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notes_11 - CHAPTER 11 THE GRAVITATIONAL FIELD(GRAVITY...

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Unformatted text preview: CHAPTER 11 THE GRAVITATIONAL FIELD (GRAVITY) GRAVITATIONAL FIELD: The groundwork for Newton’s great contribution to understanding gravity was laid by three majors players: • Newton’s Law of Gravitation o gravitational and inertial mass Copernicus (1473-1543) • Gravitational potential energy o satellites o Kepler’s 3rd Law o escape speed Galileo (1564-1642) • Gravitational field inside spheres* o hollow sphere o solid sphere • shape of planetary orbits Kepler (1571-1630) • speed around an orbit • relating time around orbit to distance from * You study using notes provided. the Sun 1 m2 m1 Consider the apple that allegedly fell in Newton’s orchard. r mm F = G 12 2 r distance between cm’s mE r FAE mA r FEA A value for G, the universal gravitational constant, can be obtained from measurements by Henry Cavendish (1731-1810) of the specific gravity of the Earth, as we shall show shortly. Cavendish’s value for the specific gravity of the Earth was 5.48, i.e., the density of the Earth is 5.48 times the density of water: ∴ ρEarth = 5.48 × 1000 kg ⋅ m−3 = 5480 kg ⋅ m−3. 4 ∴ mE = volume × density = πR E 3ρE 3 3 4 = π 6.40 × 106 × 5480 = 6.02 × 1024 kg. 3 ( ) So, Cavendish’s experiment allowed the first Newton’s 3rd Law tells us that: r r FEA = −FAE . Using the 2nd Law we have: r r mA a A = mE a E . But mA ≈ 0.1kg and mE ≈ 6 × 1024 kg. ∴ a E ≈ 1.67 × 10 −25 a A . Since near the Earth’s surface, a A = g, a E = 1.64 × 10 −24 m/s2. determination of the mass of the Earth; it is within 1% of So, the Earth also experiences an acceleration towards today’s accepted value. the apple, but it is insignificant! 2 Re-arranging the expression for the universal Let’s take a closer look gravitational constant, gives us a method for determining mA mE m g=G 2 R r = h + RE the acceleration at the surface of a massive object, is valid universally. So, if we know the mass ( mp) and radius ( rp) of a planet we can determine the The force on the apple (i.e., its weight) mm mE mA FEA = G E 2 A = G , (R E + h )2 r acceleration due to gravity at the surface because G is a universal constant! and the acceleration of the apple is F mE a = EA = G . mA (R E + h )2 gp = G R E2 g mE 2 where R E is the Earth’s radius. We know g = 9.81m/s , mE = 6.02 × 1024 kg and R E ≈ 6.4 × 106 m, R p2 Here are some examples ... Very close to Earth’s surface, h << R E and a ≡ g , so G= mp Earth mass = 6 × 10 24 kg radius = 6378km g = 9.8m/s 2 ∴ G = 6.67 × 10 −11 N ⋅ m2 /kg2. 3 Rp But look at the Earth! Mars mass = 6.42 × 10 23 kg radius = 3393km g Mars = 3.7 m/s 2 = 0.38 g R eq g=G mE RE2 ... it’s really a “flattened” sphere with R eq > R p. Moon mass = 7.35 × 10 22 kg radius = 1738km g Moon = 1.6m/s 2 = 0.165g ∴ geq < g p. In fact, g p ~ 9.83m/s2 and geq ~ 9.78m/s2 . Since ones weight is mg, one weighs less • up a mountain, • in an airplane, In deriving the expression for g we have made an assumption ... that the Earth and planets are spherical ! than at sea-level! At 30,000ft: ∆w ~ −0.3%. w 4 Gravitational and inertial mass Gravitational potential energy So far, without stating it explicitly we have used the The acceleration of gravity same “mass” (m) in Newton’s Law of gravitation as was defined by Newton’s 2nd Law in chapter 4. Note that in r = h + RE the definition of the inertial mass of an object, F inertial mass = minert = , a at a distance h above the Earth’s surface is: mE mE g=G 2 =G . ( h + R E )2 r there is no mention of gravity. The mass used in (Note: it is never zero so you are never completely Newton’s Law of gravitation is called the gravitational weightless!!) mass, which can be determined by measuring the attractive force exerted on it by another mass (M) a 9.81 distance r away, viz: r 2FMm gravitational mass = mgrav = . GM 8.0 g ( m/s 2 ) R E = 6.38 × 10 6 m 6.0 There is no mention of acceleration in this expression. 