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Barry Latham Honors Physics Bloom High School. 7.1: Kepler’s Laws of Motion  Based on years of observations, Kepler devised a series of laws known to.

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Presentation on theme: "Barry Latham Honors Physics Bloom High School. 7.1: Kepler’s Laws of Motion  Based on years of observations, Kepler devised a series of laws known to."— Presentation transcript:

1 Barry Latham Honors Physics Bloom High School

2 7.1: Kepler’s Laws of Motion  Based on years of observations, Kepler devised a series of laws known to us as his “Laws of Planetary Motion.”  1 st Law: The path of each planet about the sun is an ellipse with the sun at the focus.  2 nd Law: Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal periods of time.  3 rd Law: The ratio of the squares of the periods T of any two planets revolving about the sun is equal to the ratio of the cubes of their mean distances s to the sun: (T 1 /T 2 ) 2 =(s 1 /s 2 ) 3.

3 Kepler’s Laws

4 Problem Solving  As always, picture/data/formula/algebra!  Keep the unknown as a numerator  Rearrange the formula BEFORE you plug the values in.  Watch parenthesis, squares, cubes and their roots.  Recall that:x ½ = 2 √(x)x 1/3 = 3 √(x) x 2/3 = 3 √(x 2 ) x 3/2 = 2 √(x 3 )

5 Distance to Mars?  According to Kepler, Mars’ “year” is 687 Earth-days. Using what we know about the distance and period of the earth, we can find Mars’ distance. The mean distance of the Earth to the Sun is 1.50x10 11 m.  Picture/Data/Formula/Algebra  (T 1 /T 2 ) 2 =(s 1 /s 2 ) 3  T 1 =Mars=687 dayss 1 =Mars’ orbital radius  T 2 =Earth= dayss 2 =Earth orbital radius= 1.50x10 11 m  (687 days/ days) 2 =(s 1 /1.50x10 11 m) 3  (1.5x10 11 m)(687 days/ days) 2/3 =s 1  s 1 =2.28x10 11 m

6 Newton’s Law of Universal Gravitation  Every particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. This force acts along the line joining the two particles.  F=Gm 1 m 2 /r 2

7 Universal Gravitation Equation  F=Gm 1 m 2 /r 2  F= the force of gravity (in Newton’s)  m 1 & m 2 = the two individual masses that we are measuring  r= the distance between the masses, measured in a straight line and using their center of mass  G= the universal gravitation constant  6.67x N m 2 /kg 2  F=(N m 2 /kg 2 )(kg)(kg)/(m) 2  F=N (big surprise)

8 The Romance of Gravity  Find the force of attraction between two people. A 74 kg person is sitting 1.2 m away from a 84 kg person at a lab bench. Estimate the force attracting them together.  Picture/Data/Formula/Algebra  F=Gm 1 m 2 /r 2  G= 6.67x N m 2 /kg 2  m 1 =74 kg  m 2 =84 kg  r=1.2 m  F=G(74 kg)(84 kg)/(1.2 m) 2  F= 2.879x10 -7 N(equal to a 0.03 mg object)

9 Less Romantic Gravity, part 1  As you get far away from Earth, does your weight decrease?  What is the weight of a 2000kg satellite when it is in the lab at NASA?  F g =mg  F g =(2000 kg)(9.80 m/s 2 )  F g =19,600 N

10 Less Romantic Gravity, part 2  What is the force of gravity acting on a 2000 kg satellite when it orbits two Earth radii from the Earth’s center (r Earth = 6.38x10 6 m).  F=Gm 1 m 2 /r 2  F=G(2000 kg)(5.97x10 24 kg)/(2x6.38x10 6 m) 2  F g =4,891 N  This is 1/4th of what it was in the lab!

