Universal Gravitational Constant ex-5550
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Universal Gravitational Constant EX-5550 Page of 1Universal Gravitational Constant Equipment
** Anti-vibration pads are designed to use with a telescope and are available from Celestron, Meade, or Orion Telescopes for $50-$70. Look on-line. Using the pads decreases the amount of vibrations that the Torsion Balance picks up from the surroundings. Introduction The Gravitational Torsion Balance reprises one of the great experiments in the history of physics—the measurement of the gravitational constant, as performed by Henry Cavendish in 1798. Data collection requires less than 45 minutes. The data may then be transferred to each lab group using flash drives. This makes it possible to perform this classic experiment in a normal laboratory period with an accuracy of better than 3%. The Gravitational Torsion Balance consists of two 38.3 gram masses suspended from a highly sensitive torsion ribbon and two 1.5 kilogram masses that can be positioned as required. The Gravitational Torsion Balance is oriented so the force of gravity between the small balls and the earth is negated (the pendulum is nearly perfectly aligned vertically and horizontally). The large masses are brought near the smaller masses, and the gravitational force between the large and small masses is measured by observing the twist of the torsion ribbon. An optical lever, produced by a laser light source and a mirror affixed to the torsion pendulum, is used to accurately measure the small twist of the ribbon. Historical Overview The gravitational attraction of all objects toward the Earth is obvious. The gravitational attraction of every object to every other object, however, is anything but obvious. Despite the lack of direct evidence for any such attraction between everyday objects, Isaac Newton was able to deduce his law of universal gravitation. Newton’s law of universal gravitation: F = GmM/r2 where m and M are the masses of the objects, r is the distance between them, and G = 6.67 x 10-11 Nm2/kg2 However, in Newton's time, every measurable example of this gravitational force included the Earth as one of the masses. It was therefore impossible to measure the constant, G, without first knowing the mass of the Earth (or vice versa). The answer to this problem came from Henry Cavendish in 1798, when he performed experiments with a torsion balance, measuring the gravitational attraction between relatively small objects in the laboratory. The value he determined for G allowed the mass and density of the Earth to be determined. Cavendish's experiment was so well constructed that it was a hundred years before more accurate measurements were made. The gravitational attraction between a 38 gram mass and a 1.5 kg mass when their centers are separated by a distance of approximately 46.5 mm (a situation similar to that of the Gravitational Torsion Balance) is about 1.8 x 10-9 newtons. If this doesn’t seem like a small quantity to measure, consider that the weight of the small mass is more than two hundred million times this amount. T  he enormous strength of the Earth's attraction for the small masses, in comparison with their attraction for the large masses, is what originally made the measurement of the gravitational constant such a difficult task. The torsion balance (invented by Charles Coulomb) provides a means of negating the otherwise overwhelming effects of the Earth's attraction in this experiment. It also provides a force delicate enough to counterbalance the tiny gravitational force that exists between the large and small masses. This force is provided by twisting a very thin beryllium copper ribbon. Cavendish Balance: the large masses are first arranged in Position I, as shown in Figure 1, and the balance is allowed to come to equilibrium. The swivel support that holds the large masses is then rotated, so the large masses are moved to Position II, forcing the system into disequilibrium. The resulting oscillatory rotation of the system is then observed by watching the movement of the light spot on the scale, as the light beam is deflected by the mirror. Theory
F = GmM/b2 Eq. (1) w  here m and M are the masses of two spherical objects, b is the center-to-center distance between them, F is the force of attraction between the two masses, and G is a constant to be measured in this experiment. The force (F) is extremely small (< 2 nN) and will be measured using a very sensitive torsion balance. The torsion balance consists of two tungsten spheres, each of mass m = 38.5 g and diameter = 1.635(5) cm, attached to a thin aluminum rod that yields a center-to-center distance for the two balls of d =10.0(1) cm. The small ball dumbbell is suspended by a 40 cm beryllium copper ribbon. The large balls are made of tungsten and each have a mass of M = 1.500(5) kg and diameter D. When the two large masses (M) are in the center position (or removed), if the system is perfectly aligned, the two small masses (m) will be in the position shown in black in Figure 2 so that they are parallel to the screen. The reflection from the laser would then fall at point 2 on the screen. We will assume this is the case since it makes the derivation easier. If the small masses differ from parallel by less than 0.1 radians (5o), the errors introduced (by assuming tan = (rad)) are less than 0.5%. When the large masses (M) are moved to the position shown in Figure 2, the gravitational force (F) causes a torque () about the center of the dumbbell formed by the small masses and they rotate through an angle to the equilibrium position shown in blue. The spot on the wall due to the laser will shift by an angle of 2 or a linear distance of S/2. We then reverse the positions of the large masses and the laser spot moves to position 1. From symmetry, the distance between 1&2 is the same as between 2&3, so the total distance between 1&3 is S and this is what we will actually measure. T
