= 4 3 π( m) 3 (5480 kg m 3 ) = kg.

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1 CHAPTER 11 THE GRAVITATIONAL FIELD Newton s Law of Gavitation m 1 m A foce of attaction occus between two masses given by Newton s Law of Gavitation Inetial mass and gavitational mass Gavitational potential enegy Satellites Keple s 3 d Law Escape speed F = G m 1 m A value fo G, the univesal gavitational constant, can be obtained fom measuements by Heny Cavendish ( ) of the specific gavity of the Eath He found that the density of the Eath is 548 times the density of wate, ie, ρ Eath = kg m 3 = 5480 kg m 3 m E = volume density = 4 3 π 3 ρ E = 4 3 π( m) 3 (5480 kg m 3 ) = kg Cavendish s measuements allowed the fist detemination of the mass of the Eath; it is within 1% of today s accepted value

m E m A = h + The foce on the apple (ie, its weight) is When eleased the acceleation of the apple is Vey close to the Eath s suface, >> h and a g, so whee is the Eath s adius We know g = 981 m/s, m E

m /kg g() = G m This expession povides us a way of detemining the acceleation of gavity at the suface of a massive object like a planet If the mass is m and the adius is R p then p Examples: Mas:

2 m E m A = h + The foce on the apple (ie, its weight) is When eleased the acceleation of the apple is Vey close to the Eath s suface, >> h and a g, so whee is the Eath s adius We know g = 981 m/s, m E = kg and = m, Fom above, the acceleation due to gavity a distance fom a mass m is: F EA = m A g() = G m E m A a = F EA m = G E m A (h + ) G = m E g, = G m E m A (h + ) G = N m /kg g() = G m This expession povides us a way of detemining the acceleation of gavity at the suface of a massive object like a planet If the mass is m and the adius is R p then p Examples: Mas: Moon: In deiving these expessions we have assumed that the Eath and planets ae spheical In fact, the Eath is eally a flattened sphee g p = G m p R p mass = kg adius = m g Mas = 37 m/s mass = kg adius = m g Moon = 16 m/s

Gavitational and inetial mass R P R eq So fa, without stating it explicitly we have used the same mass (m) in Newton s Law of gavitation as was defined by

of gavity The mass used in Newton s Law of gavitation is called the gavitational mass, which can be detemined by measuing the attactive foce exeted on it by

3 Gavitational and inetial mass R P R eq So fa, without stating it explicitly we have used the same mass (m) in Newton s Law of gavitation as was defined by Newton s nd Law in chapte 4 Note that in the definition of the inetial mass of an object, Fo the Eath R eq > R p In fact, g = G m E weight is mg, one weighs less than at sea-level! up a mountain, in an aiplane, At 30,000 ft: g eq < g P g eq ~ 983 m/s and g P ~ 978 m/s Δw w ~ 03% Since ones inetial mass = m inet = F a, thee is no mention of gavity The mass used in Newton s Law of gavitation is called the gavitational mass, which can be detemined by measuing the attactive foce exeted on it by anothe mass (M) a distance away, viz: gavitational mass = m gav = F Mm GM Thee is no mention of acceleation in this expession Although these ae two diffeent concepts of mass, Newton asseted that, fo the same object, these masses ae identical, ie, m gav = m ine This is known as the pinciple of equivalence

4 Examples of gavitational and inetial mass: Gavitational mass: time it takes an object to fall fom a cetain height the weight of an object on a sping scale Inetial mass: the acceleation given to an object by a compessed sping In fact, g is a vecto ( g)! and is often efeed to as the gavitational field, ie, the egion ove which one mass influences anothe It is analogous to a magnetic field and! an electic field Since g is a vecto, the esultant gavitational field of a collection of mass is the vecto sum of the individual contibutions Question 111: Thee 50 kg masses ae located at the positions shown in the x-y plane What is the esultant foce on the mass at x = 040 m, y = 0 caused by the othe two masses?

F Rx = F 1x + F x = ( 104 10 8 N) + ( 534 10 9 N) = 157 10 8 N, and F Ry = F y = 040 10 8 N F 1x = G m

F R = ( 157 10 8 N)î + (040 10 8 N) ĵ, and!

