Transfer Orbit from Earth to Mars

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III.Transit
The shape of the orbit is going to be an ellipse. This is caused by the gravitational force acting on the object and explained by Kepler’s First law. Any object travelling in space and orbiting a planet is going to be an ellipse. If the orbit is a circle, then is just a special case of an ellipse. The transfer from Earth to Mars is no exception. Relative to the sun, the rocket is going to launch from the earth radius altitude of 1 AU, and arrive at the Mars altitude of 1.55 AU. In this case, our team considered the one tangent orbit. The one tangent portion of the name means that the orbit will only be tangent to only the Earth orbit. When payload arrives at Mars, it will have a different direction of travel than the planet. This will require an addition burn to change the angle between the planet and payload.

A. Time of Flight - Hohmann
To calculate time of flight for the payload travelling between Earth and Mars, some assumptions are made. It will be assumed that Earth and Mars have circular orbits. In actuality, they have elliptical orbits, but with a very low eccentricity. Eccentricity is a ratio that measures how circular an ellipse is. An eccentricity value of 0 denotes a perfect circle. In this report, it is assumed that Earth and Mars have an eccentricity of 0. It is also assumed that Earth and Mars travel on the same orbital plane. The numerical presentation of this is that the inclination of both planets is 0 degrees. An orbit that is commonly used to transfer between two altitudes (relative to sun) is the Hohmann orbit. Unlike the one tangent orbit, the Hohmann has two tangents; It therefore only needs to change velocity at the arrival orbit and not direction. Because of this reason, it is very energy conservative, but not time. Kepler’s Third Law (eq. 9) can be reduced to a much simpler form. Using Earth’s radius (1 year) and semi-major axis (1 AU) a new equation is formed (eq. 10). The time of flight for the Hohmann orbit is 0.71 years or about 8.5 months. The time of flight is half the time period because only half the orbit is completed. One issue arises with this orbit. It is the alignment of the planets. It requires Mars to be approximately 45° ahead of Earth’s launch position. This orbit is not feasible for a time critical mission.

B. Time of Flight - Calculation Method
For the one tangent transfer orbit, several calculations must be made in order to assure that Mars and the payload rendezvous at the same location and same time. Given an eccentricity value, an angular position (Θ) location must be calculated for where the payload intersects the Mars orbit (r = 1.55 AU). When the angular position is divided by 360°, the orbit completed ratio is calculated. Now the time period for the orbit must be calculated. This was done using the simplified Kepler’s Third Law (eq. 10). The time period was then multiplied by the orbit percent completed; it results in the time of flight in years. Mars must also be at the same angular position when the payload intersects it. Since a time of flight has already been determined and Mars initial location is known, then Mars’ final location can easily be found. This is because time and angular position are directly related (eq. 8) Multiplying the time of flight by the angular velocity, gives the final angular position of Mars. If that angular position is the same as the angular position of the payload, then the two objects have successfully intercepted each other. If not, the initial input, the eccentricity must be altered and then recalculated. It must be noted that the semi-major axis is not a variable. The SMA is dependent on the eccentricity. This was formulated using the polar equation of an ellipse (eq. 5) when the initial angular position is 0°, the radius is 1 AU. This equation (eq. 7) assures that any eccentricity value will have a corresponding semi-major axis distance that results in the payload all ways launching from Earth at the initial...
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