he torsional pendulum obeys Hooke’s Law for a rotating system, so = C Eq. (2) Where C is a constant determined by the beryllium copper ribbon supporting the small mass dumbbell. = 2(d/2)F cos Eq. (3) Actually, the angle is slightly greater than because F is along the line of centers and is not quite vertical (on the diagram). However, since with the screen at 1.5 m, S is less than 4 cm, is less that 1/2o and we may approximate this as = dF = C Eq. (4) How large a percent error does this approximation introduce? See Appendix 6 for a discussion of small corrections. The torsional pendulum will overshoot the equilibrium position and oscillate for several hours since there is very little damping in the system. Since it obeys Hooke’s law, the motion is simple harmonic (slightly damped) and the angular frequency is given by = (C/I)½ Eq. (5) where I is the rotational inertia for the small mass dumbbell about its center. Using the parallel axis theorem, I = 2[(2/5)mr2 + m(d/2)2] = m(4r2/5 + d2/2) Eq. (6) Equations 5&6 yield C = 2m(4r2/5 + d2/2) Eq. (7) And Equation 4 becomes dGMm/b2 = 2m(4r2/5 + d2/2) Eq. (8) Solving for G yields G = 2b2(4r2/5 + d2/2)/Md Eq. (9) Setup
Do Not Touch the Adjustment Wheels at the top of the Cavendish Balance!!! Many hours are required to correctly adjust them. Note to instructor: it is nice to put the plastic plate on so the students can see the small balls inside, but do it after data collection is over for all the labs using the balance. Do not do the experiment with the plastic plate since small electrostatic charges can interfere. System should be similar to Figure 3. The laser spot should move along the top of the transparent ruler. 
Figure 3: Setup Procedure
Analysis
Curve Fit
Calculation of G From the L a S values in the table, calculate and its uncertainty see Figure 4. Enter the values in the table. Use the values in the table to calculate G using Equation 9 and enter your value in the table. G = 2b2(4r2/5 + d2/2)/Md Eq. (9) Uncertainty in G For a product or quotient of terms of the form ABn/C = D where the uncertainties in A, B, and C are known, the uncertainty (D) in D is given by: D/D = [(A/A)2 + (nB/B)2 + (C/C)2]1/2 Eq. (10) Since the r2 term is much smaller than the d2 term in Equation 9, it may be ignored. The uncertainty in M can also be ignored since M/M is small compared to the other terms. Equation 10 then yields G/G = [()2 + (2)2 + (2b/b)2 + (d/d)2]1/2 Eq. (11)  Calculate the uncertainty in G and enter it in the table. Corrections to G The value you obtained for G is probably lower than the actual value by more than the uncertainty because we have ignored two significant errors.
= F d Eq. (4) the true torque is = d(F-f sin φ) Eq. (12) and the true value of G should be larger than the value you calculated by a factor of factor = F/(F-f sin f) = 1/[1- (f/F)sin φ] =1/[1- ([GmM/(b²+d²)]/[GmM/b²])sin φ] = 1/[1- (b²/(b²+d²))(b/(b²+d²)1/2) = 1/(1- [b/(b²+d²)1/2]³) Eq. (14) Calculate the (total) correction factor from step 1 and equation 14 and enter it in the table. Correct your value for G and enter it in the table. Does your value for G +/- G agree with that given in your textbook? We have ignored the rotational inertia of the aluminum rod connecting the two small balls. How would this affect our value for G? The rod is about 3 mm in diameter and 10 cm long and has a density of 2.7 g/cm3. Would it have a measurable effect? We have also ignored the gravitational attraction of the case for the small balls and the attraction of the large balls for the aluminum rod between the small masses. How would this affect our value for G?
Itrue = 2m(d/2)² + 2(2/5)mr² + (1/12)mrd² where mr = Itrue /[2m(d/2)²+ 2(2/5)mr²]= 1 + (1/6)mr/m/[1 + (8/5)(r/d)²] Correction due to attraction of Al rod 
mr = sm =
This can be directly verified by performing the experiment without the large masses in place. The data for this is included in the file called “ring only”. The results of analyzing that data is Sring/(S – Sring) = Appendix I: Initial Setup-This should be done at least four hours before the laboratory to allow time for unwanted oscillations to damp out. If you are setting up the Cavendish balance for the first time, you should begin the setup weeks in advance! P  reliminary Set Up (Note: be sure the thumbscrews on the bottom of the pendulum chamber are in the up position so the pendulum bob dumbbell is pushed up against the case (see Figure 6) and there is no tension in the beryllium copper ribbon before you move the unit! Breaking the ribbon is not a good thing!)
