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*To*: Q-list <quattro@coimbra.ans.net>*Subject*: Re: oh well!(kaboom) - a boring excursion into physics of KinematicMotion II*From*: four.rings@MCIONE.com*Date*: Mon, 27 Jul 1998 00:09:54 -0400*Sender*: owner-quattro@coimbra.ans.net

"Mark Rutherford" <mark@afnetinc.com> wrote: Because of conservation of energy laws and the restrictions, namely no friction the car would be propelled a lot further that 40 miles on 1 cup of fuel. The distance it would be propelled is infinite if frictional forces are not taken into account, Newton's law. While if the gasoline is placed below the car all of the arguments Igor provided are accurate. So a car being propelled up in the presence of a gravitational field can not be compared with a car moving in a straight line because no friction forces are acting on the car. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Mark's consideration is correct for a small horizontal travel L. However in the absence of friction the car will still not travel in a straight line. Greatly simplified the car will feel several forces applied to it: 1. The gravity force, vector pointing to the centre of Earth. 2. The reaction of the base vector pointed in the opposite direction. 3. The vector of centrifugal force in the same direction. 4. The linear component of rotational force applied as a tangential vector due to rotation of Earth. 5. The vector of Coriolice force. Also pointing to the centre of Earth but in a rather complex equation. 6. The tangential vector of force propelling the car horizontally (in our case spending the potential energy of the fuel burned). Let's call it the Propelling force. Probably some more. All except 6. are mutually compensated in our stationary sys (if we assume a moving sys of coordinates) and according to the 1st Law of Newton do not participate in the car's movement. The car however will not leave the Earth' surface and will not continue to move in a straight line tangential to the Earth's surface. For simplicity let's assume that the Earth' radius is a constant R. As the horizontal distance L increases the vector of gravity stops being normal to the vector of Propelling force. Instead we will have a right triangle of forces where the vector of gravity force is a hypotenuse H of this imaginary right angle triangle ( R, L and H). This means that the horizontal component of this gravity force becomes tangential: Ft=Fg*sinß, where ß is the angle between R and H. This component is applied in the same axis but with the opposite sign as the Propelling force, effectively slowing down the vehicle. The reality is a lot simpler, though. If the Propelling force is small, in the absence of friction (air friction, rolling friction etc.) the car will not travel in a straight line but rather will make infinite circles on the Earth' surface. However if the Propelling force is great enough once the car reaches the First Cosmic Velocity (excuse a literal translation here, I am not sure that in English it is called the same way) it will leave the Earth' surface and become it's satellite. If this force is even greater it might reach the Second Cosmic Velocity and leave the Earth' gravity for good. The car needs to overcome the gravity force which is Gamma times car's mass times Earth' mass divided by the distance between them square, where Gamma is the gravitational constant. ************************************************************ Igor Kessel '89 200TQ -- 18psi (TAP) '98 A4TQ -- mostly stock Philadelphia, PA USA http://www.geocities.com/MotorCity/Garage/8949/homepage.html ************************************************************

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