Table

Constants & Variables retaining constant value for each question

Note: Any variable NOT in this chart does not necessarily retain a constant value for each question.

Question 1

Find the Orbital Parameters of the Transfer Orbit.

a) What are the perihelion and aphelion distances of the Hohmann Transfer Orbit to Mars? Determine the semi-major axis, the eccentricity and the period of this orbit. What will the TOF (Time of Flight) be for the transfer?

a) For the transfer orbit, the perihelion distance is the Earth's orbital radius, and the aphelion distance is the orbital radius of Mars.

Rp = 1.495978 x 10^11 m

Ra = 2.27987047 x 10^11 m

The semi-major axis, a, is:

a = (Rp + Ra) / 2 = 1.887924235 x 10^11 m

The eccentricity, e, is:

e = (Ra - Rp) / (Rp + Ra) = 0.207606972

The period of the transfer orbit is given by

tau = [2Pi a^(3/2)] / Sqrt[mu sun] = 89485206 s = 517.86 days

The time of flight, TOF, is half the period of the transfer orbit:

TOF = tau/2 = 258.93 days

b) What are its velocities at aphelion and perihelion? Compare these to the circular velocity of the Earth and Mars. What is the total DV required to transfer to Mars without using patched conics?

b) Vearth = Sqrt[m sun / Rearthorbit] = 29783 m/s

Vmars = Sqrt[m sun / Rmarsorbit] = 24126 m/s

For the transfer orbit,

Vp = Sqrt[m sun (e+1) / Rp] = 32729 m/s

From conservation of angular momentum,

Va = RpVp / Ra = 21476 m/s

now,

D v1 = |Vp - Vearth| = 2946 m/s

D v2 = |Va - Vmars| = 2650 m/s

total D v = (D v1) + (D v2) = 5596 m/s

c) Make a plot of the transfer orbit, Earth and Mars.

c)

Question 2

Patched Conics is used to obtain accurate DV values..

a) Using the patched conics method described in the Orbital Mechanics supplement, calculate the actual DV required to escape earth from the ISS in a 300 km LEO.

a) Rearthsphere = Rearthorbit [(m earth / m sun)^(2/5)] = 924308408 m

Vearthsphere = |Vp - Vearth| = 2946 m/s

Vleaving = Sqrt[(Vearthsphere)^2 - 2m earth(1/Rissorbit - 1/Rearthsphere)] = 11276 m/s

Viss = Sqrt[m earth / Rissorbit] = 7722 m/s

D v1 = | Viss - Vleaving| = 3554 m/s

b) Assume that we want to enter a circular orbit around Mars at 250 km. What is the DV required to enter this orbit? Using the answer from part a), what is your total DV? For comparison, the DV required to get into orbit around the moon is 5.679 km/s.

b) Rmarssphere = Rmarsorbit [(m mars / m sun)^(2/5)] = 577332690 m

Vmarssphere = |Va - Vmars| = 2650 m/s

Varrive = Sqrt[(Vmarssphere)^2 - 2m mars(1/Rlmo - 1/Rmarssphere)] = 5515 m/s

Vlmo = Sqrt[m mars / Rlmo] = 3427 m/s

D v2 = | Vlmo - Varrive| = 2088 m/s

total D v = (D v1) + (D v2) = 5642 m/s

Question 3

If aerobraking is to be used, it would probably be done in two stages.

The first pass would be used to enter a highly elliptical orbit, and thus be captured by Mars.

The second pass would then be used to circularize the orbit or to enter a descent to the Martian surface. The DV attainable will depend on the design of the aerobrake, the amount of heat it can withstand and the thickness of the atmosphere, which depends on altitude.

a) The NASA Reference Mission is planning an elliptical parking orbit for its ERV with periapsis of 250 km and apoapsis of 33,800 km. What is the eccentricity of this orbit, its velocity at periapsis and apoapsis, and its period?

a) Rp = Rmars + (2.5 x 10^5 m) = 3647000 m

Ra = Rmars + (3.38 x 10^7 m) = 37197000 m

The eccentricity of the parking orbit is found with the equation,

e = (Ra - Rp) / (Ra + Rp) = 0.821418078

Vp = Sqrt[m mars (e+1) / Rp] = 4625 m/s

From conservation of angular momentum,

Va = RpVp / Ra = 453 m/s

The period of the parking orbit is,

t = [2p a^(3/2)] / Sqrt[m mars] = 88613 s = 1 Martian day

b) Using patched conics, what is the DV that must be effected by the aerobrake to enter this orbit from the Hohmann Transfer of Question 1? Small thrusts are used at apoapsis of this highly eccentric orbit to lower the periapsis altitude, and thus control the amount of aerobraking that will be achieved, or to enter the atmosphere. The DV required is very small, on the order of 10 m/s, and can usually be accounted for with the fuel required for small orbit corrections.

