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APS 1030 Laboratory 1 Gravity GRAVITY “To see the world in a grain of sand, And a heaven in a wild flower; Hold infinity in the palm of your hand, And eternity in an hour.” -W. Blake SYNOPSIS: By measuring the time taken for an object to fall through a distance, you will determine the acceleration due to Earth's gravity, and will use that information to calculate the mass of the Earth. You’ll also compare the Earth’s gravity with that of other objects, including black holes! EQUIPMENT: Falling body apparatus, tape measure, calculator. Introduction Acceleration is the change in speed (velocity) that a moving body experiences in a given time interval. For example, suppose that an object starts at rest and accelerates at a rate of 10 meters per second per second (written m/sec2 ). After the first second, it will be traveling at a speed of 10 m/sec. After the next second, its speed will be 10 m/sec plus 10 m/sec, or 20 m/sec, etc. So, for constant acceleration of an object starting at rest, speed = (acceleration) (time). (1) Because an accelerating object’s speed is constantly increasing, the distance it covers each second is accordingly greater. For an object starting at rest and undergoing constant acceleration, 1 distance = 2 (acceleration) (time) 2. (2) Thus, an object accelerating at 10 m/sec 2 will travel a distance of 5 meters in the first second of time; two seconds after starting, the distance covered is 20 meters, implying that the object travelled 15 meters during the second second. Every particle in the universe exerts a gravitational force on every other particle. A force gives rise to an acceleration. The combined effects of all of the particles in the Earth acting upon us produces an acceleration which we constantly experience; we refer to this acceleration simply as gravity, and we call the force on our bodies weight. If we drop an object and time how long it takes to fall a certain distance, we can use equation (2) to measure the acceleration due to gravity: acceleration = 2 (distance) (time)2 (3) APS 1030 Laboratory 2 Gravity Part I. Measuring the Acceleration due to Gravity The falling-body apparatus uses a spring-loaded clamp to hold a steel ball above a landing pad. When we release the clamp, the ball drops and the timer starts. The timer stops when the ball strikes the landing pad, and the elapsed time in seconds appears on the readout. Positioning Clamp Green Pull Knob Dropping Mechanism Timer Table Red Clamping Knob Ball Landing Pad Floor Work in a team of three individuals with one dropping apparatus. Measure the acceleration due to gravity using the small steel ball as described below. After one individual has completed his/her measurements, the other team members should independently conduct their own trial measurements (step I.2). I.1 I.2 Position the dropping mechanism over the edge of the table, and the landing pad (inside the catch basin) on the floor directly underneath it. Switch the timer on. • While holding the small steel ball in the clamp of the dropping mechanism, pull on the green knob (against the spring tension) so that the ball is captured. Tighten the red clamping knob to hold the ball in place. • Measure the drop-distance to the nearest millimeter: Use a metric tape measure to determine the distance from the bottom of the ball to the top of the landing pad (another team member may assist by holding one end of the tape, but you must actually read the distance). Record your distance measurement in meters (1 mm = 0.001 meter). • Press the reset button on the timer to initialize the display. • Release the ball by loosening the red knob. Record the elapsed time in seconds as indicated by the timer. I.3 Write down each individual's name and his/her values for distance and time in Table 1. For each individual trial, calculate the gravitational acceleration experienced by the small ball using equation (3), and record the results in the table. APS 1030 Laboratory Table 1: Experimenter Names 3 Small Ball Distance (meters) Gravity Long Distance Time (seconds) Acceleration (m/sec 2 ) Average Presumably, each of you obtained similar, but not necessarily identical, values for the drop distance, the elapsed time, and the calculated acceleration for each experimental trial. If the measurements were not identical, it does not mean that the distance, time, or acceleration actually changed between trials, nor does it necessarily imply that one or more of you made a mistake. The results simply point out that there is no such thing as a perfect