Transfer Orbit from Earth to Mars

Topics: Elliptic orbit, Orbit, Planet Pages: 4 (1332 words) Published: February 7, 2013
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...
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