Circular Motion and Gravitation:

Velocity = distance/time, so therefore - v = 2  r / T Topic 6: Circular motion and gravitation 6.1 – Circular motion a c = v 2 / r a c and F c (all three forms) a c = 4  2 r / T 2 a c = r  2 F c = mv 2 / r F c = 4  2 mr / T 2 F c = m  2 r v = r  relation between v and   =  / t (rad s -1 )

PRACTICE: Dobson is watching a 16-pound bowling ball being swung around at 50 m/s by Arnold. If the string is cut at the instant the ball is next to the ice cream, what will the ball do? (a) It will follow path A and strike Dobson's ice cream. (b) It will fly outward along curve path B. (c) It will fly tangent to the original circular path along C. Solving centripetal acceleration and force problems Topic 6: Circular motion and gravitation 6.1 – Circular motion B A C

EXAMPLE: Suppose a kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). Given that the earth has a radius of R E = m, find the speed of the ball. SOLUTION: F c = mv 2 / r The ball is traveling in a circle of radius r = m.  F c is caused by the weight of the ball so that F c = mg = (0.5)(10) = 5 n.  Since F c = mv 2 / r we have 5 = (0.5)v 2 / v = 8000 m s -1 ! Solving centripetal acceleration and force problems Topic 6: Circular motion and gravitation 6.1 – Circular motion

EXAMPLE: Suppose a kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). How long will it take the ball to return to Everest? SOLUTION: We want to find the period T.  We know that v = 8000 m s -1.  We also know that r = m.  Since v = 2  r / T (velocity = distance/time) we have: T = 2  r / v T = 2  ( )/ 8000 = (5030 s)(1 h / 3600 s) = 1.40 h. Solving centripetal acceleration and force problems Topic 6: Circular motion and gravitation 6.1 – Circular motion

EXAMPLE: Explain how an object can remain in orbit yet always be falling. SOLUTION:  Throw the ball at progressively larger speeds.  In all instances the force of gravity will draw the ball toward the center of the earth.  When the ball is finally thrown at a great enough speed, the curvature of the ball’s path will match the curvature of the earth’s surface.  The ball is effectively falling around the earth! Solving centripetal acceleration and force problems Topic 6: Circular motion and gravitation 6.1 – Circular motion

Orbital speed At any given altitude, there is only one speed for a stable circular orbit. From geometry, we can calculate what this orbital speed must be. At the earth’s surface, if an object moves 8000 meters horizontally, the surface of the earth will drop by 5 meters vertically. That is how far the object will fall vertically in one second (use the 1 st kinematic equation to show this). Therefore, an object moving at 8000 m/s will never reach the earth’s surface.

Solving centripetal acceleration and force problems Topic 6: Circular motion and gravitation 6.1 – Circular motion PRACTICE: Find the angular speed of the minute hand of a clock, and the rotation of the earth in one day. SOLUTION:  The minute hand takes 1 hour to go around one time. Thus  = 2  / T = 2  / 3600 s = rad s -1.  The earth takes 24 h for each revolution so that  = 2  / T = ( 2  / 24 h )( 1 h / 3600 s ) = rad s -1.  This small angular speed is why we can’t really feel the earth as it spins.

Essential idea: The Newtonian idea of gravitational force acting between two spherical bodies and the laws of mechanics create a model that can be used to calculate the motion of planets. F = GMm / r 2 - Newton’s Law of gravitation g = F / m – gravitational field strength g = GM / r 2 – gravitational field strength and gravitational constant Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation

Understandings: Newton’s law of gravitation Gravitational field strength Applications and skills: Describing the relationship between gravitational force and centripetal force Applying Newton’s law of gravitation to the motion of an object in circular orbit around a point mass Solving problems involving gravitational force, gravitational field strength, orbital speed and orbital period Determining the resultant gravitational field strength due to two bodies Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation

Newton’s law of gravitation  The gravitational force is the weakest of the four fundamental forces: Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation GRAVITY STRONG ELECTROMAGNETIC WEAK + + nuclear force light, heat, charge and magnets radioactivityfreefall, orbits ELECTRO-WEAK WEAKEST STRONGEST

The Force of Gravity F g = Gm 1 m 2 /r 2 –F g : Force due to gravity (N) –G: Universal gravitational constant 6.67 x N m 2 /kg 2 –m 1 and m 2 : the two masses (kg) –r: the distance between the centers of the masses (m) The Universal Law of Gravity ALWAYS works, whereas F = mg only works sometimes.