measuring instrument or procedure, and therefore there is no such thing as a perfect measurement! Since every measurement (distance, time) contains an experimental error (meaning an uncertainty, not necessarily a mistake), then any information derived from those measurements (acceleration) also contains an uncertainty. We can never know exactly what the "true" value is; we can only try to get better and better results by improving our measuring equipment or technique, and by averaging the results of our measurements to statistically reduce any random errors that may have occurred. Note: You cannot claim to know the value of the acceleration to any greater accuracy than the measurement error. It is senseless and misleading to list its value to place-values smaller than the uncertainty. For example, if your calculator comes up with a figure of 9.3487851 m/sec2 , while your individual trials ranged from 9.342 to 9.361 m/sec2 , you should list your result as 9.35 m/sec2 , with an uncertainty of about ±0.01 m/sec2 . It may not look as impressive, but it's honest! I.4 Suggest at least one way in which the measuring equipment or your measurement technique might be improved to produce a smaller experimental error. I.5 Calculate an improved measurement for the gravitational acceleration by averaging all three of your team's individual values, and enter the results in Table 1. Aristotle claimed that heavier objects fall faster than ligher objects (that is, heavy objects experienced a greater gravitational acceleration). Nearly two thousand years later, Galileo was the first to dispute this belief, claiming that all objects fall at the same rate (and asserted that air friction was responsible for the seemingly slow drop of light objects). According to lore, Galileo put his hypothesis to the test by dropping balls of different weights off the leaning tower of Pisa. I.6 Test Galileo's hypothesis. Repeat procedure I.2 using the large steel ball instead of the small one. Once again, each member should make an independent distance measurement (slightly different from the previous, since the size of the ball has changed), time measurement, and calculation of the acceleration. Record all of your teammates results in Table 2, and find the average acceleration. APS 1030 Laboratory Table 2: Experimenter Names 4 Large Ball Distance (meters) Gravity Long Distance Time (seconds) Acceleration (m/sec 2 ) Average I.7 Within your uncertainty of the measurements, does your experimental data support Galileo's contention that the acceleration due to gravity is the same for all objects? Explain. Since both balls fell nearly equal distances, your measurements haven’t tested whether gravitational acceleration was a constant value throughout the drop, but merely that it was the same for both objects (it could have changed identically for both balls during the time that they fell). I.8 Determine whether the acceleration due to gravity is a constant that is independent of the drop-distance or duration of the time-of-flight: Reposition the dropping apparatus and landing pad to perform a shorter drop to the tabletop as shown in the diagram. Using the large ball, perform independent measurements as before, and record and average your team's results in Table 3. Table 3: Experimenter Names Large Ball Distance (meters) Short Distance Time (seconds) Acceleration (m/sec 2 ) Average I.9 Within your uncertainty of the measurements, do your results support the contention that all objects experience the same, constant acceleration due to Earth's gravitational attraction? Why or why not? I.10 Average all nine of your separate measurements together (or equivalently, average the three averages) to come up with a final, best estimate of its value. APS 1030 Laboratory 5 Gravity Part II. The Mass of the Earth Isaac Newton showed that the acceleration g experienced by an object due to the gravitational attraction of a second body is directly proportional to the mass m of the attracting body, and inversely proportional to the square of the distance r between the bodies (one form of Newton's universal law of gravitation): m g = G 2 . r (4) Here G is the constant of proportionality, called the gravitational constant. (The value of G was determined very carefully by measuring the gravitational force between two large blocks of metal hung very close together from wires.) Essentially, G converts the mass and the distance to an amount of gravitational acceleration. In meters-kilograms-second (mks) units, G = 6.670 x 10-11 m3 kg· sec 2 (5) Important: When equation (4) is used