5 F Rx = F 1x + F x = ( N) + ( N) = N, and F Ry = F y = N F 1x = G m 1 m 1 = ( N m /kg (50 kg)(50 kg) ) (040 m) = N! F R = ( N)î + ( N) ĵ, and! F R = N F = G m 1 m = ( N m /kg (50 kg)(50 kg) ) (050 m) = N F x = ( N)cosθ = 080( N) = ( N), and F y = ( N)sinθ = 060( N) = ( N)

6 Question 11: Five objects, each of mass M = 300 kg ae equally spaced on the ac of a semicicle of adius R = 100 cm An object of mass m = 00 kg is located at the cente of cuvatue of the ac (a) What is the gavitational foce acting on the 00 kg mass? (b) What is the gavitational field at the cente of cuvatue of the ac? (a) The total foce components acting on m ae and F x = G Mm R + G Mm R sin 45! G Mm R sin45! G Mm R = 0, F y = G Mm R sin 45! + G Mm R + G Mm R sin45! ( ), = G Mm 1+ sin 45! R = ( N m /kg )(300 kg)(00 kg) (010 m) (414) = N! F = ( N) ĵ

7 (b) The gavitational foce acting on m is F! = m g! g!! F = m = ( N) ĵ 00 kg = ( m/s ) ĵ Question 113: The Eath otates about its N-S axis How does that affect the fee-fall acceleation measued at the equato? Assuming the Eath is a sphee, is it the same as, geate than, o less than the gavitational acceleation fo a stationay Eath, ie, g = G m E?

measued on a scale) and View of the Eath looking at the South pole and the FBD of foces acting on an object at the equato; net foce is the centipetal foce, ie, whee m g! F N!

g is the nomal foce (equal to the weight as is the gavitational foce The is the measued weight of the object Then the effective value of fee-fall acceleation is Since the Eath

8 measued on a scale) and View of the Eath looking at the South pole and the FBD of foces acting on an object at the equato; net foce is the centipetal foce, ie, whee m g! F N! F g = m! g is the nomal foce (equal to the weight as is the gavitational foce The is the measued weight of the object Then the effective value of fee-fall acceleation is Since the Eath otates though π in 4 h, π ω = ( s) = ad/s ie, Δg g = Theefoe, the effect is negligible! a F g F N F g F N = mg m g = mω, g = g ω, ie, g < g ω = ( ad/s) ( m) = 0034 m/s, Gavitational potential enegy suface was Above, we saw that the acceleation of gavity at a distance h above the Eath s g = G m E = G m E (h + )

9 U() = G m E m d = Gm E m d = G m E m + U!, whee U! is an integation constant We choose U! = 0, ie, we define the gavitational potential enegy to be zeo at = U() 0 U() = G m E m U( ) G m E m = mg We obtain a negative value fo U( ) because we defined U( ) = 0 In ealie chaptes we usually defined U( ) = 0 Howeve, it does not cause difficulties as we nomally wok with changes in potential enegy, ie, ΔU, which is independent of ou choice of U!

10 We have U() = G m E m = G m E m + h Fo small values of h, h << 1, then ( ) ( ) = G m E m 1 + h RE U() = G m E m 1 + h But G m Em 1 = G m E m 1 h = G m Em + G m Em R h E = U( ) is constant and G m E = g U() = U( ) + mgh, ie, in moving an object a height h above the Eath s suface inceases the gavitational potential enegy by ΔU = mgh, poviding h <<, a esult we used in chaptes 6 and 7 Question 114: A object is eleased fom est when it is m above the suface of a planet whose mass is M = kg and adius R = m Neglecting ai esistance, (a) what is the speed of the object just befoe it stikes the suface? (b) Compae that speed with the speed it would have achieved if the gavitational field wee constant

