***************************

The planning of interplanetary flights, such as from Earth to Mars, uses basic orbital mechanics. As long as the spacecraft speed does not
reach solar escape speed, the form of
the orbital trajectory about the sun

__will__be an ellipse. There are “min energy” trajectories, and there are faster trajectories, but__all__these trajectories__will be__ellipses.
The basics of elliptical orbits

**, including just about all the relevant analysis equations, plus a little more. An ellipse is a symmetrical closed curve containing two foci. The central body occupies one focus, the other is unoccupied. These foci are located farther from the center in the more eccentric ellipse. A circular orbit is a sort of “degenerate” ellipse with zero eccentricity, so that the foci come together at the center. I wrote this for a diverse audience of both technical and non-technical people.***are given in Figure 1*
Bear in mind that these analytical solutions apply

__only__to a 2-body problem (one object orbiting a central body). The 3-body problem (object and two bodies) requires integration of the equations of motion for an exact solution.Figure 1 – Basics of Elliptical Orbits (2-Body Problem)

For interplanetary trips,
we are talking about trajectories where the central body is the
sun. However, this same analysis applies to orbits about
the Earth or any other body. The key
features of elliptical orbits are the speeds,
which vary around the path, and
the min and max distances from the central body.

“Perihelion” is the end apex of the ellipse closest to the sun, where the speeds are highest. “Aphelion” is the end apex of the ellipse
farthest from the sun, where speeds are
lowest. The corresponding terms for an
orbit about the Earth are “perigee” and “apogee”. For the moon, they are “pericynthion” and
“apocynthion”.

*Thus the abbreviation “per” applies to the closest approach, and “apo” the farthest approach, regardless of the central body.*
When looking up tables of characteristics of astronomical
bodies, the description of their orbits
is usually cast as min and max distances from the central body. Those would be r

_{per}and r_{apo}, respectively. Their average is the semi-major axis length “a”. The eccentricity “e” and the period “P” are easily computed from these distances, all**. Further, the distance of the foci from the center of the ellipse “c”, and the length of the semi-minor axis “b”, are also easily computed, and given in the figure, as is the equation for x-y coordinates all along the ellipse, where these x-y coordinate are centered at the center of the ellipse.***as given in Figure 1*
The equation giving velocity at any radius “r” from the
central body is actually quite simple,

**. Note that “r” is bounded: r***as shown in the figure above*_{per}< r < r_{apo}. There is no simple equation giving time at any particular point around the elliptical path. Centuries ago, it was said “equal areas are swept out by the radius vector from the central body in equal amounts of time”.
Today, we would say
that the time along a segment is proportional to the area swept out by the
radius vector from the central body to the object, moving along that segment. You can obtain that time by integrating the
area under the segment from one point to another, and adding or subtracting the appropriate
triangle area. All of this should be
evident upon

**, particularly noting the shaded area.***inspection of Figure 1 above*
The “extras”

**are the equations for calculating the escape velocity and low circular orbit velocity of any celestial body, plus a form of mechanical energy conservation as you approach (or depart from) a body in an unpropelled state.***in Figure 1***Hohmann Min-Energy Transfer**

These trajectories are the ones with the minimum velocity
requirements for travel. The perihelion
of the transfer ellipse is located at the orbit of the Earth on one side of the
sun. The apohelion of the transfer
ellipse is located at the orbit of Mars on the other side of the sun.

**Because the orbit of Earth is slightly elliptical, and Mars more so, these distances vary somewhat. Using the average values gets you into the ballpark, but you should really use the worst case to size your spacecraft’s propulsion capability!***See Figure 2.*
Note that because the Hohmann transfer ellipse is tangent to
a very-nearly-circular planetary orbit at each end, then the planetary velocity and transfer
orbit velocity vectors are essentially parallel at each end. What is ordinarily a

__vector subtraction__devolves to a__simple scalar subtraction__, for determining the velocities with respect to the planet, at each end. That is__only__true when the perihelion or the apohelion of the orbit are located at the appropriate planetary orbit distance from the sun.
Specifically, the
velocity vector of the spacecraft with respect to the planet is the velocity vector
of the spacecraft with respect to the sun,
minus the velocity vector of the planet with respect to the sun:

spacecraft
V

_{wrt planet}= V_{wrt sun}– V_{planet wrt sun}where V is a__vector__velocity**An Approximation to the 3-Body Problem**

The trouble the above evaluations right at perihelion and
apohelion is that these are 2-body (spacecraft and sun) analyses, and the close-vicinity dynamics of departure
and arrival are fundamentally a 3-body problem (spacecraft-sun-planet). As already stated, it takes computerized 3-body analysis to get
velocity requirements and detailed localized trajectories exactly right.

