PuGZ Posted June 3, 2007 Report Share Posted June 3, 2007 Hi guys, I've just recently got the results back for a Physics trial exam and I'm curious about something. I'll first show you the question and then my queries in regards to it. The Keyte I satellite has a mass of 83.6kg. It orbits at a height of 690km with a period of 98.6 minutes. The graph below shows how the gravitational field strength depends on distance from the centre of the Earth. Hence there is a graph with a curve in the shape of the right side of a hyperbolic graph. On the horizontal axis is distance from the centre of the Earth (m) and on the vertical axis is gravitational field strength (N/kg). Two points are labelled: (6.38 * 10^6, 9.81) and (7.07 * 10^6, 7.98). These are the values for the Earth's surface and the radius of the orbit, respectively. The first question is such: Calculate the kinetic energy of Keyte I in orbit (assume a circular orbit Kinetic Energy [math] =\; \frac{1}{2}mv^{2}[/math] [math]m\; =\; 8.36\; \cdot \; 10^{1}[/math] [math]r\; =\; 6.38\; \cdot \; 10^{6}\; +\; 6.9\; \cdot \; 10^{5}\; =\; 7.07\; \cdot \; 10^{6}[/math] [math]G\; =\; 6.67\; \cdot \; 10^{-11}[/math] [math]T\; =\; 5.916\; \cdot \; 10^{3}[/math] [math]v\; =\; \frac{2\pi r}{T}\; =\; \frac{2\pi \; \cdot \; 7.07\; \cdot \; 10^{6}}{5.916\; \cdot \; 10^{3}}\; =\; 7.5\; \cdot \; 10^{3}[/math] Kinetic Energy [math]\; =\; \frac{1}{2}\; \cdot \; 83.6\; \cdot \; \left( 7.5\; \cdot \; 10^{3} \right)^{2}\; =\; 2.35\; \cdot \; 10^{9}\; J[/math] The second question is such: Minimum amount of energy required to place Keyte I in its orbit Area under graph = N * m / kg = J / kgTherefore, if we multiply the area under the graph (estimated to be 6.13 * 10^6 using trapezium) by the mass of the satellite (83.6 kg) we should end up with 5.13 * 10^8 J of energy. Apparently this is the wrong answer because we forgot to add the energy from the previous question. Why would this be the case? My textbook has a very similar question, but the question is worded "how much work should be done against gravity to launch the satellite?" Thanks for the pointers, Hypography. :D Quote Link to comment Share on other sites More sharing options...

sanctus Posted June 3, 2007 Report Share Posted June 3, 2007 I would say on the spot (ie without pondering a long time if I'm not about to say something stupid), is that you got only the energy to put the satellite where it is, but without the kinetic energy it would fall down right away, ie you don't have the orbit only the height... Quote Link to comment Share on other sites More sharing options...

PuGZ Posted June 3, 2007 Author Report Share Posted June 3, 2007 Yeah - d'oh - that's what I thought of this morning as I was in the shower. :) (I posted it last night). I'm almost certain that is the case now (and it makes sense, for the reason you stated) but I'll leave the question open in any case anyone wants to confirm it. EDIT: The reason I dismissed it at first is because I know satellites are only ever in free fall, but due to the curvature of the Earth they never land? Or do they have a horizontal velocity component? (Relative to the surface of the Earth). The more I try to justify a particular reason, the more confused I get. Quote Link to comment Share on other sites More sharing options...

Tormod Posted June 3, 2007 Report Share Posted June 3, 2007 Although not directly related to your question, this is related to things I have to answer when I give presentations to kids so let me try (although I don't master the math at least let me chip in!) ;) I don't think you need to know the exact orbit since height above the ground is what dictates the speed required to stay in orbit. A geostationary satellite does not move relative to the surface at about 35,786 kilometers above the ground, yet it does move much faster than the point above which it hovers because it has to travel much faster as the Earth spins. To stay in that position it has to travel [math]2*Pi*(35,786+6,378)[/math] = (roughly) 265,000 kilometers per day, or 11,042 km/h. A satellite in polar orbit 860 kilometers above the Earth (a typical height for weather satellites) travels [math]2*Pi*(860+6,378)[/math] = 45,469 km per *orbit* which means a speed of 1894 km/h. That means it takes 102 minutes to orbit once, and gives us roughly 14,5 orbits per day. That's a huge difference in kinetic energy. (Orbits are calculated from the center of the Earth, thus 6,378 in the equations above). After launch the speed of the surface is irrelevant for the satellite - it is simply necessary to calculate the speed required to stay in orbit. However, for satellites in low orbits (probably below 800 km or so) there will be a larger drag due to atmospheric resistance and this should probably be factored in. As for the freefall issue...satellites will always be in a state of microgravity. They are influenced by the Earth's gravity but only very minutely. Thus it takes a very long time for the gravitational pull to slow them down. This is corrected with thrusters if necessary. If the satellite slows down continually it will fall down (or at least go into orbit at a lower height), if it is accelerated for a long enough time it will orbit further up or even escape the Earth's gravity well and leave. To reach escape velocity, a separate engine would be necessary though. Quote Link to comment Share on other sites More sharing options...

PuGZ Posted June 4, 2007 Author Report Share Posted June 4, 2007 I understand now - thanks guys and my Physics teacher who will probably never read this. :bow_flowers: Quote Link to comment Share on other sites More sharing options...

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