b) for this question, Varrive is the same as in 2.b)

D v = Varrive - Vp = 890 m/s

c) What DV is required of the aerobrake to circularize the orbit of part a) at 250 km? This is approximately what the Mars Global Surveyor achieved during its aerobraking process.

c) the answer to this question is the difference between the speed of the circular orbit (Vlmo), and the periapsis speed of the elliptical orbit (Vp)

D v = Vp - Vlmo = 1198 m/s

Question 4

Once in a circular orbit, the minimum energy descent to the surface can be thought of as a Hohmann Transfer to the surface of the planet, not accounting for atmospheric drag.

a) What is the total DV required to land on the planet from a 250 km circular orbit? What effect will the rotation of the planet have on this DV?

In reality the Martian atmosphere ( 1% the thickness of earth's - but gravity is only 3/8ths! ) will provide most of this DV. If the spacecraft enters the atmosphere at ~6 km/s, the heatshield would reduce this to approximately 400 m/s through aerobraking. Parachutes would then be deployed which would further reduce this to 50-100 m/s, depending on the size of the chute and the mass of the spacecraft. The rest of the deceleration will have to be provided for by the spacecraft. In the NASA mission, a deceleration burn DV of 500 m/s is provided for to include a factor of safety.

a) The landing on Mars is just a Hohmann transfer from LMO to the surface of the planet. It must be noted that the planet is rotating, so the final burn must slow down the orbit to the rotational velocity of Mars.

for the transfer orbit,

Ra = Rlmo

Rp = Rmars

e = (Ra - Rp) / (Ra + Rp) = 0.035491198

Va = Sqrt[ m mars (1-e) / Ra] = 3365 m/s

D v1 = | Va - Vlmo| = 62 m/s

from conservation of angular momentum,

Vp = RaVa / Rp = 3613 m/s

Mars' rotational period = 88613 s, so

Mars equatorial velocity, Mev = 2p Rmars / t rot = 241 m/s

D v2 = | Vp - Mev | = 3372 m/s

total D v = D v1 + D v2 = 3434 m/s

The rotational velocity of Mars determines the final speed that the second burn must bring the craft's velocity to, in order to land.

b) What is the total DV to land on Mars if aerobraking and parachutes are used for capture and landing?

Assume a small DV of 500 m/s is required for the deceleration burn and maneuvering at the surface. What is the total TOF including one orbit in the elliptic capture orbit?

b) Here, we must determine the total D v needed to land on Mars, when leaving from the orbit of the ISS around Earth. Let's summarize the maneuvers needed:

burn to leave ISS = 3554 m/s

aerobrake to Mars orbital capture = 890 m/s

aerobrake to circularization = 1198 m/s

landing sequence (aerobrake, chutes) = 3434 m/s

additional surface maneuvers (burn) = 500 m/s

Note that this assumes that the 500 m/s burn specified in the assignment is separate from the 3434 m/s required to land.

so, the sum of all of these, total D v = 9576 m/s

but the amount required from burns is only 4054 m/s

Now, we need the total time of flight, including the Hohmann transfer from Earth to Mars, and assuming one orbit in the elliptical Mars parking orbit. All other times are considered insignificant.

TOF = (Hohmann transfer time) + ( 1 Mars elliptical orbit time) = 22459915 m/s = 259.95 days

Question 5

On the departure from Mars, aerobraking will not be of much use. All of the DV required will have to come from the ascent vehicle or ERV.

a) What is the DV required to rendezvous with the ERV which is in a 250 km circular orbit? Neglect any atmospheric effects and assume it is a simple Hohmann Transfer. What DV is required if the ERV was in the highly elliptic parking orbit of Question 3 a)?

a) This maneuver is a Hohmann transfer, with apoapsis in the circular low Mars orbit, and the periapsis on the surface of the planet

Ra = Rlmo

Rp = Rmars

Vp = Sqrt[ m mars (e+1) / Rmars] = 3613 m/s

D v1 = | Vp - Vmarsrot| = 3372 m/s

from conservation of angular momentum,

Va = RpVp / Ra = 3365 m/s

D v2 = | Va - Vlmo | = 62 m/s

total D v = D v1 + D v2 = 3434 m/s

Note that this is the same velocity change required for the landing maneuver from low Mars orbit, which is what we would expect from the symmetry of orbits.

