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Transcript
Rotational Motion Reading: pp. 194 – 203 (sections 8.1 – 8.3) HW #1 p. 217, question #1 p. 219, problem #1, 4, 5, 6, 15, 16, 17, 19 I. Introduction: A rotating object is one that spins on a fixed axis. The position and direction of the rotation axis will remain constant. The position of some part of the object can be specified with standard cartesian coordinates, (x,y). All objects will be assumed to rotate in a circular path of constant radius. y Both coordinates, (x, y), change over time as the object rotates. (x, y) R The object’s position can also be specified with polar coordinates, (r, q). q x For this last coordinate system, only the angle changes. The radius stays constant. II. Definitions: Since the polar coordinates only have one changing variable, the angle, we will use this to simplify analysis of motion. position is defined as where an object is in space. Here, we A. The ___________ only need to specify the angle of the object with respect to some origin or reference line. The position of the object is measured by an angle, q, measured counterclockwise from the positive x – axis. The angle will be measured in radians rather than units of ___________ q degrees . ___________ Radians are defined as the ratio of the length along the arc of a circle to the position of an object divided by the radius of the circle. s q R R s q s = length along the arc measured counterclockwise from the +x – axis. R = radius of the circular path. Since q is the ratio of two lengths, the angle measurement really does not have any units. The term “radians” is just used to specify how the angle is measured. 1 revolution = 360 degrees = 2p radians displacement of the object is just the difference in its position. B. The _____________ angular displacement of the object as the difference in We define the ____________________ its angular position. q q f q i q q o A positive q shows a counterclockwise {ccw} rotation, while a negative q shows a clockwise {cw} rotation. velocity C. Motion can also be measured through a rate of rotation, a __________. angular velocity The ____________________ of an object is defined as the amount of rotation of an object per time. The angular velocity is represented by the greek letter w (lower case omega). average angular velocity: t = elapsed time. q q q o w t t rad units s D. Another measure of motion is the rate of change of velocity, called an acceleration ________________. angular acceleration of an object is defined as the amount of The ____________________ change of the angular velocity of an object per time. The angular acceleration is represented by the greek letter a (lower case alpha). average angular acceleration: w w wo a t t rad units 2 s In general, the motion may be complex, but we will again look at constant angular acceleration cases, exactly the same way as we did back in Ch. 2. The same equations of motion can be derived for circular motion. Ch. 2 Ch. 8 Linear Motion Rotational Motion v vo at w wo a t x vot 12 at 2 q wot 12 a t 2 v vo 2ax w wo 2aq 2 2 v vo x 2 t This is only true for constant accelerations. Problem solving is the same as before. 2 2 w wo 2 q t E. The motion of an object around a circle can also be represented as actual distances along the circle and speeds tangent to the circle. R vt tangential speed , v , of an object is defined as the angular The _________________ t velocity times the radius of the circle. vt Rw The tangential speed measures the actual speed of the object as it travels around the circle. length along arc average vt = elapsed time tangential acceleration The _________________________ , at, of an object is defined as the angular acceleration times the radius of the circle. at Ra The tangential acceleration measures how the tangential speed increases or decreases over time. Example #1: A disk rotates from rest to an angular speed of 78.00 rpm in a time of 1.300 seconds. a. What is the angular acceleration of the disk? wo = 0, w = 78.00 revolutions per minute, t = 1.300 seconds. a = ? w wo at w 78.00 a rev min 2p rad 1 min 8.168 rad s 1 rev 60 sec w wo t 8.168 rad s 0 6.283 rad s 2 1.300 s b. Through what angle does the disk turn? q wot 12 at 2 q 0 rad s 1.300 s 6.283 1 2 q 5.309 rad