Question

A rocket that is in deep space and initially at rest relative to an inertial reference frame has a mass of 2.60 ✕ 105 kg, of which 1.92 ✕ 105 kg is fuel. The rocket engine is then fired for 250 s while fuel is consumed at the rate of 480 kg/s. The speed of the exhaust product relative to the rocket is 3.15 km/s.

After the 250 s firing, what is the speed of the rocket in m/s?

Sample submission: 1230

Answer 1

Solution- Using the formula

m(t) = m0 - m_dot * t

m(250 s) = (2.60 x 10^5 kg) - (480 kg/s)(250 s) = 1.40 x 10^5 kg

Now final speed

delta_v = g * Isp * ln(m0/m), here Isp = specific impulse = T / (m_dot * g)

As rocket started from rest

Therefore final speed

vf = delta_v

vf = g * [T / (m_dot * g)] * ln(m0/m)

m0 = initial mass of rocket = 2.60 x 10^5 kg

m_dot = rate of fuel consumption = 480 kg/s

ve = exhaust speed relative to rocket = 3.15 x 10^3 m/s

T = thrust

T = m_dot * ve = (480 kg/s)(3.15 x 10^3 m/s) = 1.51 x 10^6 N

vf = (T / m_dot) * ln(m0/m) = [1.51 x 10^6 N / (480 kg/s)] * ln(2.60 x 10^5 kg / 1.40 x 10^5 kg)

vf = 1947.39 m/s