Question
Calculate the length of the Hohmann Transfer (HT) semi-major axis, (a) pictured above. The HT semi-major axis is the distance to the center of the
Calculate the length of the Hohmann Transfer (HT) semi-major axis, (a) pictured above. The HT semi-major axis is the distance to the center of the ellipse along the major axis. The center of the ellipse is halfway between the two foci (one of which is the sun). To determine the length of the HT semi-major axis, first determine the length of the HT "major axis" and divide it in half. The Hohmann Transfer major axis length is determined by adding the aphelion length of Earth's orbit to the aphelion length of Mars' orbit.
Step 2:
Where T is the time in seconds,"a" is the ellipsesemi-major axis (calculated in Step 1), and= 1.327124x10^11km^3/s^2, which is referred to as the standard gravitational parameter of the Sun. The gravitational parameter of the Sun is used due to the Sun's gravity defining the ellipse of this orbital motion. For pi, use at least 4 decimal places: 3.1416
The orbital period calculated above; however, is the time (in seconds) for an entire elliptical orbit. As this spacecraft is going to be captured by Mars' gravity after only half of a rotation, you must nowdivide the Orbital Period time by 2to determine the actual time taken by this maneuver.
Your answer to this calculation will be represented in seconds.
Part 3:
You are tasked with determining the length of time, measured in days, that it will take for a spacecraft to journey from LEO to LMO using this maneuver.
Using the answer from Step 2, calculate the time taken by this maneuver, measured in days. You may assume exact lengths of seconds in a minute, minutes in an hour, and hours in a day for conversion purposes.
Step by Step Solution
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Step: 1
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