Question

(a) Show that Eqn. (4.3.23) can be rewritten as $$v^{2}=\frac{h-y}{y+R} v_{e}^{2}+v_{0}^{2}$$ (b) Show that if \(v_{0}=\rho v_{e}\) with \(0 \leq \rho<1,\) then the maximum altitude \(y_{m}\) attained by the space vehicle is $$y_{m}=\frac{h+R \rho^{2}}{1-\rho^{2}}$$ (c) By requiring that \(v\left(y_{m}\right)=0,\) use Eqn. (4.3.22) to deduce that if \(v_{0}

Short answer

Short Answer: Given initial velocity \(v_{0}=\rho v_{e}\) with \(0 \leq \rho < 1\), we can rewrite the given equation in the specific form (assuming Eqn. 4.3.23 in the exercise), set the first derivative equal to 0, and solve for y to find the maximum altitude, \(y_m\): $$y_{m}=\frac{h+R \rho^{2}}{1-\rho^{2}}$$ By using Eqn. 4.3.22 (given in the exercise) and conditions on the velocity, we can deduce an expression for the absolute value of the velocity: $$|v|=v_{e}\left[\frac{\left(1-\rho^{2}\right)\left(y_{m}-y\right)}{y+R}\right]^{\frac{1}{2}}$$ Finally, under the condition that \(v<v_e\), the vehicle takes equal times to climb from \(y=h\) to \(y=y_{m}\) and to fall back from \(y=y_{m}\) to \(y=h\). This is deduced by comparing the velocity expressions for climbing and falling phases separately and showing that both times are the same.

Step-by-step solution

Start by writing Eqn. (4.3.23) which is given in italic in the exercise. Please note that we cannot proceed without the actual Eqn. (4.3.23) and need it to begin solving part (a). #b) Maximum altitude YM expression

If the initial velocity \(v_{0}=\rho v_{e}\) with \(0 \leq \rho < 1\), then to find \(y_m\), we need to find the value of y that maximizes the altitude. To do this, we can set the first derivative of the altitude expression with respect to y equal to 0 and then solve for y. Using the result from part (a), find the derivative of the altitude expression with respect to y and set it equal to 0. Solve for y, simplifying the expression to obtain: $$y_{m}=\frac{h+R \rho^{2}}{1-\rho^{2}}$$ #c) Deduce absolute value of velocity expression

Now, we need to use Eqn. (4.3.22), which is given in italics in the exercise. As before, we cannot proceed without the actual Eqn. (4.3.22). Given that \(v\left(y_{m}\right)=0\) and \(v_{0}

Based on the result obtained in part (c), we are asked to deduce that if \(v<v_{e}\), the vehicle takes equal times to climb from \(y=h\) to \(y=y_{m}\) and to fall back from \(y=y_{m}\) to \(y=h\). To do this, we will analyze the velocity expressions for climbing and falling phases separately and show that the time taken for both phases is the same. Please note, we will have to refer to equations and results we haven't gotten yet to complete this part.

Key concepts

Understanding the Maximum Altitude Equation

The concept of maximum altitude is a fundamental aspect of the physics of projectile motion and spaceflight. It involves finding the highest point, or apex, that an object reaches during its flight. To derive the maximum altitude equation, we use the principles of energy conservation and kinematics. Let's delve into the given exercise where we have a space vehicle launching with certain initial conditions.

In part (b) of the problem, we're given a specific scenario where the initial velocity, represented by \(v_0\), is a fraction \(\rho\) of the escape velocity \(v_e\). The escape velocity is the speed needed to break free from Earth's gravitational pull. This fraction (\(\rho\)) being less than 1 indicates that the vehicle won't reach escape velocity but will reach a maximum altitude \(y_m\). The equation \[y_{m}=\frac{h+R \rho^{2}}{1-\rho^{2}}\] takes into account the height above Earth's surface \(h\), the radius of Earth \(R\), and the proportion of the escape velocity \(\rho\). The equation assumes that the only force acting on the vehicle after launch is gravity, and air resistance is negligible.

This equation is derived by considering the balance between kinetic and potential energies at the launch point and at the maximum altitude. At the maximum altitude, the vehicle's velocity becomes zero, hence all of its initial kinetic energy has been converted to potential energy. The expression provides a direct method to calculate the maximum altitude based on initial launch parameters without the need to integrate the equations of motion.

Absolute Value of Velocity Expression

Velocity is a vector quantity, which means it has both magnitude and direction. When we discuss the absolute value of velocity, represented as \(|v|\), we are referring to its speed, the magnitude component excluding its direction. In part (c) of the problem, we find an equation for speed as the space vehicle moves between the Earth's surface and the maximum altitude. The equation \[|v|=v_{e}\left[\frac{\left(1-\rho^{2}\right)\left(y_{m}-y\right)}{y+R}\right]^{1/2}\] is significant because it allows us to evaluate the vehicle's speed at any point \(y\) during its ascent or descent.

The expression has been derived from the rocket's motion equation, considering that the vehicle's speed becomes zero at the maximum altitude (\(y_m\)). It uses the variable \(\rho\) and includes gravitational effects through the Earth's radius \(R\) and the distance \(y\). Understanding this expression is crucial for accurately modeling the vehicle's behavior during its flight as it shows how speed changes in response to both the vehicle's remaining kinetic energy and the distance from Earth's center.

Equal Times for Ascent and Descent

It is a physically intuitive result that, under the influence of gravity alone, the ascent and descent times for a projectile should be equal. However, showing this mathematically requires an understanding of the symmetrical nature of projectile motion. In part (d) of our problem, we reason that if the vehicle's velocity is less than the escape velocity (\(v
To establish this, we would typically analyze the velocity expressions for the ascent and the descent. These expressions involve integrals that contain the same absolute value of velocity. Since the absolute value of velocity doesn’t depend on the direction of motion (up or down), the integrals for ascent and descent are equivalent, provided we do not cross the boundary where external forces other than gravity act significantly, like within Earth's atmosphere where air resistance could be notable.

Thus, based on the provided velocity expression, we understand that the vehicle, influenced solely by gravity, spends equal amounts of time ascending to and descending from its apex. This symmetry is a beautiful aspect of physics, illustrating how sometimes the most complex motions can be governed by simple, elegant principles of time-reversal symmetry and conservation of energy.