Episode 401: Newton s law of univesal gavitation This episode intoduces Newton s law of univesal gavitation fo point masses, and fo spheical masses, and gets students pactising calculations of the foce between objects. The meaning of invese squae law is discussed. Summay Discussion: Intoduction to Newton s law of univesal gavitation (5 minutes) Discussion: Newton s law of univesal gavitation: F = Gm 1 m / (10 minutes) Woked examples: Using F = Gm 1 m /. (5 minutes) Student questions: Moe pactice with F = Gm 1 m /. (0 minutes) Not to scale Discussion: Intoduction to Newton s law of univesal gavitation Hee ae some questions and answes which lead towads Newton s Law of Univesal Gavitation. What causes the weight that each student feels? (gavitational attaction by the Eath.) What affects the size of the Eath s pull on you? Why would you weigh a diffeent amount on the Moon? (You mass, and its mass.) If the Eath is pulling down on you, then what else must be occuing, by Newton s 3d Law? (You must be pulling up on the Eath with a foce equal to you weight.) What happens to the stength of the pull of the Eath as you go futhe away fom it? (It gets weake most students guess this coectly fom the incoect assumption that in space, astonauts ae weightless!) So, in summay the foce depends upon the masses of the Eath and you, and weakens with distance. This is all embodied in Newton s law of univesal gavitation Discussion: Newton s law of univesal gavitation Pesent the equation which epesents Newton s Law of Univesal Gavitation. F = Gm 1 m / F = gavitational foce of attaction (N) m 1, m ae the inteacting masses (kg) is the sepaation of the masses (m) 1
G is known as the univesal gavitational constant (NOT to be confused with little g). It sets the stength of the gavitational inteaction in the sense that if it wee doubled, so would all the gavitational foces. G = 6.67 10-11 N m kg - Show how the units can be woked out by eaanging the oiginal equation. This law applies between point masses, but spheical masses can be teated as though they wee point masses with all thei mass concentated at thei cente. This foce is ALWAYS attactive. In some texts you will see a minus sign in the equation, so that F = -Gm 1 m /. This minus sign is thee puely to indicate that the foce is attactive (it s a elic fom the moe coect, but well beyond the syllabus, vecto equation expessing Newton s Law of univesal gavitation). It s simplest to calculate the magnitude of the foce using F = Gm 1 m /, and the diection is given by the fact that the foce is always attactive. Evey object with a mass in the univese attacts evey othe accoding to this law. But the actual size of the foce becomes vey small fo objects vey fa away. Fo example, the Sun is about one million times moe massive than the Eath, but because it s so fa away, the pull on us fom the Sun is dwafed by the pull on us fom the Eath (which is aound 1650 times geate). As the sepaation of two objects inceases, the sepaation inceases even moe, damatically. The gavitational foce will decease by the same facto (since sepaation appeas in the denominato of the equation). This is an example of an invese squae law, so called because the foce of attaction vaies in invese popotion to the squae of the sepaation. Woked examples: Using F = Gm 1 m / You can wok though these examples, o you can set them as a task fo you students if you feel they will be able to tackle them. TAP 401-1: Woked examples; F = Gm 1 m / Student questions: Moe pactice with F = Gm 1 m / TAP 401-: Newton s gavitational law
