Friday, March 5, 2021

Fundamentals of Elliptic Orbits

What I know about orbital mechanics is restricted to the fundamentals of elliptic orbits as simple two-body problems,  out of a topic that is far wider.  These can be computed from a few equations rather easily.  Solutions can be combined with an “energy trick” to estimate interplanetary delta-vee requirements for vehicle design,  but this is most definitely inadequate for navigation purposes.

The two-body problem involves a “primary” body whose mass far-and-away overwhelms the mass of the “secondary” body,  by multiple orders of magnitude.  An example would be satellite moving about the Earth,  or a moon moving about its planet.  Those are closed orbital paths.

The shape of the closed orbital path is the ellipse from high school geometry,  with the primary located at one focus,  and the other focus unoccupied.  The secondary body is located somewhere along that ellipse,  at a radius vector r from center-of-primary to center-of-secondary.  See Figure 1.

Figure 1 – Basic Elliptic Orbit Geometry

The longer semi-major axis length is “a”,  the shorter semi-minor axis length is “b”,  and the focal distance from ellipse center to focus is “c”,  just as in high school geometry.  The ratio of c to a is the eccentricity “e” of the ellipse,  which cannot quite reach 1 and still be an ellipse,  and which becomes zero as the path becomes a circle,  with both foci located exactly at the center. The figure is symmetrical in both directions.  There is a relationship among these variables that holds for this geometry: 

a2 = b2 + c2

At nonzero eccentricity e,  the ellipse is located off-center,  with the primary at one focus.  Thus there is a larger radius “rapo” to the “apoapsis” (farthest point) of the ellipse from the primary,  and a smaller radius “rper” to the “periapsis” (nearest point) of the ellipse from the primary.  These are related to known values of a and e as follows:

rapo = a(1 + e)

rper = a(1 – e)

If instead you know the max and min orbit radii,  you can find the a and e values as follows:

a = (rapo + rper)/2

e = (rapo – rper)/(rapo + rper)

Note also that altitude h of the secondary body above the primary body surface is the radius length to the secondary,  minus the radius of the primary itself.  For an orbit in (or very near) the equatorial plane of the primary,  you would use its equatorial radius.  For a strongly inclined orbit,  the average radius would be more representative.

h = r – Rprimary

The speed V of the secondary body (with respect to the primary) varies around this orbit,  with the radius distance,  in a well-defined way.  The orbit also has a definite and repeatable period T,  meaning the time required to go fully around the orbit once.  Knowing that G is the universal gravitation constant,  and M is the mass of the primary body,  those quantities (using SI metric units) are governed by:

V2 = GM (2/r – 1/a)

T = 2 pi (a3/GM)0.5  where pi is the familiar approximate 3.141592654 value

Position around the orbit as a function of time is not an easy thing to calculate.  The fundamental notion dates to Kepler’s time:  equal areas are swept out by the radius vector r in equal intervals of time.  In the old pencil-and-paper days,  we would draw the orbit to scale on paper,  and measure the desired swept-out area with a planimeter.  We would calculate the whole ellipse area (or measure it with the planimeter).  The ratio of swept-out area to ellipse area,  multiplied by the period,  is the time to the location of the radius vector,  usually measured from the periapsis or apoapsis point.

Today one would use the equation of the ellipse in appropriate coordinates,  and integrate the appropriate areas under the curve,  plus subtracting the appropriate triangle area,  to accomplish precisely the same thing as was done in the old days with a planimeter.  The equation of an ellipse,  with its major axis aligned with the coordinate system x axis,  is known from high school geometry to be:

[(x-h)/a]2 + [(y-k)/b]2 = 1  where (h,k) is the set of coordinates of the center of the ellipse

The other two rather-useful things also easily calculated are the circular orbit (about the primary) and escape velocity speeds,  evaluated at any desired radius r from center-of-primary.  The speeds differ by a factor of square root of two.  Those governing equations use SI metric units,  and are as follows:

Vcirc2 = GM/r

Vesc2 = 2 GM/r

For reference,  here is the SI-units value of G:  6.6732 x 10-11 N-m2/kg2


Noting that apoapsis is the farthest point from the primary (as denoted by the “apo”) and periapsis is the nearest point to the primary (as denoted by “peri”),  be aware that the nomenclature “apsis” is generic.  For orbits about the Earth,  we customarily use “gee” for “apsis”,  resulting in “apogee” for the farthest point and “perigee” for the nearest point.  For orbits about the sun we use “helion” for “apsis”,  resulting in “apohelion” and “perihelion”.  For orbits about the moon,  we use “lune” for “apsis”,  resulting in “apolune” and “perilune”.   I have seen “jove” used for “apsis” for orbits about Jupiter.  But the generic “apsis” cannot be misunderstood,  regardless of context.

