In this exploration the trajectory of a spacecraft from Earth to Mars will be calculated. The aim is to determine the geometrical shape of the trajectory and what type of curve it is. Finding the interplanetary trajectory for a spacecraft is a real life problem, especially the one between Mars and Earth. It is currently solved by sophisticated computers but I am interested in learning how it is done. To do this Newton’s law of Gravitation has to be taken into account. It states that any two objects exert a gravitational force of attraction between each other. This force is directly proportional to the masses of both the objects, M1 and M2, and inversely dependent on the square of the distance, d that separates their centers. As the mass
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When a spacecraft leaves the Earth’s orbit, the gravitational force of the Earth is no longer significant and Mars is too far away so its gravitational pull has no influence either. The only object that exerts a substantial force of gravitational attraction with the spacecraft is the Sun. Because of this it is called the 2 Body Problem. It can be demonstrated using Newton’s Law of Gravitation that the orbit in a 2 Body Problem is a conic.
What could be asked now is, what is the shape of the planets orbit? Kepler in its first law states that the orbit of the planets around the sun is an elliptical shape, with the centre of the sun being located in one of the foci. An ellipse is a curve in which the sum of the distance between any point to another two points is a constant. These two points are called foci; the closer they are to each other, the more the orbit resembles a circle.
The first assumption that has to be made is that Mars and Earth have circular orbits. This is because their eccentricity, how similar to a circle it is, is very …show more content…
Iterations will be done in other to find an ellipse that contains θ1 and θ2, that complies with the time of flight and in which mars will be in the desired position after the duration of the trip.
For each iteration we will assign a value for the angle of inclination Knowing this we can write the equation for both the point of arrival and of departure. r_1=p/(1+e×cos〖(θ_1-ω)〗 ) r_2=p/(1+e×cos〖(θ_2-ω)〗 )
If we divide both equations together the variable p can be eliminated and after some rearrangements e can be found. r_1/r_2 =p/(1+e×cos〖(θ_1-ω)〗 ) × (1+e×cos〖(θ_2-ω)〗)/p r_1×(1+e×cos〖(θ_1-ω))〗=r_2 ×(1+e×cos(θ_2-ω)) r_1×e cos〖(θ_1-ω0)+r_1 〗=r_2 ×e cos〖(θ_2-ω)〗+r_2 r_1×e×cos〖(θ_1-ω)〗-r_2 ×e cos〖(θ_2-ω)〗=r_2-r_1 e×((r_1×cos〖(θ_1-ω))〗- (r_2×cos(θ_2-ω)
When a spacecraft leaves the Earth’s orbit, the gravitational force of the Earth is no longer significant and Mars is too far away so its gravitational pull has no influence either. The only object that exerts a substantial force of gravitational attraction with the spacecraft is the Sun. Because of this it is called the 2 Body Problem. It can be demonstrated using Newton’s Law of Gravitation that the orbit in a 2 Body Problem is a conic.
What could be asked now is, what is the shape of the planets orbit? Kepler in its first law states that the orbit of the planets around the sun is an elliptical shape, with the centre of the sun being located in one of the foci. An ellipse is a curve in which the sum of the distance between any point to another two points is a constant. These two points are called foci; the closer they are to each other, the more the orbit resembles a circle.
The first assumption that has to be made is that Mars and Earth have circular orbits. This is because their eccentricity, how similar to a circle it is, is very …show more content…
Iterations will be done in other to find an ellipse that contains θ1 and θ2, that complies with the time of flight and in which mars will be in the desired position after the duration of the trip.
For each iteration we will assign a value for the angle of inclination Knowing this we can write the equation for both the point of arrival and of departure. r_1=p/(1+e×cos〖(θ_1-ω)〗 ) r_2=p/(1+e×cos〖(θ_2-ω)〗 )
If we divide both equations together the variable p can be eliminated and after some rearrangements e can be found. r_1/r_2 =p/(1+e×cos〖(θ_1-ω)〗 ) × (1+e×cos〖(θ_2-ω)〗)/p r_1×(1+e×cos〖(θ_1-ω))〗=r_2 ×(1+e×cos(θ_2-ω)) r_1×e cos〖(θ_1-ω0)+r_1 〗=r_2 ×e cos〖(θ_2-ω)〗+r_2 r_1×e×cos〖(θ_1-ω)〗-r_2 ×e cos〖(θ_2-ω)〗=r_2-r_1 e×((r_1×cos〖(θ_1-ω))〗- (r_2×cos(θ_2-ω)