TAP 401-1: Woked examples; F = Gm 1 m / Woked examples Newton s Law of univesal gavitation teache s sheet Note: Remembe that a lot of students have difficulties in using standad fom coectly on a scientific calculato. You may need to tuto them in the use of the EXP button. Data equied: G = 6.67 10-11 N m kg -, mass of the Eath = 6.0 10 4 kg, adius of the Eath = 6.4 10 6 m, mass of the Sun =.0 10 30 kg, aveage distance fom the Eath to the Sun = 1.5 10 11 m. 1) Communications satellites obit the Eath at a height of 36 000 km. How fa is this fom the cente of the Eath? If such a satellite has a mass of 50 kg, what is the foce of attaction on it fom the Eath? It is (3.6 x 10 7 m + 6.4 x 10 6 m) = 4.4 x 10 7 m fom the cente of the Eath. (They should eally give this as 4. x 10 7 m this may be an oppotunity to einfoce the ole of significant figues in physical calculations). The foce is F = Gm 1 m / = (6.67 x 10-11 x 6.0 x 10 4 x 50)/ (4.4 x 10 7 ). This gives an answe of about 56 N, which fo infomation is about less than the weight as a one yea old toddle. ) What is the foce of attaction fom the Eath on you? What do we call this foce? What is the foce of attaction on the Eath fom you? They will need to estimate thei own mass in kg. If they need to convet, 1 stone is 6.4kg (and thee ae 14 pounds in a stone). They then use F=Gm 1 m / whee is the adius of the Eath. This foce is usually called thei weight. The foce on the Eath fom the student is exactly the same as thei fist answe, but in the opposite diection. 3) What is the foce of attaction fom the Sun on you? How many times smalle is this than the foce of attaction fom the Eath on you? Again, they will need to use thei own mass, and the equation F=Gm 1 m /, but this time is the aveage distance fom the Sun to the Eath. This foce should be about 1650 times less than thei weight, of the ode of 0.3-0.5 N. Small, but not negligible. 3
4) The aveage foce of attaction on the Moon fom the Sun is 4.4 10 0 N. Taking the distance fom the Sun to the Moon to be about the same as that fom the Sun to the Eath, what value of mass does this give fo the Moon? m = F /Gm 1 = (4.4 x 10 0 x (1.5 x 10 11 ) )/(6.67 x 10-11 x.0 x 10 30 ) = 7.4 x 10 kg 5) Using the mass of the Moon you calculated in question 4, what is the pull of the Eath on the Moon, if the Moon is 380 000 km away? How does this compae with the pull of the Sun on the Moon? F = Gm 1 m / = (6.67 x 10-11 x 6.0 x 10 4 x 7.4 x 10 )/ (3.8 x 10 8 ) =.1 x 10 0 N This is actually smalle than the pull of the Sun on the Moon. You could discuss whethe that means the Moon is obiting the Sun athe than the Eath. In fact, it depends on the most useful fame of efeence in a paticula situation fom the Sun s point of view, the Moon and the Eath obit the Sun, in a way that is affected by the pesence of the othe; fom the Moon s point of view, both the Sun and the Eath obit the Moon, in a way that is affected by the pesence of the othe, etc. 6. What is the foce of attaction between two people, one of mass 80 kg and the othe 100 kg if they ae 0.5m apat? F = Gm 1 m / F = G x 100 x 80 / 0.5 =.14 x 10-6 N. This is a vey small foce but it does incease as the people get close togethe! Actually this example is not accuate because Newton's law eally only applies to spheical objects, o at least objects so fa apat that they can be effectively consideed as spheical. 7. What is the foce of attaction between the Eath and the Sun? Mass of the Sun = x 10 30 kg, mass of the Eath = 6 x 10 4 kg, distance fom the Eath to the Sun = 1.5 x 10 11 m F = Gm 1 m / F = G x x 10 30 x 6 x 10 4 / [1.5 x 10 11 ] = 6.7 x 10 11 N an enomous foce! 4