with the above value for G, the mass m must be in kilograms and r in meters; otherwise, any numerical values you obtain will have nonsense units. Identical Gravitational Acceleration One Earth Mass M E Earth Mass = M E One Earth Radius R E Using calculus (which he invented to enable him to solve his mathematical problems), Newton also showed that the mass of a spherically-symmetric object, such as the Earth, gravitationally attracts other objects as if all of its mass were concentrated at its center. Thus, the acceleration produced by Earth's gravity is the same as that produced by a tiny object having the same mass as the Earth (ME), and which is located at the center of the Earth a distance of one Earth radius (RE) away. The Earth's radius is RE = 6,400 km = 6.4 x 106 meters. Since you have already measured the acceleration due to gravity, g, we can re-write equation (4) to solve for the mass of the Earth: 2 g RE ME = G . (6) II.1 Calculate the mass of the Earth using equation (6) and your team's average value for the acceleration of gravity. II.2 Compare your result with the accepted value for the mass of the Earth: ME = 5.976 x 1024 kg. Is your result accurate to 50%? 10%? 1%? Is your result too large or too small? II.3 If you were to repeat this experiment on a mountaintop, would the acceleration due to Earth's gravity be less than, equal to, or greater than that which you have measured? Explain (hint: which parameters would be different in equation 4)? APS 1030 Laboratory II.4 6 Gravity Can you think of any reason(s) why your measurement made in Boulder might have given results that were higher or lower than the "official value"? Part III. Gravity on Other Solar Sytem Objects Newton's law of gravity (Equation 4) applies to everything and everywhere in the universe, not just on the Earth (that’s why they call it the “universal” law of gravitation!). III.1 Calculate the gravity (ie. acceleration) felt by the Apollo astronauts as they walked on the Moon. To do this you'll need the mass of the Moon, 7.35 x 1022 kg, and its radius, 1738 km. Compare this acceleration to that here on Earth (don’t forget to convert from kilometers to meters!). Does this explain why the astronauts "bounced" as they walked? III.2 Now, find the “surface” gravity for Jupiter. Jupiter's mass is 1.9 x 1027 kg, and its radius is 71,398 km. Aside from being squashed into a pancake in mere seconds, what other difficulties would we face if we tried to stand on Jupiter's “surface”? III.3 Find the “surface” gravity at the outermost visible layer of the Sun, the photosphere. The Sun’s mass is 1.99 x 1030 kg, and its radius is 6.96 x 105 kilometers. Part IV. Escape Velocity When vehicles are launched into space, the rocket must have a minimum speed at launch to escape the Earth's gravity. We call this velocity the “escape velocity” (vesc). If the rocket is moving slower than this velocity, it will fall back to the Earth. From the equations given above, we can calculate the escape velocity from any object, given by the following equation: vesc = 2GM R (7) “G” is the gravitational constant given in equation (5), and M and R are the mass and radius of the object you are trying to “escape.” IV.1 Calculate the velocity needed for a spacecraft to escape from the Earth. (Note that the escape velocity does not depend on the mass of the spacecraft itself!) IV.2 Now calculate the escape velocity for the Moon. (This should explain why the Apollo astronauts needed only a small vehicle to leave the Moon’s surface and return to lunar orbit, while the Saturn-V rocket used to lift everything off of the face of the Earth was so huge!) IV.3 How fast would your spacecraft have to travel to escape from Jupiter? Individual atoms and molecules of gas in a planet’s atmosphere also have to reach these same escape velocities in order to “leak off” into space; otherwise, they stay trapped close to the planet’s surface, just like us. IV.4 Use the concept of escape velocity to explain why the Moon has virtually no atmosphere, Earth has a moderate amount, and Jupiter has so much. IV.5 Streams of charged particles, known as the solar wind, are constantly escaping from the Sun and blow out into interstellar space. How fast must these particles be travelling in order to escape the gravitational grip of the Sun? APS 1030 Laboratory 7 Gravity Part V. Black Holes When a star begins to die (ie. when it runs out of hydrogen to convert into helium), it cools off and begins to collapse due to its self-gravity. If it is more massive than our Sun, it will eventually explode in what is