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#21c     Flight to Mars: Calculations


        In the preceding section the route to Mars was identified, along a Hohmann Transfer Ellipse. The time required was derived, about 8.5 months, as well as the position of Mars at the time of launch, about 45° past closest approach.

        This section calculates two essential details: the velocity boost needed to inject the Mars spaceship into the transfer orbit, and the arrival velocity at the orbit of Mars. Will the spaceship be overtaking the planet or be overtaken by it, and what will be the velocity mismatch between ship and planet--a mismatch which will probably require an additional rocket firing? Keep reading, and if you can handle elementary algebra, you may find out.



19.Motion in a Circle

20. Newton's Gravity

21. Kepler's 3rd Law

  21a.Applying 3rd Law

21b. Fly to Mars! (1)

21c. Fly to Mars! (2)

21d. Fly to Mars! (3)

22.Reference Frames

22a.Starlight Aberration

22b. Relativity

22c. Flight (1)

22d. Flight (2)

23. Inertial Forces

23a. The Centrifugal Force

  23b. Loop-the-Loop

  24a.The Rotating Earth

24b. Rotating Frames

Notation and Escape Velocities

    Before starting out, it helps to establish a notation for the quantities involved here. Even though some of them are vectors, only their magnitudes are handled. Bold face is only used to accentuate, never to indicate vector character.

    As before, r1 = 1 AU is the distance of Earth from the Sun, r2 = 1.523691 AU that of Mars, and (as an approximation) both planets are assumed to move in circles.

    Velocity V will be measured in kilometers per second (km/s), and different velocities are identified by subscripts. Lower-case v identifies velocities associated with orbits around Earth rather than around the Sun.

    The preceding section already introduced the orbital velocity V0 of Earth around the Sun, amounting to about 30 km/s, much larger than v0 ~ 8 km/s (approx.) required by a satellite circling the Earth just above the surface(ignoring the atmosphere!). In section #21 it was noted that the escape velocity ve from such a low orbit is obtained by multiplying v0 by the square root of 2, equal to 1.41421356.... approximated here by 1.414. This gives

ve = 1.414 v0 = (1.414)(8) = 11.312 km/sec

    Such a spacecraft, however, is still not free to move to any point in space. The velocity ve has purchased its freedom from the Earth's gravity, but not freedom from the pull of the Sun, around which it continues to move in an orbit similar to Earth's, at approximately V0 =30 km/s.

    The situation is now completely analogous to escape from a low altitude Earth orbit (only the cost is higher!). To break free from a circular orbit around the Sun and leave the Solar system, the spacecraft needs to boost its velocity to a "second escape velocity"

Ve = 1.414 V0 = (1.414)(30) = 42.42 km/s   (approx.)

    To reach Ve it must somehow increase its velocity by an additional 12.42 km/s--more than is needed to escape the Earth's gravity, starting from rest on the surface! Luckily, ways exist (discussed in Section #35) of making the motion of planets (or of the Moon) provide part of this boost.

    Other velocities entering the calculation are the velocity V1 with which the spaceship starts from near Earth and enters the Hohmann ellipse (distance r1 from the Sun), and the velocity V2 with which it reaches the orbit of Mars (distance r2). Also, V3 will be the velocity of Mars in its orbit, assuming it has a constant magnitude (i.e assuming the orbit of Mars is circular).

   If V2 > V3, the spaceship overtakes Mars, while with V2 < V3 it is being overtaken.

Required Equations

    (1) Kepler's Laws

  •     The first law is "Planets move in ellipses, with the Sun at one focus. " That is already being used, e.g. the transfer ellipse is one such orbit.

  •     The second law is "The line connecting a planet to the Sun sweeps equal areas in equal times. " Let us try to extract from this a useful equation.
Application of Kepler's 2nd law
Application of Kepler's 2nd law

    The drawing here shows the orbit of Mars (solid) and the transfer ellipse (broken line), with radiuses (r1, r2) to the (perigee, apogee) points, at which the spacecraft velocity is (V1, V2). The short segments drawn at these locations represent the distance covered by the spacecraft in the next second after passing perigee or apogee, and by the definition of velocity ("distance per second") they also equal V1 and V2. Actually, those segments should be curved like the orbit, but being so short they differ negligibly from straight lines We then complete the long thin triangles, which have these lines as bases.

    Note that each of these triangles has a right angle at its bottom, because at apogee and at perigee (and nowhere else), the line to the Sun is perpendicular to the orbit.

    At perigee, the height of the triangle is r1, the length of its base is V1, so by the equation for the area A1 of a triangle

A = (1/2) (height) (base)

we get
A1 = (1/2) r1 V1

    At apogee, the height is r2, the base V2, and the area is
A2 = (1/2) r2 V2

    Each of these triangles is swept in one second, so by Kepler's 2nd law their areas can be set as equal. Multiplying both sides of that equality by 2 yields

                        r1 V1 = r2 V2                     (1)

    The equation is numbered to help refer to it later. Please note this relation only holds between apogee and perigee. At other points of the orbit, the angle between the radius and the orbit is not 900, and the area also depends on its exact value.

    Kepler's third law was already used in determining the orbital period. It will be needed again at the end.

    (2) The Energy Equation

    In section #12 it was stated that the energy E of a satellite of mass m orbiting Earth, at any point in its orbit, is

E = (1/2) mV2   –   km / r                     (2)

where r is the point's distance from the center of Earth, V is the satellite's velocity at that point, and k is some constant, related to the gravitational acceleration g. Because the energy E is conserved, the right-hand expression has the same value anywhere on the orbit. A similar relation holds for orbits around the Sun, although the value of k is different. We can express k in that case by using a simple trick, based on the escape velocity.

