We’ve argued the force of gravity is also directly proportional to the masses involved: G is a universal constant measured to be 6.67 10 -11 N·m 2 /kg.
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We’ve argued the force of gravity is also directly proportional to the masses involved:
221
R
mmF
grav∝
221
R
mmGF
grav=
G is a universal constant measured to be6.67 10-11 N·m2/kg2
0.000 000 000 066 7 N·m2/kg2
Henry Cavendish (1731 – 1810)
How irresistible is the gravitational force of attraction between a pair of us when 1 meter (center-to-center) apart?
G(80 kg) (70 kg)
(1 meter)2 = G 5600kg2
m2
= 0. 000 000 32 N
Fgrav
R
2R
R
Rm mF F
R
R/2
Two objects of mass, m, separated by a center-to-center distance R are mutually
attracted to one another by a force F.How strong is the attractive force between
the other pairs of objects shown?
A. ¼ F C. F E. 4FB. ½ F D. 2F F. other
m m
m m
m 2m
2m 2m
Weight
How hard does earth’sgravity pull on me?
2R
MmGF earthDC=
distance to the center of the earth
If unbalanced by any normal force of support how fast would I be accelerated?
2R
MmGam earthDC
DC=
F = ma, so:
How fast does the earthaccelerate me?
2R
MGa earth=
How fast does the earthaccelerate YOU?
2R
MmGam earthYOU
YOUYOU=
How fast does the earthaccelerate an apple?
2R
MmGam earthapple
appleapple=
2R
MGa earth=
The earth accelerates ALL objects near its surface at exactly the same rate:
a = (6.67 10-11 N·m2/kg2) 5.98 1024 kg(6.38 106 m)2
= 9.800 m/sec2
Actually we’ve mixed up the historical order of events somewhat.
Newton’s ideas do in fact explain whyall objects fall to earth at the same rate
independent of their mass.
But notice calculating that the acceleration rate is exactly 9.8 m/sec2 required knowing the mass of the earth! How is THAT known?
Rearth = 6.38 106 meters
Navigators and surveyors had established a size for the earth:
3
3
4R V π= = 1.09 1021 m3
Density of some STUFF in the earth:
Granite 2.7 103 kg/m3
Iron 7.8 103 kg/m3
Water 1.0 103 kg/m3
Which would suggest the mass of the earth lies somewhere between
1 – 8 1024 kg
2R
MGa earth=
It was of course Cavendish who could take:
G
RmM
earth
22 )sec/8.9(=
and for any moon orbiting any planet:
2
24
T
Ra
π=
For any planet (or star or asteroid)
2R
MGa planet=
the acceleration down toward its center is
which means:
2
2
2
4
T
R
R
MG planet π=
2
324
GT
RM
planet
π=
Which is how we’ve known the massof Mars, Jupiter and Saturn without
even going there!
How do we know the mass of Venus? Mercury?
How do we know the mass of the sun?
Johannes Kepler (1571-1630)
All planets move in ellipses, with the sun at one focus.
The undefined concepts in GEOMETRY
collection of all points in a plane equidistant from a selected point
called the center.
collection of every point in a plane whose distances from two fixed focii sums to the same constant total.
5 inches
3.5 in
ches
To build an ellipse, pick any two pointsand a number bigger than their distance apart.
3.5 inches
7 > 5
1 inch
collection of every point in a plane whose distances from two fixed focii sums to the same constant total.
To build an ellipse, pick any two pointsand a number bigger than their distance apart.
7 > 5
3 + 4 = 7
3 inc
h
4 inch
2 + 5 = 7
collection of every point in a plane whose distances from two fixed focii sums to the same constant total.
To build an ellipse, pick any two pointsand a number bigger than their distance apart.
7 > 5
2 inch
5 inch
2 + 5 = 7
speed just right
speed too highspeed too
low
The second (far) focus is empty!
The second (near) focus is empty!
Oops! Speedway too slow!
A popgun fires a standard 40mm (2.7 gram) ping pong ball at the back of a stationary wagon.
mv
mv
The ball ricochets back with almost the same speed.
A. zero momentum D. momentum 2mv.B. momentum ½mv. E. momentum 3mv.C. momentum mv. F. momentum 4mv.
The wagon must recoil with
mv
mv
Now compare the case where the tennis ballsolidly rebounds
mv
…to the case where the tennis ball smashes into but imbeds itself
mv
?
?
Which collision sends the wagon rolling forward with the greater speed?
A.
B.
C. Either collision gives the wagon the same speed.
mv
?
?
60 gram tennis ball initially at 10 m/sec60 kg wagon initially at rest
wagonwagonballballballballvmvmvm ′+′=+0
wagonvkgsmkgsmkg ′+⋅−=⋅ )60(/60.0/60.0
wagonvkgsmkg ′=⋅ )60(/20.1
wagonvsm ′=/02.0
vmmsmkgwagonball
′+=⋅ )(/60.0vkgsmkg ′=⋅ )060.60/()/60.0(
vsm ′=/00999.0
A spray of rapid fire will provide a steady pressure that can levitate this block, holding it back in place.
v = 500 m/sec
A ball rebounds from the ground straight up at 500 m/sec.
How high can it climb? (How high had it been dropped from?)
mghmv =221
gtvv −=0
tsmsm )/8.9(/5000 2−=
2/8.9
/500
sm
smt
−−
= sec51=
2
21 gtd =
22
21 )51)(/8.9( ssm=
So:
meters 745,12=Or just:
hg
v =2
2
meters h 745,12=
v = 500 m/sec
A 500 m/sec projectile traveling straight up, is in the air 102 seconds
and reaches a 12.745 km altitude.
v = 500 m/sec
A 500 m/sec projectile traveling horizontally can cross a 12 meter lecture hall in:
ssmmt 004.0)/500/(12 ==
dropping only2
21 atd =
22
21 )004.0)(/8.9( ssm=
meters 0000784.0=mm 0784.0=
Crosses the room in nearly a straight line!
][ 2
21 mvmgh Δ=Rising h=1 meter:
reduces its speed by ~4 cm/sec.
CO2 molecules of a refrigerator-cold bubble in a can of soda pop 887 mph
Room temperature nitrogen N2 molecules
1160 mph
An enormous number of these invisibly small particles exert a ~constant pressure outward, against all the surfaces
of their container:
The round shapecharacteristic of a balloonconfirms that this pressure is exerted in all directions.
A. just as hard as
Answers to Clicker QuestionsJust another example of Newton’s law: Interactions always involve mutually equal but opposite forces.
Slide 2 presents an argument as to why.
R/2m m
2Rm m
Rm 2m
R2m 2m
Compare each to: 2
2
R
mG
RR
mmGF =
××
=
⎥⎦
⎤⎢⎣
⎡×== 2
2
4/22
2
42
)2/( R
mGmG
R
mG
R
⎥⎦
⎤⎢⎣
⎡×== 2
2
22
2
4
14
2
)2( R
mG
RmG
R
mG
⎥⎦
⎤⎢⎣
⎡×=×
2
2
2 2)2(
R
mG
R
mmG
⎥⎦
⎤⎢⎣
⎡×=×
2
2
2 4)2()2(
R
mG
R
mmG
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