rad 1.300 s 2 s2 c. Through what angle will the disk turn if it were to maintain the same angular acceleration up to 254.0 rad/s? w wo 2aq 2 2 w 2 wo 2 254.0 rad s 2 02 q 2a 26.283 rad s 2 q 5134 rad d. The disk has a diameter of 12.00 inches. What is the tangential speed and acceleration at the edge of the disk the moment the disk reaches 78.00 rpm? 12.00 in 2.54 cm 1m 0.1524 m R 2 1in 100 cm vt Rw 0.1524 m8.168 rad s 1.245 m s at Ra 0.1524 m 6.283 rad s 2 0.9576 m s 2 Centripetal Force HW: See Schedule III. Centripetal Acceleration and Force. When an object moves in a circle, the direction of its velocity is always changing. This means the object is always accelerating! For an object rotating at a constant angular speed, the acceleration of the mass is always towards the center of the motion. This kind of acceleration is called centripetal {“center seeking”} acceleration, and is represented as ac. a ___________ The amount of acceleration depends on the radius of the circular path and the speed around the circle. v2 ac w 2r r w = angular speed, v = tangential speed, r = radius of circular path. Example #2: A wheel of a car has a diameter of 32.0 inches. A rock is wedged into the grooves of the tire. a. What is the centripetal acceleration on the rock if the wheel turns a rate equal to 70.0 mph? v 70.0 mile hour 1609.344 m 1 hour 31.29 m s 1 mile 3600 s 32.0 in 2.54 cm 1 m 0.4064 m r 2 1in 100 cm v 31.29 s ac 2410 m s 2 r 0.4064 m 2 m 2 b. What is the rotation rate of the tire, in rpm? v 31.29 m s w 77.0 rad s r 0.4064 m w 77.0 rad 1 rev 60 s s 2p rad 1 min 735 rpm If there is a centripetal acceleration making an object move in a circle, then there must be an unbalanced force creating this acceleration. centripetal force, and it also points towards This force is called the _____________ the center of the circular motion. This force must be made by real forces acting on an object. The centripetal force will be the sum of the radial components of the forces acting on the object. A radial component points center towards the ___________ of the circle. mv2 Fc mac mw 2 r r Example #3: A 0.475 kg mass is tied to the end of a 0.750 meter string and the mass is spun in a horizontal circle. If the mass makes 22.0 revolutions in a time of 2.50 seconds, what is the tension in the string holding the mass to the circular motion? rotation R T q 22.0 rev 2p rad 55.3 rad s w t 2.50 s 1 rev Fc mw 2 r 0.475 kg55.3 rad s 0.750 m 2 Fc 1090 N m Centripetal Force HW #2: See Schedule Last Chance for Make-up Tests! Wednesday after school! Example #4: A penny sits at the edge of a 12.00 inch diameter record. If the coefficient of static friction is 0.222 between the penny and the record, what is the maximum rotation rate of the record that will allow the penny to remain on the record? n rotation vertical forces balance. n mg Fs by definition friction is: mg Fs ,max s n Fs ,max s mg set the friction force equal to the centripetal force w s g r Fc mw 2 r Fs ,max s mg 0.2229.80 m s 2 6.00 in 0.0254 m 1in w 3.78 rad s Example #5: A mass m on a frictionless table is attached to a hanging mass M by a cord through a hole in the table. Find the speed which m must move in order for M to stay at rest. Evaluate the speed for m = 2.00 kg, M = 15.0 kg, and r = 0.863 m. Since M is at rest, the tension force lifting it is equal to the weight of M: T Mg This tension is also the centripetal force on the mass m, causing it to spin in a circular path: mv2 T r Set the two equations equal to one another: mv 2 T Mg r v v rMg m 0.863 m15.0 kg 9.80 m s 2.00 kg 2 v 7.96 ms Example #6: A common amusement park ride involves a spinning cylinder with a floor that drops away. When a high enough rotation speed is achieved, the people in the ride will stay on the side of the wall. A static friction force holds each person up. Solve for the rotation rate of the room, given the coefficient of friction for the wall and the radius of the room. Example #6: A common amusement park ride involves a spinning cylinder with a floor that drops away. When a high enough rotation speed is achieved, the people in the ride will stay on the side of the wall. A static friction force holds each person up. Solve for the rotation rate of the room, given the coefficient of friction for the wall