PHYS 100: Lecture 11

Transcription

1 PHYS 00: Lecture UNIVSAL GAVITATION and SPINGS F S x 0 F S x F x F F S = G S F = -k(x x 0 ) PHYS 00 Lecture, Slide

2 usic Who are the Artists? A) Whitney Houston and Tina Turner B) Nina Simone and Patti LaBelle C) tta James and Bonnie aitt D) The Dixie Cups ) arcia Ball, Irma Thomas, Tracy Nelson BB Why? Three Weeks Until Jazzfest!!!!! But who s counting? New Orleans Jazzfest Poster from year s past: The Sweet Soul Queen of New Orleans: Irma Thomas PHYS 00 Lecture, Slide

3 WHAT DID YOU FIND DIFFICULT? I find all of this confusing... please help me decipher physics :( OK We will touch on everything today!! I would like to talk more about the "heavenly bodies" and what we can use the more general gravity equations for. The new quations looks are a lot tougher to interpret especially Kepler's Third law quation in the context of gravitational force. The first job is to understand ) What is really new ) What are the consequences of the really new PHYS 00 Lecture, Slide 3

4 TH BIG IDAS NOT: TH BIG IDAS A ALWAYS GIVN IN TH LAST SLID We ll now try to create some structure here by starting with what s new PHYS 00 Lecture, Slide 4

5 What s New? There are TWO new PHYSICS things in this Prelecture The Universal Gravitational Force Law in mks units G = 6.67 X 0 - Nm /kg The Spring Force Law PHYS 00 Lecture, Slide 5

6 What s New? There is ON new ATH thing in this Prelecture The Gravitational force at the surface of the To find the gravitational force on an object at the surface of the : Add up gravitational forces from every piece of, r F r = df esult: Same as if all mass of were concentrated at the center of m F = G verything else in the prelecture is a consequence of using this new information in the context of Newton s Laws PHYS 00 Lecture, Slide 6

7 (Some of) The Consequences Newton s Leap to the Stars TH BIG IDA:. The moon orbits the. Therefore, some real force on the moon must be responsible for its acceleration. That force is the gravitational force exerted by the.. The weight of any object on is really the gravitational force exerted by the on that object. 3. Therefore, we can relate the acceleration due to gravity on to the acceleration of the moon in its orbit. m F = G F = mg Object on : oon: g = G a = G moon moon F = G moon moon g = a moon moon PHYS 00 Lecture, Slide 7

8 (Some of) The Consequences Kepler s Third Law TH BIG IDA:. All planets orbit the Sun. Therefore, some real force on each planet must be responsible for its acceleration. That force is the gravitational force exerted by the Sun.. Newton s Second Law can be used to determine a single relationship among the orbit parameters that must hold for all planets. G A word on the constant S : 4π Numerical value depends on units: in AU P in years Sm F = G Splanet a Splanet = v planet Splanet planet Splanet To get usual form of Kepler s Third Law, we just need to do some algebra: Period π P = v PHYSICS v = G planet Splanet This equation relates the speed of the planet to its orbital radius. Splanet Splanet G S = G S = 4π 3 P = 4π PHYS 00 Lecture, Slide 8 S

9 CheckPoint Which is a greater gravitational force, the force of gravity from the Sun on the or the force of gravity from the on the Sun? (A) Force of gravity from the Sun on the (B) Force of gravity from the on the Sun (C) They are equal A B C Great Job!! They are equal because of Newtons Third Law. The forces are equal and opposite. PHYS 00 Lecture, Slide 9

10 Acceleration An apple of mass m falls from rest from a tree. It hits the ground with velocity v. Just before it hits the ground, what is its acceleration? a = G (B) a (C) a = G (A) v m = a = g (D) () a = BB mg TWO COCT ANSWS!! (D) a = g We know all objects fall with acceleration g F = W = mg a = G (A) Acceleration is due to gravitational F = G force that exerts on mass m m Therefore we know g, the acceleration due to gravity on, in terms of G, the universal gravitation constant, and parameters (, ) g = G PHYS 00 Lecture, Slide 0

11 Weight Suppose the radius of the were half of what it is now, but its mass was the same. How would your weight change? (A) Decrease by factor of 4 (B) Decrease by factor of (C) Stay the Same (D) Increase by factor of () Increase by factor of 4 We know g, the acceleration due to gravity on, in terms of G, the universal gravitation constant, and parameters (, ) g = G We know your weight is determined by your mass and g: W = mg If mass of (and your mass) do not change: W Wnew = Wold = 4W old new old BB PHYS 00 Lecture, Slide

