Question

A 1970 kg rocket has 4620 kg of fuel on board. The rocket is coasting through space at 95.7 m/s and needs to boost its speed to 319 m/s; it does this by firing its engines and ejecting fuel at a relative speed of 727 m/s until the desired speed is reached. How much fuel is left on board after this maneuver

Answer 1

The remaining fuel on the rocket after the maneuver is 3613 kg.

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of momentum. The initial momentum of the rocket and fuel is equal to the final momentum after the maneuver. We can calculate the initial momentum using the mass of the rocket and its velocity.

We can then calculate the mass of the fuel ejected by subtracting the final velocity of the rocket from the initial velocity, dividing it by the relative speed of the fuel, and multiplying it by the mass of the rocket. Finally, we can subtract the mass of the ejected fuel from the initial mass of the fuel to find the remaining fuel on board.

Initial momentum = (mass of rocket + mass of fuel) × initial velocity = (1970 kg + 4620 kg) × 95.7 m/s.

Final momentum = (mass of rocket + remaining fuel) × final velocity = 1970 kg × 319 m/s.

Setting the initial and final momentum equations equal to each other, we can solve for the remaining fuel: (1970 kg + 4620 kg) × 95.7 m/s = 1970 kg × 319 m/s + remaining fuel × 319 m/s

By solving this equation, we find that there is 3613 kg of fuel remaining on board after the maneuver.

Answer 2

To solve this problem we will apply the propulsion equations given by Tsiolkovsky. The Tsiolkovsky rocket equation, classical rocket equation, or ideal rocket equation is a mathematical equation that describes the motion of vehicles that follow the basic principle of a rocket: a device that can apply acceleration to itself using thrust by expelling part of its mass with high velocity can thereby move due to the conservation of momentum. This is,

`\Delta V = V_e*ln((m_0)/(m_f))`

Where:

`m_0` = initial total mass = 1970 kg + 4620 kg = 6590 kg

`m_f` = final total mass

`V_e` = effective exhaust velocity

`\Delta V` = change in rocket velocity

Replacing the values we have that,

`(319-95.7) = (727) ln((6590)/(m_f))`

`223.3 = (727) ln((6590)/(m_f))`

`(223.3 )/((727)) = ln((6590)/(m_f))`

`0.3071 = ln((6590)/(m_f))`

`m_f = 4847.45kg`

Since this is the mass of both the ship and the remaining propellant, the remaining propellant mass,
`m_f`, is given by

`m_f = (4847.45 - 4620 ) kg`

`m_f = 227.45 kg`