Changing the Orbit You Are In

There are four things of general interest:  (1) changing the shape and size of the ellipse you are in,  (2) changing the plane of your orbit,  (3) deorbiting from an elliptical orbit,  and (4) deorbiting from an already-circular orbit.  The first 3 are covered by Figure 2,  and the last one by Figure 3. 

If you are in an elliptical orbit,  the best (most efficient) places to do a burn in order to change the shape or size of the orbit,  are the apoapsis or periapsis.  The radius to the point where you make the burn does not change,  the other one does,  as illustrated in the figure.  In general,  you know the changes in apoapsis or periapsis radii that you desire,  so you refigure the apoapsis and periapsis speeds for the new orbit from the known and new radius values,  as appropriate.  The change in speed at the location of your burn,  is the unfactored kinematic delta-vee (dV) for making that orbital change.

Figure 2 – Changing Your Elliptic Orbit

To emphasize:  if you want to raise or lower your apoapsis,  you make the orbit-changing burn at the periapsis of your current orbit.  If you want to raise or lower your periapsis,  you make the orbit-changing burn at the apoapsis of your current orbit. 

If you burn posigrade (in the direction of current motion),  you will raise the altitude of the other end of the orbit.  If you burn retrograde (against the direction of current motion),  you will lower the altitude of the other end of the orbit.

You have new rapo and rper values (one unchanged where you burn,  the other the desired new value).  You refigure “a” and “e” from these rapo and rper values using the equations already given.  Then you refigure the speeds at apoapsis and periapsis,  using the velocity equation already given.  Then you refigure period,  using the equation already given.  The difference in velocity values at your burn point is the delta-vee (dV) for the burn,  assuming it is “impulsive” (which for early 21st century electric propulsion,  it is not!!!).  If the burn is “impulsive”,  then it does not require factoring up for any gravity and drag losses.  That impulsive-burn presumption is the default for this article.

Orbital plane changes are a bit different.  This is merely a vector problem such that speed is unchanged,  but direction is changed.  The delta-vee (dV) calculated for a plane change angle of Δi depends directly upon the magnitude of the orbital velocity at the point where you make the burn.  Therefore,  this minimizes when your orbital velocity is minimum.  That happens at apoapsis,  so the best,  most efficient location to make a plane change in an elliptic orbit is an impulsive burn at apoapsis. 

The formula is given in the figure,  where the illustration shows initial and final velocities of equal magnitude,  and an angle bisector that is perpendicular to the dV vector.  Whether you measure angles in degrees or radians is irrelevant,  as long as you are consistent with how you figure the sine function of the indicated half-angle.  Be aware that plane changes through significant angles always cost large dV.

Deorbiting from an elliptical orbit may (or may not) match up very well with a desired location for landing.  What you are doing is lowering the periapsis altitude to one that more-or-less grazes the surface of the primary.  It does not have to be exactly a surface-grazing transfer orbit,  just close.  If the primary has an atmosphere,  your transfer periapsis needs to be at,  or under,  about half the accepted “interface altitude” for entry.  If the primary is airless,  it needs to be at,  or under,  the surface-grazing point. 

The calculation procedure is the same as the other in-plane orbital changes.  You are going to make your burn at apoapsis,  with the surface-graze point at the new periapsis.  You know the original apoapsis radius,  which will not change.  The new periapsis radius is the surface graze point,  or thereabouts,  as applicable.  You refigure the new “a” and “e” values,  then the new Vapo and Vper values,  using the equations already given.  The change in apoapsis velocities is the deorbit burn dV value,  which should need no factoring,  if an “impulsive” burn. 

