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Physics 2101 Physics 2101 Section 3Section 3
March 22March 22rdrd : Ch.13: Ch.13--1414
Announcements:Announcements:•• More on gravity.More on gravity.•• Start Ch. 14Start Ch. 14Start Ch. 14Start Ch. 14•• Exam. 3 on Mar. 24Exam. 3 on Mar. 24thth
Class Website: http://www.phys.lsu.edu/classes/spring2010/phys2101http://www.phys.lsu.edu/classes/spring2010/phys2101--3/3/http://www.phys.lsu.edu/~jzhang/teaching.htmlhttp://www.phys.lsu.edu/~jzhang/teaching.html
ChaptChapt. 13: review. 13: review
F 12 = G m1m2
r 2ˆ r 12Force on 1 due to 2:
n
r12
G = 6.67 ×10−11 N ⋅ m 2 /kg2
F 1,net = F 12 + F 13 + F 14 + ...+ F 1n = F 1i
i=1
n
∑ VECTOR ADDITION!!Force on 1 due to many:
Gravitational potential energy
If we define U = 0 at ∞, then the work done by taking mass m from R to ∞
F 12 = G m1m2
2ˆ r 12U(r) = −G m1m2
12 G
r122 12U( ) G
r12
Planets and Satellites: Planets and Satellites: Kepler’sKepler’s LawsLawsKepler Brahe Newton
1571-1630 1546-1601 1642-1727
1)1) Law of Orbits:Law of Orbits:1)1) Law of Orbits:Law of Orbits:All planets move in elliptical orbits with the Sun at one focusAll planets move in elliptical orbits with the Sun at one focus
Assumes MSun >> MEEllipse criterion:Fm + F’m = const
sun earth earth moona = semimajor axise = eccentricity
sun-earthfarthest point Ra = aphelionclosest point Rp = perihelion
earth-moon= apogee= perigee
Ra = a(1+ e)Rp = a(1− e)
Planets and Satellites: Planets and Satellites: Kepler’sKepler’s 22ndnd LawLaw2) Law of equal areas :2) Law of equal areas :
- Planets move slowest when furthest away from Sun (at Ra)Pl t f t t h l t f S ( t R )
) q) qA line that connects a planet to the sun sweeps out equal areas in equal time A line that connects a planet to the sun sweeps out equal areas in equal time
- Planets move fastest when closest away from Sun (at Rp)
ANGULAR MOMENTUM IS CONSERVED
Angular momentum L = r × p
L = rp = r mv( )= r m(ωr)( )= mr 2ω
Lm
= r 2ω
( )12
1 12 2
dA rd rdA dr r
θθ ω
=
= =
Area of “triangle”:
dA L2 2r rdt dt
ω= = dAdt
=L
2m= constant
Planets and Satellites: Planets and Satellites: Kepler’sKepler’s 33rdrd LawLaw
3) The Law of Periods : 3) The Law of Periods : The square of the period of any planet is proportional to the cube of the The square of the period of any planet is proportional to the cube of the semimajorsemimajor axis axis
of its orbitof its orbit
−Gmsat M E
2 = −msat
2πrsat
T⎛ ⎝ ⎜
⎞ ⎠ ⎟
2
⇒ rsat3 =
GM2
⎛
⎝⎜
⎞
⎠⎟ T 2
rsat2 sat rsat
sat 4π 2⎝ ⎜
⎠ ⎟
⇒ T 2 =4π 2⎛
⎜ ⎞ ⎟ r 3
NOTE: r3 ∝ T2
⇒ T =GM⎝
⎜ ⎠ ⎟ rsat
Circular OrbitCircular Orbit Elliptical OrbitElliptical OrbitCircular OrbitCircular Orbit
When used in the equations, r When used in the equations, r (radius) and a ((radius) and a (semimajorsemimajor axis) axis) are synonymous are synonymous
Example: Hot JupiterExample: Hot JupiterIn 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very close the star HD 179949 The orbit as abo t 1/9 the distance of Merc r from o r s n and it takes thestar HD 179949. The orbit was about 1/9 the distance of Mercury from our sun, 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?
3/ 25 92 1a) 2.67 10 s and 6.43 10 m
9 Mercurystar
rT r rGMπ
= = × = = ×
2 330
2
4 2.21 10 kg 1.11
star
star SunrM M
T Gπ
= = × =
52 rπ 52b) 1.51 10 m/srvTπ
= = ×
Conservation of Mechanical EnergyConservation of Mechanical EnergyEscape speed: Minimum speed (v ) required to send a mass m, from mass M andEscape 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: Emech= 0 because U = 0 and KE = 0y mechThus any other place we have:
1 GmM⎛ ⎞ 2GMEmech = KE +Ug( )= 0 ⇒ Emech =
12
mv2 −GmM
R⎛ ⎝ ⎜
⎞ ⎠ ⎟ = 0 ⇒ vescape =
2GMR
Earth = 11.2 km/s (25,000 mi/hr)( , )Escape speed: 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 ?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 13-37a) 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 f f th f ill th ti l if it l th t id’ f ith di l d fb) 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?
2a RGM 2g 2
3
a) a =3 m/s ;
2 2 2 1.732 10 m/s
g
g
a RGM MR GGM GMv a RR R
= =
= = = = ×R R
2
i f
1 mM mMb) Mechanical energy consevation: -G =-G2 R Rimv
i f
5
2
2 R R2 2.5 10 m/s2f
ii
GMR GM vR
= = ×−
i
21c) Mechanical energy consevation: 2 f
i f
mM mMG G mvR R
− = − +
21 1 1 1 1 ( ) 2 ( ) 1.4 km/s2 f f
i f i f
v GM v GMR R R R
= − ⇒ = − =
Why 2 2 ( ) ?f g g i fv a h a R R≠ = −
PressurePressure FPA
≡ “collisional force”A collisional force
Units: pascal = Pa = N/m2
1 atmosphere = 1 atm = 1.01×105 Pa = 760 torr
The collisions of gas molecules on the wall of the
760 torr = 14.7 lb/in2
molecules on the wall of the tire keep it inflated
Uniform ⊥ force on flat area
Hydrostatic Pressure: Water applies a forceHydrostatic Pressure: Water applies a force perpendicular to all of the surfaces in the pool, including the swimmer, the walls of the pool etc.