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Phy
Ch8 Universal Gravitation
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1
a] STATE KEPLER'S 1st LAW. [pg 134]
b] DRAW & LABEL AN ELLIPSE, [PRE CALC fig 9.47 pg
503,509-10]
c] STATE KEPLER'S 2nd LAW IN WORDS. [pg 134]
d] DRAW & LABEL THE DIAGRAM FOR KEPLER'S 2nd LAW. [pg
134]
e] STATE KEPLER'S 3rd LAW, DEFINE TERMS. [pg 134]
KEPLER'S LAWS
1
THE PATHS OF THE PLANETS ARE ELLIPSES, WITH THE SUN AT ONE FOCUS.
2
AN IMAGINARY LINE FROM A PLANET TO THE SUN SWEEPS OUT EQUAL AREAS
IN EQUAL TIMES.
3
R3
AVG ORBIT RADIUS CUBED
-- = ---------------------- = CONSTANT FOR ANY SYSTEM (= Primary &
Satelites)
T2
ORBIT TIME SQUARED
2
a] USE KEPLER'S 3rd LAW TO DERIVE THE EQUATION FOR ONE SATELLITE'S
ORBIT
RADIUS (Rs), IF ITS PERIOD (Ts) IS KNOWN ALONG WITH THE PERIOD AND
RADIUS FOR A
MOON (Tm & Rm).
(GIVEN Rm, Tm, & Ts, FIND Rs)
Kepler's Rs3 Rm3
Ts2
3rd Law: --- = --- Mult by Ts2-->Rs3
= Rm3--- --Cube Root-->Rs = Rm(Ts/Tm)2/3
Ts2 Tm2
Tm2
Notes: The satellite and moon can be any two
satellites, e.g. two planets
Use the same
units for Ts & Tm
Size check
(Ts/Tm)2/3 is less than Ts/Tm
b] EARTH IS 93x106
mi FROM THE SUN, WITH A PERIOD OF 365 d.
THE PERIOD OF MARS IS 687 DAYS.
DIAGRAM,
IDENTIFY EACH SYMBOL
Rm = Orbital Radius of Mars
Re = Orbital Radius of Earth (not Earth's radius).
Tm = Orbital Period of Mars
Te = Orbital Period of Earth = 365 days (not 24 hr)
DIAGRAM & FIND THE AVERAGE DISTANCE FROM MARS TO SUN IN
mi.
Rm = Re(Tm/Te)2/3
=
93x 106
mi(687 d/365 d)2/3 = 142x106
mi
PRACT 1:
Tp OF VENUS IS 225 DAYS.
PRACT 2:
Tp MERCURY IS 88 DAYS.
3
USE THE INFORMATION ABOUT A PLANET & ITS MOON TO FIND THE RADIUS
& ALTITUDE OF
A SATALLITE WITH A SYNCHRONOUS ORBIT.
EARTH Rp = 4000 mi,
MOON ORBIT RADIUS = 240,000 mi; PERIOD = 27.32 DAYS [27d 7h
43m 11.5s]
GIVEN:
Ts = 24h; Tm = 27.3d, Rm = 240,000 mi, Re = 4000 mi;
FIND Rs
R3/T2 = k --->
Ra3/Ta2
= Rb3/Tb2
-----> Ra = Rb[Ta/Tb]2/3
Tb = 27.3d*24h*3600 = 2.36 x 106,
Tb2 = 5.564 x 1012
Ta = 24h*3600s/h = 86,400 s
Ta2 = 7.465 109
.
Ra = Rb ∛7.465 109/5.564 1012]
= R2 0.1103 = 26,472 mi
ALTITUDE = Ra - Re = 26,472 - 4,000 = 22,472 mi
PRACT 1: PLUTO;
1 DAY = 153.6 hr; PLANET RADIUS [Rp] = 0.5 UNITS.
MOON'S ORBIT [Rm] = 27 UNITS; PERIOD [Tm] = 30 EARTH DAYS
PRACT 2: JUPITER,
1 DAY = 10 hr, RADIUS IS 0.5 UNITS
MOON IO: Rm =
4.2 UNITS, & Tm = 1.8 EARTH DAYS.
JUPITER HAS 4 MOONS. IO MEASURES 4.2 UNITS FROM THE CENTER OF JUPITER,
HAS A PERIOD OF 1.8 DAYS, AND GANYMEDE HAS A ORBIT OF RADIUS
10.7 UNITS.
FIND THE PERIOD OF GANYMEDE
GIVEN: FOR IO Tb
= 1.8d, & Rb = 4.2 UNITS;
FOR GANYMEDE Ta = ?,
& Ra = 10.7 UNITS.