*Yet, you can get very, very close to the velocity requirements with the following approximation technique.*
As in the above discussion,
you do the vector velocity subtraction to find the velocity of the
spacecraft with respect to the planet,
at the perihelion and apohelion conditions. For

__only Hohmann min energy transfer__, this calculation devolves to a simple scalar subtraction, because the vectors are parallel. Either way, there is a velocity magnitude involved.
The

__approximation__is to treat this relative velocity with respect to the planet as a velocity “very far from the planet”, and to approximate the pull of the planet’s gravity during the close encounter as an unpowered gravitational acceleration toward the planet, from “very far” to “very close”. This is done with__conservation of mechanical energy__, based on the “far from planet” velocity magnitude (with respect to the planet), which makes the vector direction more-or-less irrelevant, except to the trajectory details. Conservation of mechanical energy says:
0.5 m V

_{far}^{2}= 0.5 m V_{near}^{2}- change-in-PE far-to-near, where m = spacecraft mass
This approximation then makes use of the very convenient
fact that the change in potential energy from very far to very near is, in point of fact, numerically equal to the spacecraft kinetic
energy associated with the escape velocity of the planet:

0.5 m V

_{esc}^{2}= change-in-PE far-to-near
Thus, after dividing
off the “0.5 m” factors common to all three terms, we have a very simple way to estimate the
spacecraft velocity magnitude with respect to the planet, once it is “very close”. This velocity would apply for either entry
into planetary orbit, or for the initial
direct entry into the local atmosphere for aerobraking (of any type). That simple equation is:

V

_{far}^{2}= V_{near}^{2}– V_{esc}^{2}or V_{near}= (V_{far}^{2}+ V_{esc}^{2})^{0.5}
Doing the full 3-body problem on the computer refines the actual
trajectory to be flown, but does not
refine the spacecraft propulsion velocity requirements very much at all, beyond these simple estimates. These estimates are really quite good, and apply to departure as well as arrival.

This is

**, for both arrival and departure at both Earth and Mars. For arrival, V***illustrated in Figure 3*_{near}is denoted as V_{int}for “interface velocity”, and V_{near}= V_{bo}for the “burnout velocity” at departure.
Please note that Mars arrival and departure trajectories are
directed retrograde, with respect to
Mars, because the planet’s orbital
velocity about the sun exceeds the transfer ellipse apohelion velocity with
respect to the sun.

**You want to time your arrival at Mars orbital distance very slightly ahead of Mars’s arrival, so that you don’t miss closing gravitationally with the planet, or get left behind for not “leading the target” enough. Similarly, you must accelerate in the retrograde direction escaping from Mars, so that you end up at the apohelion of your return ellipse, with the appropriate slower velocity about the sun.***In effect, the planet literally runs over the spacecraft from behind upon arrival.*
The situation at Earth is different, because the transfer ellipse perihelion
velocity with respect to the sun exceeds the orbital velocity of Earth about
the sun. Thus departures and arrivals
are in the posigrade direction with respect to Earth.

**It literally runs away from the Earth in a posigrade direction upon departure.***Upon arrival, the spacecraft is literally running into the Earth from behind.***Actual Departure and Arrival “Close-In” Operations**

Arrival at, and
departure from, Mars is

**. These are similar at Earth, only the numbers are different. You are “close-in” at speed V***depicted in Figure 4*_{near}, which is V_{int}for arrival. If Mars (or Earth) were airless, this is the theoretical delta vee you have to “kill” in order to land direct. They are not airless, so your choices are entry into orbit, or some sort of direct aerobraking entry.
If you are entering Mars orbit, you want to enter it on the side of the
planet and specific location where the orbit velocity vector is fairly parallel
with your own spacecraft velocity vector.
That gives the smallest delta-vee requirement:

dV = V

_{int}– V_{orbit}on the “correct” side
If you enter orbit on the wrong side, the delta vee is much larger:

dV = V

_{int}+ V_{orbit}on the “wrong” side
Orbital entry dV values need no factoring, because these are brief impulsive burns in
space. It is not feasible to use
low-thrust long burn electric propulsion for this, at least not for manned craft. The spiral-in times are months long.

