Newton’s Law of Universal Gravitation - Boston University

1 11 Gravitation2 Definition of Weight RevisitedThe 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 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 Unit of Weight: Newton (N)3 Newton s Law of Universal Gravitation2rmMGWE=mgW=2rMGgE=where Wis the weight of an object with mass mdue to the earth s gravitational force,Gis the Universal gravitational constant = , MEis the mass of the earth, ris the distance between the object and the center of mass of the earth.

2 (1)Write Win the usual form,We getRE= = s Law of Universal Gravitation1. The gravitational acceleration, gdepends on the distance, r, between the object and the earth s center of Equation (1) can be generalized for the gravitational force between two objects with masses mand M, for which MEin eqn. (1) is replaced by M andthe distance rrepresents the distance between the centers of mass of the two By Newton s 3rdlaw, the object acts on the earth with a force having the same magnitude but pointing in the opposite FieldThe gravitational field, g, at a point is the Gravitation force an object experiences when placed at that point divided by the object s mass.

3 For gravitational field coming from the earth,mrmMGgE12 =2rMGgE=where g is in units of m/s2and ris the distance the point is from the center of mass of the earth. This result shows that the gravitational field is the same as the gravitational FieldSince the gravitational field is essential the gravitational force experienced by a unit mass at the point of interest, it should have a direction. The direction of the gravitational field is pointed towards the body that produces the other words, gravitational force always attracts the object towards the body producing the that gravitational force is a kind of interaction forces.

4 So both the object and the body involved experience the same magnitude of attractive force from each field produced by m2at a distance of r12:Gravitational Fieldm1m2F12=m1g2F21=m2g1g2= Gm2/r122g1= Gm1/r122(< g2since m1< m2)r12 Gravitational field produced by m1at a distance of r12:m1< m28 Gravitational force on the earth s surfaceSolution:Find the gravitational field gon the earth s surface. ()()()()()()2622411222262422112sm = == + EERMGg9 Gravitational force acting on the earth by a carFind the gravitational acceleration acting on the earth by a car with mass 1500 kg running on its surface.

5 Mg+mgSolution:The gravitational force, F, acting on the earth by the car is (1500kg)( ) =14700N pointed from the earth s center towards the car. The acceleration on the earth due to this force is F/ME= 14700N/( )= , which would be too small to be and MoonUsing the fact that the gravitational field at the surface of the Earth is about six times larger than that at the surface of the Moon, and the fact that the Earth s radius is about four times the Moon s radius, determine how the mass of the Earth compares to the mass of the Moon.

6 =2 Gmgr r12g1/m1= r22g2/m2m2/m1= (r2/r1)2(g2/g1) = (1/4)2(1/6) = 1/96(1: earth, 2: moon)So the mass of the moon is 1/96 times of that of the :11 Three masses on a straight line(a) Three masses, of mass 2M, M, and 3M are equally spaced along a line, as shown. The only forceseach mass experiences are the forces of gravity from theother two masses.(i) Which mass experiences the largest magnitude net force?1. 2M 2. M3. 3M4. Equal for all three5. Net force magnitude on 2M is equal to that on 3M but bigger than that on M2MM3 MRR12(ii) What s the magnitude of the net force experienced by mass 2M?

7 2MM3 MRRT hree masses on a straight lineSolutionWe can just add the forces from the other two objects. The net force on the 2M object is directed right with a magnitude of:313A triangle of massesThree point objects, 1 through 3 with identical mass, are placed at the corners of an equilateral triangle. In what direction is the net gravitational field at point A, halfway between objects 2 and 3? 23114 Net gravitational field at point AThe net gravitational field at point A comes from three sources, objects 1, 2, and 3.

8 The diagram below shows the corresponding three gravitational fields, g1, g2and g3. Obviously, g2and net field at A is due only to g1, which points Gm/riA2 1/riA2where riAis the distance between point A and object i, m is the mass of the objects. 15 Net gravitational force at point AFind the net gravitational force experienced by an object of mass 2M at point A in terms of the mass of the three source objects m and the length of each side of the equilateral triangle, the above discussion, the net gravitational field is g1, which is Gm/r1A2.

9 But r1A= Lsin60o= ( 3/2)L. So, g1= 2Gm/(3L2).The net gravitational force, F, experienced by an object with mass 2M is 2M times the field at that point. So,F = (2M)g1= 4 GmM/(3L2).Solution16 There is only one speed, v, that a satellite can have if the satellite is to remain in a circular orbit with radius, r. What is the relation between vand r ? What is vat r= 10RE?22rmMGmgrvmE==Velocity of Orbiting Satellites rGMvE= Solution:Fnet= mgFor the satellite to remain in a circular orbit with radius r, the acceleration of the satellite must equal ac= v2/rfor otherwise the unbalanced force, Fnet macwill cause the satellite to move away from the (circular) orbit.

10 172MM3 MRRT hree masses on a straight line(iii) Which of the following changes would cause the magnitude of the force experienced by the 2M object to increase by a factor of 4? Select all that apply.[ ] double the mass of all three objects[ ] change the mass of the 2M object to 8M, without changing the mass of the other objects[ ] double R[ ] Move the system to a parallel universe where the value of the Universal gravitationalconstant is four times larger than its value in our universe18 Velocity of Orbiting SatellitesrGMvE=Substitute r= 10 REto find v in the == km/sThis is faster than the speed of any plane ever flown on earth (which is ~2 km/s)!