For rendezvous into the elliptical Mars orbit, there are two possibilities. You can transfer into the periapsis of the ellipse, or the apoapsis of the ellipse. We ran both sets of numbers, and calculated that it is easier to rendezvous at apoapsis:

Rp = Rmars

Ra = Rmars + (3.38 x 10^7 m) = 3.7197 X 10^7 m

e = (Ra - Rp) / (Ra + Rp) = 0.832635364

Vp = Sqrt[ m mars (e+1) / Rp] = 4806 m/s

D v1 = | Vp - Vmarsrot | = 4565 m/s

from conservation of angular momentum,

Va = RpVp / Ra = 439 m/s

from question 3. a), the parking orbit velocity at apoapsis has been calculated to be 453 m/s, so:

D v2 = | Va - 453 m/s | = 14 m/s

total D v = D v1 + D v2 = 4579 m/s

Note, that we calculated a total D v of 4632 m/s for rendezvous at the parking orbit periapsis.

b) If you wanted to comlpete your Trans Earth Injection in one step and not stop in orbit, what DV is required? (Patched Conics must be used).

b) Vleaving = Sqrt[(Vmarssphere)^2 - 2m mars(1/Rmars - 1/Rmarssphere)] = 5670 m/s

D v = | Vleaving - Vmarsrot | = 5429 m/s

c) What is your total DV for the entire trip (returning to the ISS), and what is the round trip TOF?

c) In this question, we will assume that we burn to depart the ISS, capture into an elliptical Mars orbit using the aerobrake, then burn to circularize (saving time over an aerobrake circularization) . We then land using aerobrake, parachutes, and the 500 m/s burn for surface maneuvering. We then depart Mars without stopping in orbit.

However, before we calculate the total D v required, we must calculate the D v needed to arrive back in Earth orbit.

Varrive = Sqrt[(Vearthsphere)^2 - 2m earth(1/Riss - 1/Rearthsphere)] = 11276 m/s

D v = | Varrive - Viss | = 3554 m/s, exactly what we calculated to leave the Earth, as would be expected

now, the total mission D v,

burn to leave ISS = 3554 m/s

aerobrake to Mars orbital capture = 890 m/s

burn to circularization = 1198 m/s

landing sequence (aerobrake, chutes) = 3434 m/s

additional surface maneuvers (burn) = 500 m/s

direct trans-Earth injection = 5429 m/s

arrival back at ISS = 3554 m/s

total D v = 18559 m/s

however, the amount required from burns is only = 14235 m/s

total TOF = Hohmann time + 1 parking orbit + Hohmann time = 518.89 days

Question 6

A very important aspect of this mission is knowing where the Earth and Mars are to determine when are the best launch opportunities. Because both orbits are assumed circular, there is a constant relative motion between the two bodies.

a) What is the mean motion of Earth and Mars?

What is their synodic period, or the time between one syzygy and the next?

a) we must find the mean motions of Earth and Mars

Nearth = 2p / t earth = 1.991021278 x 10^-7 rad/s

Nmars = 2p / t mars = 1.05851434 x 10^-7 rad/s

the relative angular rate,

Nrelative = Nearth - Nmars = 9.32506938 x 10^-8 rad/s

so, the synodic period is the time required for the relative angular motion to traverse an angle of 2p

t = 2p / Nrelative = 67379502 s = 779.86 days

b) What is the angle between Earth and Mars at departure from Earth that will allow a rendezvous using the Hohmann Transfer of Question 1? Which planet "leads" the other? How often does this launch opportunity occur?

Repeat the question for the departure from Mars.

b) solve for the lead angle, q , where TOF is the Hohmann transfer time

for the trip from Earth to Mars,

q = p - (Nmars x TOF) = 0.773558256 rads

where Mars leads the Earth

for the trip from Mars to Earth,

q = (Nearth x TOF) - p = 1.312581176 rads

where Mars once again leads the Earth

opportunities for both of these transfers occurs once every synodic period, or every 779.86 days

c) Determine the three next suitable departure dates from Earth and the subsequent arrival date back on Earth based on the actual orbits of the Earth and Mars, assuming a Hohmann Transfer in both directions. Show all calculations and arrival/departure dates. These answers must be based on current data from an astronomical almanac or similar source. Quote the source.