and the radius of the room. set the friction force equal to the weight as the vertical forces balance. Fs ,max mg Fs n m mg Fs ,max s n forces balance definition of force of static friction Horizontal direction: set the normal force equal to the centripetal force. n Fc mw 2 r Combine all the information to solve for the rotation rate, w. mw r n 2 w 2 Fs , max g s mg s g w s r sr What would be the rotation rate for a room 3.00 m wide and a carpeted wall with a coefficient of friction of 0.750? w 9.80 m s2 0.7501.50 m 2.95 rad s 28.2 rpm Example #7: (Banking Angle) Determine the angle of the roadway necessary for a car to travel around the curve without relying on friction. Assume the speed of the car and the radius of the curve are given. n q q mg n cosq component of the normal force that is vertical n sin q component of the normal force that is horizontal, also becomes the centripetal force balance the vertical forces: n cos q mg set the net horizontal force equal to the centripetal force, with towards the center of the circular path as the positive direction: mv2 n sin q R substitute in: mg n cos q mg mv2 sin q cos q R v 2 Rg tan q 2 v q tan 1 Rg Example #8: Conical Pendulum. A mass of m = 1.5 kg is tied to the end of a cord whose length is L = 1.7 m. The mass whirls around a horizontal circle at a constant speed v. The cord makes an angle q = 36.9o. As the bob swings around in a circle, the cord sweeps out the surface of a cone. Find the speed v and the period of rotation T of the pendulum bob. T cos q mg mv2 T sin q R divide… T sin q mv2 T cos q mgR note: R sin q L simplify… 2 v tan q gR v v v Rg tan q L sin q g tan q 1.7 msin 36.9o 9.80 m s tan 36.9o 2 v 2.74 ms the period, T, is the time for one revolution: 2p R v T 2p R 2p L sin q T v v 2p 1.7 m sin 36.9o T 2.74 ms T 2.34 s Example #9: Another common amusement park ride is a rollercoaster with a loop. Determine the minimum speed at the top of the loop needed to pass through the top of the loop. There are two forces acting on the car: the weight pulling straight downwards and the normal force pushing perpendicular to the track. As the car goes through the loop, the normal force always points towards the center of the loop. The vector sum of the normal force and the radial component of the weight equals the centripetal force. The net force for the mass at the top of the loop is: m 2 mv Fc Fnet n mg r The faster the car goes, the greater the normal force to push the car into a circular path. The minimum speed for the car at the top of the loop is where the normal force goes to zero. 2 mvmin mg r vmin rg n mg r Example #9: (b) Use energy conservation to find the speed of the mass at the bottom of the loop. Etop 12 mv 2 min Etop KE PE mg2r 2 Ebottom 12 mvbot 0 Ebottom KE PE Set the two energies equal: 1 2 2 2 mvbot 12 mvmin mg 2r Divide by m and multiply by 2: 2 2 vbot vmin 4 gr rg 4rg 5rg vbot 5rg Example #9: (c) What is the necessary starting height of the mass if sliding from rest down the ramp? Etopramp KE PE Etoploop KE PE Etop ramp 0 mgh 2 Etoploop 12 mvmin mg2r Set the two energies equal: 2 mgh 12 mvmin 2mgr 12 mrg 2mgr 52 mgr Divide both sides by m and g: h 52 r 54 d AP Summary Problems! Example #10: {p. 135, problem #93} A circular curve of radius R in a new highway is designed so that a car traveling at speed vo can negotiate the turn safely on glare ice (zero friction). If the car travels too slowly, then it will slip toward the center of the circle. If it travels too fast, then it will slip away from the center of the circle. If the coefficient of static friction is increased (from zero), a car can stay on the road while traveling at any speed within a range from vmin to vmax. Derive formulas for vmin and vmax as functions of s, vo, g and R. Start with the too fast scenario. The car will want to slide up the ramp, so friction will point to the bottom of the ramp. n sin q F f cos q F f sin q Find horizontal and vertical components of the forces: n cosq Vertical components of the forces balance: n cos q mg Ff sin q Definition of friction: Ff s n n cos q mg s n sin q mg n cos q s sin q mg n cos q s sin q Horizontal components of the forces make the centripetal force. Take towards the left as the positive direction: mv n sin q Ff cos q r 2 mv 2 n sin