The only way to make the desired location of the landing match up,  is to wait until the location of the apoapsis of you initial elliptic orbit matches up with the rotating geometry of the primary body beneath,  such that the desired landing location will rotate under your arrival trajectory (your periapsis),  just as you get there.  This is a window-timing problem that does not exist with circular orbits,  which do not have a definite apoapsis or periapsis.  From circular,  you just time the burn to hit the target.

Deorbiting from a circular orbit is far simpler.  Because the velocity is constant around the orbit,  you can easily time your deorbit burn to hit exactly what target you desire on the surface,  there is no need to wait for the apoapsis to be in the “right place”.  The transfer orbit is merely an approximately-surface-grazing ellipse from your circular orbit,  as shown in Figure 3.  

Figure 3 – Deorbiting From a Circular Orbit

If the primary has an atmosphere,  your transfer orbit periapsis needs to be at or under about half the accepted entry interface altitude.  If the primary is airless,  the transfer orbit periapsis needs to be at or at most slightly under the surface-grazing ellipse periapsis.  The transfer orbit apoapsis is the distance at which your circular orbit already exists.

Again,  with known apoapsis and periapsis radii for the transfer orbit,  you use the equations already given to figure “a” and “e”,  then Vapo and Vper,  and the period T.  The difference between your circular orbit speed and the transfer orbit Vapo is the unfactored dV for your deorbit burn.  If “impulsive”,  that dV needs no factoring. 

Half the period of the surface-grazing transfer orbit is a pretty good estimate of the time from deorbit burn to landing.  You simply time your deorbit burn to make the surface graze point your desired landing location.  If the body has an atmosphere,  your trajectory will decelerate-and-“droop” to landing a bit early.

Interplanetary Transfer Orbits

There are two basic classes of interplanetary transfer orbits:  min-energy Hohmann ellipses,  and faster transfer ellipses.  The delta-vee requirements are lower with min-energy Hohmann,  and higher with faster transfer ellipses.  Min-energy Hohmann transfer ellipses are tangent to both the departure planetary orbit,  and to the arrival planetary orbit.  That tangency assures that planetary motion and spacecraft motion are parallel,  such that arithmetic calculations suffice to estimate delta-vees. 

The faster trajectories to outer planets all have perihelion points located at the Earth’s orbit,  and so are tangent there,  for an arithmetic calculation of departure delta-vee.  The transfer aphelion points are beyond the orbit of the destination planet,  so that full vector subtraction is necessary to determine delta-vees at arrival.  This involves determination of both speed and direction at arrival. 

The point of a faster transfer trajectory is to shorten travel time.  Therefore it makes little sense to put the outer planet transfer trajectory’s perihelion inward of Earth,  as that puts some of the fastest travel speeds out of consideration.

The faster trajectories to inner planets all have perihelion points located at the orbit of the destination planet.  The transfer apohelion is located outward of Earth’s orbit.  That puts all the higher available speeds on the transfer orbit,  for minimum travel time.  Arrival delta-vees become arithmetic calculations due to the tangency,  while departure delta-vees become full vector calculations because of the lack of tangency. 

               Min-energy Hohmann Transfer

The fundamental idea for planetary destinations outward of Earth is to put the transfer orbit perihelion at the distance of the Earth from the sun,  and to put the transfer orbit apohelion at the destination planet’s distance from the sun.  For destinations inward of Earth (those being Venus and Mercury),  one puts the transfer apohelion at Earth’s distance,  and its perihelion distance at the destination planet’s distance.  The orbit always has Earth-at-departure on one side of the transfer orbit,  and arrival-at-destination on the other side of the transfer orbit.  That is inherent to Hohmann transfer.  Thus the one-way trip time is always half the transfer orbit period.  See Figure 4.  

Figure 4 – Min Energy Hohmann Transfer Orbits

While simple enough in concept,  there is a complication:  planetary orbits about the sun are elliptical,  not circular.  Some are more eccentric ellipses than others.  Accordingly,  there are minimum and maximum distances from the sun,  for Earth,  and for any destination planet.  The higher delta-vee requirements are generally associated with the larger transfer orbits,  in turn associated with max distances of Earth and destination planet from the sun. 

Minimum delta-vee is associated with min planetary distances from the sun.  “Average” conditions prevail when both planets are near their average distances from the sun,  that being “a” for their orbits.   There is no guarantee that Earth will be at its min distance,  when the destination planet is also at its min or max distance.  All you can do is bound the delta-vee problem between the max values at max planetary distances from the sun,  and min values at min planetary distances from the sun.  Orbital calculations made at average planetary distances are merely “typical values” between these bounds. 