4.0 Although these are two different concepts of mass, mE = 5.98 × 10 24 kg 2.0 Newton asserted that, for the same object, these masses are identical, i.e., mgrav = miner . This is known as the principle of equivalence. 0 5 10 15 20 Height above Earth’s surface h ( 25 ×10 3 km ) 5 Near the Earth’s surface the gravitational potential energy of an object of mass m is Ug = mgh = mg( r − R E ) , where Ug = 0 at the Earth’s surface. Far from the Earth we must take into account the fact that g ∝ r −2. In chapter 6 we defined the change in potential energy as rr dU = −F • ds , r where F is a (conservative) force acting on the object r and d s is a general displacement. ˆ r mE r dr Fg m mm dr mm ∴ U( r ) = ∫ G E dr = GmE m∫ 2 = −G E + Uo, 2 r r r where Uo is an integration constant. We choose Uo = 0, i.e., we define the gravitational potential energy to be zero at r = ∞ . U( r ) U( r ) = −G RE φ r ds r r ˆ dr = ds cos φ = r • ds r 0 U( R E ) ← −G mE m r mE m = − mgR E RE We obtain a negative value for U(R E ) because we defined U(∞) = 0 . In earlier chapters we usually defined For the gravitational force, r mm r r r ˆ) ˆ dU = −Fg • ds = −( −Fg r • ds = Fg r • ds = G E dr . r2 U(R E ) = 0. However, it does not cause difficulties as we normally work with changes in potential energy, i.e., ∆U, which is independent of our choice of Uo. 6 We have mm mE m mE m U( r ) = −G E = −G = −G r (R E + h ) RE 1 + h R ( E ) . r Since g is a vector field, the resultant gravitational field of a system of masses is a vector sum of the individual contributions. For small values of h, h R << 1, then E m m h U( r ) = −G E 1 + RE RE −1 m m h = − G E 1 − RE RE mm mm = −G E + G E 2 h . RE RE mm m But G E = U(R E ) is constant and G E = g. RE R E2 ∴ U( r ) = U(R E ) + mgh , Question 1: Five objects, each of mass 3.00kg are are equally spaced on the arc of a semicircle of radius 10.0cm. An object of mass 2.00kg is located at the center of curvature of the arc. (a) What is the gravitational force on the 2.00kg mass? (b) What is the gravitational field at the center of curvature of the arc? i.e., in moving an object a height h above the Earth’s surface increases the gravitational potential energy by 3.00 kg ∆U = mgh , providing h << R E , a result we used in chapters 6 and 7. r r r Note that the weight of an object is Fg = mg , i.e., g is a vector and is called the gravitational field. 3.00kg 3.00 kg 10.0cm 3.00 kg 2.00kg 3.00kg 7 y M ˆ j M M ˆ i R M m M x (a) The force acting on m due to each of the masses M is mM mM mM mM Fx = G 2 + G 2 sin 45o − G 2 sin 45o − G 2 R R R R = 0, and mM mM mM Fy = G 2 sin 45o + G 2 + G 2 sin 45o R R R mM = G 2 1 + 2 sin 45o R ( r r (b) The gravitational force acting on m is F = mg . r i j j r F Fx ˆ + Fy ˆ 9.66 × 10 −8 ˆ ∴g = = = 2.00 m m = 4.83 × 10 −8 ˆ m/s2 . j ) 6.67 × 10 −11 × 2.00 × 3.00 = × 2.414 (0.100)2 = 9.66 × 10 −8 N j (in the ˆ direction). 8 We will model the system y as a sphere of mass 4 M = πR 3ρo 3 x r R Question 2: A solid sphere of radius R and density ρo has its center at the origin. There is a hollow, spherical R cavity of radius r = within the sphere and centered at 2 R x = , as shown. Find the gravitational field at points on 2 and a sphere of radius R 2 with negative mass ( − m). Then g( x ) = gsolid ( x ) + gcavity ( x ) M ( − m) =G 2 +G 2 x x − R2 ( the x-axis for x > R . y = 4 Gρo πR 3 3 x2 ) 3 4 Gρo − π R 2 3 + 2 x − R2 () ) ( x R r 4πρoR 3 1 1 = G 2 − 3 x 8 x − R2 ( 2. ) When x = R g(R ) = G 2πρoR . 