Differential Equations

First Order Equations

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Rocket Motion

Solved Problems

Example 1.

Estimate the fuel mass needed to launch a small "nanosatellite" with mass of \(50\;\text{kg}\) to a low orbit using a single-stage rocket. The specific impulse of the rocket is \(3000\,\frac{\text{m}}{\text{s}}.\)

Solution.

We can use the ideal rocket equation for rough estimates:

\[v = u\ln \frac{{{m_0}}}{m}.\]

The exhaust velocity is approximately equal to the specific impulse, so we can set:

\[u = 3000\,\frac{\text{m}}{\text{s}}.\]

The final mass of the satellite is \(m = 50\,\text{kg}.\) The initial mass \({m_0}\) includes the mass \(m\) of the satellite itself and the fuel mass \({m_p}:\)

\[{m_0} = m + {m_p}.\]

Suppose that the satellite velocity \(v\) on the low orbit is equal to the first space velocity \(7.91\,\frac{\text{km}}{\text{s}} = 7910\,\frac{\text{m}}{\text{s}}.\) We express the necessary fuel mass \({m_p}\) in terms of the remaining parameters and calculate its value:

\[v = u\ln \frac{{m + {m_p}}}{m},\;\; \Rightarrow \frac{v}{u} = \ln \left( {1 + \frac{{{m_p}}}{m}} \right),\;\; \Rightarrow 1 + \frac{{{m_p}}}{m} = {e^{\frac{v}{u}}},\;\; \Rightarrow {m_p} = m\left( {{e^{\frac{v}{u}}} - 1} \right) = 50\left( {{e^{\frac{{7910}}{{3000}}}} - 1} \right) \approx 50\left( {13.97 - 1} \right) \approx 700\left[ \text{kg} \right].\]

Thus, the fuel mass needed to launch the given satellite is \(700\,\text{kg}\) \(\left(14\right.\) times more than the mass of the satellite itself). Of course, this is an estimation of the lower mass bound, because it's based on the ideal rocket equation.

Example 2.

Estimate the acceleration of a rocket at the moment when the spacecraft reaches the orbit. Take the following parameters: the specific impulse (exhaust velocity) is \(u = 3000\,\frac{\text{m}}{\text{s}},\) the mass of the spacecraft on the orbit is \(m = 5000\,\text{kg},\) the fuel burn rate is \(\mu = 100\,\frac{\text{kg}}{\text{s}}.\)

Solution.

We again apply the ideal rocket equation:

\[v = u\ln \frac{{{m_0}}}{m}.\]

In this formula the rocket velocity \(v\) depends on the rocket mass \(m\left( t \right),\) which decreases as fuel is burning. We suppose for simplicity that the fuel burn rate is constant, so that the dependence of the mass on time is described by the linear function:

\[m\left( t \right) = {m_0} - \mu t.\]

The final mass \(m\) of the spacecraft is known in the given problem. Assuming that the exhaust velocity is constant, we can write the ideal rocket equation in the form:

\[v\left( t \right) = u\ln \frac{{{m_0}}}{{{m_0} - \mu t}}.\]

By differentiating with respect to \(t,\) we find the acceleration of the rocket:

\[\frac{{dv}}{{dt}} = a\left( t \right) = u\frac{1}{{\frac{{{m_0}}}{{{m_0} - \mu t}}}} \cdot \frac{{\left( { - {m_0}} \right)\left( { - \mu } \right)}}{{{{\left( {{m_0} - \mu t} \right)}^2}}} = \frac{{u\cancel{\left( {{m_0} - \mu t} \right)}\cancel{m_0}\mu }}{{\cancel{m_0}{{\left( {{m_0} - \mu t} \right)}^{\cancel{2}}}}} = \frac{{u\mu }}{{{m_0} - \mu t}}.\]

Thus, we have obtained the rocket acceleration formula in the form:

\[a\left( t \right) = \frac{{u\mu }}{{{m_0} - \mu t}}.\]

Notice that the acceleration \(a\) increases if any of the three parameters (the time \(t,\) the fuel burn rate \(\mu\) and the exhaust velocity \(u\)) is increased. This can be easily proved by calculating the partial derivatives with respect to each of the variables. For example, the partial derivative in \(t\) is determined by the expression:

\[\frac{{\partial a}}{{\partial t}} = - \frac{{u\mu }}{{{{\left( {{m_0} - \mu t} \right)}^2}}} \cdot \left( { - \mu } \right) = \frac{{u{\mu ^2}}}{{{{\left( {{m_0} - \mu t} \right)}^2}}} \gt 0.\]

Similarly, the partial derivative in \(\mu\) is given by

\[\frac{{\partial a}}{{\partial \mu }} = \frac{{u\left( {{m_0} - \mu t} \right) - u\mu \left( { - t} \right)}}{{{{\left( {{m_0} - \mu t} \right)}^2}}} = \frac{{u{m_0} - \cancel{u\mu t} + \cancel{u\mu t}}}{{{{\left( {{m_0} - \mu t} \right)}^2}}} = \frac{{u{m_0}}}{{{{\left( {{m_0} - \mu t} \right)}^2}}} \gt 0.\]

We can estimate the acceleration of the rocket when it reaches the orbit, assuming that the final mass is \(m = 5000\,\text{kg}:\)

\[a = \frac{{u\mu }}{{{m_0} - \mu t}} = \frac{{u\mu }}{m} = \frac{{3000 \cdot 100}}{{5000}} = 60\,\frac{\text{m}}{{{\text{s}^2}}} \approx 6\,g.\]

Hence, the rocket at the final stage of the trajectory can experience a big acceleration. Since it is a noninertial frame of reference, the force of inertia of the same magnitude will affect the astronauts in the opposite direction (that is directed towards the Earth). As it can be seen, the positive \(g\)-forces the astronauts can experience at take off can reach several \(g.\)

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