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A rocket at rest on the ground with initial mass 20,000 kg, 80% of which is fuel, burns 200kg/s as it flies

directly upwards. The exhaust gas exits the rocket at a relative speed of 1.80 km/s. Find (a) the thrust of
the rocket, (b) how long it takes to exhaust its fuel, and (c) the rocket’s speed at the end of its engine burn.
Assume g is constant and neglect air resistance.

User Manuels
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1 Answer

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Answer:

(a) The thrust of the rocket can be found using the formula:

Thrust = (mass flow rate) x (exhaust velocity)

The mass flow rate is the rate at which mass is expelled from the rocket, which is equal to the burn rate times the mass fraction of fuel:

mass flow rate = burn rate x (mass of rocket) x (mass fraction of fuel)

mass flow rate = 200 kg/s x 20,000 kg x 0.8

mass flow rate = 32,000 kg/s

The exhaust velocity is given as 1.80 km/s. Therefore, the thrust of the rocket is:

Thrust = 32,000 kg/s x 1.80 km/s

Thrust = 57,600 kN

(b) The time it takes to exhaust the fuel can be found using the formula:

time = (mass of fuel) / (burn rate)

The mass of fuel is 80% of the initial mass of the rocket:

mass of fuel = 0.8 x 20,000 kg

mass of fuel = 16,000 kg

Therefore, the time it takes to exhaust the fuel is:

time = 16,000 kg / 200 kg/s

time = 80 seconds

(c) The rocket's speed at the end of its engine burn can be found using the rocket equation:

Δv = (exhaust velocity) x ln[(initial mass) / (final mass)]

The final mass is the mass of the rocket after it has burned all of its fuel, which is:

final mass = (mass of rocket) - (mass of fuel)

final mass = 20,000 kg - 16,000 kg

final mass = 4,000 kg

The initial mass is 20,000 kg. Therefore, the change in velocity is:

Δv = 1.80 km/s x ln[(20,000 kg) / (4,000 kg)]

Δv = 1.80 km/s x ln(5)

Δv = 3.89 km/s

Since the rocket was initially at rest, its final speed is equal to the change in velocity:

final speed = 3.89 km/s

Therefore, the rocket's speed at the end of its engine burn is 3.89 km/s.

User GentryRiggen
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