Question

At Earth's surface the acceleration due to gravity is approximately \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) (with local variations). However, the acceleration decreases with distance from the surface according to Newton's law of gravitation. At a distance of \(y\) meters from Earth's surface, the acceleration is given by $$a(y)=-\frac{g}{(1+y / R)^{2}},$$ where \(R=6.4 \times 10^{6} \mathrm{m}\) is the radius of Earth. a. Suppose a projectile is launched upward with an initial velocity of \(v_{0} \mathrm{m} / \mathrm{s} .\) Let \(v(t)\) be its velocity and \(y(t)\) its height (in meters) above the surface \(t\) seconds after the launch. Neglecting forces such as air resistance, explain why \(\frac{d v}{d t}=a(y)\) and \(\frac{d y}{d t}=v(t).\) b. Use the Chain Rule to show that \(\frac{d v}{d t}=\frac{1}{2} \frac{d}{d y}\left(v^{2}\right)\). c. Show that the equation of motion for the projectile is \(\frac{1}{2} \frac{d}{d y}\left(v^{2}\right)=a(y),\) where \(a(y)\) is given previously. d. Integrate both sides of the equation in part (c) with respect to \(y\) using the fact that when \(y=0, v=v_{0} .\) Show that $$\frac{1}{2}\left(v^{2}-v_{0}^{2}\right)=g R\left(\frac{1}{1+y / R}-1\right).$$ e. When the projectile reaches its maximum height, \(v=0\) Use this fact to determine that the maximum height is $$y_{\max }=\frac{R v_{0}^{2}}{2 g R-v_{0}^{2}}.$$ f. Graph \(y_{\max }\) as a function of \(v_{0} .\) What is the maximum height when \(v_{0}=500 \mathrm{m} / \mathrm{s}, 1500 \mathrm{m} / \mathrm{s},\) and \(5 \mathrm{km} / \mathrm{s} ?\) g. Show that the value of \(v_{0}\) needed to put the projectile into orbit (called the escape velocity) is \(\sqrt{2 g R}\).

Short answer

Answer: The escape velocity is the minimum initial velocity required for a projectile to escape Earth's gravitational pull and not return to the surface. To find the escape velocity, we determined the initial velocity when the maximum height of the projectile approaches infinity. In this case, the denominator of the expression for \(y_{\max}\) should be zero (\(2 g R - v_0^2 = 0\)). By solving this equation, we obtain the escape velocity as \(v_0 = \sqrt{2 g R}\), ensuring that the projectile can reach infinitely far from Earth and escape its gravity.

Step-by-step solution

a. Explain why \(\frac{d v}{d t}=a(y)\) and \(\frac{d y}{d t}=v(t).\)

Let \(v(t)\) be the velocity of the projectile and \(y(t)\) be its height above the surface. The acceleration \(a(y)\) is the rate of change of the velocity with respect to time, so we have \(\frac{dv}{dt} = a(y)\). Since the velocity \(v(t)\) is the rate of change of the height with respect to time, we have \(\frac{dy}{dt} = v(t)\).

b. Use the Chain Rule to find \(\frac{d v}{d t}=\frac{1}{2} \frac{d}{d y}\left(v^{2}\right).\)

Using the chain rule, we have \(\frac{dv}{dt} = \frac{dv}{dy} \cdot \frac{dy}{dt}\). Since \(\frac{dy}{dt} = v(t)\), we can write \(\frac{dv}{dt} = v \cdot \frac{dv}{dy}\). We can now rewrite this equation as \(\frac{dv}{dt} = \frac{1}{2} \frac{d}{dy}(v^2)\).

c. Find the equation of motion for the projectile \(\frac{1}{2} \frac{d}{d y}\left(v^{2}\right) = a(y).\)

Since in part (a) we showed that \(\frac{dv}{dt} = a(y)\), and in part (b) we found that \(\frac{dv}{dt} = \frac{1}{2} \frac{d}{dy}(v^2)\), we can equate these two expressions to obtain the equation of motion for the projectile: \(\frac{1}{2} \frac{d}{dy}(v^2) = a(y)\).

d. Integrate both sides of the equation from part (c) with respect to \(y\) to find the relationship between \(v\) and \(y.\)

Integrating both sides of the equation \(\frac{1}{2} \frac{d}{dy}(v^2) = a(y)\) with respect to \(y\), we get: $$\int \frac{1}{2} \frac{d}{dy}(v^2) dy = \int a(y) dy$$ By the fundamental theorem of calculus, the left-hand side of the equation integrates to \(\frac{1}{2}(v^2 - v_0^2)\). Integration of the right-hand side with the initial condition \(y=0, v=v_0\) gives: $$g R \left(\frac{1}{1+\frac{y}{R}} - 1\right)$$ Thus, we get: $$\frac{1}{2}(v^2 - v_0^2) = g R\left(\frac{1}{1+\frac{y}{R}} - 1\right)$$

e. Determine the maximum height of the projectile \(y_{\max}.\)

At the maximum height of the projectile, its velocity is zero, \(v = 0\). Using the equation from part (d) and setting \(v = 0\), we obtain: $$-\frac{1}{2} v_0^2 = g R \left(\frac{1}{1 + \frac{y_{\max}}{R}} - 1\right)$$ Solving for \(y_{\max}\) gives: $$y_{\max} = \frac{R v_0^2}{2 g R - v_0^2}$$

f. Calculate the maximum height for different initial velocities.

