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Gravitation Sections 9.1 – 9.3. Reminders Reading quiz for Chapter 11 (Sections 1-2) due by start of class on Tuesday. Lab this week: Lab A7-CE – Conservation.

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Presentation on theme: "Gravitation Sections 9.1 – 9.3. Reminders Reading quiz for Chapter 11 (Sections 1-2) due by start of class on Tuesday. Lab this week: Lab A7-CE – Conservation."— Presentation transcript:

1 Gravitation Sections 9.1 – 9.3

2 Reminders Reading quiz for Chapter 11 (Sections 1-2) due by start of class on Tuesday. Lab this week: Lab A7-CE – Conservation of Energy – due by Friday at 4 p.m. In-class Quiz #4 on Thursday, October 23 – addresses chapters 8 and 9.

3 Newton’s Apple Perhaps in contemplating the fall of an apple, Newton realized that its acceleration was (in modern metric terms). He realized that the moon’s acceleration toward Earth was

4 Working… Earth acts as though all its matter is concentrated in a point at its center.

5 Working…

6 Newton Finds Kepler’s Third Law P 2 = r 3 (P in years; r in astronomical units) P 2 = kr 3 (arbitrary units) Newton derives this relationship

7 A Note about Dark Matter

8 Relationship Weight versus Mass Weight is a force defined by (kg. m/s 2 ). Recall that 1kg. m/s 2 is a Newton, N What weighs 1 N? – A medium apple – A cooked quarter-pounder burger (meat only) Mass is the amount of matter in a body. To find relationship between weight and mass make a graph of weight versus mass.

9 Time for an Experiment! Let’s graph gravitational force (weight) versus mass to find the relationship.

10 Gravitational Field Strength W t = km where k = g; ergo, W t = mg g = gravitational field strength in units of N/kg g = -9.81N/kg When simplified, Hence, g = -9.81m/s 2 Gravitational field strength has the units of acceleration and is directed downward.

11 Warning! Do not confuse g with G. g = -9.81N/kg = -9.81m/s 2 G = 6.67x10 -11 Nm 2 /kg 2 W t = mg = -GMm/r 2 g = -GM/r 2 = -(6.67x10 -11 Nm 2 /kg 2 ) * (5.97x10 24 kg)/(6,371,000m) 2 = -9.81m/s 2

12 Gravitational PE on Universal Scale PE = mgh = -GMm/r At cosmic distances, – E total = KE + PE (where PE = 0 at ∞) – E total = ½mv 2 – GMm/r

13 Black Holes Recall, E total = KE + PE = ½mv 2 – GMm/r If E total = 0, then v escape = √(2GM/r) If E total > 0, then v > v escape If E total < 0, then v < v escape In a black hole, M is “normal” but r can be vanishingly small. Hence, when something gets close enough to a black hole it can not escape – even light!

14 Sample Problems #1 and #2 #1 Calculate the weight of a 0.454kg loaf of bread on the surface of Earth. Give the magnitude and direction of that force. #2 Show that the local value of “g” (the gravitational field strength) on the surface of the moon is approximately 1/6 that on the surface of Earth. Note that g Earth = -9.81N/kg

15 Sample Problems #3 and #4 #3 How much gravitational force exists between two 1kg lead spheres whose centers are 1m apart from one another? #4 Show mathematically that the acceleration due to gravity in a 300km high orbit is 91.2% that on the surface of Earth. Assume a radius of 6,371km for Earth.


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