Escape speed on the earth is approximately 112 kms If a 2500

Escape speed on the earth is approximately 11.2 km/s. If a 25,000 kg spacecraft must reach this speed within 4.2 minutes of liftoff in order to reach space, determine a) the trust necessary to reach this speed and b) the rate at which the fuel must burn if it is expelled at 3000 m/s. Please show work and provide explanation.

Solution

Escape speed v = 11.2 km/s

                        =11.2 x10 ³ m/s

Mass of spacecraft m = 25000 kg

Time t = 4.2 min = 4.2 x 60 s

Accleration a = (v - u ) / t

Where u = initial speed = 0

Substitute values you get a = (11200-0)/(4.2 x60)

                                         = 44.444 m/s ²

(a). Let the necessary thrust be T.

We know T - mg = ma

From this T = ma+mg

                  = m(a+g)

                  = 25000 (44.444 +9.8)

                 = 1.356 x10 ⁶ N

(b).We know T = vo (dM/dt)

Where vo = velocity of expelled gasses

               = 3000 m/s

          T = 1.356 x10 ⁶ N

So, Rate of burn of fuel dM/dt = T/vo

                                           = (1.356 x10 ⁶ N)/3000

                                           = 452 kg/s