What is the minimum stay time on Mars using the Hohmann Transfer trajectories?

c) from "The Astronomical Almanac" 2000 edition, on January 17, 2000, at midnight Universal time, the mean longitudes of Earth and Mars are:

Earth - Mars = 1.828698665 rads, with Earth leading

We need Mars to be leading by 0.773558256 rads for the launch window to occur, so we need to wait a time, t, defined by

t = [2p - (Jan 1, 2000 relative angular distance) - 0.773558256] / Nrelative

= 456.87 days, using January 1, 2000 as the zero-point of time

Each subsequent launch window occurs one synodic period, or 779.86 days, after the previous window.

Earth departure 1: April 1, 2001 day 456.87

Earth departure 2: May 21, 2003 day 1236.73

Earth departure 3: July 9, 2005 day 2016.59

Now, the TOF is just the Hohmann transfer time from question 1, so:

Mars arrival 1: December 16, 2001 day 715.80

Mars arrival 2: February 4, 2004 day 1495.66

Mars arrival 3: March 25, 2006 day 2275.52

Using the same method as we used to calculate t for the trip from Earth to Mars, we can calculate the first departure window from Mars:

t = [2p - (Jan 1, 2000 relative angular distance) - 1.312581176] / Nrelative

= 389.97 days, using January 1, 2000 as the zero-point of time

However, at this time, the first mission will not have yet arrived, so this is too early a window for departure. We know that the opportunities occur every synodic period, so the next window is at:

t = 389.97 + 779.86 days = 1169.83 days

This is the first return launch window. Once again, each launch opportunity occurs every synodic period, so:

Mars departure 1: March 15, 2003 day 1169.83

Mars departure 2: May 3, 2005 day 1949.69

Mars departure 3: June 22, 2007 day 2729.55

Once again, the return trip time is just the Hohmann transfer time, so:

Earth arrival 1: November 29, 2003 day 1428.76

Earth arrival 2: January 17, 2006 day 2208.62

Earth arrival 3: March 7, 2008 day 2988.48

The minimum stay time on Mars is just the time between the first Mars arrival, and the first Mars departure,

t= 1169.83 days - 715.80 days = 454.03 days

d) What is you total mission time (from departure at ISS to arrival at ISS)?

d) the total mission time is just the time between the first Earth departure, and the first Earth arrival:

t= 1428.76 days - 456.87 days = 971.89 days

Question 7

A quicker transit may be desirable since it will keep the crew in a zero G environment for less time. While this can be done, it will be at a cost in DV. Keep in mind that the Hohmann Transfer that you have just calculated is the lowest energy trajectory to get to Mars.

This next question will demonstrate a faster transfer time but at an increased DV.

a) Take the aphelion distance of the transfer orbit to be 1.2 times the orbital radius of Mars. What is the True Anomaly of the rendezvous position with Mars and what is the TOF to get to Mars.

a) We now use a transfer ellipse with perihelion at Earth's orbit, and aphelion at 1.2 times Mars' orbit

Rp = 1.495978 x 10^11 m

Ra = 2.735844564 x 10^11 m

a = (Rp + Ra) / 2 = 2.115911282 x 10^11 m

e = (Ra - Rp) / (Ra + Rp) = 0.292986424

We want to solve for the true anomaly, f, when the transfer orbit radius, R, is equal to the radius of Mars' orbit, where the interception will occur Cos[f] = [a(1 - e^2) - R] / (R x e) solve for f,

f = 2.114576402 rads

the equation for TOF is as follows:

TOF = [[a^(3/2)] / Sqrt[m sun]] x [2 arctan[ Sqrt[(1-e) / (1+e)] tan (f/2)] [ e Sqrt[1 - e^2] Sin[f] / [1 + e Cos[f]]]]

= 13250778 s = 153.37 days

 

b) Using Patched Conics, what is the DV required to depart LEO on this trajectory? You should find that a small increase in DV will cut the TOF quite drastically.

Vp = Sqrt[ m sun (e+1) / Rp] = 33866 m/s

Vsphere = | Vp - Vearth | = 4083 m/s

Vleaving = Sqrt[(Vsphere)^2 - 2m earth(1/Rissorbit - 1/Rearthsphere)] = 11625 m/s

D v = | Vleaving - Viss | = 3903 m/s

 

c) At Mars you must calculate the relative velocity of the spacecraft wrt Mars prior to using patched conics. This is a little more difficult since the velocities of the spacecraft and Mars are not tangential. You must reduce the spacecraft's velocity to radial and tangential components and then it is a straightforward vector sum. (Note, since angular momentum is conserved, the tangential velocity can be found quite easily).

