91Ó°ÊÓ

(a) Show that if a projectile is launched straight upward from the surface of the earth with initial velocity \(v_{0}\) less than escape velocity \(\sqrt{2 G M / R}\), then the maximum distance from the center of the earth attained by the projectile is $$ r_{\max }=\frac{2 G M R}{2 G M-R v_{0}^{2}} $$ where \(M\) and \(R\) are the mass and radius of the earth, respectively. (b) With what initial velocity \(v_{0}\) must such a projectile be launched to yield a maximum altitude of 100 kilometers above the surface of the earth? (c) Find the maximum distance from the center of the earth, expressed in terms of earth radii, attained by a projectile launched from the surface of the earth with \(90 \%\) of escape velocity.

Step by step solution

01

Understand Energy Conservation

The conservation of mechanical energy for the projectile can be expressed as the sum of kinetic energy and gravitational potential energy. Initially, let the kinetic energy be \(\frac{1}{2}mv_0^2\) and gravitational potential energy be \(-\frac{GMm}{R}\). As the projectile reaches its maximum height, the kinetic energy becomes zero, and potential energy is \(-\frac{GMm}{r_{\max}}\).

02

Apply Energy Conservation Equation

From energy conservation, we have: \( \frac{1}{2}mv_0^2 - \frac{GMm}{R} = 0 - \frac{GMm}{r_{\max}} \), leading to \( \frac{1}{2}v_0^2 = \frac{GM}{R} - \frac{GM}{r_{\max}} \).

03

Solve for Maximum Distance

Rearrange the equation to solve for \(r_{\max}\): \( \frac{GM}{r_{\max}} = \frac{GM}{R} - \frac{1}{2}v_0^2 \). Simplifying further: \( r_{\max} = \frac{GM}{\frac{GM}{R} - \frac{1}{2}v_0^2} = \frac{2GMR}{2GM - Rv_0^2} \). This matches the given formula.

04

Calculate Initial Velocity for Maximum Altitude of 100 km

Set \(r_{\max} = R + 100,000 \text{ m}\). Plug this into the rearranged equation \(r_{\max} = \frac{2GMR}{2GM - Rv_0^2} \) and solve for \(v_0\).

05

Calculate Maximum Distance with 90% of Escape Velocity

Plug \(v_0 = 0.9\sqrt{\frac{2GM}{R}}\) into the maximum distance formula \(r_{\max} = \frac{2GMR}{2GM - Rv_0^2} \). Simplify to find \(r_{max}\) in terms of Earth's radius.

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Key Concepts

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

Escape Velocity

Escape velocity is the minimum speed needed for an object to break free from the gravitational pull of a celestial body without any additional propulsion. For Earth, this is approximately 11.2 km/s.
When a projectile is launched with less than the escape velocity, it won't leave Earth permanently; instead, it will reach a maximum height and then fall back down.

• Escape velocity depends on the mass and radius of the planet.
• It can be calculated using the formula: \( v_{e} = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant, \( M \) is the Earth's mass, and \( R \) is the Earth's radius.

Understanding this concept is crucial for certain calculations in projectile motion, especially when determining how high or far a rocket can travel.

Energy Conservation

The principle of energy conservation states that the total energy of an isolated system remains constant over time. For a projectile, this means its initial kinetic energy and gravitational potential energy will equal its energy at any other point in its trajectory.
Initially, a projectile's energy consists of:

• Kinetic energy: \( \frac{1}{2}mv_0^2 \)
• Gravitational potential energy: \(-\frac{GMm}{R} \)

At maximum height, the kinetic energy is zero because velocity becomes zero, and the remaining energy is all gravitational potential energy. Conserving energy helps us find the maximum distance the projectile will attain from Earth's center, as shown in calculations.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. For a projectile close to a massive body like Earth, GPE is negative, indicating it's bound by gravity.
The formula for GPE is:\[ U = -\frac{GMm}{r} \]where \( r \) is the distance from Earth's center.

• As the projectile rises, its GPE becomes less negative (it increases).
• At the maximum height, GPE is maximized which contributes to understanding how far up the object will go when launched with a particular velocity.

Knowing how GPE changes helps us understand what happens when the projectile reaches its highest point in the journey.

Kinetic Energy

Kinetic energy (KE) is the energy of motion. When a projectile is launched, its initial kinetic energy is a crucial component in determining how high it will rise.
The formula for kinetic energy is given by:\[ KE = \frac{1}{2} mv^2 \]where \(m\) is the mass and \(v\) is the velocity of the projectile.

• As the projectile moves upward, its kinetic energy decreases because it slows down due to gravity.
• At maximum height, the velocity is zero, so kinetic energy also becomes zero, with all energy accounted for by gravitational potential energy.

Understanding kinetic energy allows us to calculate initial velocities needed to reach certain altitudes, like the one required to ascend 100 km above Earth's surface.