Question

The space shuttle, with an initial mass M = 2.41×106 kg, is launched from the surface of the earth with an initial net acceleration a = 26.1 m/s^2. The rate of fuel consumption is R = 6.90×103 kg/s. The shuttle reaches outer space with a velocity of vo = 4632. m/s, and a mass of Mo = 1.45×106 kg. How much fuel must be burned after this time to reach a velocity vf = 5495. m/s?

*Possible to solve without calculus and considering the force of gravity?*

Answer 1

first we have to find thrust, which is F=ma, or Thrust = (26.1 m/s^2 + 9.81 m/s^2)(2.4e6 kg). (If you don't understand why to add gravity, then draw the FBD.)

Thrust = 86543100 N

which is also equal to Vemission(delta mass/delta time) (which in this question is given as R.) So,

86543100 = Vemission(6.9e3)
Vemission = 12542.5 m/s

Assuming that the velocity of the emitted fuel remains constant throughout, you can now sub Vemission into the equation Vf - Vi = Vemission*ln(Mi/Mf) and solve for Mf.

5495 - 4632 = 12542.5*ln(1.45e6/Mf)
Mf=1.353e6 kg

Now subtract your final mass from your initial mass, and this should give you the amount of fuel burned!

1.45e6 kg - 1.353e6 kg = 96413.82 kg