For any given distances of Earth and destination planet from the sun,  you figure “a” from the sum of the two planetary distances,  and “e” from the difference divided by the sum,  as per the equations already given.  From these,  you can figure everything else,  including transfer orbit perihelion and apohelion speeds,  and the period,  per the equations already given. 

Bear in mind that these orbital speeds for spacecraft and planets are measured with respect to the sun.  For the departure,  the difference between the spacecraft speed and Earth’s speed is effectively the spacecraft speed with respect to the Earth,  something required of the spacecraft when “far” from Earth,  meaning outside the Earth’s gravitational influence.  That speed is tangent to both Earth’s orbit and the transfer orbit, which is why an arithmetic difference is adequate. 

For the arrival,  the difference between spacecraft speed and the destination planet’s speed is the spacecraft speed with respect to the destination planet,  something required of the spacecraft when “far” from the destination planet,  meaning outside its gravitational influence.  That speed is tangent to the destination planet’s orbit,  and the transfer orbit,  which is why an arithmetic difference is adequate. 

For destinations outward of Earth,  the departure speed is larger than Earth’s orbital speed,  requiring a posigrade-direction burn.  For destinations inward of Earth,  the departure speed is smaller than Earth’s orbital speed,  requiring a retrograde-direction burn. 

For destinations outward of Earth,  the arrival speed is less than the destination’s orbital speed,  so that the destination planet wants to “run over” the spacecraft “from behind”.  For destinations inward of Earth,  the arrival speed is greater than the destination’s orbital speed,  so that the spacecraft overtakes the destination,  and wants to “run into it from behind”. 

               Faster Ellipse Trajectories

The situation for faster ellipse trajectories is similar to Hohmann,  except that the transfer ellipse is only tangent to the inward planet.  The case of Earth to Mars is illustrated in Figure 5,  including the area-under-the-curve concept for calculating 1-way transfer time (1-way trip time).    For the example shown,  the difference between transfer perihelion speed and Earth orbital speed can be computed arithmetically,  precisely because of the parallel vectors,  which is due to the tangency.  That value is the velocity “far” from Earth (after escape),  which is required in order to be on the transfer trajectory.

Again for the example shown,  the apohelion of the transfer orbit lies far beyond the orbit of Mars.  Arrival is at the point where the two orbits cross.  The spacecraft velocity vector with respect to the sun is not parallel to the velocity vector of Mars with respect to the sun.  The velocity of the spacecraft with respect to Mars is its velocity with respect to the sun,  minus the velocity of Mars with respect to the sun,  done as a vector calculation. 

Because of the nonparallel geometry,  this subtraction is inherently a vector subtraction,  as indicated in the figure.   The angle between the two vectors is the difference of the angles of the tangent lines of the two orbits.  Because it is a vector subtraction,  the negative of the Mars velocity is added graphically to the spacecraft velocity.  This can be done by the method of resolved components,  in any convenient coordinates,  such as those which are radial and tangent to Mars’s orbit.

That vector velocity with respect to Mars is the speed and direction of the spacecraft “far” from Mars,  before Mars’s gravity can act to pull the spacecraft both closer and faster.

Figure 5 – Faster Ellipse Trajectories (Mars as Example)

Besides the faster trip time,  there is a second possibly-beneficial characteristic of the faster orbit.  If its semi-major axis length is correctly chosen (solve the period equation for “a”,  given a desired period value),  the period of the transfer orbit becomes an integer multiple of Earth’s 1-year orbital period.  What that means is the transfer orbit can serve as an abort orbit,  taking the spacecraft back to Earth,  if the Mars arrival is not made.  The period must be an integer multiple of 1 year,  so that Earth will actually be there,  as the spacecraft arrives at perihelion.

There is another difference between this abort-orbit situation and a simple Hohmann transfer:  it is possible to support a shorter stay upon (or near) Mars.  The time available in the example is the transfer period minus two of the 1-way trip times.  With Hohmann at average conditions,  there is a little over a year’s stay on Mars required,  before the orbits are “right” for the Hohmann trip home to Earth.