If instead you are aerobraking (whether one pass or
multi-pass), your initial

__entry interface__speed is essentially V_{near}= V_{int}. From there deceleration is by drag, not propulsion, until touchdown. On Mars, retropropulsive touchdown is required at one level or another, since terminal parachute speeds are high subsonic at best. Whatever the terminal velocity is, that is the theoretical delta-vee you need to “kill” for touchdown. I recommend that theoretical value be increased by factor 1.5 to cover maneuver and hover allowances.
Departure from Mars is the exact reverse, but in the same direction as arrival. You want to end up at V

_{near}= V_{bo}still close to the planet, so that your V_{far}is the aphelion velocity of the transfer ellipse back to Earth. If departing from orbit, you burn on the side where the orbital motion is locally retrograde, so that the delta-vee required is lower:
dV = V

_{bo}– V_{orbit}
Departing Mars direct from the surface, your theoretical dV is V

_{bo}, and it must be directed retrograde. This dV needs to be factored-up for small drag and gravity losses. Recommendations are given in the figure.
While

**, departing Earth is the same process and analysis, just with everything oriented in the posigrade direction. If you leave orbit, you do it on the side where orbital motion is posigrade, and end up at V***not shown in the figure*_{bo}in a posigrade direction. That delta vee is:
dV = V

_{bo}– V_{orbit}
If you depart directly from the surface, your theoretical delta vee is V

_{bo}, which must be factored-up for gravity and drag losses. Recommendations are**.***in the figure*
Arriving at Earth has exactly the same values. If you enter into orbit, you do it on the side where orbit motion is
posigrade, so that the delta vee is:

dV = V

_{int}– V_{orbit}
If you enter the atmosphere for aerobraking, your entry interface speed is V

_{int}. Depending upon the design of your spacecraft, there may or may not be a touchdown burn. If there is, it is some terminal speed to “kill”. I recommend factoring that up by 1.5 for hover and maneuver allowances.**Faster Trajectories**

There is

__no such thing__as a “direct flight to Mars”. All fast trajectories are still elliptical about the sun, unless your spacecraft propulsion is capable of far greater than solar escape speed. None are, at this time in history.
Faster trajectories reduce the travel time at the expense of
higher required delta-vees.

__There is no way around that bit of physics__. What you want to do is employ the higher-speed end of your transfer ellipse as your trajectory, and arrive at your destination__before__you reach the lower speed portion of your ellipse. Thus for Earth-Mars, your transfer ellipse perihelion will still be at Earth’s orbital distance, while your transfer ellipse apohelion will be well beyond the orbit of Mars.
There is no point to putting the transfer ellipse perihelion
inward of Earth’s orbit, because that
moves your trip segment towards the slower end of your transfer ellipse. You would thus average lower speeds over
about the same path length.

For any such faster transfer ellipse, the situation is

**. Your perihelion velocity is higher, so your departure delta vee is higher. But the same departure V***as depicted in Figure 5*_{bo}and V_{far}calculations apply, as for the Hohmann ellipse.
You will “get off” the transfer ellipse when your distance
from the sun is at Mars’s distance. You
need to time your arrival to be just as Mars gets there. It will be a real vector subtraction to
determine your velocity with respect to Mars at arrival. Its magnitude is your V

_{far}. The geometry of closure with the planet is more complicated because of the nonparallel angle between the vectors to be subtracted. Even so, just use the “kinetic energy thing” on V_{far}and V_{esc}to get V_{int}. From there, it’s the same basic choices for orbital entry or direct landing, even though the detailed geometries are changed.
Departing Mars is the reverse. Whether from orbit or direct from the
surface, you will end up at V