11 Universal Gravitation & Kepler’s 3 rd Law  If we take Newton’s law of universal gravitation,  F=Gm 1 m 2 /r 2  And Kepler’s 3 rd Law (with centripetal acceleration included)  F=m4  2 r/T 2  m=mass of central body  T=period of orbit (time of one revolution)  Set them equal (they are both F) and solve for T  T=√(r 3 (4  2 )/(Gm))  This represents the time it takes for an object to go around a central body one time.

12 Cavendish’s Device  Using a device shown to the right, Cavendish was able to accurately measure the force of gravity.  By twisting the string slightly, the scale displayed the force of attraction between the sphere on the rod and sphere A.

13 Geophysical Applications  When discussing objects on the earth’s surface, m 1 becomes m E and r is the radius of the earth (r E ).  Original formulaF=Gm 1 m 2 /r 2  substituted m 2 g for Fm 2 g=Gm E m 2 /r E 2  cancel m 2 g=Gm E /r E 2  So the acceleration due to gravity of an object is dependant upon the mass of the object and the radius of the object (in this case, the earth)  In lieu of Earth, we can use the data from any object!

14 What’s the mass of Earth?  Unless you have a HUGE scale, we need another way to determine it.  g=Gm E /r E 2 but solve for m E  m E =g r E 2 /G  m E =(9.80 m/s 2 )(6.38x10 6 m) 2 /(6.67x N m 2 /kg 2 )  m E =5.98x10 24 kg

15 Is “g” really 9.8?  Find g at sea level on Earth.  g=Gm E /r E 2  g=(6.67x N m 2 /kg 2 )(5.98x10 24 kg)/(6.38x10 6 m) 2  g= m/s 2  So, yes, it’s basically 9.8.

16 Is “g” really 9.8?  Gravity at the top of Mt. Everest. Mt. Everest has a height of 8,850 m above sea level.  g=Gm E /r E 2  g=(6.67x N m 2 /kg 2 )(5.98x10 24 kg)/(6.38x10 6 m + 8,850 m) 2  g= m/s 2  So, yes, it’s basically still 9.8.

17 Is g 9.8 everywhere on Earth?  Explain the losses or gains in gravity.  HW p. 178 q. 6-10

18 7.2: Satellites  Artificial satellites are put into orbit with three different final velocities.  Circular orbit (space shuttles)  27,000 km/h  Elliptical orbit  30,000 km/h  If an object needs to leave earth’s orbit, such as solar probes, and extra-solar objects  40,000 km/h

19 Satellite Orbital Velocity  Geostationary satellite (communications & weather)  Stays above the same point on earth

20 Satellite Equation Derivation  To remain above the same point, the satellite must have a period (T) of one day (one cycle per day).  Apply Newton’s Second law and a new formula (centripetal acceleration)  Assume the orbit is circular (which it pretty much is)  The only force on the satellite is the force of universal gravitation. Plug our new gravity equation and centripetal acceleration into F=ma.  F=maF=Gm 1 m 2 /r 2 a c =v 2 /r  Gm sat m E /r 2 =(m sat )(v 2 /r)

21 Speed (v) of a Satellite Orbiting a Central Body in Circular Orbit  Since the satellite revolves around the earth at the same rate the earth turns, v=2  r/T  T is the period of the movement, 1 day (86,400 s)  Gm sat m E /r 2 =m sat v 2 /rcancel m sat and r  Gm E /r=v 2 solve for v  v=√(Gm E /r)m E is the mass of the Earth but can represent the central body  √((6.67x )(5.98x10 24 )/(4.23x10 7 ))=3070 m/s  We find the same result if we use, v=2  r/T.  v=2  (4.23x10 7 )/(86,400)=3076 m/s

22 Period (T) of a Satellite Orbiting a Central Body in Circular Orbit  We saw this on Slide 11!  T=√(r 3 (4  2 )/(Gm))  This represents the time it takes for an object to go around a central body one time.  If the mass (m) is Earth, then m=m E