Appendix II. Leveling the Gravitational Torsion Balance
Figure 7: Using the Leveling Sight Figure 8: Adjusting the Height of the Pendulum Vertical Adjustment of the Pendulum (generally only required the first time the Cavendish balance is used.) The base of the pendulum should be flush with the floor of the pendulum chamber. If it is not, adjust the height of the pendulum:
Appendix III: Setting Up the Laser and the Screen
 
Figure 9: Laser Setup Figure 10: Making Cav. Box Parallel to Screen  
Figure 11: Polarizer Figure 12: Attaching the ground wire Appendix IV. Rotational Alignment of the Pendulum Bob Arms - Zeroing (Generally only required the first time the Cavendish balance is used. May take many hours to do.) Do Appendices I-III first. The pendulum bob arms must be centered rotationally in the case — that is, equidistant from each side of the case (Figure 13). Figures 14a and 14b illustrate what the goal is. The lower spot is the reflection off the glass. It does not move. The middle spot is the reflection off the mirror. The upper spot is reflected off the mirror, reflected back off the glass, and reflected a second time off the mirror. If the balls are centered in the case (assuming the mirror is parallel to the rod connecting the two small masses), then the three spots should be aligned vertically (as in Figure 14a) when the system is at rest. This is difficult to achieve. A more realistic case is shown in Figure 14b where the center spot is 1.5 cm to the right. The screen is at 1.5 m so the system is off by only 1/100 radian (0.50). This introduces a negligible error. To get these images, the video camera and polarizer have been removed.   
Figure 13: Centering the Pendulum Bob Figure 14a: Figure 14b: Perfect Alignment Realistic Alignment Alignment procedure
Zero Adjust Thumbscrew 
Figure 15: Refining the Rotational Alignment of the Pendulum Bob Appendix V: Small Corrections Since we have claimed an uncertainty of only a couple of percent, we must consider small effects that might become important at this level of precision. We have identified seven effects that could be significant:
These effects are all small and partially canceling so the net effect is less than 0.1% and can be safely ignored. For those who are interested, details of the calculations are included in the remaining indices. We have ignored any interaction between the various corrections (for example, we calculated the force on the aluminum rod holding the small balls assuming it was on the symmetry plane of the case. Such small corrections to small corrections are surely ignorable. 1. The laser is aimed slightly upwards so the beam goes over the laser head to hit the screen. The laser must be tipped up by about 20 for the reflected beam to clear the laser head at 50 cm. Doesn't matter. Horizontal deviation, θ, depends only on the rotation of the mirror around a vertical axis. Only difference is that it means that the perpendicular distance to the screen is L (which is what was called for in the Procedure) and not the distance to the line where the spot moves. Really doesn't matter much since cos (tip angle) = cos 2⁰ = 0.9994. Correction = 0.00% (or +0.06% if measured distance to spot rather than perpendicular distance) 2. The mirror is not exactly vertical. This angle can be determined by comparing the reflected angle measured from horizontal to the incoming angle measured from horizontal. For the two PASCO Cavendish balances I checked, the mirror was tipped forward by 1.40 and 2.30. I take 20 as an average figure. Angular velocity is a vector so the rotation speed about the tipped axis, ω', is given by ω' = ω cos 2⁰ = 0.9994 ω → Correction = +0.06 % 3. Mirror horizontal position is off due to not achieving perfect initial alignment and/or the mirror not being perfectly aligned with the axis of the small ball dumbbell. The mirror error can be measured by noting where the small masses hit the case (see Appendix IV, step 1). For the two units I checked, the angles were 0.70 and 0.90.  Figure 16: Angles
Here the mirror is rotated by θ₀ from the ideal position shown in black to the positions shown in blue. The laser beam comes in along the red vertical line and goes out along the red line. When the mirror rotates to the right the normal shifts by θ to the dotted green line and the laser to the solid green line. When the mirror rotates to the left the normal shifts θ (same torque) to the left to the dotted purple line and the laser to the solid purple line. θ = θ2 + θ0 & θ = θ1 – θ0 → 2θ = θ2 + θ1 S1 + S2 = (L tan 2θ1 - L tan 2θ0) + (L tan 2θ2 + L tan 2θ0) = L tan 2θ1 + L tan 2θ2 This is the same as before provided we make the small angle approximation tan 2α = 2α. The maximum angle (hits case) is about 50 for which tan is 0.25% larger than . More realistically, assuming a mirror angle error of 10, an alignment error of 0.750 (2 cm at 150 cm) in the same direction plus the 0.670 (the shift from equilibrium when the large masses are shifted), the angles should be less than 2.50 and the using the small angle approximation makes our value for G about 0.06% low. Actually only about half this since if one of the above angles is 2.50, the other is only 1.20 (since shifts in the opposite direction). Correction < +0.03% 4. Unless the initial alignment is perfect, the small balls are not initially centered in the case. We assume that the small mass dumbbell is 0.380 off of the center of the case (laser spot off 2 cm at 150 cm). 
