Topic 6: Circular motion and gravitation 6.2 Newton s law of gravitation

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1 Topic 6: Circular motion and gravitation 6.2 Newton s law of gravitation Essential idea: The Newtonian idea of gravitational force acting between two spherical bodies and the laws of mechanics create a model that can be used to calculate the motion of planets. Nature of science: Laws: Newton s law of gravitation and the laws of mechanics are the foundation for deterministic classical physics. These can be used to make predictions but do not explain why the observed phenomena exist.

2 Topic 6: Circular motion and gravitation 6.2 Newton s law of gravitation Understandings: Newton s law of gravitation Gravitational field strength Applications and skills: Describing the relationship between gravitational force and centripetal force Applying Newton s law of gravitation to the motion of an object in circular orbit around a point mass Solving problems involving gravitational force, gravitational field strength, orbital speed and orbital period Determining the resultant gravitational field strength due to two bodies

3 Topic 6: Circular motion and gravitation 6.2 Newton s law of gravitation Guidance: Newton s law of gravitation should be extended to spherical masses of uniform density by assuming that their mass is concentrated at their centre Gravitational field strength at a point is the force per unit mass experienced by a small point mass at that point Calculations of the resultant gravitational field strength due to two bodies will be restricted to points along the straight line joining the bodies

4 Topic 6: Circular motion and gravitation 6.2 Newton s law of gravitation Data booklet reference: F = G Mm r 2 g = F m F = G M r 2 Theory of knowledge: The laws of mechanics along with the law of gravitation create the deterministic nature of classical physics. Are classical physics and modern physics compatible? Do other areas of knowledge also have a similar division between classical and modern in their historical development?

5 Topic 6: Circular motion and gravitation 6.2 Newton s law of gravitation Utilization: The law of gravitation is essential in describing the motion of satellites, planets, moons and entire galaxies Comparison to Coulomb s law (see Physics sub-topic 5.1) Aims: Aim 4: the theory of gravitation when combined and synthesized with the rest of the laws of mechanics allows detailed predictions about the future position and motion of planets

6 Field (Fundamental) Forces Gravitational Force Gravitational force is cumulative and extended to infinity. The gravitational force between two object is due to the cumulative effect of billions of billions of the atoms made up both bodies. This means that the larger the body (contain more matter), the stronger the force. But on the scale of individual particles, the force is extremely small, only in the order of times that of the strong force. Electromagnetic Force The force is long range, in principle extending over infinite distance. However, the strength can quickly diminishes due to shielding effect. Many everyday experiences such as friction and air resistance are due to this force. This is also the resistant force that we feel, for example, when pressing our palm against a wall. This is originated from the fact that no two atoms can occupy the same space. However, its strength is about 100 times weaker within the range of 1 fm, where the strong force dominates. But because there is no shielding within the nucleus, the force can be cumulative and can compete with the strong force. This competition determines the stability structure of nuclei. Weak Nuclear Force Responsible for nuclear beta decay and other similar decay processes involving fundamental particles. The range of this force is smaller than 1 fm and is 10-7 weaker than the strong force Strong Nuclear Force This force is responsible for binding of nuclei. It is the dominant one in reactions and decays of most of the fundamental particles. This force is so strong that it binds and stabilize the protons of similar charges within a nucleus. However, it is very short range. No such force will be felt beyond the order of 1 fm (femtometer or m).

7 One of the most significant intellectual achievements in the history of thought. It is universal it applies to all objects regardless of their location anywhere in the Universe. Every object in the universe attracts every other object. Gravitational force between two point masses m 1 and m 2 is proportional to their product, and inversely proportional to the square of their separation r. mm F = G r G = 6.67x10-11 Nm 2 / kg 2 Universal gravitational constant the same value anywhere in the universe - very small value no significant forces of attraction between ordinary sized objects. The actual value of G, the universal gravitational constant, was not known until Henry Cavendish conducted a tricky experiment in 1798 to find it. r

8 Newton s law of gravitation The earth, planets, moons, and even the sun, have layers kind of like an onion: In other words, NONE of the celestial bodies we observe are point masses. many Given that the law is called the universal law of gravitation, how do we use it for planets and such?

9 Newton s law of gravitation Newton spent much time developing integral calculus to prove that A spherically symmetric shell of mass M acts as if all of its mass is located at its center. M -Newton s shell theorem. m Thus F = Gm 1 m 2 / r 2 works not only for point masses, which have no radii, but for any spherically symmetric distribution of mass at any radius like planets and stars. r

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11 r is the distance between the centers of the masses. m FYI The radius of each 1 m 2 F 12 F 21 r EXAMPLE: The earth has a mass of M = kg and the moon has a mass of m = kg. The mean distance between the earth and the moon is m. What is the gravitational force between them? SOLUTION: F = GMm / r 2. F = ( )( )( ) / ( ) 2 F = N. mass is immaterial. What is the speed of the moon in its orbit about earth? SOLUTION: F C = F G = mv 2 / r = ( ) v 2 / v = ms -1. What is the period of the moon (in days) in its orbit about earth? SOLUTION: v = 2 r / T. T = 2 r / v = 2 ( ) / = ( s)(1 h / 3600 s)(1 d / 24 h) = 27.2 d.