Getting to “Near” Velocities from “Far” Velocities,  and Back

The following “trick” is a way to correct spacecraft velocities with respect to a planet which were computed “far” from a planet (per above),  to the velocities “near” the planet,  after its gravity has pulled the spacecraft to faster speeds.  This is based on conservation of mechanical energy,  and works well for figuring delta-vee values for spacecraft design.  It is totally inadequate for navigation!  For that,  you must run a real 3-body problem in a computer program.  Here is the estimated correction:

Vnear2 = Vfar2 + Vesc2  where Vesc is figured at the appropriate distance r from center-of-planet

For departure,  Vnear is the vehicle burnout speed (with respect to the planet) that must be reached,  which is larger than Vesc,  such that “far” from the planet,  it will still traveling at the lower value Vfar with respect to the planet,  which when corrected to be with respect to the sun,  is the transfer trajectory velocity with respect to the sun that is required at departure.

For arrival,  spacecraft velocity with respect to the sun gets refigured by vector subtraction to velocity with respect to the planet,  but before the planet’s gravity has effect,  here denoted as Vfar.  This gets corrected to an appropriate Vnear,  which could then be the direct atmospheric entry speed,  the speed which must be “killed” to land directly on an airless planet,  or the speed which must be reduced by a suitable delta-vee,  in order to enter orbit about the planet.

What Delta-Vee Values Will You Need?

There are at least three practical things to consider,  sometimes four:  (1) departure delta-vee,  (2) correcting the inclination of the transfer orbit plane,  (3) other course corrections (some are always required for a precision approach),  and (4) the delta-vee or delta-vees required to enter orbit or to land directly.  Be aware that the same basic considerations apply for return trips,  although the detailed numerical circumstances can be quite different.

               Departure Delta-Vee

Departure delta-vee has already been addressed in the sense that there is a “near” velocity greater-than-escape (with respect to the planet) that the spacecraft must attain.  This could be from orbit,  or from the surface.  The details of calculating delta-vee from that information vary with exactly how you ascend and depart. 

Regardless,  there is a “far” velocity larger than escape that the vehicle must achieve,  oriented parallel to the Earth orbit velocity if Hohmann,  and also parallel to the Earth orbit velocity if a faster ellipse headed outward of Earth.   If a faster trajectory headed inward of Earth,  these velocities are not parallel,  requiring vector addition processes.

Figure 6 shows what must happen,  for departures from low Earth orbit onto Hohmann trajectories to other planets.  The transfers outward of Earth require Vfar values exceeding Earth escape,  directed in the same direction as Earth’s orbital velocity about the sun,  and oriented parallel to the Earth orbital velocity.  This parallelism is true whether Hohmann or faster.  Arithmetic calculations work. 

Transfers inward of Earth require Vfar values exceeding Earth escape,  directed opposite to the Earth’s orbital velocity about the sun.  The direction is parallel to Earth’s orbital velocity if Hohmann transfer,  and not parallel if a faster trajectory.  Not parallel requires vector addition processes.

Estimates of this type are adequate for the delta-vee sizing of vehicles,  but they are totally inadequate for navigation.  That requires a 3-body trajectory program.  The departure delta-vee is the difference between Vnear and the circular orbit velocity.  If the burn is impulsive,  no factoring-up is needed:

dVfrom orbit = Vnear – Vcirc   for factor = 1.00 if impulsive

Figure 6 – Departure From Low Earth Orbit

Departures from the Earth’s surface can be either to Earth orbit,  or direct to escape onto the transfer trajectory.  Either way,  the initial sharply-rising portion of the trajectory is subject to both gravity and drag losses.  The nearly-horizontal but exoatmospheric portion is subject to a minimal gravity loss,  but no drag loss.  That gets you to low circular orbit. 

You can try to model the higher losses on the early portion of the ascent if you want to,  but just using 5% each for gravity and drag losses against low orbit velocity works just as well as any other estimate.  Which is why I generally just do that simpler easier estimate.  See Figure 7,  which shows this.

Departure can then be direct,  or from circular orbit,  but either way,  that portion of the delta-vee needs no factoring,  because from orbit,  there are no drag and gravity losses for impulsive burns. 

These numbers and recommendations are for eastward launches from Earth.  The desired orbital velocity has both kinetic and potential energy values associated,  so that the “more representative” orbit velocity in this book-keeping would actually be surface circular.  