Extenal efeence Questions 6 and 7 taken fom Resouceful Physics 5
TAP 401-: Newton s gavitational law These questions ae intended to give you pactice in using the gavitational law. They will give you a feeling fo typical foces with a ange of masses and also how sensitive foce is to distance. Useful data G = 6.67 10 11 N m kg Eath s mass = 5.97 10 4 kg Moon s mass = 7.34 10 kg Sun s mass is.0 10 30 kg Radius of the Moon = 1.64 10 6 m Radius of the Eath = 6.37 10 6 m Eath Moon distance = 3.8 10 5 km Eath Sun distance = 1.5 10 8 km 1. You may sometimes find it difficult to get up fom the sofa afte watching a TV pogamme. Assuming the foce of gavity acts between the cente of you body and the cente of the sofa, estimate the attaction between you and you sofa.. Calculate the size of the gavitational pull of a sphee of mass 10 kg on a mass.0 kg when thei centes ae 00 mm apat. What is the foce of the.0 kg mass on the 10 kg mass? 3. At what distance apat would two equal masses of 150 kg need to be placed fo the foce between them to be.0 10 5 N? 6
4. Calculate the gavitational pull of the Eath on each of the following bodies: the Moon; satellite A with mass 100 kg at a distance fom the Eath s cente 4. 10 7 m; and satellite B mass 80 kg at a distance fom the Eath s cente 8.0 10 6 m. 5. Show that the unit fo G, the univesal gavitational constant, can be expessed as m 3 s kg 1. 6. Calculate the weight of an astonaut whose mass (including spacesuit) is 7 kg on the Moon? What is the astonaut's weight on Eath? 7
Comment on the diffeence. 7. Show that pull of the Sun on the Moon is about. times lage than the pull of the Eath on the Moon. 8. Why then does the Moon obit the Eath? Fomatted: Bullets and Numbeing The Ameican space agency, NASA, plans to send a manned mission to Mas late this centuy. Mas has a mass 6.4 x 10 3 kg and a adius 3.38 x 10 6 m. G = 6.67 x 10 11 N m kg - 9 (a) The mass of a typical astonaut plus spacesuit is 80 kg. What would be the gavitational foce acting on such an astonaut standing on the suface of Mas? (b) State whethe an astonaut on Mas would feel lighte o heavie than on Eath. 8
Pactical advice This esouce could be used fo eithe class wok o homewok. Answes and woked solutions 1. Fo the values estimated in the answes: GMm ( 6.67 10 N kg m ) 60 kg 100 kg 6 = 1.6 10 ( 0.5 m) N.. Pull on the.0 kg mass GMm ( 6.67 10 N kg m ) 10 kg.0 kg 9 = 3.3 10 ( 0.00 m) The pull on the 10 kg mass will be equal but opposite in diection. N. 3. = Gm F = m G F = 150 kg 6.67 10.0 10 Nkg 5 N m = 0.7 m. 4. GMm ( 6.67 10 Nkg m ) (5.97 10 kg) (7.34 10 kg) 0 =.0 10 N. GMm (6.67 10 Nkg 8 ( 3.8 10 m) m ) (5.97 10 7 ( 4. 10 m) 6 ( 8.0 10 m) 4 4 kg) 100 kg = 3 N. GMm ( 6.67 10 Nkg m ) (5.97 10 kg) 80 kg = 5.0 10 N. 4 3 1 5. N kg m = kg m s kg m = m s kg. 6. Moon GMm ( 6.67 10 Nkg m ) 7 kg (7.34 10 kg) = 1.3 10 N. 6 ( 1.64 10 m) 9
Eath 4 GMm ( 6.67 10 N kg m ) 7 kg (5.97 10 kg) = 7.1 10 7. Sun Moon 6 ( 6.37 10 m) N. GMm ( 6.67 10 Nkg m ) (.0 10 kg) (7.34 10 kg) 0 = 4.4 10 N. Eath Moon 30 11 ( 1.5 10 m) 4 GMm 6.67 10 N kg m ) (5.97 10 kg) (7.34 10 kg) =.0 10 ( 0 8 ( 3.8 10 m) N. 4.4 10 atio of attactions =.0 10 0 0 N =. N 8. The Moon does of couse obit the Sun, as pat of the Eath Moon system. You can think of the Moon s obit of the Eath as supeimposed on its obit of the Sun. 9 F = (Gm astom x M mas ) / F = (6.67 x 10 11 N m kg - ) x 80 kg x (6.4 x 10 3 kg) / (3.38 x 10 6 m) = 300 N (b) Would feel lighte. Extenal efeences Questions 1-8: This is taken fom Advancing Physics Chapte 11, 80W Question 9: This is taken fom Saltes Hones Advanced Physics, section STA, additional sheet 8 and 9 10