11 h K i = 0 : U(R + h) K f = 1 mv f : U(R) (a) Using the consevation of mechanical enegy, ie, K f + U f = K i + U i, we obtain 1 mv f G Mm R = 0 G Mm (R + h) v f = ( N m /kg )( kg) m m = (m/s) v f = m/s (b) Assuming g is constant, g at the suface is g = G M R = ( N m /kg ) ( kg) ( m) = 107 m/s Question 115: When fathest fom Eath, ie, at apogee, the Moon-Eath distance is 406,400 km When closest to Eath, ie, at peigee, the Moon-Eath distance is 357,600 km If the mass of the Eath is kg, what ae the obital speeds of the Moon at apogee and peigee? We have ΔK = ΔU 1 m v = mgh, ie, v = gh = (107 m/s )( m) = m/s

12 In the pevious chapte we saw that poviding thee ae no extenal foces/toques, then the angula momentum of a system like the Eath-Moon is conseved The angula momentum of the Moon is L! =! m v! v p p a va But at peigee and apogee! v!, so L = mv mv a a = mv p p, ie, v a = v p p a Also, mechanical enegy is conseved, ie, 1 mv a G M E m = 1 a mv p G M E m p Substituting fo the given values Fom ealie ( ) = (m/s) v p 057 v p = (m/s) 057 = 1080 m/s v a = v p p a = (1080 m/s) ( m) ( m) = 950 m/s v a G M E a = v p G M E p Hence v p p a G M E a = v p G M E p, ie, v p 1 p 1 = GM E 1 a p a

Satellites M E v m constant T = 4π Gm E 3 ie, T 3 Conside a satellite, mass m, in a cicula obit, adius, aound

centipetal acceleation So, in a stable obit G m Em v = Gm E The time fo one obit T = π v ie, v = 4π T = mv = G

~ v = G m E ie, the velocity of a satellite in a stable obit does not depend on its mass only the mass of the

13 Satellites M E v m constant T = 4π Gm E 3 ie, T 3 Conside a satellite, mass m, in a cicula obit, adius, aound the Eath (o a planet aound the Sun) The gavitational foce between the satellite and the Eath povides the centipetal acceleation So, in a stable obit G m Em v = Gm E The time fo one obit T = π v ie, v = 4π T = mv = G m E = g, whee g is the gavitational field acting on the satellite Note also ~ Keple s 3d Law of planetay motion ~ v = G m E ie, the velocity of a satellite in a stable obit does not depend on its mass only the mass of the cental object! so what? well, it povides us a way of woking out the mass of a planet Take the Eath-Moon system fo example

= v G = (103 103 m/s)(384 10 8 m) 667 10 11 N m /kg = 589 10 4 kg Note: we didn t need to know the mass of the Moon, only its distance fom the

14 Teating the Moon as a satellite aound the Eath v = G m E Using a simila appoach we can calculate the mass of the Sun by consideing the Eath as a satellite The speed of the Moon is cicumfeence of obit v = time of obit = Then, fom above = π T π( m) = 103 km/s ( s) m E = v G = ( m/s)( m) N m /kg = kg Note: we didn t need to know the mass of the Moon, only its distance fom the Eath and its obital peiod v = π( m) ( s) = m/s m S = v G = ( m m/s) N m /kg = kg

15 (a) The fee-fall acceleation at the suface of a planet is g = G M R, Question 116: A 00 kg satellite cicles the planet Roton evey 8 h in an obit of adius m If the adius of Roton is m, what is (a) the magnitude of the fee-fall acceleation at the suface of Roton, and (b) the kinetic enegy of the satellite? whee M is its mass and R is its adius Howeve, we must find M Using Keple s 3 d Law we have whee R s is the adius of the obit and T is obital time Now T = (8 h)(3600 s/h) = s, Hence M = 4π G R s 3 T, 4π ( m) 3 M = ( N m /kg ) ( s) = kg g = ( N m /kg ) kg = 69 m/s ( m)

16 (b) Fo a satellite in obit, the centipetal foce mv = G Mm R s R s ie, K = 1 mv = G Mm R s = ( N m /kg ) ( kg)(00 kg) ( m) = J Question 117: Euopa obits Jupite with a peiod of 355 days at an aveage distance of m fom Jupite s cente (a) Assuming the obit is cicula, what is the mass of Jupite? (b) Anothe moon, Callisto, obits Jupite evey 167 days What is its aveage distance fom Jupite?