For polar launches,  you would need to add to that ideal velocity requirement the tangential velocity of your launch point as a function of its latitude.  The eastward velocity of a point on the Earth’s surface at its equator is 1670 km/hr,  or 0.463 km/s.  Multiply that by the cosine of latitude for sites off the equator.  If launching retrograde,  add two of those tangential velocities to the ideal surface circular orbit velocity. 

For launches from other worlds of different atmospheres and surface gravities using the simpler factored orbital velocity method,  reduce the 0.05 values by the ratio of surface density and gravity to Earth standard values,  as indicated in the figure.  Such is adequate for vehicle design delta-vee values,  but totally inadequate for navigation estimates.  

Figure 7 – Factored Delta-Vee Recommendations for Earth Ascent and Elsewhere

Navigation-grade estimates require a full trajectory program incorporating appropriate models for gravity and atmosphere,  and for all the vehicle characteristics. 3 degrees of freedom is adequate,  but 6 degrees of freedom is better,  although it costs a lot more effort to set a 6-dof model up.

               Correcting to the Inclination of the Transfer Orbit at Departure

Up to this point,  the entire solar system has been treated as if co-planar.  It is not.  Each planet has an orbit slightly differently-inclined (and oriented) than the Earth’s orbital plane,  which is customarily used as the reference plane for describing Solar System orbit inclinations.  The difference in inclination between the destination planet’s orbit and the Earth’s orbit is the maximum plane change angle magnitude that you need to “cover” in your design.  It can even be zero. 

This needs to be done as you depart,  so that the error does not build-up into something out-of-reach later.  It can be book-kept as part of the general course correction budget,  but it really needs to be part of the departure burn process.

If done after escape,  this gets figured with the plane change equation,  at the spacecraft transfer orbit departure velocity with respect to the sun:  i.e.,  V = Vper for the transfer orbit to Mars.  That is because you are adjusting the plane of an orbit about the sun.  That equation is:

dV = 2 V sin(Δi/2) ~ V Δi for small-angle Δi measured in radians (“SRT rule”)

One should understand that this equation applies to a 2-body problem,  being originally developed for satellite orbits about the Earth.  For Δi < 10 degrees,  the “SRT rule” applies,  such that the sine and tangent values for Δi are numerically the same as the value of Δi itself measured in radians.  The speed V is the current speed in the two-body orbit at the point you make the burn to change the inclination.  Whatever it is you are orbitingis what you measure V “with respect to”

The change in angle Δi for our transfer orbit relative to the plane of Earth’s orbit about the sun,  is the difference in planetary inclinations,  as a maximum value,  usually in the neighborhood of a few degrees (the inclination of Mars’s orbit is 1.85 degrees,  Earth as a reference is zero by definition).  

For direct surface launch onto the interplanetary transfer trajectory,  you simply aim your ascent trajectory such that its final post-escape residual “Vfar” is aligned for the transfer trajectory.  You make this aiming well before you escapewhile the 2-body problem with Earth as primary still applies!  That makes V in the inclination equation the relatively small number of your flight speed relative to the Earth.  The customary factoring-up for gravity and drag losses during the ascent covers all the mostly-eastward launches well enough,  including this case.    When estimated this way,  no additional departure delta-vee for inclination correction is needed. 

If you launch into low Earth orbit,  and then depart onto the transfer trajectory from there,  you want the inclination of that orbit about the Earth to match the inclination of your transfer trajectory.  If your Earth orbit is elliptical,  you also want its orientation to match your transfer trajectory.  If circular,  that second requirement is deleted.  No additional departure delta-vee for inclination correction is needed,  since you already corrected it early in your ascent. 

Many mission concepts have called for a space station in orbit about the Earth,  at which to assemble vehicles,  and from which to depart to other planets.  Such a station should be in a circular orbit whose inclination matches that of Earth’s orbit about the sun,  not equatorial about the Earth. 

From there,  you correct to the transfer orbit inclination with a plane change burn,  while still in orbit about the Earth!  Then you burn for departure,  with no further course correction required until later.  For this case,  the worst-case angle change for Mars is 1.85 degrees,  and your velocity in orbit about the Earth is about 7.9 km/s.  The resulting inclination-correcting dV is about 0.26 km/s,  worst case. 