3 9 We use conservation of energy, i.e., Ki + Ui = K f + Uf . Using the expression for Kf = 0 : U( R E + h) gravitational potential h 1 K i = mv i 2 : U( R E ) 2 energy, we get 1 mm mv i2 − G E 2 RE =0−G Question 3: An object is fired upward from the surface of the Earth with an initial speed of 4.0km/s. Find the maximum height it achieves. Compare that height with the height it would have achieved if the gravitational field were constant. mE m . (R E + h ) Solving for h we find RE RE h= = , 2GmE 2 gR E − 1 − 1 R E v i2 v i2 GmE where g = is the gravitational field at the R E2 Earths’s surface. ∴h = 6370 × 103 2 × 9.81 × 6370 × 103 − 1 (4.0 × 103 )2 = 9.35 × 105 m (935km). 10 Satellites If the gravitational field (g) were constant, then v m ∆K = − ∆U = mgh ′ , r v 2 (4.0 × 103 )2 = 815.5km. i.e., h ′ = i = 2g 2 × 9.81 In the first part we found the “correct” height was RE h= . 2 gR E − 1 v i2 v2 But, from above, i = h ′ , where h ′ is the height for a 2g constant gravitational field. RE h ′R E = ∴h = . R E − 1 (R E − h ′ ) h′ Check h ( km) = 815.5 × 6370 = 935km. (6370 − 815.5) ME Consider a satellite, mass m, in a circular orbit, radius r, around the Earth (or a planet around the Sun). The gravitational force between the satellite and the Earth provides the centripetal acceleration. mE m mv 2 ∴G 2 = . r r So, in a stable orbit GmE . r 2 πr The time for one orbit T = v v2 = i.e., v 2 = 4π2 r 2 mE 2 = G r = g′ r , T where g′ is the gravitational field acting on the satellite. 11 constant ∴ T2 = Consider the Moon as the satellite around the Earth. 4π2 3 r i.e., T2 ∝ r 3. GmE v 3840 km v= G mE r ~ Kepler’s 3rd Law of planetary motion ~ Note also m v= G E. r i.e., the velocity of a satellite in a stable orbit does not depend on its mass only the mass of the central object! so what .. ? ... well, it provides us a way of working out the mass of a planet. Take the Earth-Moon system for example ... The speed of the Moon is distance around orbit 2πr v= = time of orbit T = 2π × 3.84 × 108 = 1023m/s. 27.3 × 24 × 60 × 60 From above we have: mE = v 2 r G 10232 × 3.84 × 108 = = 6.0 × 1024 kg , −11 6.67 × 10 which the the same as the result we got before. Note: we didn’t need to know the mass of the Moon, only its distance from the Earth and its orbital period. 12 Using a similar approach we can determine the mass of the Sun by thinking of the Earth as a satellite! Look ... Question 4: If K is the kinetic energy of a satellite in a 1.5 × 10 8 km 2 πr m v= =GS T r stable Earth orbit and Ug is the potential energy of the Earth-satellite system, what is the relationship between K and Ug? ∴v = So mS = v 2 2π × 1.5 × 1011 = 2.99 × 10 4 m/s. 365 × 24 × 60 × 60 r 1.5 × 1011 = (2.99 × 10 4 )2 × G 6.67 × 10 −11 = 2.0 × 1030 kg. 13 We saw that the centripetal force of an Earth-satellite system is a result of the gravitational attraction between the Earth and the satellite, i.e., m m mv 2 , Fg = G E = r r2 1 1 mm ∴ K = mv 2 = G E . 