10a-0 GMm U (r) - r = how high does it go? Satellite Motion

F
GMm
R2
Satellite Motion
GMm
U (r)  r
Ei  Ef
10a-0
U (r   )  0
how high does it go?
Escape velocity
Kepler’s 2nd Law ::= Areas  Angular Mom. Conservation !!!!
Newton’s Universal Law of Gravity
10a-1
M
F
r
Fm
F= force of gravity on m (M) due to M (m)
1) F acts along the line connecting the centers of objects
“Central Force” (for spherical objects)
2) F points toward M (attractive force)
3) Newton says:
GMm
F=
r2
G=6.67x10-11 [m3/Kg s2]
for special case on the Earth’s surface,
this acceleration is: g = 9.81 m/s2
G is a very small number!
10a-1a
10a-1
Recall
e
10a-1b
Properties of Gravity
Gravitation from a Sphere
Gravitational force between
an object and a sphere is the
same as if all the mass of
the sphere was concentrated
at its center.
10a-2
10a-3
Calculate amoon-in-orbit for later
(circular orbit)
What is amoon in orbit ?
(2 )2 R
a
T2
(2[3.14])2 3.84(10)8 m

[2.3(10)6 s ]2
(10)8 m
 28.6
(10)12 s 2
aMoon
 R  3.84(10)8 m
m
 0.00286 2
s
cannon fired horizontally
Re
low velocity- ~parabolic path
RMoon dist
higher velocity
- earth curves away under falling object
For special ”orbital” velocity
the ball falls
at the same rate that the earth curves away
“a” of object in orbit (at surface) is same as dropped object
Re
RMoon dist
6.378(10)6 m

3.84(10)8 m
 .0166 
1
60.2
Newton’s connection afalling-apple=aorbiting-apple=g !!!!!!
earth
moon
GMm
F
 ma
2
R
GM
 amoon
2
R
2
aMoon
 Re


 aapple
 RMoon dist 
aMoon
m
 1 

9.8

s2
 60.2 
Same idea for moon in orbit- moon “falling” toward Earth
2
aapple  g
amoon
Re2
 aapple 2
R
a
m
 0.0027 2
s
GM
Moon
2
2
amoon
R
2
 R  e2
R
aapple GM R
amoon  g e2
dead right Newt. nailed it !!!
2
R
10a-4 R e
apples-moon –everything ruled by 1/r2 gravity
http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/newt/newtmtn.html
Recall energy conservation near earth's surface
Throw an object up (very slow) – how high does it go?
Ei  Ef
1
mv 2  mgh
2
10a-5
2
vi
h
2g
2
2
y= h
vtop = 0 : know
100
h
2(9.8)
E f  0  mgh
h  510 m
vi = 100 m/s
U=0
y=0
1
2
E i  mv i  0
2
m
 
s
m
( 2)
s
Universal Gravitational Potential Energy
GMm
:::: U(r =
r
U(r)
=-
Ei  Ef
vtop = 0 : know
∞)
=0
GM e m
1
2 GM e m
mvi 2
Re
r
GM e m
Ef  r
r
3 cases
E< 0 can find rmax and object
falls back
E= 0 can just get to
and stops
vi
1
2 GM e m
Ei  mvi 2
Re
Re
Earth
10a-6
E > 0 gets to r  
And keeps going
r
Universal Gravitational Potential Energy
GMm
U (r)  :::: U (r  )  0
r
Throw an object up (very fast)– how high does it go?
GMe m
GMe m
1
2
mvi  2
Re
r
Ei  Ef
vtop = 0 : know
GM e m
Ef  r
E< 0 can find rmax and
object falls back
r
vi
1
2 GM e m
Ei  mvi 2
Re
Re
Earth
recall
mg=m
so gR e =
GMe
R 2e
GMe
Re
10a-6a
GM e m
r  GM e m
1
2
mvi 2
Re
Re
r 
Re
1
vi 2
2GMe
rmax
Re

vi 2
{1}
2 gRe
vi 2
note {1- 2 gR }  0  rmax  
e
Universal Gravitational Potential Energy
10a-6b
GMm
U (r)  :::: U (r  )  0
r
mg=m
so
Escape velocity : vf=0 and rmax=  E=0
Ei  Ef
r
1
2 GM e m
Ei  mvi 2
Re
Earth
gR e =
1
mvesc 2
2
2
2GMe