q s n cos q r mv 2 n sin q s cos q r Ff s n mg mv 2 sin q s cos q r cos q s sin q sin q s cos q v rg cos q s sin q 2 Next is the too slow scenario. The car will want to slide down the ramp, so friction will point to the top of the ramp. n sin q n cosq F f sin q Find horizontal and vertical components of the forces: Vertical components of the forces balance: n cos q Ff sin q mg Definition of friction: F f cos q Ff s n n cos q s n sin q mg mg n cos q s sin q mg n cos q s sin q Horizontal components of the forces make the centripetal force. Take towards the left as the positive direction: mv 2 n sin q Ff cos q r mv 2 n sin q s n cos q r mv 2 n sin q s cos q r Ff s n mg mv 2 sin q s cos q r cos q s sin q sin q s cos q v rg cos q s sin q 2 Example #11: A mass is tied to the end of a string and spun in a vertical circle. What is the difference in tension in the string between top and bottom of the circle? mvtop 2 mg r Ttop mg Ttop mvbottom 2 Tbottom mg r energy conservation: Tbottom 1 2 mg mvbottom2 12 mvtop 2 mg 2r mvbottom 2 Tbottom mg r mvtop 2 r Tbottom mg Ttop mvbottom 2 mg r Ttop mvtop 2 r mg subtract: Tbottom Ttop Tbottom Ttop 2 mvbottom mg r mvbottom 2 mvtop 2 r mvtop 2 2mg r mg now use energy conservation to eliminate speeds Tbottom Ttop 1 2 mvbottom 2 mvtop 2 r 2mg mvbottom2 12 mvtop 2 mg 2r mvbottom2 mvtop 2 2mg 2r Tbottom Ttop 4mgr 2mg r Tbottom Ttop 6mg Centripetal Force HW check tomorrow! Example #12: Sonam is seated on the top of a frictionless hemispherical mound of ice of radius R. Why she is there, I do not know. A small breeze upsets the equilibrium and she starts sliding down the ice. At what vertical height from the ground does she leave the ice surface? Give the answer in terms of R. When Sonam is at the side, as shown, the sum of the radial components gives the centripetal force: mv mg cos q n R 0 Sonam leaves the surface when the normal force becomes zero. mg cos q h = R cos q 2 n mv 2 mgR cos q mv 2 mgh q mg Now use energy conservation to find the speed of Sonam if she starts at rest at the top. Etop mgR Eside 12 mvside 2 mgh From the previous slide: mv 2 mgh mgR 12 mgh mgh 32 mgh h 23 R Newton’s Universal Law of Gravity HW #3 on schedule Turn in Rotary Motion Lab Newton’s Law of Universal Gravity: Any two objects are attracted to each other through the force of gravity. The force is proportional to each mass. F m1 F m2 The force is inversely proportional to the square of the center to center distance between the masses. 1 F 2 r “inverse square law” These two equations can be combined into a single equation, and a proportionality constant can be introduced to make an equality: Gm1m2 F r2 2 By Newton’s 3rd Law, N m 11 G 6 . 67384 10 forces are equal and kg 2 opposite! Example #1: Two students, each with mass 70 kg, sit 80 cm apart. (a) What is the force of attraction between the two students? m1 m2 70 kg r 80 cm 0.80 m Gm1m2 F r2 2 N m 11 6.67384 10 70 kg 70 kg 2 kg 2 0.80 m F 5.110 7 N (b) How does this compare to the weight of the students? weight m1 g 70 kg 9.80 m s 2 weight 686 N 7 F 5.110 N 10 7 10 weight 686 N The force between you and your neighbor is less than 1 part in a billion of your normal body weight. Example #2: What is the direction of the net force on the mass at the center of the image? Why? All forces balance out pairwise except for one: Force on central mass points upwards. Example #3: What is the direction of the net force on the mass at the center of the image? Why? All forces balance out pairwise except for one: Force on central mass points towards the left. II. Compare the force of gravity to weight: The weight of an object near the surface of the Earth is just the force of gravity exerted on the object by the Earth. GM E m mg 2 RE Remember, r is the center to center (of the earth) distance. M E 5.98 1024 kg RE 6.378 106 m The mass of the object cancels out, and what is left is a theoretical value for the acceleration due to gravity near the surface of the Earth. GM E g 2 RE Example #4: Estimate the value of ‘g’ for the Earth. GM E g 2 RE 2 N m 11 24 6.67384 10 5.98 10 kg 2 kg 2 6 6.378 10 m g 9.81 m s 2 Example #5: Estimate the value of ‘g’ near the surface of Peter Griffin. Example #5: Estimate the value of ‘g’ near the surface of Peter Griffin. width 1.4 m Estimate volume as sphere 1.4 m in diameter. 