Figure 17: Motion of m Figure 17 shows the ideal case where the small masses are initially centered in the case and then Δb = (d/2) sin θ = (10.1 cm/2) sin 0.3500 = 0.0308 cm If the initial angle is off by 0.760 (laser spot off 2 cm at 10 cm), the masses are displaced by b0 where b0 = (d/2)sin 0.760 = 0.033 cm, and the m to M distance becomes (b – [b + b0]) for one position of the large masses and (b – [b – b0]) for the other position. The forces in the two positions then become: F1 = GmM/(b-[Δb+b0])² F2 = GmM/(b-[Δb-b0])² 1 = 2F1d = kθ1 τ2 = 2F2d = kθ2 2d(F1+F2) = k(θ1+θ2) = k(2θ) Thus the angle that the small masses swing through is directly proportional to the sum of F1+F2. Let F be the force when b0 is zero (perfect alignment), then (F1+F2)/(F+F) = [(b-[Δb])² + (b-[Δb])²]/[(b-[Δb+b0])² + (b-[Δb-b0])²] = [2(1-[Δb]/b)²]/[(1-[Δb+b0]/b)² + (1-[Δb-b0]/b)²] (Δb = 0.0308 cm, b = 4.27 cm) = [2(1-0.0308/4.27)²]/[(1-[0.031+0.033]/4.27)² + (1-[0.031-0.033]/4.27)²] = 1.0012 So the G value is too large by 0.12%. Correction due to b0 < -0.1 5. The angle between M and m tilts slightly as m moves. Refer to Figure 17. Δb = (d/2) sin θ = 0.0308 cm b₁ = b - Δb = 4.27(1) cm - 0.03 cm = 4.24(2) cm c = d - d cos θ = 9.4 x 10-5 cm new distance = (b₁² + c²)1/2 = b₁ (we already assumed that) α = tan-1(c/b₁) = 1.3x10-3 degrees perpendicular component of F = F cos = F cos 0.00130= F, so we can ignore the change in angle. Correction = 0 6. The mass of the aluminum plates on the side of the case attracts the small balls. Assume the small mass m moves from halfway between the plates to a point b to the right. The right hand plate will then exert a force F+ which is slightly greater than the force exerted by the left hand plate, F-. We start by treating the plates as circles of radius R. df is the differential force exerted on m by the mass element dm. dF is the differential force exerted on m by the mass, dM, of the ring of radius r and lies along the axis of the plates. The plates are separated by 2w. 
df cos = (Gm dm /[r2 + (w-b)2])([b-b]/ [r2 + (w-b)2]1/2) dF = df cos = Gm dm (w-b)/[r2 + (w-b)2]3/2) where the integral is evaluated around the ring of radius r. dF = Gm dM (w-b)/[r2 + (w-b)2]3/2) = Gm(2rdr)(w-b)/[r2 + (w-b)2]3/2) where is the mass/area of the aluminum plate. = (2.7 g/cm3)(0.20 cm) = 0.54 g/cm2 F+ = dF = 2Gm(w-b)rdr/[r2 + (w-b)2]-3/2 from r=0 to r=R. F+ = 2Gm(1 - (w-b)/[r2 + (w-b)2]1/2) Similiarly F- = 2Gm(1 - (w+b)/[r2 + (w+b)2]1/2) Fnet /F = (F+ - F-)/F = 2(b2/M)((w+b)/[r2 + (w+b)2]1/2 - (w-b)/[r2 + (w-b)2]1/2) where F is the force between m and M. The maximum circle that could be cut from the plate (centered on m) would have R = 2.5 cm. For this value of R Fnet /F = 0.0007 This is actually an upper bound since the function Fnet /F is maximum for R = 2.1 cm. This may seem counterintuitive, but consider if the plates were really large. Then the force would be independent of the distance from the plate. Correction < 0.07%. The motion is not truly simple harmonic. It is clear from the goodness of fit of the damped sine wave to the data that the motion is very nearly simple harmonic, except that it is damped. This changes the frequency slightly. For the damped harmonic case: ω = [ω₀² - B²]1/2 = [(0.01289/s)² - (~3x10-4/s)²]1/2 = 0.0128865/s Correction = 0 Written by Chuck Hunt 
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rient the Gravitational Torsion Balance so the mirror on the pendulum bob faces a screen or wall that is at least 1.5 meters away.
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