    As noted earlier, for an object in the Earth's orbit to completely escape the Sun (but just barely!), it needs a velocity Ve = 1.414..V0 = 42.42 km/s. Let E0 be the energy of such an object. Then since

Ve2 = 2 V02
we get (at the Earth's orbit)
E0 = m V02   –   km / r1

    Because it has escape velocity, if we wait a long, long time, this object will be extremely far from Earth, and, having exhausted practically all of its kinetic energy, its velocity will be very close to zero. Then both terms on the right side of equation (2) tend to zero, suggesting

E0 = 0

            The above fits the meaning of the sign of E:
    Negative E always identifies elliptic orbits bound to the Sun, like those of planets. Positive E characterizes unbound "hyperbolic" trajectories, like that of the space probe Voyager 2, on its way out of the solar system and into the wide black yonder, always maintaining an appreciable velocity. And motions with E=0 are "parabolic" and lie on the boundary between the two groups, not limited in distance but slowing down to a crawl as that distance grows. Orbits of non-periodic comets come close to this behavior.

So we have
m V02   –   km / r1 = 0

    Dividing by m and shifting the negative Ve term to the right

V02 = k / r1

    from which the value of k may be written

                                  k = V02 r1                     (3)        


In earlier sections we approximated

r1 = 150,000,000 km                 V0 = 30 km/sec

In what follow more precision will be used, taking the astronomical unit as

r1 = 149,598,000 km

A circular orbit around the sun with this radius has length

2 π r1 = 938,952,000 km

and assuming a Julian year of 365.25 days of 86400 seconds each gives

V0 = 939,952,000 / [(365.25)(86400)] = 29.785 km/s
(Rather than 30 km/s)
      V02 = 887.163 (km/s)2                 k = 1.32818 1011

    Returning now to the spaceship in a Mars transfer orbit, its energy should be the same at the perigee P and the apogee A, so by equation (2)

1/2 m V12   –   km / r1   =   1/2 m V22  –   km / r2

    Divide both sides by m ("cancel m") and multiply both sides by 2

V12   –   2 k / r1   =   V22   –   2 k / r2

Transferring terms (by adding suitable quantities to both sides) and substituting (3) gives

V12   –   V22   =   2 V02 r1 (1/r1   –   1/r2)  

                                        =   2 V02 (1   –   (r1/r2))                     (4)

In astronomical units
r1 = 1                               r1 = 1.523691
r1 / r2 = 0.656301                     1 – (r1 / r2) = 0.343699

We therefore have all we need to calculate the right side of equation (4).

          2 V02 (1  –  (r1/r2))   =   2 (887.163) (0.343699)   =   609.834 (km/s)2

On the left side we have two unknown quantities V1 and V2, but we can use (1)
r1 V1 = r2 V2
to express one in term of the other.
                    V2 = V1 (r1/r2)                     (5)

V22 = V12(r12/r22)

    Substituting this on the left side of (4)

        V12 – V22   =   V12 (1   –   (r12/r22)) =                            

                  =   V12 (1 – (0.656301)2   =   0.569269 V12   =   609.834 (km/s)2

Dividing both side by 0.569269

V12 = 1071.26 (km/s)2

    Extracting square root
V1 = 32.730 km/s
Applying now equation (1)

V2   =   V1 (r1 / r2) = (32.730)(0.656301)   =   21.481 km/s

showing we need add just 2.945 km/s, a shade short of 3 km/s or 10% of the orbital velocity.

Arrival at Mars

The velocity V2 at which the spaceship arrives at Mars is found from (5)

V2 = V1 (r1/r2) = (1 / 1.523691)( 32.730) km/s

= 21.4807 km/s

    It has given up some of its kinetic energy to overcome the pull of the Sun and moVe further outward from the Sun. The big question now is--how does this compare with the velocity V3 of Mars in its orbit?

    To derive a velocity in km/s, distances must be ultimately rendered in kilometers and times in seconds, but "for the benefit of pedestrians," we break up the calculation, avoiding big numbers and scientific notation. We start with Kepler's 3rd law for circular orbits, with distance r in AU and orbital period T in years. As derived in the preceding section (and also in section #10), in these units

T2 = r3

    For Mars, r = 1.523691, T2 = (1.523691)3 = 3.53745

T = 1.8808 years

Assuming 365.25 days per (Julian) year:

T = 1.8808 years = 686.96 days

    During that time the spaceship covers

2 π r = (6.2832) (1.523691) (149,598,000) km = (1432.2) (1,000,000) km

    Dividing by T, this comes to

(1432.2 / 686.96) (1,000,000) = (2.08484) (1,000,000) km/day

                       = 2,084 840 km/day

Each day has (24)(3600) = 86400 seconds, so the orbital distance covered by Mars each second is

2,084,840 / 86400 = 24.130 km

    Distance-per-second is of course the definition of velocity. Therefore

V3 = 24.130 km/s

Comparing to
V2 = 21.481 km/s

we see that Mars is the one moving faster, and will be overtaking the spaceship. To match velocities with Mars, the spaceship must generate an extra boost of 2.649 km/s .

Next Stop: #21d. Flight to Mars: the Return Trip

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Author and Curator:   Dr. David P. Stern
     Mail to Dr.Stern:   stargaze("at" symbol) .

Last updated: 9-22-2004
Reformatted 24 March 2006