4p 3 volume 0.70 m 3 1.44 m3 Multiply by density to get mass: m 900 m3 1.44 m 1293 kg kg 3 2800lbs! Estimate ‘g’ at surface: GM g 2 R 2 N m 11 6.67384 10 1293 kg 2 kg 2 0.70 m g 1.8 10 7 m s2 III. Satellite Motion. A satellite is an object that orbits (travels in a circular path around) some gravity source The force of gravity provides the centripetal force to keep the satellite moving in a circular path. The equation of motion is: 2 mv GMm 2 r r m = satellite mass (divides out…) M = mass of the gravity source r = radius of the orbit (center to center distance!) v = the tangential speed of the orbiting object Example #6: The space shuttle orbits the Earth at an altitude of 400 km above the surface of the Earth. (a) Determine the speed of the space station around the Earth. mv2 GMm 2 r r r 6.378 106 m 4.00 105 m 2 N m 11 24 6.67384 10 5 . 98 10 kg 2 GM kg v 6 5 r 6.378 10 m 4.00 10 m v 7,673 ms (b) How much time will it take for the space station to orbit the Earth? dist 2p r T time rate v 2p 6.778 106 m T 7673 ms T 5550 s 92.5 min Example #7: Write an equation that gives the period (time for one revolution) of an orbit in terms of the radius of the orbit. 2p r v T 2 mv GMm 2 r r GM 2p r 2 v r T 2 r 3 GM 2 2 T 4p Example #8: Determine the altitude (height above the surface) for a geosynchronous satellite. The period of the satellite matches the length of the Earth’s day, 24 hours. 3600 s 8.64 104 s T 24 h 1h GM 2 T r T GM 2 4p 2p 2 3 11 N m 3 r 6.67384 10 2 kg 2 8.64 10 s 24 5.98 10 kg 2p 4 2 11 N m r 3 6.67384 10 2 kg 2 8.64 10 s 24 5.98 10 kg 2p 4 r 4.23 107 m This is the height from the center of the Earth. Subtract the radius of the Earth to get the height from the surface. 7 6 h 4.23 10 m 6.378 10 m h 3.59 10 m 7 2 Example #9: {on your own} Determine the orbital velocity of Grog’s golf ball. r REarth 6.378 106 m Example #9: Determine the orbital velocity and period of Grog’s golf ball. mv2 GMm 2 r r 2 11 N m 24 6.67384 10 5.98 10 kg 2 kg GM v 6 6.378 10 m r v 7,910 ms Example #10: What is your perceived weight when you stand at the equator of the Earth? n Example #10: What is your perceived weight when you stand at the equator of the Earth? mw 2 r mg n m mg Your perceived weight is the normal force pushing on your feet to support you. r n mg mw r 2 Some of the true gravity force is used to make you move in a circle. The remainder of that force is what you feel as your weight. The percentage your weight is reduced is: mw 2 r mg 2 2p rad 6 6.378 10 m s 24 h 3600 h 9.80 m s2 0.003 0.3% Example #11: {on your own!} What is maximum rotation rate for the Earth so that objects at the equator will stay on the surface and not “fly off”? Hint: The normal force goes to zero at this maximum rotation rate. The objects are essentially in a low orbit. mw r mg n 2 w 0 9.80 m s2 g 3 rad 1.24 10 s 6 r 6.378 10 m T 2p w 5070 s 84.5 min Potential Energy and Newton’s Law of Gravity ? Newton’s Law of Universal Gravity: Any two objects are attracted to each other through the force of gravity. Gm1m2 F r2 2 N m 11 G 6.67384 10 kg 2 This force also stores energy between the two masses. The potential energy stored in the two masses is: Gm1m2 PE r Example #12: What is the minimum speed that a spaceship at the surface of the Earth must have to completely escape Earth’s gravity? m m Total mechanical energy is conserved. Esurface KE PE 1 2 E far away 0 mvsurface PE mv far away 2 1 2 1 2 mvsurface 2 0 KE PE 2 The least speed at the surface corresponds to the least speed far away. GM Earth m 0 r 1 2 mvsurface vsurface vsurface 2 GM Earth m 0 r 2GM Earth r 2 N m 11 24 2 6.67384 10 5.98 10 kg 2 kg 6 6.378 10 m vsurface 1.12 10 4 m s 11.2 kms Kepler’s Laws of Satellite Motion Handout HW #4 IV. Kepler’s Laws. Johannes Kepler was a German mathematician, astronomer and astrologer. He is best known for his eponymous laws of planetary motion, which he was able to put together by painstakingly analyzing the volumes of data collected by Tycho Brahe of the planets motions in the sky. Kepler Brahe Kepler’s 1st Law: All planets (satellites) orbit the sun (or gravity source) in elliptical orbits. The sun sits at one focus of the ellipse. The planet will be closer to the sun for part of its “year”, and farther away for the