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Gravitational Force That the Earth Exerts on the Object

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Definition of Weight Revisited The weight of an object on or above the earth is the gravitational force that the earth exerts on the object. The weight always points toward the center of mass of the Gravitation earth. On or above another astronomical body, the weight is the gravitational force exerted on the object by that body. The direction of the weight (or gravitational force) points towards the center of mass of that body. SI Unit of Weight: Newton (N) 1 2 Newton’s Law of Universal Gravitation Newton’s Law of Universal Gravitation M E m 1. The gravitational acceleration, g depends on the W = G (1) r 2 distance, r, between the object and the earth’s center of mass. where W is the weight of an object with 2. Equation (1) can be generalized for the gravitational mass m due to the earth’s gravitational force between two objects with masses m and M, for force, G is the universal gravitational which M in eqn. (1) is replaced by M and the constant = 6.67x10-11 m2/kg2, M is the E E distance r represents the distance between the mass of the earth, r is the distance centers of mass of the two objects. between the object and the center of rd mass of the earth. 3. By Newton’s 3 law, the object acts on the earth with a force having the same magnitude but pointing in 6 the opposite direction. Write W in the usual form, W = mg RE = 6.38x10 m M = 5.97x1024 kg M E E We get g = G -11 2 -2 r 2 G = 6.67x10 m kg 3 4 Gravitational Field Gravitational Field The gravitational field, g, at a point is the gravitation Since the gravitational field is essential the force an object experiences when placed at that point gravitational force experienced by a unit mass at the divided by the object’s mass. For gravitational field point of interest, it should have a direction. The coming from the earth, direction of the gravitational field is pointed towards mM 1 the body that produces the field. In other words, g = G E ⋅ r 2 m gravitational force always attracts the object towards the body producing the field. M Note that gravitational force is a kind of g = G E r 2 interaction forces. So both the object and the body involved experience the same magnitude of where g is in units of m/s2 and r is the distance the attractive force from each other. point is from the center of mass of the earth. This result shows that the gravitational field is the same as the gravitational acceleration. 5 6 1 Gravitational Field Gravitational force on the earth’s surface Find the gravitational field g on the earth’s F12=m1g2 surface. m1 m2 F21=m2g1 Solution: M E m1 < m2 g = G 2 r12 RE 24 −11 2 2 ()5.98×10 kg = 6.67 ×10 N ⋅ m kg Gravitational field produced by m at a distance ()2 2 ()6.38×106 m of r12: g = Gm /r 2 2 2 12 ()5.98kg = ()6.67 N ⋅ m 2 kg 2 ×10−11+24−2×6 6.38 m 2 Gravitational field produced by m1 at a distance () 2 of r12: 2 = 9.80 m s g1 = Gm1/r12 (< g2 since m1 < m2) 7 8 Earth and Moon Gravitational force acting on the earth by a car Using the fact that the gravitational field at the surface of the Earth is about six times larger than that at the Find the gravitational acceleration acting on the earth by surface of the Moon, and the fact that the Earth’s radius a car with mass 1500 kg running on its surface. is about four times the Moon’s radius, determine how the mass of the Earth compares to the mass of the Moon. Solution: The gravitational force, F, acting on the Solution: 2 earth by the car is (1500kg)(9.8m/s ) −mg Gm =14700N pointed from the earth’s g = r 2 center towards the car. +mg ⇒ r 2g /m = r 2g /m (1: earth, 2: moon) The acceleration on the earth due to 1 1 1 2 2 2 this force is F/ME 2 2 m2/m1 = (r2/r1) (g2/g1) = (1/4) (1/6) = 1/96 = 14700N/(5.98x1024kg) = 2.46x10-21m/s2, which would be too small to be detected. So the mass of the moon is 1/96 times of that of the 9 earth. 