12 Gravitational field strength Suppose a mass m is located a distance r from a another mass M. The gravitational field strength g is the force per unit mass acting on m due to the presence of M. g = F m The units are newtons per kilogram (N kg -1 = m s -2 ). Suppose a mass m is located on the surface of a planet of radius R. We know that it s weight is F = mg. But from the law of universal gravitation, the weight of m is equal to its attraction to the planet s mass M and equals F = GMm / R 2. mg = GMm / R 2. g = GM R 2 gravitational field strength at surface of a planet of mass M and radius R Double the distance from the centre, g is 4 times less, and so is weight

13 PRACTICE: A 525-kg satellite is launched from the earth s surface to a height of one earth radius above the surface. What is its weight (a) at the surface, and (b) at altitude? SOLUTION: (a) AT SURFACE: g surface = 9.8 m s -2. F = mg = (525)(9.8) = 5145 N. W=5100 N (b) AT ALTITUDE: g surface+r = 2.45 m s -2. F = mg = (525)(2.46) = N W=1300N

14 Gravitational field strength Compare the gravitational force formula F = GMm / r 2 (Force) with the gravitational field formula g = GM / r 2 (Field) Note that the force formula has two masses, and the force is the result of their interaction at a distance r. Note that the field formula has just one mass namely the mass that sets up the local field in the space surrounding it. It curves it. The field view of the universe (spatial disruption by a single mass) is currently preferred over the force view (action at a distance) as the next slides will show.

15 Consider the force view. In the force view, the masses know the locations of each other at all times, and the force is instantaneously felt by both masses at all times. This requires the force signal to be transferred between the masses instantaneously. Einstein s special theory of relativity states unequivocally that the fastest any signal can travel is at the (finite) speed of light c. SUN

16 Thus the action at a distance force signal will be slightly mdelayed in telling the orbital mass when to turn. The end result would have to be an expanding spiral mmotion, as illustrated in the following animation: We do not observe planets leaving mtheir orbits as they travel around the msun. Thus force as action at a distance doesn t work if we are to believe special relativity. And all current evidence points to mthe correctness of special relativity. SUN

17 So how does the field view take care of this signal lag problem? Simply put the gravitational field distorts the space around the mass that is causing it so that any other mass placed at any position in the field will know how to respond immediately. Think of space as a stretched rubber sheet like a drum head. Bigger masses curve the rubber sheet more than smaller masses. The next slide illustrates this gravitational curvature of the space surrounding, for example, the sun.

18 each mass feels a different slope and must travel mat a particular speed to stay in orbit. The field view eliminates the need for long distance msignaling between two masses. Rather, it distorts the mspace about one mass.

19 In the space surrounding the mass M which sets up mthe field we can release test masses m 1 and m 2 mas shown to determine the strength of the field. m 2 g 1 g 2 m 1 (a) Because g = GM / r 2. It varies as 1 / r 2. M (b) Because the gravitational force is attractive. (a) The field arrow is bigger for m 2 than m 1. Why? (b) The field arrow always points to M. Why?

20 By placing a series of test masses about a larger mmass, we can map out its gravitational field: M The field arrows of the inner ring are longer than the field arrows of the outer ring and all field arrows point to the centerline.

21 Gravitational field strength Simplification: Instead of having arrows at every point we take the convention of drawing field lines as a single arrow. Density/concentration of the field lines convey us comparable strengths. The closer together the field lines, the stronger the field. SUN In the red region the field lines are closer together than in the green region. Thus the red field is stronger than the green field. SUN

22 PRACTICE: Sketch the gravitational field about the earth (a) as viewed from far away, and (b) as viewed locally (at the surface). SOLUTION: (a) (b) or The closer to the surface we are, the more uniform the field concentration.

23 EXAMPLE: Find the gravitational field strength at a point between the earth and the moon that is right between their centers. M = kg m = kg SOLUTION: Make a sketch. Note that r = d / 2 = / 2 = m. g m = Gm / r 2 g m = ( )( )/( ) 2 = N. g M = GM / r 2 g M = ( )( )/( ) 2 = N. g m Finally, g = g M g m = N,. g M d = m

24 PRACTICE: Two spheres of equal mass and different radii are held a distance d apart. The gravitational strength is measured on the the two masses at position x which varies. Which graph shows the variation of g with x correctly? field line joining There is a point between M and m where g = 0. Since g = Gm / R 2 and R left < R right, then g left > g right at the surfaces of the masses.

25 EXAMPLE: Derive Kepler s law, which states that the period T of an object in a circular orbit about a body of mass M is given by T 2 = 4π2 GM r3 SOLUTION: F c = mv2 r = G Mm r 2 v 2 = G M r 2πr T 2 = G M r T 2 = 4π2 GM r3

26 EXAMPLE: A satellite in geosynchronous orbit takes 24 hours to orbit the earth. Thus, it can be above the same point of the earth s surface at all times, if desired. Find the necessary orbital radius, and express it in terms of earth radii. R E = m. SOLUTION: T = (24 h)(3600 s h 1 ) = s mv 2 r = G Mm r 2 v 2 = G M r 2πr T 2 = G M r r 3 = GM 4π 2 T2 r 3 = π 2 (86400) 2 r = m = 6.63 R E

27 mv 2 r = G Mm r 2 v 2 = G M r 2πr T 2 = G M r r 3 = GM 4π 2 T2 FYI Kepler s third law originally said that the square of the period was proportional to the cube of the radius and nothing at all about what the constant of proportionality was. Newton s law of gravitation was needed for that!

