Explanation: One dimensional, straight-line acceleration path is followed in such an environment as the only force acting on the rocket is its thrust and it acts in the flight direction. Under the influence of gravity, the flight path will become curved.

Explanation: Both mass flow and thrust remain constant in such an environment. Then for a given burning time tb, total mass expelled from the rocket can be determined from mass flow rate m = m/tb equation.

Explanation: We account for residual propellant mass in inert mass section. Inert mass includes the masses of the engine system, like that of the nozzles, tanks, cases or unused, residual propellants. It doesn’t come under the useful propellant mass consumed for propulsion.

up = ∆u = -c ln(1-ζ).

Explanation: Because doing any of these except better nozzle cooling will result in better Isp. Velocity increment is directly proportional to Isp, so more Isp means more velocity increment.

Explanation: 0.9 is the correct answer. Because of the relation up = ∆u = -c ln(1-ζ), we can see that higher ζ gives higher up.

Explanation: Isp increases because of the relation up = c ln(mo/mb). Here c is the exhaust velocity of the vehicle and ln denotes natural logarithm.

Explanation: This is because the effective propellant fraction is the ratio of the mass of propellants to the initial mass of the system. If inert mass(mf) increases, then the total initial mass(mo) will also increase as mo = mp + mf.

Explanation: Single-stage rocket vehicles can have mass ratio up to about 20. To minimize weights and lateral loads, spherical shape of the rocket is desirable, although it may not be the easiest to manufacture.

Explanation: Multistage vehicles can have mass ratios that exceed 200. In this case, the rocket is segmented into multiple segments in which each of the segment is a separate propulsive system.