#### Transcript Astronomy101.sept29

Astronomy 101 The Solar System Tuesday, Thursday 2:30-3:45 pm Hasbrouck 20 Tom Burbine [email protected] Course • Course Website: – http://blogs.umass.edu/astron101-tburbine/ • Textbook: – Pathways to Astronomy (2nd Edition) by Stephen Schneider and Thomas Arny. • You also will need a calculator. Office Hours • Mine • Tuesday, Thursday - 1:15-2:15pm • Lederle Graduate Research Tower C 632 • Neil • Tuesday, Thursday - 11 am-noon • Lederle Graduate Research Tower B 619-O Homework • We will use Spark • https://spark.oit.umass.edu/webct/logonDisplay.d owebct • Homework will be due approximately twice a week Exam #1 • Average was 85 • Grades ranged from 40s to 100s HW #5 • Due Thursday • A hypothesis is an educated guess, based on observation. Usually, a hypothesis can be supported or refuted through experimentation or more observation. A hypothesis can be disproven, but not proven to be true. • A scientific theory summarizes a hypothesis or group of hypotheses that have been supported with repeated testing. A theory is valid as long as there is no evidence to dispute it. Therefore, theories can be disproven. • A law generalizes a body of observations. At the time it is made, no exceptions have been found to a law. assume all mass is concentrated in the center of a body F = G M1 M2 r2 • The value of G was determined by Henry Cavendish between 1797-1798 • G = 6.67 x 10-11 m3/(kgs2) • http://blogs.howstuffworks.com/2009/04/13/diycalculate-the-gravitational-constant-like-cavendishdid/ http://www.makingthemodernworld.org.uk/learning_modules/maths/06.TU.02/illustrations/06.IL.09.gif What is the attraction of two people in this room? F = G M1 M2 r2 • • • • • Say their masses are both 100 kg Their distances are 10 meters apart F = 6.67 x 10-11 m3/(kgs2) * 100*100 kg2/(10*10 m2) F = 6.67 x 10-9 N = 0.0000000067 N Remember the person weighs 980 N F = G M1 M2 r2 • How would the force between the two people change if they were only 5 meters apart instead of 10 meters? • A) Stay the same • B) Double (Increase by a Factor of 2) • C) Quadrupul (Increase by a Factor of 4) • D) halve (decrease by a factor of 2) F = G M1 M2 = G M1 M2 (r/2)2 r2/4 =4 G M1 M2 r2 • How would the force between the two people change if they were only 5 meters apart instead of 10 meters? • A) Stay the same • B) Double (Increase by a Factor of 2) • C) Quadrupul (Increase by a Factor of 4) • D) halve (decrease by a factor of 2) Acceleration of gravity (g) F = ma = G Mm r2 a =GM r2 g=a=GM r2 M is the Earth’s mass r is the Earth’s radius m is the mass of an object F is the force a is the acceleration Acceleration of gravity (g) M is the Earth’s mass r is the Earth’s radius g=GM r2 g = 6.67 x 10-11 m3/(kgs2) * (6.0 x 1024 kg) (6.4 x 106 m) * (6.4 x 106 m) g = 9.8 m/s2 Gravitational acceleration • Gravitational acceleration is different on different planets because they have different sizes and masses • Gravitational acceleration (on Moon) = 1.6 m/s² (0.165 g) • Gravitational acceleration (on Jupiter) = 24.8 m/s² (2.53 g) Experiment on the Moon • http://www.youtube.com/watch?v=5C5_dOEyAfk How things fall • Heavy and light objects fall at the same rate • The heavy object does not fall faster (as long as there is no air resistance) g=GM r2 (does not depend on mass of object) How does gravity work? • Gravity distort space-time • http://www.hulu.com/watch/19766/spaceripeinsteins-messengers Escape velocity • Velocity above this will allow an object to escape a planet’s gravity v For Earth: v = square root[(2 x 6.67 x 10-11 m3/(kgs2) x (6.0 x 1024 kg)] (6.4 x 106 m) v = square root [1.25 x 108 m2/s2] v = 11.2 x 103 m/s = 11.2 km/s Escape velocity • Escape velocity is different on different planets because they have different sizes and masses • Escape velocity (on Moon) = 2.4 km/s • Escape velocity (on Jupiter) = 59.5 km/s What causes tides on earth? • Moon pulls on different parts of the Earth with different strengths • http://www.youtube.com/watch?v=Rn_ycVcyxlY • http://www.youtube.com/watch?v=aN2RM5wa1e k Forces on Water • Average Force on 1 kg water on Earth from Moon F = G M m = 6.67 x 10-11 m3/(kgs2) * (7.35 x 1022 kg) * (1 kg) r2 (3.84 x 108 m) 2 • F = 3.33 x 10-5 N • Force of 1 kg on water on near-side of Earth from Sun F = G M m = 6.67 x 10-11 m3/(kgs2) * (7.35 x 1022 kg) * (1 kg) r2 (3.84 x 108 m -6.37 x 106 m) 2 • F = 3.44 x 10-5 N • Difference in forces is 1.1 x 10-6 N • Called Tidal Force • Tidal force arises because the gravitational force exerted on one body by a second body is not constant across its diameter • Water flows so this tidal force causes the tides that are seen on Earth Effects on tides due to Sun • Sun exerts a stronger gravitational force on the Earth • But since farther away, the differential force from one side of the Earth to the other is smaller • Sun’s tidal effect is about one-half that of the Moon Forces on Water • Average Force on 1 kg water on Earth from Sun F = G M m = 6.67 x 10-11 m3/(kgs2) * (2 x 1030 kg) * (1 kg) r2 (1.5 x 1011 m) 2 • F = 5.928889 x 10-3 N • Force of 1 kg on water on near-side of Earth from Sun F = G M m = 6.67 x 10-11 m3/(kgs2) * (2 x 1030 kg) * (1 kg) r2 (1.5 x 1011 m -6.37 x 106 m) 2 • F = 5.929392 x 10-3 N • Difference in forces is 5.0 x 10-7 N due to Sun • Difference in forces is 1.1 x 10-6 N due to Moon Remember • Force downwards is 9 Newtons on 1 kg of water • Water won’t be pulled off Earth • Water can flow Shoemaker-Levy 9 • • • • • Comet that hit Jupiter Jupiter-orbiting comet Broken apart by tidal forces Discovered in 1993 Hit Jupiter in 1994 Roche Limit • The smallest distance at which a natural satellite can orbit a celestial body without being torn apart by the larger body's gravitational force (tidal forces). The distance depends on the densities of the two bodies and the orbit of the satellite. • If a planet and a satellite have identical densities, then the Roche limit is 2.446 times the radius of the planet. • Jupiter's moon Metis and Saturn's moon Pan are examples of natural satellites that survive despite being within their Roche limits Why is the Roche Limit important? • Comet Shoemaker-Levy 9's decaying orbit around Jupiter passed within its Roche limit in July, 1992, causing it to break into a number of smaller pieces. • All known planetary rings are located within the Roche limit • The first impact occurred at 20:15 UTC on July 16, 1994 • Fragment A of the nucleus slammed into Jupiter's southern hemisphere at a speed of about 60 km/s. • Instruments on Galileo detected a fireball which reached a peak temperature of about 24,000 K, compared to the typical Jovian cloudtop temperature of about 130 K, before expanding and cooling rapidly to about 1500 K after 40 s. • http://www.youtube.com/watch?v=tbhT6KbHvZ8 Has this happened before? Ganymede Any Questions?