28 F c = mv2 r = G Mm r 2 v 2 = G M r 2πr T 2 = G M r T 2 = 4π2 GM r3 T = 4π2 GM 3/2 r 3/2

29 mv 2 r = G Mm r 2 r = G M v 2 r X r y = G M v x 2 G M v Y 2 = v Y 2 v x 2 = 2πr Y T Y 2πr X T X 2 = T Xr Y T Y r X 2 = 8r Y r X 2 r X r y 3 = 64 r X r y = 4

30 Solving problems involving gravitational field Consider Dobson inside an elevator which is not moving If he drops a ball, it will accelerate downward at 10 ms -2 as expected. PRACTICE: If the elevator is accelerating upward at 2 ms -2, what will Dobson observe the dropped ball s acceleration to be? SOLUTION: Since the elevator is accelerating upward at 2 ms -2 to meet the ball which is accelerating downward at 10 ms -2, Dobson would observe an acceleration of 12 ms -2. If the elevator were accelerating downward at 2, he would observe an acceleration of 8 ms -2.

31 PRACTICE: If the elevator were to accelerate downward at 10 ms -2, what would Dobson observe the dropped ball s acceleration to be? SOLUTION: He would observe the acceleration of the ball to be zero! He would think that the ball was weightless! FYI The ball is NOT weightless, obviously. It is merely accelerating at the same rate as Dobson! How could you get Dobson to accelerate downward at 10 ms -2? Cut the cable!

32 The Vomit Comet

33 PRACTICE: We have all seen astronauts experiencing weightlessness. Explain why it only appears that they are weightless. SOLUTION: The astronaut, the spacecraft, and the tomatoes, are all accelerating at a c = g. They all fall together and appear to be weightless. International Space Station

34 PRACTICE: Discuss the concept of weightlessness in deep space. SOLUTION: Only in deep space which is defined to be far, far away from all masses will a mass be truly weightless. In deep space, the r in F = GMm / r 2 is so large for every m that F, the force of gravity, is for all intents and purposes, zero.

35 Since the satellite is in circular orbit F C = mv 2 / r. Since the satellite s weight is holding it in orbit, F C = mg. Thus mv 2 / r = mg. Finally g = v 2 / r.

36 A satellite of mass m orbits a planet of mass M and radius R as shown. The radius of the circular orbit of the satellite is x. The planet may be assumed to behave as a point mass with its mass concentrated at its centre. (a) Deduce that the linear speed v of the satellite in its R x orbit is given by expression v = GM x g = GM x 2 a c = v2 x = GM x 2 (a c = g in circular orbits). v 2 = GM x v = GM x (b) Derive an expression, in terms of m, G, M and x, for the kinetic energy of the satellite. From a v 2 = GM x E K = 1 2 mv2 = GMm 2x

37 This question is about gravitation. A binary star consists of two stars that each follow circular orbits about a fixed point P as shown. The stars have the same orbital period T. Each star may be considered to act as a point mass with its mass concentrated at its centre. The stars, of masses M 1 and M 2, orbit at distances R 1 and R 2 respectively from point P. M 1 R 1 R 2 P M 2 (a) State the name of the force that provides the centripetal force for the motion of the stars. It is the gravitational force. (b) By considering the force acting on one of the stars, deduce that the orbital period T is given by the expression M 1 experiences F C = M 1 v 12 / R 1. Since v 1 = 2 R 1 / T, then v 12 = 4 2 R 12 / T 2. Thus F C = F G M 1 v 12 / R 1 = GM 1 M 2 / (R 1 +R 2 ) 2. M 1 (4 2 R 12 / T 2 ) / R 1 = GM 1 M 2 / (R 1 +R 2 ) R 1 (R 1 +R 2 ) 2 = GM 2 T 2 T 2 4 = 2 R 1 (R 1 +R 2 ) GM 2 2 Note that F G = GM 1 M 2 / (R 1 +R 2 ) 2.

38 (c) The star of mass M 1 is closer to the point P than the star of mass M 2. Using the answer in (b), state and explain which star has the larger mass. From (b) T 2 = (4 2 / GM 2 )R 1 (R 1 +R 2 ) 2. M 1 R 1 R 2 P M 2 From symmetry T 2 = (4 2 / GM 1 )R 2 (R 1 +R 2 ) 2. (4 2 / GM 2 )R 1 (R 1 +R 2 ) 2 = (4 2 / GM 1 )R 2 (R 1 +R 2 ) 2 (1 / M 2 )R 1 = (1 / M 1 )R 2 M 1 / M 2 = R 2 / R 1 Since R 2 > R 1, we see that M 1 > M 2.