Monday, December 14, 2015

Manned Mercury Mission Part 02a - Trajectory Analysis Part 1

One of the most important parts of the development of any interplanetary mission is trajectory planning. The exact trajectory the spacecraft takes affects several factors, including travel time, delta-V needed, thermal and radiation environment, and more. Given that Mercury is quite far from Earth, and in a very different orbit, trajectory analysis will be especially important for this mission.

Previous unmanned missions to Mercury, such as Mariner 10 and MESSENGER, used multiple flybys of Venus and Earth to reduce the amount of delta-V needed. However, I won't be using them for this mission. First, using flybys, while reducing delta-V, usually increases the travel time greatly; the MESSENGER probe took about four years to arrive at Mercury. Since this is a manned mission, we want to ensure that our astronauts do not spend any longer in space than they have to. Second, NASA has some of the best supercomputers in the world available to do the highly complex calculations needed for planning flyby trajectories. I have an old laptop and a graphing calculator. So, instead, our trajectory will be as direct as possible of a transfer from Low Earth Orbit to Mercury's surface (and back).

Some of the orbital parameters of Mercury and Earth can be found here: http://nssdc.gsfc.nasa.gov/planetary/factsheet/mercuryfact.html. 

It becomes immediately apparent that there are several problems. Mercury's orbit is inclined about 7 degrees, compared to virtually nil for Earth. This means that a planar change will be required, significantly increasing the delta-V requirements. Mercury's orbit is also highly eccentric, with the perihelion and aphelion more than 20 million kilometers apart. In order to minimize the delta-V required, we should try to intercept Mercury near its aphelion.

According to this site, roughly 17 km/s of delta-V is required for an Earth to Mercury transfer. However, the results given there are approximate, and don't account for plane changes or any other factors. I'd prefer to do the numbers myself. First, we'll calculate the delta-V needed to transfer from Earth's orbit to Mercury (neglecting the velocity needed to escape from Earth's gravitational pull). This is known as the "excess hyperbolic velocity", it's called that because the escape trajectory looks like a hyperbola relative to Earth (orbits with an eccentricity greater than 1 are hyperbolas, while those with e<1 are ellipses). We'll assume that a single burn is done that changes our orbit from one that is the same as Earth's to one that intersects Mercury's at aphelion (which will be our new perihelion).

From that fact sheet I linked earlier, Earth has a mean orbital velocity of 29.78 km/s (our planet's orbit is nearly circular, the min and max are only about a kilometer apart). The orbital radius of Earth's orbit is 1.496*10^8 kilometers, while Mercury's orbital distance at aphelion is 6.98*10^7 kilometers. Since both bodies (and our hypothetical spacecraft) are in solar orbit, the value of  μ (the standard gravitational parameter) is 1.327*10^20 m^3/s^2. Substituting these into the equation for the delta-v of the first burn for a Hohmann transfer orbit
we see that a delta-v of 6026 m/s is needed. Note that since we're moving inward from Earth, we actually have to slow down our craft by that much. This value of 6026 m/s is our excess hyperbolic velocity. We'll call it 6.1 km/s, since there might be a few uncertainties and simplifications I've done that could add a couple m/s.

Next, we have to account for the velocity needed to escape Earth. That's given by the following formula;

Computing this for an escape from Earth's surface will give a value of about 11 km/s. However, our mission will start in Low Earth Orbit, a few hundred kilometers above Earth's surface. Since it's in orbit, it will already be traveling at very high speed. This means that our delta-V needed will be less than what is given by the formula. Assuming a 300 km parking orbit; our escape velocity will be approximately 10.93 km/s. However, at a 300 km orbit, our orbital velocity is about 7.73 km/s, so we only need ~3.2 km/s to escape from Earth's gravitational pull. If we simply added this to our excess hyperbolic velocity, we would get that we need a delta V of almost 10 km/s at Earth departure. Fortunately, this is not the case;

Computing the value for "v" given our known escape velocity and excess hyperbolic velocity, we get a delta V of  roughly 6.9 km/s.

Our calculations for Earth departure are complete except for one step; accounting for the difference in inclination between Earth's orbit an Mercury's orbit. Mercury's orbit is inclined 7 degrees relative to Earth's; failing to account for this would result in our spacecraft missing its destination by millions of kilometers.

If we perform the inclination change immediately after our burn to leave Earth, our needed delta V would be given by the following formula;

Changing our inclination to completely match Mercury's would require an additional delta V of about 2.9 km/s. On the other hand, if we were able to time our launch such that we only had to change our inclination by 3 degrees to intercept Mercury, only 1.25 km/s of delta V would be required. In the second case, our total Earth departure delta V would come to about 8.2 km/s.


The procedure to compute the delta V upon arrival at Mercury is essentially the same as the one for Earth departure, only in reverse.

We will plan on entering a 100 km parking orbit around Mercury. In order to become captured by Mercury's gravity, we must eliminate all velocity relative to the planet, less our orbit speed in the parking orbit (we will assume the capture burn takes place at our point of closest approach to Mercury). In a 100 km parking orbit around Mercury, our orbital speed will be 2.94 km/s. Looking at the Hohmann transfer orbit equations, we see that the velocity needed to match Mercury at aphelion from our transfer orbit is given by the following;

Running the numbers, we get that a delta V of  7.31 km/s is needed. It would appear then, that our capture velocity is given by (7.31-2.94) = 4.37 km/s. However, while becoming captured, we will be going deeper into Mercury's gravity well, trading gravitational potential energy for kinetic energy. This will require additional delta V to cancel, equal to the escape velocity from an altitude of 100 km above Mercury; 4.16 km/s. Therefore, our total deltaV at capture is 8.51 km/s. Combining this with our Earth departure deltaV, we see that our total delta V needed to travel from Low Earth Orbit to Low Mercury Orbit is approximately 16.7 km/s, which we can round up to 17 km/s.





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