Your spacecraft is in a circular, geostationary orbit around earth and you would like to fly to the moon. Draw a suitable Hohmann transfer orbit that will get you there. How long does the journey take? (Neglect the gravitational fields of the moon and the sun for this problem, approximate the lunar orbit by a circle and consider an ideal transfer orbit.)
A geosycnchronous orbit is one with a period equal to the earth's rotational period, which, contrary to popular belief, is 23hr-56min-4.09sec., not 24 hours. Thus, the required altltude providing this period is ~22,238.64 miles, or ~35,787.875 kilometers. The orbital velocity of satellites in this orbit is ~10,088.25 feet per second or ~6,877 MPH. The point on the orbit where the circular velocity of the launching rocket reaches 10,088.25 fps, and shuts down, is the point where the separated satellite will remain.
The required Hohman Transfer velocity to transfer from the geostationary orbit to the lunar orbit, a mean distance of 239,000 miles away, is 13,544fps. The spacecraft will arrive at the lunar orbit with a velocity of 3340fps, taking ~136 hours to get there.
Sufficient propulsion capability will be needed to execute an orbital plane change of from ~18.5º to 28.5º.
To draw a suitable Hohmann transfer orbit from a circular, geostationary orbit around Earth to the Moon, follow these steps:
1. Start by drawing a circle to represent Earth's orbit, with the radius equal to the distance from the center of the Earth to the spacecraft in its geostationary orbit.
2. Draw another circle to represent the orbit of the Moon, with a radius equal to the average distance from the center of the Moon to the center of the Earth.
3. Identify two points on the Earth's orbit that are 180 degrees apart and are located on the opposite side of the Earth from the Moon. These points will be the departure and arrival points of the transfer orbit.
4. Draw an ellipse that is tangent to both the departure and arrival points on Earth's orbit. The major axis of the ellipse should be aligned with the line connecting these points.
5. The transfer orbit is the portion of the ellipse that lies beyond Earth's orbit and intersects with the Moon's orbit.
Now, let's calculate the time it takes for the spacecraft to complete the journey using the Hohmann transfer orbit.
The time for the spacecraft to complete the transfer orbit can be approximated using the following formula:
T = π * √((a^3) / (μ))
Where:
T is the transfer time
π is a mathematical constant approximately equal to 3.14159
a is the length of the semi-major axis of the transfer ellipse
μ is the gravitational parameter of Earth (approximately 3.986 × 10^5 km^3 / s^2)
To find the semi-major axis (a) of the transfer ellipse, use the formula:
a = (r1 + r2) / 2
Where:
r1 is the radius of the initial circular orbit around Earth
r2 is the radius of the final circular orbit around the Moon
Using the average Earth radius of 6,371 km and the average Moon radius of 1,737 km (ignoring the radius of the spacecraft's geostationary orbit), we can calculate the values:
r1 = 6,371 km
r2 = 6,371 km + 384,400 km
a = (6,371 km + (6,371 km + 384,400 km)) / 2
Now, substitute the values into the transfer time formula to find T:
T = π * √((a^3) / (μ))
By solving this equation, you can find the approximate time it takes for the spacecraft to complete the journey from the circular, geostationary orbit around Earth to the Moon using the Hohmann transfer orbit.