R3/T2 = k ---> Ra3/Ta2 = Rb3/Tb2 -----> Ta =
TbSQRT[Ra3/Rb3]
Ta = Tb[Ra/Rb]3/2
Ra = 10.7, Rb = 4.2 ----> Ra/Rb = 2.5476
Ta = 1.8d[2.5476]3/2 = 1.8d*4.066 = 7.319 d --sd--> 7.3 d
4
a] DESCRIBE NEWTON'S 3 CONTRIBUTIONS TO THE THEORY OF GRAVITY. [see b]
A HE POSTULATED
THAT GRAVITY WAS NOT UNIQUE TO THE EARTH--ANY TWO
BODIES WILL PULL
ON EACH OTHER.
B HE POSTULATED
THAT THE MASS OF A BODY CAN BE TREATED AS IF IT WAS
ALL CONCENTRATED
AT ONE POINT--THE CENTER OF GRAVITY.
C HE ANALYZED THE
MOON & USED KEPLER'S 3rd LAW TO DERIVE THE
EQUATION FOR
GRAVITY.
D HE SAW THE MOON
(PLANETS) AS FALLING AROUND THE EARTH (SUN).
E HE TESTED HIS
EQUATION FOR GRAVITY BY COMPARING EARTH'S g AT
240,000 mi WITH V2/R FOR THE MOON.
b] WRITE THE EQUATION FOR THE LAW OF GRAVITY, DEFINE THE
SYMBOLS.
DRAW & LABEL THE
DIAGRAM.
Fg = GM1M2 Fg = Force of Gravity
d2
G = Universal Gravitational Constant = 6.67 10-11Nm2/kg2
M1 & M2 = Masses
d = Distance between the centers of gravity for M1 & M2
c] ANALYZE THE MOON, & SUBSTITUTE KEPLER'S 3rd LAW INTO
THE EXPRESSION FOR
CENTRIPETAL FORCE TO FIND THE EQUATION FOR GRAVITY.
Gravity = Centrepital Force
Fg = mV2/Ro
Ro & To = Radius & Period of
Orbit
V = 2pRo/To
m = Mass of Moon
Substituting for V ---> Fg = m4p2Ro2/To2
= m4p2Ro/To2
-----------
Ro
Kepler's 3rd Law: To2/Ro3
= k ---> To2 = kRo3
Substituting for To2 --->Fg = m4p2Ro/kRo3
Simplifying: Fg = m4p2/kRo2
Let 4p2/k
= a new constant K ---> Fg = mK/Ro2
Since every object must pull on every other object K must
include the mass of
the Earth, i.e., K = GMe, Me = Earth's Mass.---> Fg = GMem/Ro2
Generalizing this equation to all objects we have:
Fg = GM1M2/d2 , d = the distance between their
centers of gravity.
5
a] DRAW THE DIAGRAM FOR CAVENDISH'S EXPERIMENT [pg 162]
b] TWO LEAD SPHERES [20 kg & 30 kg] ARE SUSPENDED CLOSE
TO EACH OTHER.
THEIR CENTERS ARE 0.25 m APART. THE
FORCE BETWEEN THEM IS 6.4 10-7
N. FIND
G.
Given: M1 = 20
kg
Fg
= GM1M2/d2
M2 = 30
kg
6.4 10-7
N = G(20kg)(30Kg)/(0.25m)2
d =
0.25
m 6.4 10-7
N = G 9600 kg2/m2
Fg = 6.4 10-7
N G = 6.4 10-7
N/9600 kg2/m2
Find G =
?
= 6.67 10-11
N/kg2/m2
6
A 300,000 kg AND A 150,000 kg LEAD SPHERE ARE SUSPENDED CLOSE TO EACH OTHER.
THEIR CENTERS ARE 3.3 m APART.