12 CheckPoint On your weight is W. Suppose you are now on Planet X whose mass is twice the mass of the s but whose diameter is also twice the s diameter. 4 BB Compare W X, your weight on Planet X to your weight on. (A) W X = ¼ W (B) W X = / W (C) W X = W (D) W X = W () W X = 4 W You said: When you double the mass of the planet, you increase the numerator by a factor of. Doubling the radius will increase the denominator by a factor of 4, thus WX = /W When you double the mass of the planet, you increase the numerator by a factor of. Doubling the radius will increase the denominator by a factor of 4, thus WX = /W The diameter of is two times that of Planet X. Thus, one's weight is times the weight of one's weight on A B C D PHYS 00 Lecture, Slide

13 CheckPoint On your weight is W. Suppose you are now on Planet X whose mass is twice the mass of the s but whose diameter is also twice the s diameter. Compare W X, your weight on Planet X to your weight on. (A) W X = ¼ W (B) W X = / W (C) W X = W (D) W X = W () W X = 4 W We know your weight is determined by your mass and g: W = mg We know g, the acceleration due to gravity on a planet, in terms of G, the universal gravitation constant, and planet parameters ( P, P ) g = G planet planet planet Therefore your weight is proportional to g planet W planet planet planet W X = W X X W W X X = W = W ( ) () = W PHYS 00 Lecture, Slide 3

14 Speed Two satellites are in circular orbits about the. Satellite is twice as far from the as is Satellite. How is the speed of Satellite (v ) related to the speed of Satellite (v )? BB (A) v < v (B) v = v (C) v > v (D) Cannot determine: depends on their masses (D) Is NOT COCT: that is the point of Kepler s Third Law!! WHY?? ) Both accelerations (a = v /) are caused by the same force!! ) That force (gravitation) is proportional to the satellite s mass F m = G a PHYSICS BUT F = m = G v a = v = G Satellites in larger orbits have smaller speeds. v PHYS 00 Lecture, Slide 4

15 CheckPoint 3 Case I: a satellite orbits the once every days at an orbital distance of o. Case II: a satellite travels around the at an orbital distance of 4o. BB 3 How long would take the satellite in Case II to make one complete revolution around the? 3 (A) day (B) days (C) 4 days (D) 8 days () 6 days (F) 3 days You said: Period is directly proportional to radius. Therefore, if the radius increases by a factor of 4, so does the period. the period is in the denominator, and is a squared term. for simplicity we can say that the radius cubed / period squared for case one is /4. so, the second term having a radius 4 times larger will need a period 6 days. If 0 takes two days, 4 squared is, and multiply by the original days, to get 3 days A B C D F PHYS 00 Lecture, Slide 5

16 CheckPoint 3 Case I: a satellite orbits the once every days at an orbital distance of o. Case II: a satellite travels around the at an orbital distance of 4o. How long would take the satellite in Case II to make one complete revolution around the? 3 (A) day (B) days (C) 4 days (D) 8 days () 6 days (F) 3 days ecall the speed question: larger orbits correspond to slower speeds v = G Using definition of period (P = π/v), we get Kepler s Third Law: Case I Case II = P 3 3 (4 ) 0 0 P 4 P = P 3 = constant = P 64P 3 P = 8P 0 PHYS 00 Lecture, Slide 6

17 Directions BB The springs have equal spring constants (k) and unstretched lengths (x 0 ) in the two cases Compare the directions of the forces that the spring exerts on the mass in the two cases (A) I II (B) I II (C) I II (D) I II Springs ALWAYS exert a STOING FOC Force direction ALWAYS towards UNSTTCHD POSITION PHYS 00 Lecture, Slide 7

18 CheckPoint 4 The springs have equal spring constants (k) and unstretched lengths (x 0 ) in the two cases BB Compare X I and X II, the distances the srings are stretched/compressed in the two cases (A) X I < X II (B) X I = X II (C) X I > X II You said: Draw a freebody diagram. The forces are the same. F spring=mg sinx equals k abs(x - x0) = mg sinx X is greater because the string is at the top of the incline and the box is moving down so the distance between the base of the spring and the box are increasing so it is stretching. X is smaller because the spring is placed at the base of the incline and the box is going to slide that way naturally due to gravity so it is compressing the spring and making it smaller A B C PHYS 00 Lecture, Slide 8

19 CheckPoint 4 The springs have equal spring constants (k) and unstretched lengths (x 0 ) in the two cases Compare X I and X II, the distances the srings are stretched/compressed in the two cases (A) X I < X II (B) X I = X II (C) X I > X II TH KY: F BODY DIAGAS They are the SA!! F spring = mg sinθ m x x0 = g sinθ k x x 0 = mg sinθ k PHYS 00 Lecture, Slide 9

20 INDS Homework on Universal Gravitation & Springs Due Tuesday Final xam onday ay 7 at :30pm Transportation Building Contact me if you have a conflict PHYS 00 Lecture, Slide 0