Download Transparancies for Gravity & Circular Motion Section

Handout III : Gravitation and Circular Motion
EE1 Particle Kinematics :
Newton’s Legacy
“I frame no hypotheses; for whatever is
not deduced from the phenomena is to be
called a hypothesis; and hypotheses,
whether metaphysical or physical, whether
of occult qualities or mechanical, have no
place in experimental philosophy.”
October 2004
http://ppewww.ph.gla.ac.uk/~parkes/teaching/PK/PK.html
Chris Parkes
Gravitational Force
Myth of Newton & apple.
He realised gravity is universal
same for planets and apples
•Any two masses m1,m2 attract each other
with a gravitational force:
F
F
m1m2
F G 2
r
r
m2
m1
Newton’s law of Gravity
Inverse square law 1/r2, r distance between masses
The gravitational constant G = 6.67 x 10-11 Nm2/kg2
•Explains motion of planets, moons and tides
24kg,
m
=5.97x10


mE m
GmE
E
Gravity on
m
F G
 
RE=6378km
2
2

earth’s surface
RE
RE 
Mass, radius of earth

GmE
1
 9.81ms
Or F  mg Hence, g 
2
RE
Circular Motion
360o = 2 radians
180o =  radians
90o = /2 radians
• Rotate in circle with constant angular speed 
R – radius of circle
s – distance moved along circumference
=t, angle  (radians) = s/R
• Co-ordinates
x= R cos  = R cos t
y= R sin  = R sin t
• Velocity
•Acceleration
d
v x  ( R cos t )   R sin t
dt
d
v y  ( R sin t )  R cos t
dt
d
d
a x  (v x )  ( R sin t )   R 2 cos t
dt
dt
d
d
a y  (v y )  ( R cos wt )   R 2 sin t
dt
dt
y
R
s
=t
x
t=0
Magnitude and direction of motion
•Velocity
v 2  vx  v y  R 2 w2 sin 2 t  R 2 2 cos 2 t   2 R 2
2
2
v=R
tan  
And direction of velocity vector v
Is tangential to the circle
vy
vx

cos t
1

 sin t
tan 
    90o
•Acceleration
2

a
a  ax  a y 
2
v
2

R 2 w4 cos 2 t  R 2 4 sin 2 t   4 R 2
a= 2R=(R)2/R=v2/R
And direction of acceleration vector a
a= -2r
a x   2 x
a y   2 y
Acceleration is towards centre of circle
Angular Momentum
• For a body moving in a circle of radius r at speed
v, the angular momentum is
L=(mv)r
= mr2= I 
(using v=R)
I is called moment of inertia
The rate of change of angular momentum is
dL d
dv
 dt (mvr)  mr
 mra
dt
dt
 r  ma  r  F

r
s
– The product rF is called the torque of the Force
• Work done by force is Fs =(Fr)(s/r)
= Torque  angle in radians
Power
d
 Torque 
 Torque  
dt
= rate of doing work
= Torque  Angular velocity
Force towards centre of circle
•
Particle is accelerating
–
•
•
1.
2.
3.
So must be a Force
Accelerating towards centre of circle
– So force is towards centre of circle
F=ma= mv2/R in direction –r 2
v
or using unit vector F  m rˆ
r
Examples of central Force
Tension in a rope
Banked Corner
Gravity acting on a satellite
N.B. general solution is an ellipse not a circle - planets travel in ellipses around sun
Satellites
•Centripetal Force provided by Gravity
Mm mv2
F G 2 
R
R
M
2
M
v G
v

G
R
R
m
R
M
Distance in one revolution s = 2R, in time period T, v=s/T
R
T  2R / v  2R
GM
T2R3 , Kepler’s 3rd Law
•Special case of satellites – Geostationary orbit
•Stay above same point on earth T=24 hours
3
24  60  60  2
R  42,000km
R2
GM E
Gravitational Potential Energy
• How much work must we do to move
m1 from r to infinity ?
m1m2
F G 2
R
– When distance R
r
m2
m1
– Work done in moving dR dW=FdR

– Total work done
m1m2
W  G
r
R
2
dR
W  Gm1m2 ( R1 )r  Gm1m2 ( 1 ) Gm1m2 ( 1r )

W
Gm1m2
r
Choose Potential energy (PE) to be zero when at infinity
m1m2
Then stored energy when at r is –W
 PE  G
r
-ve as attractive force, so PE must be maximal at 
Compare Gravitational P.E.
• Relate to other expression that you know
Potential Energy falling distance h to earth’s surface = mgh
Uses:
1) Expression for g from earlier
• g=GME/RE2

M m 
M m
   G E     G E 
RE  
RE  h 

 GM E 

1




 m
1


1  h / RE 
2) Binomial expansion given h<<RE
 RE 
• (1+)-1 = 1- +…..smaller terms…  mgRE (1  1  h / RE  smaller ..terms...)
 mgh
•Compare with Electrostatics:
Same form, but watch signs: attractive or repulsive force
attract
m1m2
F G 2
r
repel
Q1Q2
F k 2
r
Maximal at 
m1m2
PE  G
r
Minimal at
Q1Q2
PE  k
r
A final complication:
what do we mean by mass ?
• Newton’s 2nd law F = mI a
mI is inertial mass
• Law of Gravity
mG M G
F G
rˆ
2
r
mG, MG is gravitational mass
- like electric charge for gravity
Are these the same ?
•Yes, but that took another 250 years till
Einstein’s theory of relativity to explain!