10 Three masses on a straight line Three masses on a straight line 2M M 3M (a) Three masses, of mass 2M M 3M 2M, M, and 3M are equally R R spaced along a line, as R R shown. The only forces (ii) What’s the magnitude of the net force experienced by each mass experiences are the forces of gravity from the mass 2M? other two masses. (i) Which mass experiences the largest magnitude net force? Solution We can just add the forces from the other two objects. The 1. 2M net force on the 2M object is directed right with a magnitude 2. M of: 3. 3M 4. Equal for all three 5. Net force magnitude on 2M is equal to that on 3M but bigger than that on M 11 12 2 A triangle of masses Net gravitational field at point A The net gravitational field at point A comes from three Three point objects, 1 through 1 sources, objects 1, 2, and 3. The diagram below shows the 3 with identical mass, are corresponding three gravitational fields, g1, g2 and g3. placed at the corners Obviously, g2 and g3 cancel. of an equilateral triangle. The net field at A is due only to g1, which points up. 3 1 2 2 gi = Gm/riA 2 ∝ 1/riA In what direction is the net gravitational field at point A, where riA is the distance halfway between objects 2 and 3? g1 between point A and 2 g2 g3 3 object i, m is the mass of the objects. 13 14 Net gravitational force at point A Velocity of Orbiting Satellites There is only one speed, v, that a satellite can have if the Find the net gravitational force experienced by an object of satellite is to remain in a circular orbit with radius, r. What is mass 2M at point A in terms of the mass of the three source the relation between v and r ? What is v at r = 10R ? objects m and the length of each side of the equilateral E triangle, L. Solution: Solution For the satellite to remain in a circular orbit with radius r, the From the above discussion, the net gravitational field is g1, acceleration of the satellite must which is Gm/r 2. But r = Lsin60o = (√3/2)L. So, 2 1A 1A equal ac = v /r for otherwise the Fnet = mg unbalanced force, F – ma will g = 2Gm/(3L2). net c 1 cause the satellite to move away The net gravitational force, F, experienced by an object with from the (circular) orbit. mass 2M is 2M times the field at that point. So, v2 mM m = mg = G E r r 2 2 F = (2M)g1 = 4GmM/(3L ). GM 15 ⇒ v = E 16 r Three masses on a straight line Velocity of Orbiting Satellites 2M M 3M Substitute r = 10RE to find v in the orbit: GM R R v = E r (iii) Which of the following changes would cause the magnitude of the force experienced by the 2M object to 6.67x10−11m2kg-2x5.97x1024 kg increase by a factor of 4? Select all that apply. = 10x6.38x106 m [ ] double the mass of all three objects [ ] change the mass of the 2M object to 8M, without 6.67m 2kg -2 ×5.97kg = ×10()−11+24−1−6 ×1/ 2 changing the mass of the other objects 6.38m [ ] double R [ ] Move the system to a parallel universe where the value of = 2.50 km/s the universal gravitational constant is four times larger than its value in our universe This is faster than the speed of any plane ever flown on 17 earth (which is ~2 km/s)! 18 3 The Orbital Radius for Synchronous The Orbital Radius for Synchronous satellites satellites In many applications of satellites such as digital satellite Solution: GM system television, it is desirable that the motion of the satellite From last example, v = E follows a circular orbit and be synchronized with the r 2π r earth’s self rotation (so that the But, for uniform circular motion, v = τ satellite is always at the same R = 6.38x106 m location above the earth’s surface). 2πr GM E ⇒ = E This requires that the period, τ, 24 τ r ME = 5.97x10 kg of the satellite be exactly one day, G = 6.67x10-11 m2kg-2 4 τ GM i.e., 8.64 x 10 s. What is the height, ⇒ r 3/ 2 = E (2) H, of the satellite above the earth’s 2π surface? ⇒ r = 4.22 x 107 m 7 So, H = r – RE = 3.58 x 10 m 19 20 General Orbital Motions of satellites Gravitational potential energy 1. They often trace out an ellipse. Therefore, the The gravitational interaction or potential energy of two objects gravitational force is not always perpendicular to with masses m and M and separation r is: the satellite’s velocity. GmM 2. K + U is conserved throughout the orbit U =− g r 3.
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