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A satellite in force free space sweeps stationary interplanetary dust at a rate \(d M / d t=\alpha w\), where \(M\) is the mass, \(v\) is the velocity of the satellite and \(\alpha\) is a constant. What is the deacceleration of the satellite? (a) \(-2 \mathrm{~cm}^{2} / \mathrm{M}\) (b) \(-\alpha v^{2} / M\) \((\mathbf{c})+\alpha_{\underline{ }^{2}} / M\) (d) \(-\alpha v^{2}\)

Short Answer

Expert verified
The deacceleration is \(-\frac{\alpha v^2}{M}\), matching option (b).

Step by step solution

01

Analyze the problem

We are given that the satellite's mass changes at a rate described by \(\frac{dM}{dt} = \alpha v\), where \(\alpha\) is a constant and \(v\) is the velocity of the satellite. We are tasked with determining the deceleration, or negative acceleration, caused by this mass change.
02

Apply conservation of momentum

Consider the conservation of momentum for the satellite. The force exerted on the satellite is related to the change in momentum by \(F = \frac{d}{dt}(Mv) = \frac{dM}{dt} \cdot v + M \cdot \frac{dv}{dt}\). Since it's a force-free space, the external force \(F\) should be zero.
03

Substitute given rate of mass change into momentum equation

Substitute \(\frac{dM}{dt} = \alpha v\) into the momentum equation from step 2: \[0 = \alpha v \cdot v + M \cdot \frac{dv}{dt}\]. This simplifies to \[0 = \alpha v^2 + M \cdot \frac{dv}{dt}\].
04

Solve for deceleration \(\frac{dv}{dt}\)

Rearrange the simplified momentum equation to solve for \(\frac{dv}{dt}\): \[M \cdot \frac{dv}{dt} = -\alpha v^2\]. Divide both sides by \(M\) to isolate \(\frac{dv}{dt}\): \[\frac{dv}{dt} = -\frac{\alpha v^2}{M}\].
05

Select the correct option

The expression obtained for deceleration matches option (b): \(-\frac{\alpha v^2}{M}\).

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Key Concepts

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

Momentum Conservation
Imagine a satellite moving through a quiet part of space, where no external forces are acting on it. In these conditions, the principle of momentum conservation becomes key to understanding how the satellite behaves as it sweeps up interplanetary dust. Momentum conservation states that in an isolated system, the total momentum remains constant over time. For the satellite, this means any change in momentum caused by collecting dust must be balanced internally.In mathematical terms, momentum (...MV) is the product of mass and velocity. As the satellite accumulates dust, its mass changes at a rate of \(\frac{dM}{dt} = \alpha v\), where \(\alpha\) is a constant, and \(v\) is the satellite's velocity. Using momentum conservation, we derive the relationship that connects mass change, velocity, and acceleration: \[F = \frac{d}{dt}(Mv) = \frac{dM}{dt} \cdot v + M \cdot \frac{dv}{dt}\].In a force-free environment, \(F\) is zero. This powerful relationship allows us to find the deceleration of the satellite, which is essential to understand its behavior in space.
Deceleration
Deceleration, or negative acceleration, is what happens when something slows down. For our satellite, as it picks up interplanetary dust, its momentum changes, causing it to decelerate. In force-free space, deceleration is influenced purely by changes in mass and velocity.After substituting the rate of mass change \(\frac{dM}{dt} = \alpha v\) into the conservation of momentum equation, we obtain:\[0 = \alpha v^2 + M \cdot \frac{dv}{dt}\]. This equation equates the mass change term with the acceleration term, hence solving it for deceleration gives us:\[\frac{dv}{dt} = -\frac{\alpha v^2}{M}\]. This formula tells us that as the velocity \(v\) increases or mass \(M\) decreases, the satellite slows down more quickly due to its interaction with the dust. The negative sign indicates that this is indeed deceleration.
Mass Change Rate
The mass change rate \(\frac{dM}{dt}\) is a critical component when considering satellite dynamics, especially when it collects dust or other particles in space. This rate determines how much mass is added to or subtracted from the satellite over time.In the given scenario, the rate is described by the equation \(\frac{dM}{dt} = \alpha v\). Here, \(\alpha\) is a constant that signifies how efficiently the satellite gathers mass as a function of its velocity \(v\). This makes intuitive sense: the faster the satellite moves, the more dust it collects per unit time.The mass change directly influences both the momentum and the velocity of the satellite. This relationship is vital because it allows scientists and engineers to predict changes in the satellite's trajectory and ensure its mission objectives are met efficiently.

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Most popular questions from this chapter

A body of mass \(40 \mathrm{~kg}\) resting on a rough horizontal surface is subjected to a force \(P\) which is just enough to start the motion of the body. If \(\mu_{n}=0.5, \mu_{k}=0.4\), \(g=10 \mathrm{~ms}^{-2}\) and the force \(P\) is continuously applied on the body, then the acceleration of the body is (a) \(2 \mathrm{rro}\) (b) \(1 \mathrm{~ms}^{-2}\) (c) \(2 \mathrm{~ms}^{-2}\) (d) \(2.4 \mathrm{~ms}^{-2}\)

A \(24 \mathrm{~kg}\) block resting on a floor has a rope tied to its top. The maximum tension, the rope can withstand without breaking is \(310 \mathrm{~N}\). The minimum time in which the block can be lifted a vertical distance of \(4.6 \mathrm{~m}\) by pulling on the rope is (a) \(1.2 \mathrm{~s}\) (b) \(1.3 \mathrm{~s}\) (c) \(1.7 \mathrm{~s}\) (d) \(2.3 \mathrm{~s}\)

A body of mass \(2 \mathrm{~kg}\) travels according to law \(x(t)=p t+q t^{2}+r t^{3}\), where \(p=3 \mathrm{~ms}^{-1}, q=4 \mathrm{~ms}^{-1}\) and \(r=5 \mathrm{~ms}^{-3}\). The force acting on the body at \(t=2 \mathrm{~s}\) is [NCFRT Exemplat] (a) \(136 \mathrm{~N}\) (b) \(134 \mathrm{~N}\) (c) \(158 \mathrm{~N}\) (d) \(68 \mathrm{~N}\)

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