EXAMPLE: The earth has a mass of M = 5.98  kg and the moon has a mass of m = 7.36  kg. The mean distance between the earth and the moon is 3.82  10 8 m. What is the gravitational force between them? SOLUTION: Use F = GMm / r 2. F = (6.67×10 −11 )(5.98  )(7.36  ) / (3.82  10 8 ) 2 F = 2.01  N. Solving problems involving gravitational force  Be very clear that r is the distance between the centers of the masses. Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation m1m1 m2m2 r F 12 F 21 FYI  The radius of each mass is immaterial.

FYI  For circular orbits, the gravitational force is the centripetal force. Thus F C = F G. EXAMPLE: The moon has a mass of m = 7.36  kg. The mean distance between the earth and the moon is 3.82  10 8 m. What is the speed of the moon in its orbit about earth? SOLUTION: Use F C = F G = mv 2 / r.  From the previous slide F G = 2.01  N. Then 2.01  = ( 7.36  ) v 2 / 3.82  10 8  Then v = 1.02  10 3 ms -1. Solving problems involving gravitational force Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation

FYI  For circular orbits, the gravitational force is the centripetal force. Thus F C = F G. EXAMPLE: The moon has a mass of m = 7.36  kg. The mean distance between the earth and the moon is 3.82  10 8 m. What is the period of the moon (in days) in its orbit about earth? SOLUTION: Use v = d / t = 2  r / T.  From the previous slide v = 1.02  10 3 ms -1. Then T = 2  r / v = 2  ( 3.82  10 8 ) / 1.02  10 3 = (2.35  10 6 s)(1 h / 3600 s)(1 d / 24 h) = 27.2 d. Solving problems involving gravitational force Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation

Solving problems involving gravitational field strength Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation g = GM / R 2 gravitational field strength at surface of a planet of mass M and radius R g = GM / r 2 gravitational field strength at distance r from center of a planet of mass M

Solving problems involving gravitational field strength Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation PRACTICE: The mass of the earth is M = 5.98  kg and the radius of the earth is R = 6.37  10 6 m. Find the gravitational field strength at the surface of the earth, and at a distance of one earth radius above its surface. SOLUTION:  For r = R: g = GM / R 2 g = (6.67×10 −11 )(5.98  )/(6.37  10 6 ) 2 g = 9.83 N kg -1 (m s -2 ).  For r = 2R: Since r is squared…just divide by 2 2 = 4. Thus g = 9.83 / 4 = 2.46 m s -2.

Solving problems involving gravitational field strength Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation PRACTICE: A 525-kg satellite is launched from the earth’s surface to a height of one earth radius above the surface. What is its weight (a) at the surface, and (b) at altitude? SOLUTION: Use information from the previous slide: (a) AT SURFACE: g surface = 9.83 m s -2.  Then from F = mg we get F = (525)(9.83) = 5160 N. (b) AT ALTITUDE: g surface+R = 2.46 m s -2.  Then from F = mg we get F = (525)(2.46) = 1290 N.

Gravitational field strength  Compare the gravitational force formula F = GMm / r 2 (Force – action at a distance) with the gravitational field formula g = GM / r 2 (Field – local curvature of space)  Note that the force formula has two masses, and the force is the result of their interaction at a distance r.  Note that the field formula has just one mass – namely the mass that “sets up” the local field in the space surrounding it. It “curves” it.  The field view of the universe (spatial disruption by a single mass) is currently preferred over the force view (action at a distance) as the next slides will show. Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation

Gravitational field strength  The gravitational field distorts the space around the mass that is causing it so that any other mass placed at any position in the field will “know” how to respond immediately.  Think of space as a stretched rubber sheet – like a drum head.  Bigger masses “curve” the rubber sheet more than smaller masses.  The next slide illustrates this gravitational “curvature” of the space surrounding, for example, the sun. Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation

Gravitational field strength  Note that each mass “feels” a different “slope” and must travel at a particular speed to stay in orbit. Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation FYI  The field view eliminates the need for long distance signaling between two masses. Rather, it distorts the space about one mass.

