I saw that this weeks Space Access '09 conference,
Space tourism survey targets cost factor. Online results hint at future price points for suborbital and orbital flights. By Leonard David Senior space writer updated 4:53 p.m. ET, Tues., Oct. 3, 2006 "Pricey seats. "So far, orbital space tourism has been the propelled province of well-heeled millionaires. Even for projected suborbital jaunts =97 up to the edge of space and return to Earth =97 the price tag for a Virgin Galactic spaceliner seat slaps your purse or wallet for roughly $200,000. Several key results of the space tourism survey point out: The prices of current space treks into suborbital and orbital are generally too high at present, with only 7 percent registering for a suborbital flight and 4 percent for an orbital adventure at current price levels. Suborbital flights would really take off at $25,000, and orbital flights at $500,000, if such price levels were compatible with an operator=92s business plan. If price were not an issue, nearly two-thirds of the respondents would want to go on a round-the-moon adventure."
You can make a calculation for how much fuel you would need for a rocket flying horizontally to reach a certain distance by using the rocket equation for velocity:
Vf -Vi =3D Ve*ln(Mi/Mf), where Vf, Mf are the final velocity and final mass, and Vi, Mi are the initial velocity and initial mass, and Ve is the exhaust velocity. The formula still works for intermediate points in the trip where you burned only a portion of the fuel, where Vf and Mf are the values at these intermediate times. Let's say you're burning propellant at a rate r kgs/sec. Then the mass of the vehicle at time t will be Mf =3D Mi-rt. I'll say the initial velocity Vi is zero, and let the velocity at time t be V(t). Then the formula becomes:
V(t) =3D Ve*ln[Mi/(Mi-rt)]. Then we can integrate this formula for velocity to get the distance traveled, S(t):
S(t) =3D Ve*t - (Ve/r)*(Mi-rt)*ln[Mi/(Mi-rt)]
This formula is for the case of constant thrust, where the acceleration will gradually increase since the mass is decreasing as the fuel is used up. It might be more comfortable for the passengers if instead we used a constant acceleration flight. This would be accomplished by making the fuel flow rate, and therefore thrust, decrease as the weight decreases. The formulas for this case can be constructed in an analogous fashion to those of the classic rocket equation. I haven't calculated it but my guess is the total fuel usage would be the same as for using the fuel at a constant rate. In any case, I will assume that just as for SpaceShipOne it will have aerodynamic shape to allow lift so that most of this propulsion can go towards providing horizontal thrust. I didn't include the drag in this first order calculation of the constant fuel rate case, but it can be added in a more detail examination. You can reduce the drag by having the craft undergo the hypersonic flight at high altitude. You can save fuel to reach this altitude by using a carrier craft such as the White Knight for SpaceShipOne. Note that you don't have to use the fuel on the carrier craft or suborbital vehicle to get to a height of say 100 km, but only to get to high enough altitude to reduce the drag and heating on the vehicle at the hypersonic velocities. XCOR is planning on using kerosene and LOX for their engines so I'll use this type of engine for getting the Ve number. Kerosene/LOX engines can have Isp of 360 s at high altitude, which I am assuming will be the only time the rocket will be used. So Ve will be in the range of 3600 m/s at high altitude. First let's say you want to go across the continental U.S., 4500 km. For a first generation transport vehicle let's say it's comparable in size to SpaceShipOne about 1,000 kg empty and 3,000 kg fully loaded with fuel to carry one pilot and two passengers. Let's put in some numbers in order to calculate the distance, S(t): say t =3D 2500 s, about 42 minutes, r =3D 1 kg/s, and Mi consists of a
1000 kg vehicle with passengers and 2500 kg fuel, for a total of 3500 kg. Then we calculate: S(t) =3D 3600*2500 - (3600/1)*(1000)*ln (3500/1000) =3D 4,490,000 meters, or 4,490 km. The time of 42 minutes compares to about 6 hours for a normal passenger jet to travel this distance. The maximum speed would be Vf =3D 3600*ln(3500/1000) =3D 4500 m/s, or Mach 15, quite a high speed. The X-15 was able to reach Mach 6.7 and was planned on being able to reach Mach 8. It had an Inconel skin with a titanium frame to resist the heat loads at these high Mach numbers. Still for Mach 15 you might need materials even more heat resistant. In this article Burt Rutan says SpaceShipOne's carbon composite structure would not be sufficient for even the Mach 6.7 speeds of the X-15:X-15 and today=92s spaceplanes. by Sam Dinkin Monday, August 9, 2004
Microturbo TRS-18-1 Engine Specifications.
Pratt & Whitney Canada PW600.
Bob Clark