2 2 r mm But, the potential energy is Ug = −G E , r U ∴K = − g 2 . m so v 2 = G E . r Therefore, the total mechanical energy is U = − 1 G mE m . E = K + Ug = g 2 2 r Question 5: Europa orbits Jupiter with a period of 3.55d at an average distance of 6.71 × 108 m from Jupiter’s center. (a) Assuming the orbit is circlar, what is the mass of Jupiter? (b) Another moon, Callisto, orbits Jupiter with a period of 16.7d. What is its average distance from Jupiter? Note that E < 0 so the satellite is “bound”. Energy K r E Ug 14 (a) In algebraic form, Kepler’s 3rd Law states that Te2 = Work done to put a satellite in orbit 4π2 R e3, GMJ m where Te and R e are the orbital period and the orbital radius of Europa, respectively, and MJ is the mass of ME r r r dr Jupiter. ∴ MJ = 4π2 4π2 × (6.71 × 108 )3 R e3 = GTe2 6.67 × 10 −11 × (3.55 × 24 × 3600)2 = 1.90 × 1027 kg. (b) Also, Tc2 = 4π2 R c3, where subscript c refers to the GMJ The work done against the gravitational force in moving r an object a distance d r is rr rr Mm dW = F • dr = G E dr ( F r ). r2 (Note, if dr ( = h ) is very small compared with r, e.g., taking r as the Earth’s radius (6400 km) and h ≈ a few moon Callisto. 2 3 T R T ∴ c = c , i.e., R c = R e 3 c Te Te Re 2 = 6.71 × 108 × 3 16.7 3.55 = 1.88 × 109 m. ( ) 2 meters, so g is constant, then m dW = m G E h = mgh , R E2 a result we obtained before for the work done in lifting an object a short distance.) 15 Hence, the work done lifting a satellite to a height h is W= RE +h ∫ RE R +h G ME mdr = GM m E 1 dr ∫ 2 E r2 r R The escape speed ( v e) is defined as the minimum speed required for an object to “escape” the Earth’s gravitational field. Then Ki + Ui = K f + Uf , 1 i.e., mv e2 − U(R E ) = 0. 2 Kf = 0 : U(∞) = 0 E R +h 1E = GME m− r R E 1 1 = −GME m − R E + h R E GME hR E R − (R E + h ) = −GME m E = m R 2 (R + h ) E E R E (R E + h ) hR E = mg . (R E + h ) 1 K i = mv i 2 : U( R E ) 2 so, v e = hR E 12 mv = mg , (R E + h ) 2 hR E i.e., v = 2 g . (R E + h ) 2GME = 2 gR E . RE For the Earth, we find v e = 2 × 9.81 × 6370 × 103 = 1.12 × 10 4 m/s By the work-energy theorem, the launch speed required for the satellite to reach a height h is given by 1 Mm ∴ mv e2 = G E , 2 RE = 11.2 km/s. For an object launched from Earth, • if v e < 11.2 km/s the object will be “bound” and will orbit the Earth in a circle or ellipse, • if v e > 11.2 km/s the object will be “unbound” and will follow a hyperbolic path and leave the Earth and never return. 16 Question 6: One way of thinking about a black hole is to consider a spherical object whose density is so large that the escape speed at the surface is greater than the speed of light, c. The critical radius for the formation of a black hole is called the Schwarzschild radius, R S. (a) 2GM Show that R S = 2 , where M is the mass of the black c hole. (b) Calculate the Schwarzschild radius for a black hole whose mass is equal to 10 solar masses. (a) The escape speed is given by 2GM ve = . R For a black hole, v e = c and R = R S, i.e., c = 2GM 2GM ⇒ RS = 2 . RS c (b) If M = 10MSun = 10 × 1.99 × 1030 = 1.99 × 1031 kg, ∴ RS = 2 × 6.67 × 1011 × 1.99 × 1031 (3 × 108 )2 = 2.95 × 10 4 m (29.5km). 17 Gravitational field inside a hollow sphere In fact, the result is true for any thickness of shell. We The gravitational field inside a hollow sphere is zero, i.e., r g = 0 ( r < R ). can apply the previous result by dividing the shell into a We can show this result by and summing the fields at the the considering a small (point) point of interest. Thus, r g = 0, A1 m1 ∆r r1 mo r 2 ∆φ A2 m2 mass mo situated at some point inside the sphere. If continuum of concentric shells inside all hollow spheres. the distance from the mass to points on the shell diametrically opposite are r1 and r2, then the areas A1 and A2 and the volumes V1 = A1∆r and V2 = A2 ∆r are proportional to r12 and r22, respectively. If the density of the shell is ρ, then m1 = V1ρ ∝ r12 and m2 = V2ρ ∝ r22, m1 m2 mm mm i.e., 2 = 2 and G o 2 1 = G o 2 2 . r1 