Transcript PHSX 114, Wednesday, September 18, 2002

PHSX 114, Wednesday,
September 17, 2003
• Reading for today: Chapter 5 (5-6 -- 5-8)
• Reading for next lecture (Fri.): Chapter 5
(5-8 -- 5-10)
• Homework for today's lecture: Chapter 5,
question 13; problems 25, 33, 62, 69
Other announcements
• Exam #2 is one week from today, covers
Chapters 4 and 5
• Review and equation sheets are posted on
the web
• Monday's class will be review
Newton’s Law of Universal
Gravitation
•perhaps the most important equation in
the history of science
•Historical context will be discussed
Friday
•The first great “unification” in physics
– motion of a falling apple and motion
of the moon about the Earth explained
by the same theory
The falling apple
• FG=mg
• What direction does the apple
fall?
•Down? (see video clip)
•toward the center of the Earth
An orbiting body is a
falling body
• the Earth is not flat
• curvature is such that the ground drops
about 5 m for every 8000 m traveled
horizontally
• it takes about one second to fall one meter
• if your horizontal velocity is about 8000 m/s
(18,000 mph), you are always at the same
height above the ground
Orbiting at the Earth’s
surface
• r is the Earth’s radius, 6.38 x 106 m
• centripetal acceleration is g (9.8 m/s2)
• aR= v2/r =
(8000 m/s)2/(6.38 x 106 m)= 9.8 m/s2
Is the moon’s acceleration 9.8 m/s2?
• No, it’s 0.0027 m/s2
• g (9.8 m/s2) is 3600 times bigger (602)
• Moon’s period is 27.3 days=2.36 x 106 s
• Earth-moon distance is 3.84 x 108 m
• v=2πr/T=2π(3.84 x 108 m)/(2.36 x 106 s) =
1020 m/s
• aR=v2/r=(1020 m/s)2/(3.84 x 108 m) =
0.0027 m/s2
• (3.84 x 108 m) /(6.38 x 106 m)= 60
• force of gravity decreases as the distance
squared
Newton's law of
universal gravitation
r
m1
• F = Gm1m2/r2
• r is the distance separating the two objects
• G is the gravitational constant,
G= 6.67 x 10-11 N-m2/kg2
• Force is attractive, direction is along line
joining the two objects
• Valid not just on Earth, but everywhere
m2
Example: Find the gravitational force
between the Earth and the Moon
• F = GmEmM/r2
• mE= 5.97 x 1024 kg; mM= 7.35 x 1022 kg;
r=3.84 x 108 m
• F = (6.67 x 10-11 N-m2/kg2)(5.97 x 1024 kg)(7.35
x 1022 kg)/(3.84 x 108 m)2 = 1.98 x 1020 N
• aM=F/mM = (1.98 x 1020 N)/(7.35 x 1022 kg)
=0.0027 m/s2
• Note: Newton's third law applies, aE=F/mE is
much smaller
Your turn
Estimate the force of gravity between you and
the person sitting next to you.
• Answer: approximately 3 x 10-7 N
• F = Gm1m2/r2
• m1= m2= 70 kg (150 lb); r=1 m
• F = (6.67 x 10-11 N-m2/kg2)(70kg)(70kg)/(1m)2
= 3 x 10-7 N
Cavendish experiment (1798)
• In 1665, Newton didn't have the technology
to measure G
• Henry Cavendish uses a clever "torsion
pendulum" apparatus to find G
• Video clip
• Cavendish is said to have measured the
mass of the Earth
G and g
• two expressions for the force of gravity at
the Earth’s surface
•FG = GmmE/rE2
• FG = mg
• mg = GmmE/rE2
• g = GmE/rE2
• g and rE well known in Cavendish’s time
• by measuring G, mass of Earth could now
be determined by mE = g rE2/G
• example
Your turn
If g=3.72 m/s2 on the surface of Mars and the
radius of Mars is 3400 km, find the mass of
Mars.
• Answer: 6.4 x 1023 kg (about 1/10 that of Earth)
• mM = g rM 2/G=(3.72 m/s2 )(3.4 x 106 m)2/
(6.67 x 10-11 N-m2/kg2)
Weightlessness and
Apparent weight
•Astronauts in orbit are “weightless”,
yet they are in Earth’s gravity
• in orbit, they are falling with
downward acceleration g
• “apparent weight” is zero
Apparent weight
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Scale reads the force w
ΣF=ma= w – FG
Elevator example
FG=mg doesn't change with elevator's acceleration
If a=0 (elevator has constant speed), w = FG
If a>0 (elevator accelerating upward), w > FG
If a<0 (elevator accelerating downward), w < FG
If a=-g (freefall), w=0 (-mg = w – mg)
example
Your turn
Find the apparent weight of a 100 kg object if
the elevator is accelerating upward with
a=3.0 m/s2.
• Answer: 1280 N
(in kg: 131 kg=1280N / 9.8 m/s2)
• ma= w – FG; w=ma+mg=m(a+g)=
(100kg)(3.0 + 9.8 m/s2)=1280 N