! Plugging in the values gives a=42,300 km, or a height of about 36,000 km above the earth’s surface. Easy calculation for you to try: what is the orbital speed of a geosynchronous satellite? Compare your result with the answer on page 130 of Giancoli.

14 Weightlessness You have probably felt “weightless” on a carnival ride, roller coaster, or rapidly-accelerating elevator. Were you truly weightless? One could only wish. You simply experienced free fall. An astronaut in the space station is not weightless either, but experiences the weightless sensation because he/she is constantly falling around the earth. The effective value of g at the orbital radius of the space station is not zero! Nevertheless, it is probably valid to claim that the apparent weightlessness of free fall simulates true zero-g weightlessness in outer space.


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