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How come this term is not shown in the mathematical expression, -kv ? what does 'k' mean ? How can we determine this value ? what kind of factors are involved in determining k value (Actually, the shape of paraschute or falling body can be a very important factors to determine the k value). The resistance force ? Judging from our common sense, the shape/size of the paraschute definately influence The explanation said the air resistance can be expressed as -kv, but how did you know that ? Jugding from the expression, it seems that the air resistance is determined only by the velocity ? How did you guarantee that the resistance is not influenced by other factors like acceleration, mass of the falling body ? How about the shape of the falling body or paraschute ? Doesn't this influence I hope you would have a lot of questions that has not clearly been explained in my explanation above and that has not been treated in your math course.įor example, my own questions are like this. Note : This can be one of the simplest modeling example, but it doesn't mean that you should be able to understand everything here in 5 minutes and you are dumb if you don't understand this so quickly. Time_elapsed, time_elapsed/3600.Total Force applied to a body = Motion of the body Vel_next = vel_now + acc_now * Time_interval Īcc_next = g * (Radius / total_distance) * (Radius / total_distance) įprintf(stdout," Total time = %f seconds or %f hours.\n Final speed = %f m/s\n", 5 * acc_now * Time_interval * Time_interval Total_distance = Number_of_radius*Radius Īcc_next = g / (Number_of_radius*Number_of_radius) Number_of_radius += 1.0 /* Distance from the Earth's center */
FREEFALL EQUATION FREE
* Initial distance: Number_of_radius*Radius, where Radius = Earth's Radius*/ĭouble Number_of_radius,Time_interval,Radius,time_elapsed,total_distance įprintf(stderr,"\n\nUse: %s InitialDistance\n", argv) įprintf(stderr,"\nEval free fall time given the initial height(in Earth's Radius units)\n\n") This means the max distance will be 52.047 * 6370 km, or about 86% of the Earth Moon distance.ĭouble acc_now,vel_now,acc_next,vel_next,step
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Use 52.047 as parameter to get 96 hours as the free fall time. I used a numerical method (simplified program below). The point is: since the max distance is a small distance, the object and the Earth will be reasonably at the same orbit around the Sun all along, so you don't need to worry about the orbit.Ĭould you back this up with a calculation? After 4 days the object will reach its max distance from the Earth (a fraction of Earth-Moon distance), and 4 days later, the object will return back to you. Suppose you are at north pole, just to simplify the problem (otherwise, we should add a horizontal velocity component to the launching speed in order to shoot the object in a real radial direction). So with a little bit of luck it will still land on the surface of the earth. This amounts to 0,637 of the earth diameter. The difference between the two is 5,43*10^-5 R.
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R is 1AU and \alpha is the fraction of the orbit traveled, so this is 4/365,25. The rocket would have traveled a distance 2\pi R \alpha while the earth would have moved a distance Rsin(2\pi \alpha ) in the same direction.