Astronomy101.sept29

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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/(kgs2)
• 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/(kgs2) * 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/(kgs2) * (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/(kgs2) 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/(kgs2) * (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/(kgs2) * (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/(kgs2) * (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/(kgs2) * (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?