Using the formula from part (e), we can find the maximum height of the projectile for different initial velocities \(v_0\): 1. For \(v_0 = 500 \mathrm{m/s}\): $$y_{\max} = \frac{R (500)^2}{2 g R - (500)^2} \approx 28,962 \mathrm{m}$$ 2. For \(v_0 = 1500 \mathrm{m/s}\): $$y_{\max} = \frac{R (1500)^2}{2 g R - (1500)^2} \approx 906,859 \mathrm{m}$$ 3. For \(v_0 = 5000 \mathrm{m/s}\): $$y_{\max} = \frac{R (5000)^2}{2 g R - (5000)^2} \approx 6,388,548 \mathrm{m}$$

g. Determine the escape velocity needed to put the projectile into orbit.

To find the escape velocity, we need to determine the initial velocity \(v_0\) at which the projectile can reach infinitely far from Earth. In other words, when \(y_{\max} \to \infty\), the denominator of the expression for \(y_{\max}\) should be zero: $$2 g R - v_0^2 = 0$$ Solving for \(v_0\), we get the escape velocity: $$v_0 = \sqrt{2 g R}$$

Key concepts

Acceleration Due to Gravity

Acceleration due to gravity is a critical concept when discussing projectile motion on Earth or any other celestial body. On Earth's surface, this acceleration is approximately 9.8 meters per second squared (\( g = 9.8 \, \text{m/s}^2 \)), though slight variations can occur based on altitude and local geology.
As a projectile moves upwards and away from the Earth's surface, the pull of gravity weakens. This decrease follows Newton’s law of gravitation, leading to a variation in gravitational acceleration based on the distance from the Earth's center. At a distance \( y \) from the surface, the acceleration is provided by the formula: \[ a(y) = -\frac{g}{(1 + \frac{y}{R})^2} \] where \( R \) is the Earth's radius (approximately 6.4 million meters). This negative sign indicates that gravity acts in the opposite direction of the projectile's upward motion. Understanding this decreasing acceleration is crucial in calculating a projectile's motion and trajectory over large distances from the Earth's surface.

Newton's Law of Gravitation

Newton’s Law of Gravitation lays the foundation for understanding how bodies in space interact. It states that every point mass attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. For Earth and a projectile, the formula simplifies to:
\[ F = \frac{G \cdot M \cdot m}{r^2} \]
Here, \( F \) is the force of attraction, \( G \) is the gravitational constant, \( M \) is the Earth's mass, \( m \) is the mass of the projectile, and \( r \) is the distance between their centers. On a practical level, this law aids in deriving the gravitational pull experienced by a projectile as it moves from Earth's surface, represented by the acceleration equation \( a(y) \) discussed earlier. Such understanding allows for precise predictions of trajectories and helps calculate escape velocities needed for orbit or interplanetary travel.

Chain Rule

In calculus, the chain rule is a technique used to differentiate composite functions. When applied to problems of motion, like those encountered with projectiles, the chain rule becomes extremely useful. To exemplify its application in this context, consider the velocity \( v(t) \) in relation to height \( y(t) \) above Earth. Here, the chain rule allows us to link the rate of change of velocity with respect to time to its rate of change with respect to height:
\[ \frac{dv}{dt} = \frac{dv}{dy} \cdot \frac{dy}{dt} \].
Given that \( \frac{dy}{dt} = v(t) \), we can substitute to derive:
\[ \frac{dv}{dt} = v \cdot \frac{dv}{dy} \].
This substitution is integral in finding the equation of motion. It helps relate changes in kinetic energy to changes in gravitational potential, thus describing the energy dynamics of a projectile effectively.

Equation of Motion

An equation of motion describes the behavior of a projectile under the influence of forces such as gravity, allowing us to predict its position and speed at any time. In our scenario, where air resistance is negligible, the equation stems from combining gravitational acceleration and the chain rule insights.
From Newton's law, the force (or acceleration \( a(y) \)) acting upon the projectile is determined. Using the chain rule, we connect this acceleration to changes in velocity with height, resulting in the equation:
\[ \frac{1}{2} \frac{d}{dy}(v^2) = a(y) \].
Integrating this differential equation provides the relationship between velocity \( v \) and height \( y \). Further, by solving for boundary conditions—like initial velocity \( v_0 \) when \( y = 0 \)—we derive the formula:
\[ \frac{1}{2}(v^2 - v_0^2) = gR \left( \frac{1}{1+ \frac{y}{R}} - 1 \right) \].
This expression helps calculate pivotal parameters such as the maximum height of a projectile or determining escape velocity requirements.