(Visualize two masses equal to 300 & 150 cars) [0.277N = 1 oz].
WHAT IS THE GRAVITATIONAL FORCE BETWEEN THEM IN lbs?
GM1M2 6.67 10-11Nm2x3x105kgx1.5x105kg 6.67 10-11Nx1010x4.5
Fg = ----- = ------------------------------ = -------------------
R2
kg2 x 3.32m2
10.89
30.02 10-1N
= -----------
= 0.2756 N = 1 oz
10.89
PRACT 1: 100 TONS & 200 TONS, 6 m.
PRACT 2: 10 kg &
20 kg, 4 m
7 a] USE THE LAW OF GRAVITY TO DERIVE THE EQUATION FOR g.
M1 = Mass of a planet (Mp)
M2 = Mass of an object (m)
d = Planet radius.
Fg = Weight
GM1M2/d2 = Weight
GMpm/Rp2 = mg
---Divide by m---> GMp/Rp2> = g
That is, g is determined by the
planets mass and radius.
b] WHAT IS THE MASS OF THE EARTH (Me)? [R = 6,380,000 m = 4,000 mi]
GMe/Re2 = g
6.67 10-11
N/kg2/m2 Me/(6,380,000
m)2 = 9.8 m/s2
9.8 m/s2 x
(6,380,000
m)2
Me = -------------------------
6.67 10-11
N/kg2/m2
Me = 59.8x1023 kg
8A
THE PLANET OF AN OBJECT IS CHANGED.
DERIVE FOR THE EQUATION FOR NEW WEIGHT.
W' = We(M'/Me)(Re/R')2
We = Weight on Earth
Wx = Weight on Planet X
Wx = G Mx m
G Mx m
Rx2
Wx Rx2
We = G Me m Divide
Wx by We ---> --- = -------
Re2
We G Me m
Re2
Cancelling G & m
Mx
Wx Rx2
-- = ----
We Me
Re2
Dividing by Me/Re2--> Wx Mx(Re)2
--
= -------
We Me(Rx)2
8B
HOW MUCH DOES A 135 lb PERSON WEIGH ON PLANET X, IF IT HAS TWICE THE
MASS OF EARTH AND
THREE TIMES THE RADIUS?
Mx = 2 Me
Rx = 3 Re
Wx = ?
Wx Mx(Re)2 Wx
2Me(Re)2
--
= ------- ---> ------ = ---------
We Me(Rx)2 135
lb Me(3Re)2
Wx 2(1)2
------ = --------- = 2/9
135 lb (3)2
Wx = 135 lb x 2/9 = 30 lb
PRACT 3: 200
lb, 3xM & 1/2xR PRACT 4: 150 lb, 4xM & 1/4xR
8C
A 180 lb BOX IS MOVED TO ALTITUDE OF 8,000 mi, [Re = 4,000 mi].
a] DRAW & LABEL THE DIAGRAM; [b]
WHAT WILL IT WEIGH?
Mx = Me
Rx = Re + Altitude = 12,000 mi
Wx = ?
Wx Me(Re)2 Wx
(4,000)2
--
= ------- ---> ------ = ---------
We Me(Rx)2 180
lb (12,000)2
Wx (1)2
------ = -------- = 1/9
180 lb (3)2
Wx = 180 lb x 1/9 = 20 lb
PRACT 1: 200
lb, 4000 mi PRACT 2: 400 lb, 12,000 mi
9
ONE CAN ANALYZE A MOON TO DERIVE
4p2Rmoon3
THE EQUATION FOR A PLANET'S MASS.
Mp = ---------
G*Tmoon2
a] IF DIST TO EARTH = 148.8 109
m, FIND THE SUN'S MASS.
b] HOW MANY EARTHS = THE MASS OF THE SUN?
PRACT 1:
IF THE CENTERS OF THE EARTH & MOON ARE 3.8 108 m APART, AND THE
MOON'S PERIOD IS 27.32 DAYS, FIND THE EARTH'S MASS.
PRACT 2:
JUPITER HAS A MOON WITH AN ORBITAL RADIUS OF 4.27 108
m,
AND A PERIOD OF 1.8 d.
a] FIND JUPITER'S MASS.
b] HOW MANY JUPITERS = THE MASS OF THE SUN?
c] HOW MANY EARTHS = THE MASS OF JUPITER?
