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Gravitation
• The Law of Universal Gravitation is based on the observed fact that all masses attract all other masses. The force of attraction decreases as the distance between the masses increases.
• This relationship is called an inverse square law since the decrease in attraction between objects is relative to the square of that distance.
• If the distance between the masses doubles, the force of gravitational attraction becomes ¼ of the original force. If the distance triples, the force becomes 1/9 of the original, as so on.
• For example, 22 = 4, 32 = 9, etc.
Weight & Mass
• Weight and mass measure different things. • Mass measures the quantity of matter which is present. It
represents the inertia property of matter meaning its ability of resist changes in motion. Mass is measured in grams, kilograms or slugs.
• Weight is a force resulting from the effect of gravity on a mass. A mass without gravity is weightless. Weight is measured in dynes, newtons or pounds.
• Gravity on the Earth’s surface is measured as – 980 cm/s2, - 9.8 m/s2 or – 32 ft/s2. (The negative sign means that gravity always acts downward)
• As we move above the Earth’s surface or to other planets, the strength of the gravitational field changes and so does the weight.
Planet
Force of gravity
Orb
ital
path
Force of gravity (weight) at the Earth’s surface
Fearth = G m1 me
re 2
Force of gravity (weight) at a point (P) above the
Earth’s surface
Fpoint P = G m1 me
rp2
w = m1g = G m1 me
r2
g = G me
r2
g r2 = Gme
gearth r2earth = Gmearth
gpoint p r2point P = Gmearth
Therefore
gearth r2earth =gpoint p r2
point P
Radius of Earth = 4000 miles
scale150 lbs
Two Radius of Earth = 8000 miles
scale37.5 lbs
Three Radius of Earth = 12000 miles
scale16.7 lbs
¼ wt
1/9 wt
Normalwt
Satellites
• When satellites orbit a planet the force which supplies the centripetal force, and thereby the circular motion, is the pull of gravity of the planet which is orbited.
• Without the force of gravity, the satellite would move in a straight line due to inertia.
• When the satellite orbits, the force of gravity must equal the centripetal force. If the force of gravity exceeded the centripetal force the satellite would spiral into the planet. If the centripetal force exceeded the force of gravity, the satellite would seek a wider orbit or move off in a straight line.
Earth
A satellite is a projectile shot from a very high elevationand is in free fall about the Earth.
Inertial position
Centripetal force
Centripetal force
Centripetal force
Centripetal force
Gravity suppliescentripetal forceinward towardsthe center of the
circular path
Fg = gravity force betweenm1 and m2 separated by
a distance r
G is the Universal Gravitational Constant
The weight of an object is its mass times g’, the
gravity value at location r
Planet
Force of gravityFg
Centripetal forceFc
Orb
ital
path
Fg = Fc
Fg = G m1 m2
r2
Fc = m v2
r
G m1 m2 = m1 v2
r2 rCanceling m1 & r on both sides
V2 = G m2
r
(1) V2 = G m2
r (2) V2 r= G m2
(3) V = ωr (4) ω= 2πf (5) T = 1/f
(6) ω = 2π / T (7) (ωr)2r = Gm2
(8) ω2 r3 = Gm2
(9) ( 2π / T)2 r3 = Gm2
(10) 4π2 r3 / T2 = Gm2
(11) T2 / r3 = 4π2/ Gm2 = a constantT2 / r3 = a constant
Kepler’sThird Law
Kepler’s Laws of Satellite Motion• (1) Satellites travel in elliptical paths. (The Earth and
the inner planets as well as the moon travel in nearly circular orbits. The orbits of the outer planets are more ellipsoid. Comets orbits are very elliptical.)
• (2) Areas swept out in equal times are equal even though the speed of the satellite varies. Satellite velocity is least when it is furthest from the central body (apogee) and greatest when it is nearest (perigee).
• (3) The period of motion squared divided by the average orbital radius cubed gives a constant for all satellites orbiting the same body. ( T2
1/ r31 = T2
2/ r32)
FORCE
(N)
DISPLACEMENT (M)
X1 X2
WORK = AREA UNDER THE CURVEW = F X (SUM OF THE BOXES)
WIDTH OF EACH BOX = X
AREA MISSED - INCREASINGTHE NUMBER BOXES WILL
REDUCE THIS ERROR!
AS THE NUMBER OF BOXESINCREASES, THE ERROR
DECREASES!