other part of its “year”. Rp = perihelion = closest distance of planet to sun Ra = aphelion = farthest distance of planet to sun Kepler's first law. An ellipse is a closed curve such that the sum of the distances from any point P on the curve to two fixed points (called the foci, F1 and F2) remains constant. That is, the sum of the distances, F1P + F2P, is the same for all points on the curve. A circle is a special case of an ellipse in which the two foci coincide, at the center of the circle. Kepler’s 2nd Law: As planets orbit around the sun, an imaginary line from the planet to the sun will sweep out equal areas in equal times. Kepler's second law. The two shaded regions have equal areas. The planet moves from point 1 to point 2 in the same time as it takes to move from point 3 to point 4. Planets move fastest in that part of their orbit where they are closest to the Sun. Exaggerated scale. Potential energy from gravity still has the same behavior on this scale: The farther the planet moves from the sun, the higher the potential energy of the planet. But total energy is still constant: E KE PE const As the planet moves farther from the sun, it slows down. The closer the planet to the sun, the faster it moves in its orbit. Measurable Application: Earth’s weather is due to the tilt of the Earth’s axis relative to its orbital plane. Earth at perihelion Earth at aphelion Since summer (for the northern hemisphere) occurs when Earth is farthest from the sun, the Earth spends slightly more time in this part of the orbit as compared to winter (again, for northern hemisphere). The effect is that summer is about 3 days longer than winter. This is sometimes called the “summer analemma”. The analemma is the location of the sun in the sky at the same time each day of the year plotted on a graph or photograph. The uneven shape is due to the elliptical orbit of the Earth around the sun. Kepler’s 3rd Law: The ratio of the cube of the length of the semi major axis, to the square of the period of revolution, is constant for all planetary satellites. r3 K 2 T r 3 GM 2 T 4p 2 r = length of semi major axis. T = period of orbit. Newton later figured out how this constant fit into his gravitational scheme. Example #13: Earth’s orbital information is sometimes used as a standard: The orbital period is one year and its semimajor axis is one AU (astronomical unit). If the orbital period for Jupiter is 11.9 years, determine the semimajor axis distance for Jupiter. r 3 GM const 2 2 T 4p 3 3 aJ aE 2 2 TJ TE TJ TJ aE 2 TE TE 2 a J aE 3 2 3 11.9 y a J 1.00 AU 1.00 y aJ 5.21 AU 2 3 Example #14: Earth’s orbit has a semimajor axis length of 1.50×108 km and a period of 365.24 days. Determine the mass of the sun. r 3 GM 2 2 T 4p 4p 2 r 3 M 2 GT 4p 1.50 10 m M 2 2 11 N m h s 6.67384 10 365.24d 24 d 3600 h 2 kg 2 11 M 2.00 10 kg 30 3 Example #15: A satellite moves in a circular orbit around Earth at a speed of 5,000 m/s. Determine (a) the satellite’s altitude above the surface of Earth and (b) the period of the satellite’s orbit. mv2 GMm 2 r r GM r 2 v 2 11 N m 24 6.67384 10 5.98 10 kg 2 kg r m 2 5000 s r 1.60 10 m 7 subtract the radius of Earth to get the altitude: h 9.59 10 m 6 How much time will it take for the space station to orbit the Earth? dist 2p r T time rate v T 2p 1.596 107 m 5000 ms T 2.01104 s 334 min 5.57 hours Example #16: Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22 × 105 km. From this data, determine the mass of Jupiter. r 3 GM 2 2 T 4p M 4p 2 r 3 M 2 GT 4p 2 4.22 10 m 8 3 2 2 N m 11 h s 1.77 d 24 3600 d h 6.67384 10 2 kg M 1.90 10 kg 27 Example #17: Assume the Earth has a uniform density of r. Determine the strength of Earth’s gravitational field as a function of its radius for the interior of the Earth. density is mass per unit volume! M M 3M r 4 3 3 V p R 4 p R 3 When inside a spherically symmetric object, the pull of gravity at that point depends only on the amount of mass contained in a ball whose radius matches that location. The shell of mass on the outside of that position does not contribute to the pull of gravity. GM inside g r r2 Determine the mass located inside: M inside 3M 4 3 rV pr 3 4p R 3 3 M inside r M 3 R Plug into the formula for the acceleration due to gravity: GM inside g r r2 G r3 2M 3 r R GM 3 r R