10
WHAT FORCE: a] HOLDS TEA IN A CUP?
b]
MAKES BUBBLES RISE?
c] MADE THE EARTH ROUND? d]
CAUSES THE NUCLEAR REACTIONS IN STARS?
11
WHAT ARE THE TWO MASSES AND THE ONE DISTANCE THAT DETERMINS YOUR WEIGHT?
12
CALCULATE YOUR WEIGHT IF YOU WERE FIVE TIMES FARTHER FROM THE CENTER OF
THE
EARTH THAN YOU ARE NOW? TEN
TIMES?
13
a] DIAGRAM A PLANETARY PERTURBATION.
b] WHAT CAUSES PLANETARY PERTURBATIONS?
14
SINCE THE PLANETS ARE PULLED TO THE SUN BY GRAVITATIONAL ATTRACTION, WHY
DON'T
THEY SIMPLY CRASH INTO THE SUN? 1
ANS: 5) 6.67 10-11 Nm2/kg2
6) 1 oz, (P1) 0.134 oz, (P2) 8.34 10-9
N
7) 59.8 1023 kg
8b) 20 lb; (P1) 50 lb; (P2) 25 lb
8c) 30 lb; (P3) 2400 lb; (P4) 9600 lb
9a) 1.96
1030
kg,
b)
327,800;
P1) 59.8 1023
kg,
P2) 1.96 1027
kg, 1032, 319
P3) 2.935 1071
kg, 1.498 1044
10) Gravity
11) Your mass, the mass of the Earth, & the Earth's radius.
12) Wx = We (1/5)2 = We/25
Wx = We (1/10)2 = We/100
13b) The pull of gravity from a near by planet.
14) A planet's tangential velocity is so great that they are falling around
the
sun.
Old Notes
1
DERIVE THE EQUATION FOR THE PRIMARY'S MASS FORM THE PERIOD & ORBIT
RADIUS OF
A SATELLITE.
2
SHOW HOW WOULD YOU DETERMINE THE ACCELERATION OF GRAVITY ON ANOTHER
PLANET
FROM EARTH--WHAT WOULD YOU MEASURE & CALCULATE--SHOW EQUATIONS.
STEP 1: MEASURE A MOON'S ORBITAL
RADIUS & PERIOD [Rm & Tm]
STEP 2: CALCULATE THE PLANET'S MASS
Mp: Mp = 4pi2Rm3/Tm2
STEP 3: MEASURE THE PLANET'S
RADIUS, Rp;
STEP 4: FINALLY, CACULATE g:
g = GMp/Rp2
DEFINE
C of G.
THE CENTER OF GRAVITY OF AN OBJECT IS THE POINT LOCATED AT THE CENTER OF AN
OBJECT'S WEIGHT DISTRIBUTION.
THERE IS A SMALL DIFFERENCE BETWEEN CENTER OF GRAVITY AND CENTER OF MASS WHEN
THE OBJECT IS LARGE ENOUGH FOR GRAVITY TO VARY FROM ONE PART TO ANOTHER.
3 THE MOON DOESN'T REVOLVE AROUND ___.
THE EARTH & MOON REVOLVE AROUND CENTER
OF MASS OF THE EARTH-MOON
SYSTEM.
4
AN OBJECT WILL REMAIN UPRIGHT IF ITS CG IS ABOVE THE AREA OF SUPPORT.
5 UNSTABLE OBJECTS WILL CHANGE POSITION UNTIL ITS CG IS THE LOWEST
POSSIBLE.
THERE
ARE 3 TYPES OF STABILITY:
A UNSTABLE--ANY
MOTION CAUSES THE CG TO BE LOWERED--NO WORK
REQUIRED. EX: A CONE ON
ITS TIP
B STABLE--ANY
MOTION CAUSES THE CG TO BE RAISED & WORK MUST BE
DONE. EX: A CONE ON
ITS BASE
C NEUTRAL--ANY
MOTION CAUSES NO CHANGE IN THE ELEVATION OF THE CG,
NO WORK IS DONE.
EX: A CONE ON ITS SIDE.
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