_{bo}near the planet. The “kinetic energy thing” gets you V_{far}. That speed and an appropriate direction must add vectorially with the planet’s orbital velocity vector, to obtain the velocity vector you want for the return trajectory (correct magnitude, and direction tangent to the transfer ellipse path). You have to time this such that Earth will be at your perihelion point when you get there.
The easiest way to get the angles for the vector additions
is to just plot the transfer ellipse,
and a circle at the Mars distance.
Draw the tangents where they cross,
and measure the angle “a” between them with a protractor. Otherwise,
when evaluating the points on the trajectory, compute the slope at the encounter point
numerically. The tan

^{-1}(slope) is numerically its angle “a1” below reference, where reference is a line parallel to the semi-major axis (a negative angle on the perihelion half of the ellipse, and positive on the apohelion side).
The encounter coordinates give you the angle a3 of the
radius vector at encounter as the value tan

^{-1}(y/(c-x)). The circle approximation for Mars’s orbit through the encounter point has a tangent normal to that radius vector. Its angle below reference “a2” is 90^{o}-radius vector angle a3. The difference in the angles is the angle “a” between the velocity vectors.**.***See Figure 6*
The easiest way to get the time from perihelion to the Mars
encounter point is (again) to plot the transfer ellipse and a circle at the
Mars orbit distance. Bound this with the
semi-major axis, and with the radius
vector from the sun to the Mars encounter point. Then

__use a planimeter__to measure the swept area.
The area of the entire ellipse corresponds to the period of
the whole transfer orbit. Or, if desired,
half the area of the ellipse corresponds to the travel time from
perihelion to apohelion. The ratio of
your planimeter area swept for the trip,
to the ellipse area, is the same
as the ratio of 1-way trip time to orbital period. Or if ratioed to half the ellipse area, the ratio of travel time to one-way trip
time.

If you don’t have a planimeter with which to measure areas
on your plot, then integrate numerically
the area under the ellipse curve (relative to semi-major axis) from the
perihelion point to the encounter point.
Add or subtract as appropriate the area of the right triangle formed by
the encounter vector from the sun to Mars as its hypotenuse. A spreadsheet would work for this. (If you have values for a and b, then you have the equation for the
ellipse, by which to generate coordinate
values for x and y.)

**Trajectories to Venus or Mercury**

Trips to Venus or Mercury work almost the same way as trips
outbound to Mars or further. The
difference is that the transfer perihelion is at the destination, not at Earth.
This is

**. If Hohmann min energy, then the apohelion is at Earth’s orbit. If a faster trajectory, the apohelion is outward from Earth’s orbit.***shown in Figure 7*
Quite frankly, as
fast as these inward-from-Earth trips are with Hohmann min energy
ellipses, there seems little point to
the added complications of a faster trajectory.
Venus is only 143 to 149 days away,
and Mercury is only 95 to 117 days,
using min energy Hohmann trajectories.
Compare that with Mars: 235 to
283 days away, and
Ceres-as-typical-of-the-asteroid-belt at 428 to 517 days away. Times outward of Earth are longer simply because
the distances are larger, and the
velocities are lower.

**Reference Data for Solar System Bodies**

The universal gravitation constant for Newtonian gravity is
G = 6.6732E-11 N-m

^{2}/kg^{2}. The masses and radius data for some selected bodies are as follows:__Body__

__mass, kg__

__eq.R, km__

__avg.R, km__

Sun 1.991E30 695950
695950

Earth 5.979E24
6378.5 6371.3

Mars 6.418E23
3386 3380

moon 7.354E22 1738.7 1738.3

moon 7.354E22 1738.7 1738.3

Phobos 2.72E16
11.3 10.4

Basic orbital data for selected bodies are as follows:

__Body__

__r__

_{per}, km__r__

_{apo}, km
Earth 1.4707E8 1.5207E8

Mars 2.0656E8 2.4912E8

Moon 363,299 405,506

Phobos a = 9408 km
--------------

**Specific Case Study Numbers for Trips from Earth to Mars**

For purposes of typical results, I presume that both Earth and Mars are at
their average distances from the sun,
meaning the radii to their orbits are their “a” values. I also set the transfer orbit perihelion
distance at the Earth orbit distance (in this case its value of “a”) for this
study.