WEIGHT
(N)
Distance Above Center of Earth (m)
r1 r2
WORK = AREA UNDER THE CURVEW = F X (SUM OF THE BOXES)
r1 & r2 are are two pointsin the gravity field
Work = - G m1 m2
r
r2 = infinityr1 = radius of the Earth
K.E. work of lift satelliteK.E. = ½ mv2
½ m1 v2 = - G m1 m2
r
Vesc = 2 G m2 1/2
r
r2
r1
|
Velocity required to leave theEarth’s gravity field
)(
Gravitation & Satellite Problems(a)What is the velocity of a satellite 5000 km above
the Earth? (b)What is it period of rotation ?
5000 km
Re
r = re + hme = 6 x 10 24 kgRe = 6.4 x 10 6 m
(a) V2 = G m2 , v = G me 1/2
r rV = (6.67 x 10-11 )(6 x 1024) 1/2
(6.4 x 106 + 5 x 106)V = 5900 m/s
(b) V = 2π r / T, T = 2π r/ v T = 2π (11.4 x 106) / 5900 = 1.2 x 104 sec = 3.4 hr.
( )( )
Gravitation & Satellite ProblemsWhat is the gravitational force between two elephants which
are 2 meters apart. Each elephant weighs 3200 lbs.
• m = 3200lbs / 2.2 lbs/kg = 1455 kg (now using MKS units and the Law of Universal Gravitation)
• Fg = G m1 m2 = 6.67 x 10-11 (1455)(1455)
r2 22
• Fg = 3.5 x 10-5 N
Fg = G m1 m2
r2
W = mg
Gravitation & Satellite ProblemsFind the escape velocity of an object leaving the Earth (escape velocity is the velocity required for an object to
leave Earth’s gravity as a projectile).
• mearth = 6 x 1024 kg, rearth = 6.4 x 106 m, G = 6.67 x 10-11 N m2 / kg2
• V = 2 (6.67 x 10-11 )(6 x 1024 ) 1/2
(6.4 x 106 )• V = 1.12 x 104 m/s or 25,000 mph
Vesc = 2 G m2 1/2
r
Escape velocity formula
( )
( )
Gravitation & Satellite ProblemsFind the height of a geosynchronous Earth satellite. (A satellite which maintains its position at a fixed point above the Earth)
• Using Kepler’s third law
• And the facts that the moon is 242,000 miles (3.9 x 108 m) from Earth and its period is 27.3 days (2.36 x 106 sec)
• (27.3)2 / (242,000)3 = (1)2 / rsat3,
• rsat = ((12 x 242,0003) / 27.32))1/3 = 26,700 miles above Earth’s center or (26,700 – radius of Earth (4000 miles)) = 22,700 miles above the Earth’s surface.
In order to maintain its position theperiod of the satellite must equal that
of the Earth (24 hours or 1 day)
T21/ r3
1 = T22/ r3
2
Gravitation & Satellite Problems(a) Find the acceleration due to gravity at 1000 km
above the Earth. (b) What is the weight of a 70 kg man at this location ?
• (a) gearth r2earth =gpoint p r2
point P
• (9.8) x (6.4 x 106 )2 = gp (6.4 x 106 + 1 x 106)2
• gp = 7.33 m/s2
• (b) g at 1000 km above the Earth is 7.33 m/s2 and weight = mass x gravity, therefore• w = 70 kg x 7.33 m/s2 = 513 N
1000 km
Re
gearth r2earth =gpoint p r2
point P
rearth = 6.4 x 106 mgearth = 9.8 m/s2
What is the force of attraction between two 1000 kg objectsSeparated by 3 meters? Express the answer in newtons.
(A) 0.13 (B) 0.0000074 (C) 0.000032 (D) 320
Find the value of gravity at 2000 km above the Earth? Express the answerin meters per second squared.
(A) 32 (B) 5.7 (C) 16.9 (D) 0
A person weighs 128 lbs on Earth. What is his weight in lbs, at 6.4 x 106
meters above the Earth ?(A) 64 (B) 10.2 (C) 32 (D) weightless
Find the radius of the orbit of a satellite above the Earth with a periodof 12 hours. Express the answer in kilometers
(A) 3200 (B) 2670 (C) 7200 (D) 320
A satellite orbits the Earth with a radius of 8000 km. Find its velocityIn meters per second.
(A) 7000 (B) 5000 (C) 550 (D) 20,000
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