called a supernova. In the explosion, the outer layers of the star are blown off and form a supernova remnant like the Crab Nebula. The inner core of the star does not explode outward but instead implodes, collapsing into a very small sphere. Einstein's theory of special relativity tells us that nothing can go faster than the speed of light: it is the ultimate speed limit. Imagine that our star has collapsed into a sphere so small (ie., its R is small) and is massive enough (ie. M is large) such that its escape velocity is greater than the speed of light. Nothing would then be able to escape its gravitational pull, not even light! This dark star is the ubitiquous “black hole.” Obviously such an object deserves quite a bit of respect and we must make sure we do not get too close to it, otherwise we will be uncontrollably pulled in. So, let's find out how close we can get. If we substitute the speed of light “C” in for vesc in equation (7) and rearrange the terms, we find: RSch = 2GM C2 (8) Rsch is called the “Schwarzschild radius.” It is the radius at which nothing can escape from the black hole; if anything gets closer than Rsch to a black hole, no force in the universe can save it from being pulled in, because the necessary escape velocity would be greater than the speed of light. Rsch is not the actual radius of the black hole; it is the “event horizon,” so-called because nothing inside this radius can be seen and we can only speculate on what is happening inside. No one really knows for certain, but it is speculated that the black hole itself collapses into what is mathematically known as a singularity - a point that has no radius, no volume, and an infinite density. V.1 Imagine that the Sun collapsed into a black hole (which, incidentally, will not happen). Using equation (8), what would its Schwarzschild radius (Rsch) be? How does this compare to the actual radius of the Sun? Compare the Sun's Rsch to the radius of the Earth. If the Sun did actually collapse into a black hole, would the Earth and the other planets be sucked in? Why or why not? V.2 Suppose that the Earth collapsed into a black hole. What would its Rsch be? Suppose that you collapsed into a black hole! How does this Rsch compare to the radius of a proton (which is around 10-15 m) ? Now, imagine that you are an astronaut that is going to bravely (i.e., foolishly) fly into a black hole. When you are 1.5 Rsch away from the black hole, you glance to your side and notice a strange sight. At this distance, beams of light orbit the black hole much like a satellite orbits the Earth. So, when you look to the side you see the back of your head! You continue inward, oblivious to the fact that you have crossed the Schwarzschild radius and now can never return. Once inside this radius, we have no idea what has happened to you since no light can escape to tell us - though we can envision what kind of gruesome death awaits you! You are now about to experience the effect of “differential gravity” - also known as the tidal effect. In plain English, your body will be stretched out and pulled apart. Your feet are closer to the black hole are therefore are being pulled towards the black hole faster than your head, causing you to be stretched apart! APS 1030 Laboratory 8 Gravity You Black Hole Schwarzschild Radius R Sch V.3 Suppose that for a brief moment that you are actually standing on the Schwarzschild radius of the black-hole Sun calculated earlier; ie. your feet are a distance Rsch from the black hole. Using equation (4), what acceleration would your feet feel? V.4 Assuming that you are 1.5 meters tall (that is, your head is Rsch+1.5 m away from the black hole), what acceleration would your head feel? V.5 Now, calculate how far this acceleration will take you in one second (presuming that you're not moving at first) by using equation (2). To get a feel for how far this really is, compare this distance to the size of Pluto's orbit, 5.9 x 1012 m. (You may be wondering how this is possible since the Schwarzshild radius of our black hole is much smaller than this. It isn't - you'll be pulled into the black hole much faster than a second!) V.6 Now, calculate the difference between the two accelerations; this is the differential acceleration. This is how quickly your head and feet are moving apart from each other! This should be on the order of 1010 m/sec2 . V.7 Finally, calculate how far your head and feet would be separated after one second. Again, use equation (2) to get a distance in meters. Compare this to the distance from the Earth to the Moon (3.8 x 108 m).