PHYS 1443 – Section 501 Lecture #15 Monday, Mar. 22, 2004 Dr. Andrew Brandt Gravitational Potential Energy and Escape Speed Power Momentum Monday, Mar. 22, 2004 PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

Announcements HW#6 on Ch. 7+part of Ch. 8 is due Weds 3/24 at midnight (note HW#7 will be due 3/29) Test 2 on ch. 6-10 will be Weds Apr. 7 Monday, Mar. 22, 2004 PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

The Gravitational Field The gravitational force is a field force. The force exists every where in space. If one were to place a test object of mass m at any point in space in the existence of another object of mass M, the test object will feel the gravitational force, , exerted by M. Therefore the gravitational field g is defined as In other words, the gravitational field at a point in space is the gravitational force experienced by a test particle placed at the point divided by the mass of the test particle. So how do we write the Earth’s gravitational field? Where is the unit vector pointing outward from the center of the Earth E Far away from the Earth’s surface Close to the Earth’s surface Monday, Mar. 22, 2004 PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

The Gravitational Potential Energy What is the potential energy of an object at the height y from the surface of the Earth? No Is this a general expression for the Earth’s Gravitational potential? Why not? Because this formula is only valid for the case where the gravitational force is constant, near the surface of the Earth and the generalized gravitational force is inversely proportional to the square of the distance. OK. Then how would we generalize the potential energy in the gravitational field? Because gravitational force is a central force, and a central force is a conservative force, the work done by the gravitational force is independent of the path. RE m ri Fg rf The path can be considered as consisting of many tangential and radial motions. Tangential motions do not contribute to work!!! Monday, Mar. 22, 2004 PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

More on The Gravitational Potential Energy Since the gravitational force is a radial force, it performs work only when the path is in the radial direction. Therefore, the work performed by the gravitational force that depends on the position becomes Potential energy is the negative change of work in the path Since the Earth’s gravitational force is So the potential energy function becomes Since only the difference of potential energy matters, by taking the infinite distance as the initial point of the potential energy, we obtain For any two particles? The energy needed to move the particles infinitely apart. For many particles? Monday, Mar. 22, 2004 PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

Example of Gravitational Potential Energy A particle of mass m is displaced through a small vertical distance Dy near the Earth’s surface. Show that in this situation the general expression for the change in gravitational potential energy is reduced to DU=mgDy. Taking the general expression of gravitational potential energy The above formula becomes Since the situation is close to the surface of the Earth Therefore, DU becomes Since on the surface of the Earth the gravitational field is The potential energy becomes Monday, Mar. 22, 2004 PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

Escape Speed RE m h ME vi vf=0 at h=rmax Consider an object of mass m projected vertically from the surface of the Earth with an initial speed vi and eventually coming to a stop (vf=0) at the distance rmax. Because the total energy is conserved Solving the above equation for vi, one obtains Therefore if the initial speed vi is known, one can use this formula to compute the final height h of the object. In order for the object to escape Earth’s gravitational field completely, the initial speed needs to be This is called the escape speed. This formula is valid for any planet or large mass objects. How does this depend on the mass of the escaping object? Independent of the mass of the escaping object Monday, Mar. 22, 2004 PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

Power Average power Instantaneous power Unit? Rate at which work is performed What is the difference for the same car with two different engines (4 cylinder and 8 cylinder) climbing the same hill? 8 cylinder car climbs up faster Is the amount of work done by the engines different? NO Then what is different? The work rate is higher for 8 cylinder than 4. Average power Instantaneous power Unit? What do power companies sell? Energy Monday, Mar. 22, 2004 PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

Energy Loss in Automobile A typical automobile uses only 13% of its fuel to propel the vehicle. 67% wasted in the engine: 1. Incomplete burning 2. Heat 3. Sound 16% in friction in mechanical parts of the car Why? 4% in operating other crucial parts such as oil and fuel pumps, etc The 13% of the energy from fuel is used for balancing energy loss related to moving vehicle, like air resistance and road friction to tire, etc Two frictional forces involved in moving vehicles Coefficient of Rolling Friction; m=0.016 Air Drag Total Resistance Total power to keep speed v=26.8m/s=60mi/h Power to overcome each component of resistance Monday, Mar. 22, 2004 PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

Example for Power A compact car has a mass of 800kg, and its efficiency is rated at 18%. Find the amount of gasoline used to accelerate the car from rest to 27m/s (~60mi/h). Use the fact that the energy equivalent of 1gal of gasoline is 1.3x108J. First let’s compute what kinetic energy is needed to accelerate the car from rest to a speed v. Since the engine is only 18% efficient we must divide the necessary kinetic energy by this efficiency in order to figure out what the total energy needed is. Then using the fact that 1gal of gasoline can put out 1.3x108J, we can compute the total volume of gasoline needed to accelerate the car to 60 mi/h. Monday, Mar. 22, 2004 PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

Linear Momentum The principle of energy conservation can be used to solve problems that are harder to solve using Newton’s laws. It is used to describe motion of an object or a system of objects. A new concept of linear momentum can also be used to solve physical problems, especially the problems involving collisions of objects. Linear momentum of an object whose mass is m and is moving at a velocity of v is defined as What can you tell from this definition about momentum? Momentum is a vector quantity. The heavier the object the higher the momentum The higher the velocity the higher the momentum Its unit is kg.m/s What else can use see from the definition? Do you see force? The change of momentum in a given time interval Monday, Mar. 22, 2004 PHYS 1443-501, Spring 2004 Dr. Andrew Brandt