where h a 8000 m is the so-called"scale height" of the atmosphere, and po is the air density at sea level. It turns out that the differential equation that results for the velocity cannot be integrated explicitly and, in prac eeds to be int Numerically. It ng to note. however. that the effect of drag losses is usually quite small, and it is often reasonable to ignore it in a first calculation In order to see the importance of D versus the the effect of gravity, we can estimate the value of the ratio D/mg. At conditions typical for maximum drag, p a 0. 25 kg/m and v=700 m/s. Considering a rocket of 12, 000 kg with a cross section of A= 1 m2 and Cp=0.2, we have pACD20.25×1×0.2×7002 2×12.000×9.8 0.021, which indicates that the drag force is only about 2% of the gra force Example Single vs. Two Stage Rockets [1] To achieve an orbital speed of v= 7600 m/s, we require an ideal Au of about 9000 m/s where the extra velocity is needed to overcome gravity and drag. Chemical rockets produce exhausts jets at velocities of A 2500-4500 m/s. Using the higher c, if we wish to place a payload in orbit with a single stage rocket re have a mass ratio(mass at burn out, mf, divided by initial mass, mo) of 9000/4500 =e-2=0.135 The mass mf must include all components of the rocket infrastructure, including the engine, empty tank guidance equipment, etc, as well as the payload. The mass of the propellant will be mo-mf =(1 0.135 )mo=0.865mo. If we assume that the tank plus engine are the main contributors to the weight of the rocket and weigh 10% of the propellant, 0.087mo, we have maod=(0.135-0.087)mo=0.048mo Clearly, there is not much margin here, and, in fact, single-stage-to-orbit vehicles have yet to be engineered ccessfully. The alternative solution is to subdivide the rocket, so that empty tanks are dropped when they are no longer needed Two stages Consider now a two stage vehicle. We shall again assume that each empty tank plus its engine weighs 10% of the propellant it carries. The required Av is now subdivided into two Aus of 4500 m/s each First stage If mo is the initial mass and mi is the mass after burn out, we have m1=c-45004500m0e-1mo=0.368m0where H ≈ 8000 m is the so-called “scale height” of the atmosphere, and ρ0 is the air density at sea level. It turns out that the differential equation that results for the velocity cannot be integrated explicitly and, in practice, needs to be integrated numerically. It is interesting to note, however, that the effect of drag losses is usually quite small, and it is often reasonable to ignore it in a first calculation. In order to see the importance of D versus the the effect of gravity, we can estimate the value of the ratio D/mg. At conditions typical for maximum drag, ρ ≈ 0.25 kg/m3 and v = 700 m/s. Considering a rocket of 12, 000 kg with a cross section of A = 1 m2 and CD = 0.2, we have, ρACDv 2 2mg = 0.25 × 1 × 0.2 × 7002 2 × 12, 000 × 9.8 = 0.021 , which indicates that the drag force is only about 2% of the gravity force. Example Single vs. Two Stage Rockets [1] Single stage To achieve an orbital speed of v = 7600 m/s, we require an ideal ∆v of about 9000 m/s where the extra velocity is needed to overcome gravity and drag. Chemical rockets produce exhausts jets at velocities of c ≈ 2500 − 4500 m/s. Using the higher c, if we wish to place a payload in orbit with a single stage rocket, we have a mass ratio (mass at burn out, mf , divided by initial mass, m0) of mf m0 = e −9000/4500 = e −2 = 0.135 . The mass mf must include all components of the rocket infrastructure, including the engine, empty tank, guidance equipment, etc., as well as the payload. The mass of the propellant will be m0 − mf = (1 − 0.135)m0 = 0.865m0. If we assume that the tank plus engine are the main contributors to the weight of the rocket and weigh 10% of the propellant, 0.087m0, we have, m 1stage payload = (0.135 − 0.087)m0 = 0.048m0 . Clearly, there is not much margin here, and, in fact, single-stage-to-orbit vehicles have yet to be engineered successfully. The alternative solution is to subdivide the rocket, so that empty tanks are dropped when they are no longer needed. Two stages Consider now a two stage vehicle. We shall again assume that each empty tank plus its engine weighs 10% of the propellant it carries. The required ∆v is now subdivided into two ∆v’s of 4500 m/s each. First stage If m0 is the initial mass and m1 is the mass after burn out, we have, m1 = e −4500/4500m0 = e −1m0 = 0.368m0 4