K
keermalec
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Hi, calculating a Hohmann transfer orbit with high-thrust acceleration at departure and arrival is pretty straightforward (http://en.wikipedia.org/wiki/Hohmann_transfer).<br /><br />However, when one tries to calculate a minimum transfer orbit using continuous low-thrust, things get much trickier.<br /><br />For example, getting from Earth to Mars using a Hohmann transfer orbit requires delta-vs of 2.93 km/s at Earth orbit and 2.64 km/s at Mars orbit. the trip takes 258 days and is a semi-ellipse with the Earth at perigee and Mars at Apogee.<br /><br />Now if I want to get to Mars using a continuous but low thrust as, for example, when using an ion drive, I suppose the total delta-v will still be 5.57 km/s but the duration and total angle accomplished around the sun depends entirely on the amount of thrust.<br /><br />Is it right to suppose that by continuously accelerating for 258 days to attain a speed of 5.57 km/s I will get to Mars, or is it more complex than that? 5.57 km/s in 258 days is about 0.00025 m/s/s acceleration, or 0.000025 Gs. Will I do a 180° turn around the sun at this thrust? If I have a higher thrust, will I do less than 180°?<br /><br />In the end the question is: if Earth is at angle A and Mars is at angle B, what is the thrust and total delta-v needed to get from Earth to Mars, sing continuous thrust? <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>