What is the relative velocity of the spacecraft wrt Mars and using patched conics what is the DV required by the aerobrake to capture into a 250 km circular orbit?

c) For this question, we must break-up the spacecraft velocity into tangential and radial components.

Ri = radius at interception = 2.27987047 x 10^11 m

solve for the total velocity of the craft at interception be equating the total energies at perigee of the transfer orbit, and interception

[(Vp^2) / 2] - (m sun / Rp) = [(Vi^2) / 2] - (msun / Ri)

this gives total Vi = 23171 m/s

TANGENTIAL COMPONENT:

from conservation of angular momentum,

tangential Vi = VpRp / Ri = 22222 m/s

Mars has a velocity of 24126 m/s, which is entirely in the tangential direction, so the relative tangential velocity is

tangential Vrel = | 24126 m/s - 22222 m/s | = 1904 m/s

RADIAL COMPONENT:

radial Vi = Sqrt[ (total Vi)^2 - (tangential Vi)^2 ] = 6563 m/s

 

Mars has no velocity component in the radial direction, so

radial Vrel = radial Vi = 6563 m/s

Now, the relative total velocity is given by:

total Vrel = Sqrt[ (radial Vrel)^2 + (tangential Vrel) ] = 6834 m/s

Since this is the relative velocity of the craft with respect to Mars, it is also the Vsphere at Mars' sphere of influence:

Vsphere = 6834 m/s

Varrive = Sqrt[(Vsphere)^2 - 2m mars(1/Rlmo - 1/Rmarssphere)] = 8373 m/s

now, the D v required by the aerobrake to enter into low Mars orbit is:

aerobrake D v = | Varrive - Vlmo | = 4946 m/s

d) What will the lead angle be now for the transfer from Earth to Mars? What will be the date of the next launch opportunity, and when is the first opportunity to return to earth? What is your stay time on Mars and when is your arrival back at Earth?

d) since the true anomaly at interception is 2.114576402 rads, the angle between the planets for the launch window to occur can be found using the formula:

q = 2.114576402 - (Nmars x TOF) = 0.711962549 rads, with Mars leading Earth

from question 6, we know that on January 1, 2000 at midnight UT, the relative position of Earth and Mars is 1.828698665 rads, with Earth leading

We need Mars to be leading by 0.711962549 rads for the launch window to occur, so we need to wait a time, t, defined by

t = [2p - (Jan 1, 2000 relative angular distance) - 0.711962549] / Nrelative

= 464.51 days, using January 1, 2000 as the zero-point of time

Earth departure: April 9, 2001 day 464.51

Now, the TOF is just the transfer time from question 7.a), so:

Mars arrival: September 9, 2001 day 617.88

The TOF on the return trip is the same as on the outbound trip.

TOF = 13250778 s = 153.37 days

solve for the lead angle, theta

theta = (Nearth x TOF) - 2.114576402 = 0.523682358 rads,

with Mars leading Earth the launch window can now be found

t = [2Pi - (Jan 1, 2000 relative angular distance) - 0.523682358] / Nrelative t = 487 days

(too early, try next launch window 1 synodic period later)

t = 1267.74 days,

using January 1, 2000 as the zero point of time

Mars Departure: June 21, 2003 day 1267.74

Using the TOF calculated above,

ISS Arrival: November 22, 2003 day 1421.11

The stay time on Mars is just the time between Mars arrival and Mars departure

t = 1267.74 days - 617.88 days = 649.86 days

e) When do you think humans will first walk on Mars?

e) The choice to send humans to Mars is largely a political one. The technology and scientific knowledge necessary to get there are available today, and so if the next President of the United States were to call for a manned mission to Mars, there would be nothing fundamental stopping us from accomplishing this goal in relatively short order. However, the political climate today, combined with NASA's ongoing struggle to keep the Space Station program alive, have put the goal of humans to Mars on the back burner for the time being. Not until the Space Station is complete (optimistically, 5 years from now, but pessimistically, never) will we see a return of humans to activity beyond low Earth orbit. Assuming a research and development rate approximately equal to that of the Apollo program, and accounting for launch windows and transit time, it is conceivable that humans could be on Mars little more than 10 years after the decision is made to go there. The only question is, who will make that decision, and when will this person make the decision to inscribe his/her name forever into the pages of human history?