You are piloting a spacecraft whose total mass is \(1000 \mathrm{~kg}\) and attempting to dock with a space station in deep space. Assume for simplicity that the station is stationary, that your spacecraft is moving at \(1.0 \mathrm{~m} / \mathrm{s}\) toward the station, and that both are perfectly aligned for docking. Your spacecraft has a small retro-rocket at its front end to slow its approach, which can burn fuel at a rate of \(1.0 \mathrm{~kg} / \mathrm{s}\) and with an exhaust velocity of \(100 \mathrm{~m} / \mathrm{s}\) relative to the rocket. Assume that your spacecraft has only \(20 \mathrm{~kg}\) of fuel left and sufficient distance for docking. a) What is the initial thrust exerted on your spacecraft by the retro-rocket? What is the thrust's direction? b) For safety in docking, NASA allows a maximum docking speed of \(0.02 \mathrm{~m} / \mathrm{s}\). Assuming you fire the retro-rocket from time \(t=0\) in one sustained burst, how much fuel (in kilograms) has to be burned to slow your spacecraft to this speed relative to the space station? c) How long should you sustain the firing of the retrorocket? d) If the space station's mass is \(500,000 \mathrm{~kg}\) (close to the value for the ISS), what is the final velocity of the station after the docking of your spacecraft, which arrives with a speed of \(0.02 \mathrm{~m} / \mathrm{s}\) ?

Step by step solution

01

a) Calculate the initial thrust exerted on the spacecraft and its direction

To determine the initial thrust exerted on the spacecraft, we can use the Thrust equation given by \(F_{thrust} = v_{exhaust} \times dm/dt\) where \(v_{exhaust} = 100 \mathrm{~m} / \mathrm{s}\) and \(dm/dt = 1.0 \mathrm{~kg} / \mathrm{s}\). Plugging the values into the equation, we have \(F_{thrust} = 100 \mathrm{~m} / \mathrm{s} \times 1.0 \mathrm{~kg} / \mathrm{s} = 100\mathrm{~N}\). Since the spacecraft is moving towards the space_station, the direction of thrust exerted by the retro-rocket should be opposite to the direction of motion.

02

b) Calculate the amount of fuel required to slow the spacecraft down to the safe docking speed

Let's assume that the retro-rocket produces a constant acceleration. In that case, we can use the following kinematic equation: \(v_{final} = v_{initial} + a \times t\). To calculate the value of acceleration, we substitute the variables, \(v_{final} = -0.02 \mathrm{~m} / \mathrm{s}\) (opposite direction), \(v_{initial} = 1.0 \mathrm{~m} / \mathrm{s}\), and \(t = t\). We know that \(F_{thrust} = m \times a\) and from part (a), we know that the initial thrust \(F_{thrust} = 100\mathrm{~N}\). Now, the mass of the spacecraft (\(m\)) will change as fuel is consumed, so we let \(m\) be \(1000\mathrm{~kg} - M\) (where \(M\) is the amount of fuel burnt). Thus: \(m \times a = F_{thrust}\) \((1000 - M)a = 100 \mathrm{~N}\) Since \(F_{thrust}(dm/dt) = v_{exhaust}(dm/dt)\), we can find acceleration as a function of fuel consumption: \(a = \dfrac{v_{exhaust}(dm/dt)}{m}\) Now substitute in the kinematic equation: \(-0.02 \mathrm{~m/s} = 1.0 \mathrm{~m/s} + \dfrac{100 \mathrm{~m/s} \times 1.0 \mathrm{~kg/s}}{(1000-M)}t\) Solve this for \(M\) and we get: \(M = 19.6 \mathrm{~kg}\) of fuel required.

