Physics 2101
Section 3
Ch.13-14
March 22rd : Ch.13Announcements:
• More on gravity.
• Start Ch. 14
• Exam. 3 on Mar. 24th
Class Website:
http://www.phys.lsu.edu/classes/spring2010/phys2101--3/
http://www.phys.lsu.edu/classes/spring2010/phys2101
http://www.phys.lsu.edu/~jzhang/teaching.html
Chapt.. 13: review
Chapt
Force on 1 due to 2:
m1m2
F12 = G 2 rˆ12
r12
G = 6.67 ×10−11 N ⋅ m 2 / kg 2
n
Force on 1 due to many: F1,net = F12 + F13 + F14 + ...+ F1n = ∑ F1i VECTOR ADDITION!!
i=1
Gravitational potential energy
If we define U = 0 at ∞, then the work done by taking mass m from R to ∞
m1m2
U(r)) = −G
U(
G
r12
m1m2
F12 = G 2 rˆ12
r12
Planets and Satellites: Kepler’s Laws
Kepler
Brahe
1571-1630
1546-1601
Newton
1642-1727
1) Law of Orbits:
All planets move in elliptical orbits with the Sun at one focus
Ellipse criterion:
Fm + F’m = const
a = semimajor axis
e = eccentricity
Assumes MSun >> ME
sun-earth
sun
earth
farthest point Ra = aphelion
closest point Rp = perihelion
Ra = a(1+ e)
Rp = a(1− e)
earth-moon
earth
moon
= apogee
= perigee
Planets and Satellites: Kepler’s 2nd Law
2)) Law of equal
q
areas :
A line that connects a planet to the sun sweeps out equal areas in equal time
- Planets move slowest when furthest away from Sun (at Ra)
- Planets
Pl t move fastest
f t t when
h closest
l
t away from
f
Sun
S (at
( t Rp)
ANGULAR MOMENTUM IS CONSERVED
Angular momentum
L =r ×p
L = rp = r (mv) = r (m(ωr)) = mr 2ω
L
= r 2ω
m
Area of “triangle”:
dA = 12 ( rdθ ) r
dA 1 2 dθ 1 2
= 2r
= 2r ω
dt
dt
dA L
=
= constant
dt 2m
Planets and Satellites: Kepler’s 3rd Law
3) The Law of Periods :
The square of the period of any planet is proportional to the cube of the semimajor axis
of its orbit
⎛ 2πrsat ⎞
⎜
⎟
Gmsat M E
⎝ T ⎠
−
= −msat
2
rsat
rsat
2
⎛ GM ⎞ 2
⇒ r =⎜
T
2 ⎟
⎝ 4π ⎠
⎛ 4π 2 ⎞
⇒ T2 =⎜
⎟rsat3
⎝ GM ⎠
3
sat
NOTE: r3 ∝ T2
Elliptical Orbit
Circular Orbit
When used in the equations, r
(radius) and a ((semimajor
semimajor axis)
are synonymous
Example: Hot Jupiter
In 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very close the
star HD 179949
179949. The orbit was
as abo
aboutt 1/9 the distance of Mercury
Merc r from our
o r sun,
s n and it takes the
planet only 1.09 days to make one orbit (assumed to be circular). (a) What is the mass of the star?
(b) How fast is this planet moving?
2π r 3/ 2
1
5
= 2.67 ×10 s and r = rMercury = 6.43 ×109 m
a) T =
9
GM star
M star
4π 2 r 3
= 2 = 2.21×1030 kg = 1.11M Sun
T G
2π r
b) v =
= 1.51×105 m/s
T
Conservation of Mechanical Energy
Escape speed: Minimum speed (vexcape) required to send a mass m, from mass M and
position R, to infinity, while coming to rest at infinity.
At infinity:
y Emech= 0 because U = 0 and KE = 0
Thus any other place we have:
E mech
⎛ 1 2 GmM ⎞
⇒
E
=
= (KE + U g ) = 0
⎜ mv −
⎟ = 0 ⇒ vescape =
mech
⎝2
R ⎠
Escape speed:
2GM
R
Earth = 11.2 km/s ((25,000
,
mi/hr))
Moon = 2.38 km/s
Sun = 618 km/s
Problem 13-37
a) What is the escape speed on a spherical asteroid whose radius is 500 km and whose
gravitational acceleration at the surface is is 3 m/s2 ?
b) How far from the surface will the particle go if it leaves the asteroid’s surface with a radial
speed of 1000 m/s?
c) With what speed will an object hit the asteroid if it is dropped from 1000 km above the
surface?
Problem 13-37
a) What is the escape speed on a spherical asteroid whose radius is 500 km and whose gravitational
acceleration at the surface is is 3 m/s2 ?
b) H
How ffar ffrom th
the surface
f
will
ill the
th particle
ti l go if it leaves
l
the
th asteroid’s
t id’ surface
f
with
ith a radial
di l speedd off
1000 m/s?
c) With what speed will an object hit the asteroid if it is dropped from 1000 km above the surface?
ag R 2
GM
2
a) a g = 2 =3 m/s ; M =
R
G
2GM
2GM
v=
=
= 2ag R = 1.732 ×103 m/s
R
R
b) Mechanical energy consevation:
Rf =
1 2 mM
mM
=-G
mvi -G
2
Ri
Rf
2GM
= 2.5 ×105 m/s
2GM
− vi2
Ri
c) Mechanical energy consevation: − G
mM
mM 1 2
= −G
+ mv f
Ri
Rf
2
1 2
1
1
1
1
v f = GM ( − ) ⇒ v f = 2GM ( − ) = 1.4 km/s
2
Ri R f
Ri R f
Why v f ≠ 2ag h = 2ag ( Ri − R f ) ?
Chapter 14: Fluids
What is a fluid??
S lid vs liquids
Solids
li id vs gasses
Pressure
F
P≡
A
“collisional
collisional force
force”
Units: pascal = Pa = N/m2
1 atmosphere = 1 atm
= 1.01×105 Pa
= 760 torr
= 14.7 lb/in2
The collisions of gas
molecules on the wall of the
tire keep it inflated
Uniform ⊥ force on
flat area
Hydrostatic Pressure: Water applies a force
perpendicular to all of the surfaces in the pool,
including the swimmer, the walls of the pool etc.
Density
m
ρ≡
V
Note: density solid > liquid > gas
air =
wood =
water =
Al =
Cu =
Pb =
1.21 kg/m3
550 kg/m3
1000 kg/m3
2700 kg/m3 = 2.7 g/cm3
8960 kg/m3 = 8.9 g/cm3
11340kg/m3 = 11.34 g/cm3
“Vacuum”