If instead you wait until after escape,  when you are already on the interplanetary trajectory,  you are now orbiting the sun,  at around 30+ km/s.  The max angle change is still 1.85 degrees,  and so the dV for the correction burn is now 0.97 km/s.  That’s a big difference,  and very unfavorable!  It is far better to correct the inclination while still in orbit about the Earth,  or even earlier,  during direct ascents.  This is because the values of velocities relative to the planet are far lower than those relative to the sun. 

               Midcourse and Terminal Course Corrections

These always take place while on the interplanetary trajectory,  but involve very tiny angle changes.  These are easier estimated as cross range distance divided by range-to-target,  which is essentially the angle change in radians.  As a credible guess,  the midcourse correction for Mars might be about 10 planet diameters (at roughly 4200 miles diameter),  when the range is close to the average distance of Mars from the sun (close to 150 million miles).  The resulting angle change is near 42,000 miles/150 million miles = 0.00028 radians.  Speeds will still be in the 25-30 km/s class,  for a dV in the 0.008 km/s class.  This rises rapidly if you wait to later in the trajectory.  But you do need enough travel to successfully determine where you are versus where you really need to be.  That’s why it is called a “midcourse” correction!

The terminal correction can be estimated similarly.  For a Mars mission,  you may be wanting to “hit” a narrow “window” for direct entry or orbital entry.  The cross-range distance correction might be on the order of 100 miles,  when you are two or three planet diameters (4200 miles diameter) away.  Velocity will be near 25 km/s with respect to the sun,  so the dV will be a bit larger,  at something like 0.24 km/s. 

Add these up and then double it for a safety factor.  For the Mars mission example,  that would be 2 x 0.248 km/s ~ 0.5 km/s budget for midcourse and terminal course corrections,  excluding any initial inclination adjustment (which should be taken care of before escape onto the interplanetary trajectory).  

               Delta-Vees for Landings or for Entering Orbit

Upon arriving at Mars or any other destination,  there are only two things you can do.  You can either enter orbit about the destination,  or you can try to land directly upon it.  The calculated arrival “Vnear” is your relative velocity to the destination,  at very close distances,  so time is very short to impact!  Your location and orientation relative to the planet is critical,  so the navigation problem is critical to success.  That is not the topic here,  but delta-vee for sizing vehicles is. 

To enter orbit,  there is an orbit speed,  usually presumed to be low circular,  with respect to the planet.  Your “Vnear” value is also with respect to the planet.  For destinations outward of Earth that rotate in more-or-less the same direction as they revolve about the sun,  you would like to orbit in that same direction,  if you intend to land later.  That is to take advantage of the rotation to reduce any delta-vee taking off.  See Figure 8.

On the sunward side of the planet,  the orbital velocity reduces the burn requirement,  while increasing it on the anti-sunward side,  as indicated in the figure.  The ideal delta-vee to enter such a sunward-side orbit is Vnear – Vorbit,  and circumstances may require vector addition instead of arithmetic computation.    If impulsive,  this delta-vee needs no factoring.  (On the anti-sunward side,  the delta-vee is far higher at Vnear + Vorbit.)

How you might land directly upon land upon the destination depends upon whether or not it has an atmosphere substantial enough to use for aerobraking.  See Figure 9.

If airless,  you must “kill” all of Vnear (as figured at the surface with the surface Vesc) with retropropulsion,  and that must be factored for gravity losses,  and a budget added for any terminal zone hover or divert effects avoiding obstructions.  Just as a rule-of-thumb,  that terminal budget should be something like the last 0.5 km/s of Vnear,  factored up by at least 1.5.  Thus your airless delta-vee estimate to land might be:  dV = Vnear * (1 + grav loss) + 0.5 km/s *(1 + 50% or more).

If there is an atmosphere substantial enough for aerobraking,  you need to do an entry estimate to find out what your altitudeangle-below-horizontal,  and speed might be at the end of braking hypersonics.  Such entry analyses are not covered here,  see Refs. 1 and 2 for that.  That end-of-hypersonics speed would be the velocity corresponding to local Mach 3 at the destination.  On Mars,  because the atmosphere is so thin,  the corresponding altitudes are invariably quite low,  compared to those experienced here in the thicker atmosphere at Earth. 

