Question

6. An interplanetary probe is launched from a space station. The specific impulse of the propulsion system is 3,000 ? ∙ ?/??. The fuel load constitutes 80% of the launch mass, and the vehicle itself is an additional 16%, leaving 4% for the rest of the payload.

   1. Relative to the space station, what will be the final speed of the probe?

   2. What final speed can be achieved if the mass of the payload is reduced by 25%?

Answer 1

let the total mass of the probe including fuel, be 'M' (kg). Fuel mass = 0.8M Vehicle mais = 0.16M Payload mass = 0.04M Specific Impulse = 3000 N.s/kg . :. Net Impulse = 3000 x full man = 3000 x 0.8 M * Net Impulse = 2400 M N. We know, Impulse = fat = AP [change in momention] => Net Impulse = (m2V2 -mirl 7 2400M = 0.2M 200 Ime = 0.16M +0.041 1 m, = M = Vo = 12000 m/s

Thus, I final speed of probe is. q = 12,000 m/el If Payload mass in decreased by 254, 1 i.e., New payload mass = 75 of 0.0um = 0.03M mg = 0.16 M +0.03M Net Impulse = marz-mini 1 => 2400 M = 0.19 M V2 - 0 . 7 V2 = 12 631.579 mle So, new final speed would be 12631.579 m/a