r2 r1 r2 Gravitational field inside a solid sphere To determine the gravitational field inside a solid sphere, a distance r from the center, we use the previous result. The mass of the part of the sphere outside r makes no contribution to r R the gravitational field at or inside r. Only the mass M′ inside radius Thus, the net force on mo is zero. That argument can be r contributes to the gravitational field at r. This mass made for all diameterically opposite areas A1 and A2 over the sphere. So, the gravitational field inside a produces a field equal to that of a point mass M′ at the center of the sphere. If M is the mass of the sphere, then hollow sphere is zero. 18 43 πr r3 3 M = M = 3 M. 43 R πR 3 Then, using an earlier expression for g, we find M′ GM r g( r ≤ R ) = G 2 = 3 r , r R Question 7: Agamemnon is a planet of radius 4850km inside the sphere. and density 5500kg/m3. A straight shaft is drilled g( r ) M gr = G 2 r M gr = G 3 r R completely through the planet Agamemnon. If a stone is dropped from the top of the shaft, what would its speed r R be (a) at the center, and (b) at a distance of 2500km above the center? (c) Would the stone reach the opening at the other end of the shaft? The magnitude of the gravitational field is zero at the origin and increases linearly with r. But, as before, it is directed towards the center of the sphere. The stipulation for this result is that the sphere is of uniform mass density. 19 (a) If we let a mass m fall down the shaft, to a radius r, the work done by the gravitational field ∆W is r rr rr ∫ Fg • dr = ∫ Fgdr = ∫ mg r dr , R r R i.e., 12 mv = 1.81 × 107 m. 2 ∴ v = 2 × 1.81 × 107 = 6014 m/s. R R where g r is the gravitational field at radius r. But M′ M gr = G 2 = G 3 r , r R (b) With R = 4850 km and r = 2500 km, ∆W = ∆K = 1.33 × 107 m (J). ∴ v = 2 × 1.33 × 107 = 5154 m/s. where M′ is the mass contained within the sphere of (c) Yes, the stone would just reach the opposite opening. radius r. Since the scenario is symmetrical, its speed at the ∴ ∆W = − GMm r GMm 2 2 R −r . ∫ rdr = 2R 3 R3 R [ 4 4 But M = πR 3ρ = π(4850 × 103 )3 × 5500 3 3 = 2.63 × 1024 kg, opening would be zero, having done ∆W = 1.81 × 107 m Joules of work against the gravitational field. v ( m/s) 7500 6014m/s 5000 and from the work-energy theorem, ∆W = ∆K . 5154m/s With R = 4850 km and r = 0, 2500 GM 6.67 × 10 −11 × 2.63 × 1024 ∆W = ∆ K = m= m (J). 2R 2 × 4850 × 103 r 1 0 R 1 20 In the previous chapter we saw that providing there are no external forces/torques, then the angular momentum of a system like the Earth-Moon is conserved. The rr r angular momentum of the Moon is L = r × mv . vp rp But at perigee and apogee rr r ⊥v , ra va so L = mvr . rp ∴ mv a ra = mv p rp, i.e., v a = v p . ra Question 8: When farthest from Earth, i.e., at apogee, the Moon-Earth distance is 406,400km. When closest to Earth, i.e., at perigee, the Moon-Earth distance is 357,600km. If the mass of the Earth is 5.89 × 1024 kg, what are the orbital speeds of the Moon at apogee and Also, mechanical energy is conserved, i.e., 1 Mm 1 Mm mv a 2 − G E = mv p2 − G E . 2 2 ra rp perigee? ∴ v a 2 − 2G Hence i.e., rp v p2 ra ME M = v p2 − 2G E . ra rp 2 − 2G 2 2 1 − rp vp 2 ra ME M = v p2 − 2G E , ra rp 1 1 = 2GME − . rp ra 21 ra 2 − rp2 ra − rp ∴ v p2 2 = 2GM E r r ap ra ra + rp 2GME so v p2 , = ra rp i.e., v p = = 2GME ra rp ( ra + rp ) 2 × 6.67 × 10 −11 × 5.98 × 1024 × 4.064 × 108 ( 3.576 × 108 × 4.064 × 108 + 3.576 × 108 = 1090m/s ) (1.090km/s). 3.576 × 108 ∴ v a = v p = 1090 × ra 4.064 × 108 rp = 959m/s (0.959km/s). 22 ...
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