17 (a) Keple s 3 d Law states that whee T eu and R eu ae the obital peiod and obital adius Of Euopa, espectively, and (b) Also Callisto M J M J = 4π R eu 4π T eu= 3 R GM eu, J is the mass of Jupite 3 GT eu 4π ( m) 3 = ( N m /kg )( s) 4π T ca = 3 R GM ca, J T ca T eu = R ca = kg R eu 3 whee subscipt ca efes to T, ie, R ca = R ca eu = ( m) 167 d 355 d = m 3 T eu 3 Escape speed The escape speed fo the Eath is defined as the minimum initial speed of a pojectile that enables it to escape the influence of the Eath s gavitational field Then K i + U i = K f + U f Fo the Eath Fo an object launched fom the Eath s suface if v i < 11 km/s, the pojectile will be bound and will obit the Eath in a cicle o ellipse if v i > 11 km/s, the pojectile will be unbound and will follow a hypebolic path and leave the Eath and neve etun K f = 0 : U( ) = 0 K i = 1 mv e : U( ) v e = ie, 1 mv e + U( ) = 0 1 mv e = G m E m, ie, v e = Gm E = g (981 m/s )( m) = 11 km/s

18 The wok-kinetic enegy theoem tells us that the minimum wok done to escape the Eath o Moon is the same as the change in kinetic enegy to achieve escape Question 118: Does it take moe fuel, less fuel o the same amount of fuel fo a ocket to tavel fom the Eath to the Moon as fom the Moon to the Eath? speed W = ΔK = K i K f But K f = 0, W = 1 mv e The escape speed is v e = gr Since > R M and g E > g M, then v e (Eath) > v e (Moon) In fact, Since the Eath s escape speed is geate than the Moon s escape speed, it equies moe wok, ie, moe fuel, fo a spacecaft to tavel fom the Eath to the Moon than vice vesa v e (Eath) = 11 km/s v e (Moon) = 36 km/s

The escape speed fom the Eath s suface is v e = whee M E and ae the mass and adius of the Eath, espectively So, the escape speed of a planet at a distance R fom the cente of the Sun is GM E, Question

19 The escape speed fom the Eath s suface is v e = whee M E and ae the mass and adius of the Eath, espectively So, the escape speed of a planet at a distance R fom the cente of the Sun is GM E, Question 119: Each planet in the sola system obits at its own paticula speed aound the Sun If the obital speeds of the planets wee doubled, which of the following is tue? Each planet would A: assume a close obit aound the Sun, B: assume a moe distant obit aound the Sun, o C: escape the sola system D: Any o all of the above depending on the planet v e = GM S R Ealie, we found that the obital speed of a planet aound the Sun is v = GM S R So, if the obital speed is doubled, v GM S R, which is geate than the escape speed Theefoe, each planet will escape the sola system (answe C) Note that the obital speed need only to be inceased by a facto of fo a planet to escape the sola system

(a) Fom ealie, we showed that the escape speed is v e = GM R Fo a black hole, v e = c and R = R S, Question 1110: A simple model of a black hole would be to conside it a spheical object whose mass is

20 (a) Fom ealie, we showed that the escape speed is v e = GM R Fo a black hole, v e = c and R = R S, Question 1110: A simple model of a black hole would be to conside it a spheical object whose mass is so lage that the escape speed at the suface is geate than the speed of light, c The citical adius fo the fomation of a black hole is called the Schwazschild adius, R S (a) Show that R S = GM c, whee M is the mass of the black hole (b) Calculate the Schwazschild adius fo a black hole whose mass is equal to 10 sola masses Take c = m/s ie, c = GM R S R S = GM c (b) M = 10M Sun = 10( kg) = kg R S = ( N m /kg )( kg) ( m/s) = m (96 km)

Chapter 13 Gravitation

Chapter 13 Gravitation Chapte 13 Gavitation In this chapte we will exploe the following topics: -Newton s law of gavitation, which descibes the attactive foce between two point masses and its application to extended objects