__What changes is the transfer orbit apohelion distance__.
I ran 4 values of transfer orbit apohelion distance: (1) Hohmann transfer at Mars “a” = 2.28E8 km
for comparison, (2) 3.21E8 km to get a
2-year orbit period so that a free return to Earth is possible, (3) 4.00E8 km near the inner edge of the
asteroid belt, and (4) 4.671E8 km to get
a 3-year orbital period so that a free return to Earth is possible from an
orbit whose apohelion is well within the asteroid belt.

For all cases,
departure is from low Earth orbit (LEO).
Arrival at Mars could be any of 3 cases:
(1) direct aerobraking entry leading to a landing, (2) entry into low Mars orbit (LMO), or (3) rendezvous with and touchdown upon
Phobos, Mars’s inner moon. Once V

_{far}is determined for each of the transfer orbit cases, then each arrival sub-case must be analyzed separately.
Finding the encounter point requires solving the equations
of the ellipse and the circle models simultaneously, unless one does this graphically. The ellipse is centered at the origin of the
x-y coordinates, the circle is not
(being centered at the positive-x focus where the central body is located, in all the figures above depicting this).

Now, for the transfer
ellipse, the semi-major (a) and
semi-minor (b) axis distances, and the
distance to the foci (c), are all known. Its center is the origin (0,0). Its equation is:

x

^{2}/a^{2}+ y^{2}/b^{2}= 1
The circle is offset from the origin, centered at (c,0) to the right of that
origin, with a radius equal to the Mars
orbit distance (call it R). Its equation
is therefore:

(x – c)

^{2}+ y^{2}= R^{2}
Solving the circle equation for the y

^{2}term gets us something we can substitute into the ellipse equation, getting us one equation in one variable (x) that we can solve:
x

^{2}/a^{2}+ [R^{2}– (x – c)^{2}]/b^{2}= 1 substitution into ellipse to eliminate y^{2}
We have to expand
the squared binomial in the second term,
and then distribute the 1/b

^{2}coefficient, followed by collection of like terms in x^{2}and x:
x

^{2}/a^{2}+ [R^{2}– (x^{2}- 2xc + c^{2})]/b^{2}= 1
x

^{2}/a^{2}+ [R^{2}– x^{2}+ 2xc – c^{2}]/b^{2}= 1
x

^{2}/a^{2}+ R^{2}/b^{2}– x^{2}/b^{2}+ 2xc/b^{2}– c^{2}/b^{2}= 1
(1/a

^{2}– 1/b^{2})x^{2}+ (2c/b^{2})x + (R^{2}/b^{2}– c^{2}/b^{2}-1) = 0
That result is a
quadratic equation in standard form Ax

^{2}+ Bx + C = 0, where:
A = 1/a

^{2}– 1/b^{2}
B = 2c/b

^{2}
C = R

^{2}/b^{2}– c^{2}/b^{2}-1
for which the most
convenient solution is by means of the quadratic formula:

x = -B/2A +/- (D^0.5)/2A, where D is the discriminant D = B

^{2}– 4AC.
For there to be one
and only one x solution, the
discriminant must be zero, so that x =
-B/2A. If the discriminant is positive, there are two real solutions for x per the
formula. If the discriminant is
negative, there are no real-number
solutions at all.

Once we have values
for solution x, the corresponding y
coordinates can be determined from either the ellipse equation or the circle
equation (we are interested in the positive-y roots for the arrival
encounter; the negative-y roots
correspond to the departure point):

y = +/- [R

^{2}– (x – c)^{2}]^{0.5}from the circle equation
y = +/- [b

^{2}(1 - x^{2}/a^{2})]^{0.5}from the ellipse equation
Once the x-location
(along the semi-major axis) is known, we
can integrate numerically under the ellipse curve (relative to the semi-major
axis) from the solution x to the perihelion x value. The area of half the ellipse (to one side of
the semi-major axis) is 0.5*pi*a*b.
There is a triangle formed by the radius vector to encounter: its height is the y coordinate at
encounter. Its base is the focus length
c minus the x coordinate. Thus the
triangle area is 0.5*y*(c-x). To create
the area swept by the radius vector, the
area of this triangle subtracts from the integral area, as long as the encounter x is less than
c. The swept area factor SAF is the swept
area divided by the ellipse half area.
This area ratio applies to half the orbit period, for the one-way trip time.

