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Tom Wilson, Department of Geology and Geography Environmental and Exploration Geophysics I tom.h.wilson Department of Geology.

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Presentation on theme: "Tom Wilson, Department of Geology and Geography Environmental and Exploration Geophysics I tom.h.wilson Department of Geology."— Presentation transcript:

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2 Tom Wilson, Department of Geology and Geography Environmental and Exploration Geophysics I tom.h.wilson Department of Geology and Geography West Virginia University Morgantown, WV Gravity Methods I

3 Tom Wilson, Department of Geology and Geography Gravity Passive source & non-invasive LaCoste Romberg GravimeterWorden Gravimeter

4 Tom Wilson, Department of Geology and Geography x spring extension m s spring mass k Young’s modulus g acceleration due to gravity Colorado School of Mines web sites - Mass and spring Pendulum measurement Hooke’s Law

5 The spring inside the gravimeter Tom Wilson, Department of Geology and Geography The spring is designed in such a way that small changes in gravity result in rather large deflections of the movable end of the beam. Early gravimeters read the mechanical movement of the spring. Today’s gravimeters use electrostatic feedback systems that hold the movable end of the beam at a fixed position between the plates of the capacitor. The voltage needed to hold the beam at a fixed position is proportional to the changes in gravity.

6 Tom Wilson, Department of Geology and Geography Newton’s Universal Law of Gravitation Newton.org.uk m1m1 m2m2 r 12 F 12 Force of gravity G Gravitational Constant

7 Tom Wilson, Department of Geology and Geography g E represents the acceleration of gravity at a particular point on the earth’s surface. The variation of g across the earth’s surface provides information about the distribution of density contrasts in the subsurface since m =  V (i.e. density x volume). m s spring mass m E mass of the earth R E radius of the earth Like apparent conductivity and resistivity g, the acceleration of gravity, is a basic physical property we measure, and from which, we infer the distribution of subsurface density contrast.

8 Tom Wilson, Department of Geology and Geography Units Most of us are familiar with the units of g as feet/sec 2 or meters/sec 2, etc. From Newton’s law of gravity g also has units of The milli Gal

9 Tom Wilson, Department of Geology and Geography Using the metric system, we usually think of g as being 9.8 meters/sec 2. This is an easy number to recall. If, however, we were on the Martian moon Phobos, g p is only about meters/sec 2. [m/sec 2 ] might not be the most useful units to use on Phobos. Some unit names used in detailed gravity applications include 9.8 m/sec Gals (or cm/sec 2 ) milli Gals (i.e. 1000th of a Gal & m/s 2 ) m/sec 2 =the gravity unit (gu) (1/10 th milliGal) We experience similar problems in geological applications, because changes of g associated with subsurface density contrasts can be quite small.

10 Tom Wilson, Department of Geology and Geography If you were to fall from a height of 100 meters on Phobos, you would hit the ground in a.10 seconds b.1 minute c.3 minutes You would hit the ground with a velocity of a.1 m/s b.5 m/s c.30 m/s How long would it take you to accelerate to that velocity on earth? a.10 seconds b.1 second c.1/10 th of a second =189s =1m/s =0.1s The velocity you would reach after jumping off a brick. 27x22x18km

11 Tom Wilson, Department of Geology and Geography 6km If you could jump up about ½ meter on earth you could probably jump up about 1.7 kilometers on Phobos. (It would be pretty hard to take a running jump on Phobos).

12 Tom Wilson, Department of Geology and Geography 6km That would give you a velocity of 4.43 m/s and on Phobos that would keep you off the surface for 26 minutes (13 up and 13 down). With a horizontal component of about 4 meters per second you’d come down on the opposite rim.

13 Tom Wilson, Department of Geology and Geography

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15 Diameter 6794 km Diameter 12,756 km 78 x 10 6 km

16 Summary relationships Tom Wilson, Department of Geology and Geography 1 milligal = 10 microns/sec 2 1 milligal equals m/sec 2 or conversely 1 m/sec 2 = 10 5 milligals. The gravity on Phobos is m/s 2 or 560 milligals. Are such small accelerations worth contemplating? Can they even be measured?

17 Spring sensitivity Tom Wilson, Department of Geology and Geography Today’s gravimeters measure changes in g in the  Gal (10 -6 cm/s 2 ) range. If spring extension in response to the Earth’s gravitational field is 1 cm, a  Gal increase in acceleration will stretch the spring by m – a length covered by 100 hydrogen atoms lined up side-by-side. The spring response in today’s modern field portable gravimeters is amplified so that detection of these small changes is possible…. for the modest price of $80,000 to $90,000

18 Tom Wilson, Department of Geology and Geography Note that the variations in g that we see associated with these large scale structures produce small but detectable anomalies that range in scale from approximately milliGals. Calculated and observed gravitational accelerations are plotted across a major structure in the Valley and Ridge Province,

19 Tom Wilson, Department of Geology and Geography We usually think of the acceleration due to gravity as being a constant m/s 2 - but as the forgoing figures suggest, this is not the case. Variations in g can be quite extreme. For example, compare the gravitational acceleration at the poles and equator. The earth is an oblate spheroid - that is, its equatorial radius is greater than its polar radius. R p = km R E = km 21.4km difference

20 Tom Wilson, Department of Geology and Geography R p = km R E = km g P = m/s 2 g E = m/s 2 This is a difference of 5186 milligals. If you weighed 200 lbs at the poles you would weigh about 1 pound less (199 lbs) at the equator. Substitute for the different values of R Difference in polar and equatorial gravity

21 Tom Wilson, Department of Geology and Geography Significant gravitational effects are also associated with earth’s topographic features. R. J. Lillie, 1999

22 Tom Wilson, Department of Geology and Geography Isostatic compensation and density distributions in the earth’s crust R. J. Lillie, 1999

23 Tom Wilson, Department of Geology and Geography Some problems to consider 1. Given that G=6.672 x m 2 kg -1 s -2, that g = 9.8 m/s 2, and that the radius of the earth is 6366km, calculate the mass of the earth. 2. At birth assume that you were delivered by an obstetrician with a mass of 75kg, and that the obstetrician’s center of mass was 0.5 meters from yours. Also assume that at that very point in time, Mars was closest to the earth or about 78 x 10 6 km from your center of mass. The mass of Mars is approximately 6.42 x kg. Determine the acceleration due to the gravitational field of the obstetrician and of Mars. Which was greater?

24 Tom Wilson, Department of Geology and Geography 3. A space traveler lands on the surface of a spherically shaped object that produces an acceleration due to gravity of m/s 2. The object has average density of 5500 kg/m 3. What is the radius of this object? How long would it take you to fall 5 meters assuming a constant g of milliGals? Due this Thursday

25 Tom Wilson, Department of Geology and Geography Start doing some background reading for the gravity lab ….

26 Tom Wilson, Department of Geology and Geography Does water flow downhill?

27 Tom Wilson, Department of Geology and Geography Keep reading Chapter 6. Look over the three problems handed out in class today. We’ll finish these up in class on Thursday Just a reminder that the gravity papers are available in the mailroom.


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