03

c) Calculate how long the retrorocket should be sustained

Now that we have the value for the fuel consumed (\(M = 19.6 \mathrm{~kg}\)), we can plug this into the formula given in part (b) to find the time it takes: \(-0.02 \mathrm{~m/s} = 1.0 \mathrm{~m/s} + \dfrac{100 \mathrm{~m/s} \times 1.0 \mathrm{~kg/s}}{(1000-19.6)}t\) Solve for \(t\) and we get: \(t = 19.8 \mathrm{~s}\)

04

d)Calculate the final velocity of the station after the docking

We can apply the conservation of momentum equation to determine the final velocity of the space station after docking: \((m_1 \times v_1) + (m_2 \times v_2) = (m_1+m_2) \times v_{final}\) In this case, \(m_1 = 500,000\mathrm{~kg}\), \(v_1 = 0\) (stationary), \(m_2 = (1000 - 19.6)\mathrm{~kg}\) (mass of the spacecraft minus fuel consumed), and \(v_2 = -0.02\mathrm{~m/s}.\) Substitute the values into the equation: \((500,000\mathrm{~kg} \times 0) + (980.4\mathrm{~kg} \times -0.02\mathrm{~m/s}) = (500,000\mathrm{~kg} + 980.4\mathrm{~kg}) \times v_{final}\) Solve for \(v_{final}\) and we get: \(v_{final} = -3.9 \times 10^{-5} \mathrm {~m/s}\). After docking, the space station will have a final velocity of \(-3.9 \times 10^{-5} \mathrm {~m/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thrust Equation

When we consider the physics of spacecraft docking, the thrust equation plays a crucial role. Thrust is a force that moves a spacecraft in space, and it's produced by the expulsion of mass from the craft, like gases from a rocket engine.

The basic thrust equation is given as:
\[ F_{thrust} = v_{exhaust} \times \frac{dm}{dt} \] where

• \( F_{thrust} \) is the force of thrust,
• \( v_{exhaust} \) is the velocity of the exhaust gases relative to the spacecraft,
• \( \frac{dm}{dt} \) is the mass flow rate, or the rate at which mass is ejected from the spacecraft.

In the exercise, the given gas exhaust velocity is \( 100 \, m/s \), and the fuel burns at a rate of \( 1.0 \, kg/s \). These values allow us to calculate the thrust exerted on the spacecraft – which is critical for understanding how the spacecraft's speed and direction can be manipulated.

Notably, the direction of the thrust is as significant as its magnitude. For a spacecraft aiming to dock with a station, thrust must be directed opposite to its motion to decelerate it effectively.

Kinematic Equations

The kinematic equations describe the motion of objects without accounting for the forces that cause this motion. These equations are fundamental in calculating the spacecraft's trajectory and final speed upon docking.

In the context of our problem, we specifically look at the equation: \[ v_{final} = v_{initial} + a \times t \] This relates the final velocity \( v_{final} \), the initial velocity \( v_{initial} \), the acceleration \( a \), and the time \( t \) over which the acceleration takes place. It's important because it helps us understand how much time and fuel are needed to decelerate the spacecraft to a safe docking speed.

However, spacecraft motion is more complex because the mass of the spacecraft decreases as fuel is burned. A varying mass complicates the acceleration calculations because the acceleration is not constant. This problem requires integrating the mass flow rate into the kinematic equations, thereby modifying the simple models we would use in an Earth-based context with constant mass systems.

Conservation of Momentum

The principle of conservation of momentum states that in an isolated system, the total momentum remains constant if no external forces are acting on it. In space, docking maneuvers often approximate such an isolated system, making this principle extremely useful.

The momentum of a system is the product of its mass and velocity. When two bodies interact, such as a spacecraft docking with a space station, their combined momentum after they connect should equal their total momentum before the interaction. This equation is given by: \[ (m_1 \times v_1) + (m_2 \times v_2) = (m_1 + m_2) \times v_{final} \] Here, \( m_1 \) and \( m_2 \) are the masses of the station and spacecraft, while \( v_1 \) and \( v_2 \) are their respective velocities before docking. The final collective velocity \( v_{final} \) is what we seek to calculate.

In our exercise, even though the space station is vastly heavier than the spacecraft, docking will slightly change the station's velocity due to the momentum brought by the spacecraft. The change is minuscule due to the station's massive size compared to the spacecraft, a testament to the conservation of momentum.