23 Orbital height (r) of a Satellite Orbiting a Central Body in Circular Orbit  Using any number of equations containing r, we can find the orbital height  Remember that height is from the center of the central object to the center of the satellite  F=Gm 1 m 2 /r 2 Newton’s Universal Gravitation  r= √( Gm 1 m 2 /F)  F=m4  2 r/T 2 Kepler’s Third Law  r=FT 2 /(m4  2 )  T= 2  √ (r 3 /(Gm))Newton’s version of Kepler’s 3 rd Law  r=  v=√(Gm E /r)v of satellite in circular orbit  r=Gm E /v 2

24 Inertial Mass  According to F=ma, the mass of an object is the relationship between the Force applied and the acceleration of the object.  Because every object with mass has inertia, we can rearrange and use subscripts.  m inertia =F net /a  More mass means less reaction to the same force  Inertial mass is independent of location or a reference body  It’s how they measure the astronauts mass in space!  Lab 7-2

25 Gravitational Mass  Unlike Inertial Mass, gravitational mass is dependant on a reference material (sometime a whole planet!)  Using Newton’s Universal Gravitation formula, we can rearrange to find the mass of an object  F grav =Gm grav m E /r 2  m grav = r 2 F grav /(Gm E )  To measure this, we use a triple beam or digital balance in the presence of gravity

26 Relativity & Gravity  According to Einstein, gravity is not a force, but an effect of space itself  Mass causes space to curve, so objects are pulled toward the “dimple” in space-time (even light)  Larger objects make larger “dimples” hence larger gravitational pull

27 Earth’s gravitational field distorting space-time

28 Blackholes, defined  Black hole- a concentration of mass so dense that nothing can escape the pull of it’s gravity, not even light  Formed by the severe gravitational pull overcoming the thermonuclear reaction pushing the material out  Black hole formation    Size? Black holes can be measured by mass or area  Mass to kg (10-1,000,000 times as massive as our Sun)  Volume- based on Schwartzchild radius or radius of the horizon  A black hole with a mass the same as the Sun would be 3 km in radius  The Sun has a radius of 700,000 km now!  A 10-solar-mass black hole would be 30km and so on  Event Horizon- the point of no return, the Schwartzchild radius

29 Black Holes evaporate?  Black Holes will not eventually suck up everything in the universe  They emit “Hawking radiation”  At the Event Horizon, a particle-antiparticle pair can appear from within the black hole  One particle gets sucked back in, the other escapes  Momentum & Energy are conserved  The emitted particle is detected  This happens so frequently, it is observed as a matter/energy stream

30 Hawking Radiation

31 Black Hole Images

32 A wormhole

33 End of Unit Review, 1  What would make you weigh more: an earth twice as massive with the same diameter or one having half its diameter with the same mass?  Solution: Find out using F=Gm 1 m 2 /r 2  F g1 =Gm E m you /r E 2 normally, what is your F g ?  F g2 =G(2m E )m you /r E 2 now, what is your F g ?  F g3 =Gm E m you /(½r E ) 2 now, what is your F g ?  How do they all compare?

34 End of Unit Review, 2  Would one need a longer rocket to take off from the moon vs. earth?  What is needed to escape the Earth? The Moon?  Gravity holds us in place, so which has less gravity?  Solution: Find out using g=Gm E /r E 2  For earth, g= m/s 2  For the Moon, g=Gm M /r M 2  g=G(7.349x10 22 kg)/(1.738x10 6 m) 2  g=1.62 m/s 2

35 End of Unit Review, 3  Based on our prior work, what would your weight be on the Moon?  F gE =m(9. 80 m/s 2 )  F gM =m(1.62 m/s 2 )  F gE =(84.1 kg)(9. 80 m/s 2 )=824 N  F gM =(84.1 kg)(1.62 m/s 2 )=136 N

36 End of Unit Review, 4  If an astronaut was to break the connection rope between him and the space shuttle, will gravity quickly pull him back to the shuttle?  F=Gm 1 m 2 /r  g=Gm/r 2  What happens to the acceleration of an asteroid as it approaches a planet?  If g based on distance?  g=Gm/r 2


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