Gravitational field strength  Drawing the the gravitational field - simplified:  In fact, we don’t even have to draw the sun – the arrows are sufficient to denote its presence.  To simplify field drawings even more, we take the convention of drawing “field lines” as a single arrow. Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation SUN

Gravitational field strength  In the first sketch the strength of the field at a point is determined by the length of the field arrows in the vicinity of that point.  The second sketch has single arrows, so how do we know how strong the field is at a particular point in the vicinity of a mass?  We simply look at the concentration of the field lines. The closer together the field lines, the stronger the field.  In the red region the field lines are closer together than in the green region.  Thus the red field is stronger than the green field. Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation SUN

PRACTICE: Sketch the gravitational field about the earth (a) as viewed from far away, and (b) as viewed “locally” (at the surface). SOLUTION: (a) FYI  Note that the closer to the surface we are, the more uniform the field concentration. Solving problems involving gravitational field strength Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation (b) or

Solving problems involving gravitational field strength Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation EXAMPLE: Find the gravitational field strength at a point between the earth and the moon that is right between their centers. SOLUTION:  Make a sketch.  Note that r = d / 2 = 3.82  10 8 / 2 = 1.91  10 8 m.  g m = Gm / r 2 g m = (6.67× )(7.36  )/(1.91  10 8 ) 2 = 1.35  N.  g M = GM / r 2 g M = (6.67× )(5.98  )/(1.91  10 8 ) 2 = 1.09  N.  Finally, g = g M – g m = 1.08  N,→. M = 5.98  kg m = 7.36  kg gMgM gmgm d = 3.82  10 8 m

PRACTICE: Jupiter’s gravitational field strength at its surface is 25 N kg -1 while its radius is 7.1  10 7 m. (b) Using the given information and g = GM / R 2 calculate Jupiter’s mass. (c) Find the weight of a 65-kg man on Jupiter. SOLUTION: (b) g = GM / R 2 M = gR 2 / G(manipulation) M = (25)(7.1  10 7 ) 2 / 6.67×10 −11 = 1.9×10 27 kg. (c) F = mg F = 65(25) = 1600 N. Solving problems involving gravitational field strength Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation

Applications: Kepler’s law states that the period T of an object in a circular orbit about a body of mass M is given by T 2 = [ (4  2 / (GM) ]r 3. Solving problems involving orbital period Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation

EXAMPLE: A satellite in geosynchronous orbit takes 24 hours to orbit the earth. Thus, it can be above the same point of the earth’s surface at all times, if desired. Find the necessary orbital radius, and express it in terms of earth radii. R E = 6.37  10 6 m. SOLUTION: T = (24 h)(3600 s h -1 ) = s.  Then from Kepler’s law T 2 = [ 4  2 / (GM) ]r 3 we have r 3 = T 2 / [ 4  2 / (GM) ] r 3 = / [ 4  2 / (6.67   5.98  ) ] = 7.54  r = ( m)(1 R E / 6.37  10 6 m) = 6.63 R E. Solving problems involving orbital period Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation

PRACTICE: If the elevator is accelerating upward at 2 ms -2, what will Dobson observe the dropped ball’s acceleration to be? SOLUTION:  Since the elevator is accelerating upward at 2 ms -2 to meet the ball which is accelerating downward at 10 ms - 2, Dobson would observe an acceleration of 12 ms -2.  If the elevator were accelerating downward at 2, he would observe an acceleration of 8 ms -2. Solving problems involving gravitational field  Consider Dobson inside an elevator which is not moving…  If he drops a ball, it will accelerate downward at 10 ms -2 as expected. Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation

PRACTICE: If the elevator were to accelerate downward at 10 ms -2, what would Dobson observe the dropped ball’s acceleration to be? SOLUTION:  He would observe the acceleration of the ball to be zero!  He would think that the ball was “weightless!” FYI  The ball is NOT weightless, obviously. It is merely accelerating at the same rate as Dobson!  How could you get Dobson to accelerate downward at 10 ms -2 ? Solving problems involving gravitational field Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation Cut the cable!