Presentation on theme: "Copyright Sautter 2003. Gravitation The Law of Universal Gravitation is based on the observed fact that all masses attract all other masses. The force."— Presentation transcript:
Copyright Sautter 2003
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, 2 2 = 4, 3 2 = 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/s 2, m/s 2 or – 32 ft/s 2. (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 Orbital path Force of gravity (weight) at the Earth’s surface F earth = G m1 m1 meme re re 2 Force of gravity (weight) at a point (P) above the Earth’s surface F point P = G m1 m1 meme rp2rp2 w = m 1 g = G m1 m1 meme r2r2 g = G meme r2r2 g r 2 = Gm e g earth r 2 earth = Gm earth g point p r 2 point P = Gm earth Therefore g earth r 2 earth =g point p r 2 point P
Radius of Earth = 4000 miles scale 150 lbs Two Radius of Earth = 8000 miles scale 37.5 lbs Three Radius of Earth = miles scale 16.7 lbs ¼ wt 1 / 9 wt Normal wt
scale 150 lbs scale 25.6 lbs scale 406 lbs g = 9.81 m/s 2 g = 1.67 m/s 2 g = 26.6 m/s 2
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.
Velocity Vectors Acceleration Vectors Force Vectors
Earth A satellite is a projectile shot from a very high elevation and is in free fall about the Earth.
Inertial position Centripetal force Gravity supplies centripetal force inward towards the center of the circular path
F g = gravity force between m 1 and m 2 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 gravity F g Centripetal force F c Orbital path Fg Fg = FcFc Fg Fg = G m1 m1 m2m2 r2r2 Fc Fc = m v2v2 r G m1 m1 m2 m2 = m1 m1 v2v2 r 2 r Canceling m1 m1 & r on both sides V2 V2 = G m2m2 r
(1) V 2 = G m 2 r (2) V 2 r= G m 2 (3) V = ωr (4) ω = 2πf (5) T = 1/f (6) ω = 2π / T (7) (ωr) 2 r = Gm 2 (8) ω 2 r 3 = Gm 2 (9) ( 2π / T) 2 r 3 = Gm 2 (10) 4π 2 r 3 / T 2 = Gm 2 (11) T 2 / r 3 = 4π 2 / Gm 2 = a constant T 2 / r 3 = a constant Kepler’s Third 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. ( T 2 1 / r 3 1 = T 2 2 / r 3 2 )
Velocity increases (perigee) Velocity decreases (apogee) Equal Areas In Equal Times
F O R C E (N) DISPLACEMENT (M) X1X1 X2X2 WORK = AREA UNDER THE CURVE W = F X (SUM OF THE BOXES) WIDTH OF EACH BOX = X AREA MISSED - INCREASING THE NUMBER BOXES WILL REDUCE THIS ERROR! AS THE NUMBER OF BOXES INCREASES, THE ERROR DECREASES!
W E I G H T (N) Distance Above Center of Earth (m) r1r1 r2r2 WORK = AREA UNDER THE CURVE W = F X (SUM OF THE BOXES) r 1 & r 2 are are two points in the gravity field
Law of Universal Gravitation Fg Fg = G m1 m1 m2m2 r2r2 Work = Fdr = G m1 m1 m 2 dr r2r2 Work = - G m1 m1 m2m2 r r2r2 r1r1 |
Work = - G m 1 m 2 r r 2 = infinity r 1 = radius of the Earth K.E. work of lift satellite K.E. = ½ mv 2 ½ m 1 v 2 = - G m 1 m 2 r V esc = 2 G m 2 1/2 r r2r2 r1r1 | Velocity required to leave the Earth’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 ReRe r = r e + h m e = 6 x kg R e = 6.4 x 10 6 m (a) V 2 = G m 2, v = G m e 1/2 r r V = (6.67 x )(6 x ) 1/2 (6.4 x x 10 6 ) V = 5900 m/s (b) V = 2π r / T, T = 2π r/ v T = 2π (11.4 x 10 6 ) / 5900 = 1.2 x 10 4 sec = 3.4 hr. () ()
Gravitation & Satellite Problems What 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) F g = G m 1 m 2 = 6.67 x (1455)(1455) r F g = 3.5 x N F g = G m 1 m 2 r 2 W = mg
Gravitation & Satellite Problems Find 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). m earth = 6 x kg, r earth = 6.4 x 10 6 m, G = 6.67 x N m 2 / kg 2 V = 2 (6.67 x )(6 x ) 1/2 (6.4 x 10 6 ) V = 1.12 x 10 4 m/s or 25,000 mph V esc = 2 G m 2 1/2 r Escape velocity formula () ( )
Gravitation & Satellite Problems Find 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 10 8 m) from Earth and its period is 27.3 days (2.36 x 10 6 sec) (27.3) 2 / (242,000) 3 = (1) 2 / r sat 3, r sat = ((1 2 x 242,000 3 ) / )) 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 the period of the satellite must equal that of the Earth (24 hours or 1 day) T 2 1 / r 3 1 = T 2 2 / r 3 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) g earth r 2 earth =g point p r 2 point P (9.8) x ( 6.4 x 10 6 ) 2 = g p (6.4 x x 10 6 ) 2 g p = 7.33 m/s 2 (b) g at 1000 km above the Earth is 7.33 m/s 2 and weight = mass x gravity, therefore w = 70 kg x 7.33 m/s 2 = 513 N 1000 km ReRe g earth r 2 earth =g point p r 2 point P r earth = 6.4 x 10 6 m g earth = 9.8 m/s 2
What is the force of attraction between two 1000 kg objects Separated by 3 meters? Express the answer in newtons. (A) 0.13 (B) (C) (D) 320 Find the value of gravity at 2000 km above the Earth? Express the answer in 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 10 6 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 period of 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 velocity In meters per second. (A) 7000 (B) 5000 (C) 550 (D) 20,000 Click here for answers