I used a spreadsheet
to do this analysis, supplemented by
hand plots of the orbits to ensure the calculations were getting the right
answers. This process required multiple
iterations before I got it “right”.

**the basic transfer orbit-related data that are independent of the exact nature of Mars arrival. The basic orbital parameters a, b, c are given (blue-highlighted values expressed as Mkm (millions of km) have the most significant figures). Average, perihelion, and apohelion velocities (with respect to the sun) are given in km/s. The period and half-period values are shown, with the half-period shown in seconds, days, and months.***Figure 8 shows*
Also included in the figure are the basics of Earth
departure velocities and Mars arrival velocities. All the Earth departure velocities are
tangent to both the transfer orbit and Earth’s orbit, since the transfer perihelion is always at
Earth’s orbit. The scalar difference
between perihelion velocity and Earth’s average orbital velocity is the V

_{inf}value, typical of “far from Earth” velocity needs, and measured with respect to Earth. Adjusted for the effects of Earth’s gravity to a “near Earth” value, this produces the V_{bo}values with respect to Earth.
Mars arrival is a little more complicated, since the velocities must add vectorially for
all but the Hohmann min energy transfer case.
In the middle group in the figure,
the angle “a” is that between the velocity of the spacecraft “V” in its
transfer orbit at Mars encounter, and
the velocity vector of Mars that is tangent to its orbit. Only for the Hohmann case is this angle
zero.

The velocity “far from Mars” with respect to Mars is the
velocity vector V at angle a relative to Mars’s vector, minus Mars’s velocity vector. This is the Mars arrival V

_{inf}in the figure. I did not include all the spreadsheet details of computing that angle, but my hand plots showed that I was indeed computing the correct values. Adjusted for Mars’s gravitational attraction, this corresponds to a higher V_{int}“near Mars”.
The bottom group in the figure relates to the swept-area
estimated 1-way trip time. Again, I didn’t include all the details of the
numerical integration, but my area
estimates were indeed confirmed by the hand-plotted orbits. The blue highlighted data are the transfer
orbit parameters expressed as Mkm, for
the most significant figures.

The principal results are plotted versus transfer orbit
apohelion distance

**. These include the half-period of the orbit, the “near Earth” departure requirement V***in Figure 9*_{bo}with respect to Earth, the “near Mars” arrival speed V_{int}with respect to Mars, and the 1-way trip time.
Note that the 1-way trip time of 8.62 months and the
half-period of the transfer orbit (8.62 months) are identical for the Hohmann
transfer case, where the apohelion
distance is the average orbital distance of Mars. This requires 11.57 km/s achieved burnout
speed to depart, and arrives close to
Mars at some 5.69 km/s, more-or less
lined up tangent to Mars’s orbit.

The next faster case investigated has a half-period of 12.00
months, for a full round trip of exactly
2 years. If the Mars encounter were to
fail, the spacecraft would arrive back
at Earth’s orbit just as Earth got there.
This offers the possibility of a free-return abort, if 2 years in space is tolerable. Otherwise,
the 1-way trip to Mars is much faster at 4.26 months. It costs more: the Earth departure burnout requirement is
12.26 km/s, and the near-Mars encounter
velocity is some 7.40 km/s, skewed
off-tangent at about 34-35 degrees.

The third case used an even 400 million km apohelion
distance. Its total transfer orbit
period is 30.3 months, which is
nonresonant with Earth’s period about the sun.
There is

__no free-return abort__using this orbit! The 1-way trip time is really fast at 3.67 months, but this costs quite a bit. The required departure burnout speed is 12.77 km/s, and the near-Mars encounter velocity is some 7.36 km/s, skewed about 38 degrees off tangential. This apohelion is actually into the inner edge of the main asteroid belt.
The final case has a half-period of 18.0 months, or a full round-trip period of 3 years, resonant with Earth. This offers the possibility of a free-return
abort, if 3 years in space is
tolerable. The 1-way trip time is the
shortest of the cases investigated, at
3.40 months. The cost is high: 13.14 km/s at departure burnout, and a near-Mars encounter speed of 6.53 km/s
to deal with. That last is skewed about
33-34 degrees off tangential.