Your choices from that point are twofold:  (1) continued aero-deceleration followed by a final retropropulsive touchdown,  or retropropulsion-to-touchdown without any further aero-deceleration.  Your ballistic coefficient and resulting end-of-hypersonics altitude will decide between those choices at any given destination,  including the Earth.  That topic is not covered here,  see Refs. 2 and 3 for that. 

Either way,  for delta-vee estimating purposes,  there is a speed-at-some-altitude which must be “killed” propulsively.  Calculate the kinetic and potential energies at that point,  add them,  and convert it to a surface speed “Vsurf”,  using Vsurf2/2 = V2/2 + g*h,  where g is the local acceleration of gravity.  That is what you have to “kill” with your propulsion.  It requires factoring for hover/divert considerations:  dV = Vsurf *(1 + at least 50%)

Figure 8 – How to Enter Orbit

Figure 9 – Paths to Landing Directly

What You Do With All These Delta-Vees

Any given stage or vehicle must deliver its share of the total delta-vee to perform the mission.  Since these are appropriately-factored delta-vees,  they are “mass ratio-effective”.  Thus you can use the rocket equation in reverse to determine required stage or vehicle mass ratios from the factored delta-vees,  given appropriate exhaust velocities for the engines and propellants you are presuming.  That topic is also out of scope here.  See Ref. 4 for the “how-to” of estimating those things. 

References (all located at

#1. 30 June 2012              Atmosphere Models for Earth,  Mars,  and Titan

#2. 14 July 2012               “Back of the Envelope” Entry Model

#3. 5 August 2012           Ballistic Entry From Low Mars Orbit

#4. 23 August 2018         Back-of-the-Envelope Rocket Propulsion Analysis

An Example One-Way Cargo Mission to Mars

As indicated earlier,  this kind of orbital mechanics analysis is suitable for sizing vehicles,  not for navigation.  The entire analysis is oriented toward determining realistic delta-vees for all the necessary things that must be done.  The concept for a one-way cargo delivery to Mars is shown in Figure 10.  Upper and lower bounds and an average for the transfer orbit are given in Figure 11.  Departure delta-vees and arrival entry interface speeds are given in Figure 12.  Ascent and landing estimates are given in Figure 13.  Figure 14 indicates the appropriate uses for this kind of data.  

Figure 10 – Mission Concept and Assumptions

Figure 11 – Average and Bounds on Transfer Trajectory

Figure 12 – Departure Delta-Vees and Arrival Entry Interface Speeds

Figure 13 – Ascent and Landing Delta-Vees

Figure 14 – Appropriate End Uses for These Data


  1. Dear Mr. Johnson,

    “The faster trajectories to outer planets all have perihelion points located at the Earth’s orbit, and so are tangent there, for an arithmetic calculation of departure delta-vee.”

    I saw a similar statement in Bate (Fundamentals of Astrodynamics, page 362). Checking empirically, using

    rTerra =149600000000 m, rMars = 227900000000 m, and Msol*G = 1.3274E20, I found

    Perihelion 149600000000 m 1480000000 m 149600000000 m 148800000000 m

    Aphelion 247000000000 m 2377250000 m 237122000000 m 232992000000 m

    Transit time
    rT to rM 180 days 180 days 200 days 200 days

    total delta V 8419 m/s 7486 m/s 7161 m/s 6836 m/s

    showing that for a required transit time, there is a savings in delta vee if perihelion is smaller than rTerra. Have I misunderstood the situation?

    I tried to check my calculations against your Figure 10. Assuming a circular orbit and deriving

    Msol*G = VE2 * rE = (29771 m/s)^2 *149570000000 m = 1.3257E20 m^3/s^2 ,

    then perihelion speed for the Holmann orbit would be 32716 m/s, not 33378 m/s. Again, have I misunderstood?

    Thank you, MBMelcon

  2. I'm not sure how to respond to your comment or question. My reasoning was very simple: orbital velocities are higher nearer periapsis than near apoapsis. Higher velocities over the same distance means shorter transit times. I was looking at shortest transit time, not min delta-vee for a fixed transit time (I have no way to do that). As for the difference in numbers, your input numbers are not identical to mine. So you should not be surprised when you answer differs from mine. -- GW