
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 450 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 planeta mismatch which will probably require an additional rocket firing? Keep reading, and if you can handle elementary algebra, you may find out. 
Index
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. LooptheLoop 24a.The Rotating Earth 24b. Rotating Frames 
Notation and Escape VelocitiesBefore 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, r_{1} = 1 AU is the distance of Earth from the Sun, r_{2} = 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. Lowercase v identifies velocities associated with orbits around Earth rather than around the Sun. The preceding section already introduced the orbital velocity V_{0} of Earth around the Sun, amounting to about 30 km/s (more precisely, 29.77 km/s), much larger than v_{0} ~ 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 v_{e} from such a low orbit is obtained by multiplying v_{0} by the square root of 2, equal to 1.41421356.... approximated here by 1.414. This gives
Such a spacecraft, however, is still not free to move to any point in space. The velocity v_{e} 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 V_{0} =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"
To reach V_{e} it must somehow increase its velocity by an additional 12.42 km/smore 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 V_{1} with which the spaceship starts from near Earth and enters the Hohmann ellipse (distance r_{1} from the Sun), and the velocity V_{2} with which it reaches the orbit of Mars (distance r_{2}). Also, V_{3} 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 V_{2} > V_{3}, the spaceship overtakes Mars, while with V_{2} < V_{3} it is being overtaken. 
Required Equations(1) Kepler's Laws

The drawing here shows the orbit of Mars (solid) and the transfer ellipse (broken line), with radiuses (r_{1}, r_{2}) to the (perigee, apogee) points, at which the spacecraft velocity is (V_{1}, V_{2}). 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 V_{1} and V_{2}. 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 r_{1}, the length of its base is V_{1}, so by the equation for the area A_{1} of a triangle
we get At apogee, the height is r_{2}, the base V_{2}, and the area is
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
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 90^{0}, 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
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 righthand 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 V_{e} = 1.414..V_{0} = 42.42 km/s. Let E_{0} be the energy of such an object. Then since
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
So we have
Dividing by m and shifting the negative V_{e} term to the right
from which the value of k may be written
CalculationReturning 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)
Divide both sides by m ("cancel m") and multiply both sides by 2
Transferring terms (by adding suitable quantities to both sides) and substituting (3) gives
= 2 V_{0}^{2} (1 – (r_{1}/r_{2})) (4) At this point it helps to step back and take a broader look. We have twoand only twounknown quantities, V_{1} and V_{2} , and we also have two separate equations that involve them, numbered (1) and (4). In mathematics, in general, two equations are enough to nail down two unknowns. It is not an ironclad guarantee, but usually two equations will do the job, and here they indeed do.
As the first step, use equation (1) to express V_{2} :
Squaring Substituting this on the left side of (4)
= V_{1}^{2} (r_{2}^{2} – r_{1}^{2}) / r_{2}^{2} This may be set equal to the right side of (4), and then we already have an equation for step (2)just one equation, with only one unknown, namely V_{1}. But rather than rushing, it pays to separate as many factors as possible. Maybe something will simplify! By a wellknown identity (see also "Identities")
so Keep eq. (6) for reference and consider next the right side of (4). We might introduce there a common denominator r_{2}: By (4), expressions (6) and (7) are equal:
Our hunch has paid off: both sides can be divided by (r_{2} r_{1}), removing that term, and both sides can also be multiplied by r_{2} , creating more simplification. If we had rushed ahead earlier, our expressions might have been much more formidable! What remains is
Multiply by r_{2}, divide by (r_{2} + r_{1}), and the unknown quantity stands alonesquared, but that's OK:
Time to plug in numbers:
so Extracting square roots from both sides (first law of algebra)
showing we need add just 2.966 km/s, a shade short of 3 km/s or 10% of the orbital velocity. Arrival at MarsThe velocity V_{2} at which the spaceship arrives at Mars is found from (5)
It has given up some of its kinetic energy to overcome the pull of the Sun and moV_{e} further outward from the Sun. The big question now ishow does this compare with the velocity V_{3} 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
For Mars, r = 1.523691, T^{2} = (1.523691)^{3} = 3.53745
Let us go with the accepted number 1.8822 (here some approximations were made). Assuming 365.25 days per (Julian) year:
During that time the spaceship covers
Dividing by T, this comes to Each day has (24)(3600) = 86400 seconds, so the orbital distance covered by Mars each second is Distancepersecond is of course the definition of velocity. Therefore
Comparing to 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 2.545 km/s . 
Next Stop: #21d. Flight to Mars: the Return Trip
Timeline Glossary Back to the Master List
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated: 9222004
Reformatted 24 March 2006