Most Mars mission designs will be leaving from orbit about
the Earth. The Spacex Starship is one of
those. Low circular Earth orbit (LEO) has
a speed about the Earth of about 7.9 km/s,
departure starts from there and must reach “V

_{bo}”. At Mars, the choices are (1) direct entry and retropropulsive touchdown, (2) entry into low Mars orbit (LMO) with possibly a separate vehicle to deorbit and enter for a retropropulsive landing, and (3) entry into Phobos’s orbit and touching down propulsively on Phobos. For that last, I simply applied a 1.5 factor for maneuver and hover to the escape speed from Phobos, to estimate a mass ratio-effective delta-vee requirement.
A few of my estimation items are simply assumptions. I assumed that any of the transits (both
outbound and inbound) have a course correction allowance of dV = 0.5 km/s. I also assumed that any vehicle returning to
LMO must rendezvous with another vehicle,
necessitating a rendezvous allowance.
I assumed that allowance to be dV = 0.1 km/s. Anything deorbiting from LMO to land upon
Mars requires an allowance for a deorbit burn.
For this I used dV = 0.05 km/s.
Finally, I assumed the terminal
speed after Mars entry to be about a Mach number, leading to an estimate of the speed to be
“killed” (and about 0.5 Mach on Earth).

A summary of the detailed transit and terminus estimates is

**. Note that the transfer ellipse influences departures and arrivals, but nothing else. Delta-vee information is highlighted blue, and entry speeds for aerobraking are highlighted green.***given in Figure 10***(see Update 11-24-19)**

This same information is rearranged into groups representing
each kind of overall mission, with the
data parametric upon the transfer trajectory used. The presumption is that the return to Earth
uses the same transfer ellipse trajectory as the outbound trip to Mars.

**the velocity requirements (blue) and entry interface speeds (green) for a mission that departs Earth orbit, makes a direct entry and landing upon Mars, then makes a direct escape from Mars, leading to a direct entry and landing at Earth. Touchdowns are assumed retropropulsive. There are many mission designs which might use this architecture. The most notable recent example is Spacex’s proposed “Starship”.**

*Figure 11 shows***the velocity requirements (blue) and entry speeds (green) for a typical orbit-to-orbit mission to Mars. This is broken down into a transit velocity requirement, and a separate landing velocity requirement, since it is likely the lander is a separate vehicle. Similarly, the takeoff is separate from the transit home. Earth return is to LEO, not a landing. That is presumed to be a different vehicle.**

*Figure 12 shows*
The Phobos (only) visit is

**. Because the landing allowance is so small, it is presumed the transit vehicle also makes the landing. Transit is from LEO to Phobos orbit, and from Phobos orbit to LEO for the return. There is no aerobraking in this scenario. Upon return to LEO, it is presumed that some other vehicle is used for the final return-to-Earth landing.***shown in Figure 13***(see Update 11-24-19 instead!)**

As mentioned above, I
utilized by-hand plots (pencil and paper) to verify correct calculation of many
items. I did not need such a plot for
the Hohmann min-energy transit, but I
did for the 3 faster trajectories. The
following two figures are photographs made of those working hand plots, two plots per page. Note that I mis-plotted one of them and had
to try again.

Figure 15 – Photo of Hand-Plotted Orbits Made for
Verification, part 2

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**Update 11-24-19: Revised Delta-Vee Estimates for Phobos Mission**

I have corrected an error in estimating Vinf for the Phobos misson. I used the surface escape speed 5 km/s for the conservation of mechanical energy estimate, when I should have used the escape velocity out at Phobos's orbital distance, some 3 km/s. This reduces the Vnear values substantially, thus reducing the estimates for delta-vee required. This changes Figure 13 entirely, and the Phobos Departure/Arrival details in Figure 10.

*The revised data are given in Figure 16 below.*Figure 16 -- Revised Estimates for the Phobos Mission

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Thanks for all the maths. It should come in handy!

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