Exercise: Fast Transfer -- Technical Notes

Illustration of a Fast Transfer

In the Fast Transfer shown here, Delta-V1 is, as in the case of a Hohmann Transfer, carried out at periapsis of the transfer trajectory. However, the radius of apoapsis of the transfer orbit is greater than the radius of the outer circular orbit, and, instead of proceeding to apoapsis, the journey of the satellite is interrupted by a second burn when it reaches the desired final radius. It can be seen that Delta-V2 changes not only the magnitude of the velocity vector but also its direction.

This discussion and the related exercise are based on Example 3-6-5 in Hale, Francis J., Introduction to Space Flight, Englewood Cliffs, N.J.: Prentice-Hall (1994), pp. 48-50.

Derivation of Values Used for Delta-Vs

The amount of Delta-V needed can be calculated using the Law of Cosines, where Vo is the velocity of the outer orbit, Vt is the velocity of the transfer orbit where it intersects the outer orbit, and α is the angle betweenand

The transfer trajectory can be a parabola or hyperbola instead of an ellipse, but an ellipse is the usual choice.

For the Fast Transfer you are using a transfer ellipse with a semimajor axis twice as large as that of the one you used in the Hohmann Transfer, which was 42,238 km. The energy of the transfer orbit is

= -4.072 x 106 m2/sec2

where a is the semimajor axis. The velocities at periapsis of the transfer trajectory and at the point where it intersects the outer orbit are, respectively,

= 10.528 km/sec = 3.276 km/sec

Recalling that Vi = 7.713 km/sec, you can calculate Delta-V1 in the same manner as for the Hohmann Transfer:

Delta-V1 = Vt0 - Vi = 2.814 km/sec

Notice that Delta-V1 is greater here than in the Hohmann Transfer, since the Fast Transfer ellipse has a greater semimajor axis and, consequently, a greater specific mechanical energy than the Hohmann ellipse.

Before you can apply the Law of Cosines to compute Delta-V2, you need to know the angle α between the velocity vector of the spacecraft in the transfer orbit and its velocity vector after transferring to the outer circular orbit (see illustration). It will help here (and in calculating time of flight) to know the eccentricity of the Fast Transfer ellipse and the true anomaly of the spacecraft's position at the intersection between the transfer trajectory and the outer circular orbit.

Since you know the radius of periapsis and semimajor axis of the transfer ellipse, you can derive its eccentricity from the relationship rp = a(1 - e):

= 0.8631

The ellipse's semilatus rectum is given by

p = a(1 - e2) = 12,482 km

With e and p, you can determine the true anomaly at ro:

= 144.7 deg

The angle between the two velocity vectors is

= 59.35 deg


= 3.148 km/sec

and the total Delta-V needed for the Fast Transfer is

Delta-V = Delta-V1 + Delta-V2 = 5.962 km/sec

which is more than 50 % greater than that needed for the Hohmann Transfer.

Hohmann and Fast Transfer Compared (TOF)

As you have seen, the Fast Transfer is more than 50% more expensive than a Hohmann Transfer that achieves the same result. However, it is also nearly twice as fast.

Since the Hohmann transfer begins at periapsis and ends at apoapsis of the transfer ellipse, the time of flight (TOF) is just half the period of the ellipse, i.e.

= 19,046 seconds = 5.29 hours

For the Fast Transfer, first calculate eccentric anomaly (E):

Then apply Kepler's Equation to derive mean anomaly (M):

    M = E - e sin E = 0.559 rad

Finally, use the mean anomaly value to calculate TOF:

= 9585 seconds = 2.66 hours

Again, the greater speed of the Fast Transfer is not surprising, given the fact that it occurs about halfway between perigee and apogee of the transfer ellipse and does not include the significant slowing down that occurs when an orbiting body approaches apoapsis.