Paramecium Brain Posted July 2, 2006 Report Share Posted July 2, 2006 Alright...so in my introduction thread I explained that I was here mostly because of a bet...(that I couldn't find the answer to this problem...) So here's the problem... And here's to hoping you can help me. :) A space shuttle releases a 470 kg communications satellite while in orbit 280kms above the surface of the earth. A rocket engine on the satellite is fired to boost it into a geosynchronus orbit. How much energy does the engine have to provide to change orbits? The total energy, E, for a satellite in circle orbit is e = 1/2U. Take the period of rotation of the earth as 24hrs. Geosynchronous orbit: Kepler's third law: T^2 = (4(pi)^2/GM)r^3 Please help. :shrug: Quote Link to comment Share on other sites More sharing options...

Jay-qu Posted July 2, 2006 Report Share Posted July 2, 2006 Geosynch orbit has raduis 42,164,000m from the center of the Earth. Energy needed to move from 6,658,000 to 42,164,000 will equal: [math]\int_{6,658,000}^{42,164,000}\frac{M_eM_sG}{r^2}dr[/math][math]M_e = 5.9742 * 10^{24} kg[/math][math]M_s = 470 kg[/math][math]G = 6.67 * 10^{-11}[/math] But this does not include the energy need to speed the satellite up to have a period of 24hrs, for that you need the initial velocity of the satellite. Quote Link to comment Share on other sites More sharing options...

CraigD Posted July 3, 2006 Report Share Posted July 3, 2006 A space shuttle releases a 470 kg communications satellite while in orbit 280kms above the surface of the earth. A rocket engine on the satellite is fired to boost it into a geosynchronus orbit. How much energy does the engine have to provide to change orbits?If you only want to know how much kinetic energy the engine must provide to the satellite, you can calculate use the following approach.Assuming the starting orbit is nearly circular, calculate the starting orbital speed [math]V[/math] by solving for centripetal = gravitational acceleration, [math]V^2 = \frac{M_{Earth} \times G}{R}[/math] for starting distance from the center of the Earth [math]R[/math] (6658000 m). Performing this calculation, you’ll discover an practical detail: the ”standard gravitation parameter”, [math]M_{Earth} \times G[/math], is known to higher precision than either [math]M_{Earth}[/math] or the gravitational constant [math]G[/math] separately, gaining a little precision, if necessary.Calculate or look up the orbital speed [math]V_{GEO}[/math] of a satellite in GEOsynchonous orbit ([math]R_{GEO}[/math] = 42164000 m)Calculate the change in speed [math]\Delta V_P[/math] and [math]\Delta V_A[/math] required at the beginning and end of a transfer orbit starting at [math]R[/math] and ending at [math]R_{GEO}[/math]Calculate the change in kinetic energy ([math]E = \frac{1}{2} \times V^2 \times M[/math]) for the satellite’s mass [math]M[/math] and the values calculated above.While technically a solution to the problem, this isn’t a very useful one, since the energy required by a rocket engine isn’t the same thing as that it gives to its payload. To get a useful answer, you have to know the engine’s specific impulse [math]v_e[/math], and use a rocket equation (rocket mechanics are discussed in several existing threads). The tradition at hypography is for people to give direction like the above, and for the person asking the question to then try solving the problem, posting their answer for checking, or indicating any trouble they encounter – because solving the problem is much more fun than having it solved for you! Quote Link to comment Share on other sites More sharing options...

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