Re
vesc 
GMe
Re
r
GMe m
 0
Re
Escape velocity
vesc
Re
GMe
R 2e
GMem
1
2
mv e =0
2
Re
vtop = 0 : know
GMem
Ef   0(at r  )
r
vi
recall
2GM e
 2 gRe
Re
recall
GMe
gR e =
Re
10a-6c
Escape velocity : vf=0 and rmax=  E=0
E=0 critical escape condition
E i  Ef  0
1
2 GM e m
mvi 0
2
Re
vtop = 0 : know and U(r=)=0
vesc
2
2GMe
=
Re
Escape velocity
Ef  0
vesc =
r
vi
Re
1
2 GM e m
Ei  mvi 2
Re
Earth
vesc = 2(9.8
v esc
2GM e
= 2gR e
Re
m
6
)(6.378(10)
m)
2
s
m
=11.1(10)
s
3
Kepler’s 1st Law
10a-7
(all objects bound to sun- Kepler figured out with Mars)
1. The orbital motion of the planets about the sun is an ellipse with the sun at one focus.
other
empty focus.
semi-minor axis rmin
ellipse = squashed circle
eccentricity ~ rmin/rmaj
rminrmaj get circle
rmaj
semi-major axis
Note in reality both Msun and mplanets-etc orbit common CM
Newton!!
- all motion in 1/r2 force law (like gravity) follows conic sections (slices of a cone)
- bound motion ellipses – unbound motion paranoia or hyperbola
- note one assumes for this Msun>>>mplanets-etc.
-Halley’s Comet coming back every 76 years nailed this
(Halley didn’t live to see it return but used Newton’s Laws to predict exactly the return)
Your chance to see Halley’s Comet
1986+76= 2062
(my father saw in 1910 with his
grandfather)
( I showed my children, in 1986)
Kepler’s 2nd Law
10a-8
(all objects bound to sun- Kepler figured out with Mars)
2. As the planet moves in its orbits it sweeps out equal areas in an equal times
planet moves faster (slower) when closer to (farther from) to the sun Msun>>>mplanets-etc.
planet
comet
r
l= vt
Newton!!
-Gravity = a central force (acts between centers of ~spherical objects)
 no torque on planet (comet…etc.)  angular momentum conserved
Area triangle= A
1
1
ΔA= r (Δl) = r (v  Δt)
2
2
ΔA 1
1
1
= r v  [
] (mr v  ) =[
] (L)
Δt 2
2m
2m
Recall !!! Ang. Mom.
L  mr v 
Kepler’s 2nd Law ::= Areas  Angular Mom. Conservation !!!!
10a-9
Kepler’s 3rd Law
P k R
2
sec
sec2
m3
P R
2
3
years
3
10a-9a
AU= R(earth) =1 AU
m
Kepler probably tried
Pn  R m
R  Pn/m
n/m=1/2 NO
curves down
n=m=1 NO
curves up
n=2: m=3 YES
linear !!
Kepler’s 3rd Law
(T2~R3)
Newton's Law
connection
how
“Newton” knew
he’d nailed it
Mass of Sun !!!
10a-10
Satellite Motion
Mm 
G 2 m
R
v2
 GM
R
Fma
v2
RE
M
R
v
ME=5.97(10)24m
R
v
m
GM
R
Satellite at surface (of Earth)
R=RE=6378(10)3m
2
GM
6 m
v
 gRE  9.8(6.378)(10) 2
RE
s
GMm
 mg
2
RE
GM
 gRE
RE
v=7.906(10)3 m/s
2 R
v
T
2 R 2(3.14159)6.378(10)6
3
T


5.07(10)
s  84.5min
3
v
7.906(10)
10a-11
Satellite Motion
Mm 
G 2 m
R
v2
Fma
v2
RE
R
 GM
R
2 R
v
T
M
GM
v
R
2
(2

)
T2 
R3
GM
Geocentric Satellite
GM 2
R 
T
2
(2 )
R
v
m
ME=5.97(10)24m
GM
 gRE
RE
4



T 24h 8.64 10 s
3
10a-12
GM 2
R3
T
2
(2 )
R=9598 (10)3m
Geocentric satellites
• Find the height (altitude) of a satellite above the Earth
surface so that it is always above the same point on the
Earth’s surface.
2
M E mS
vS

G
mS
2
(RE  h )
(RE  h )
2 (RE  h )
vS 
T
2
 2 (RE  h )


ME
T



G
(RE  h )
(RE  h )2
(RE  h )
3
GM E T 2

4 2
2
GM
T
E
(RE  h )  3
2

4
10a-12a
4



T 24 h 8 .64 10 s
6


h 3 .58 10 m
Notice, in order for the satellite to
hang above the same spot on the
Earth it has to have T = 24 hours