Question

A projectile with mass of \(500 \mathrm{kg}\) is launched straight up from the Earth's surface with an initial speed \(v_{i} .\) What magnitude of \(v_{i}\) enables the projectile to just reach a maximum height of \(5 R_{\mathrm{E}},\) measured from the center of the Earth? Ignore air friction as the projectile goes through the Earth's atmosphere.

Short answer

The initial speed \(v_i\) should be approximately 11083 m/s.

Step-by-step solution

Understand the Problem

A projectile needs an initial speed to reach a particular height above Earth's surface. This height, measured from the Earth's center, is 5 times the Earth's radius, \(R_E\). We need to calculate this initial speed, \(v_i\).

Apply Energy Conservation

The principle of conservation of mechanical energy states that the total mechanical energy (kinetic plus potential) is conserved if only conservative forces are acting. Initially, at Earth's surface, the projective has kinetic energy \(\frac{1}{2} mv_i^2\) and gravitational potential energy \(- \frac{G M m}{R_E}\). At maximum height, the kinetic energy is 0 and the potential energy is \(- \frac{G M m}{5R_E}\).

Write the Energy Conservation Equation

Set the initial total energy equal to the total energy at maximum height:\[ \frac{1}{2} mv_i^2 - \frac{G M m}{R_E} = - \frac{G M m}{5R_E} \]

Simplify the Equation

First, cancel the mass \(m\) from both sides of the equation since it appears in all terms. Then simplify:\[ \frac{1}{2} v_i^2 = \frac{G M}{5R_E} - \frac{G M}{R_E} \] Combine the potential energy terms on the right side:\[ \frac{G M}{5R_E} - \frac{G M}{R_E} = -\frac{4G M}{5R_E} \]

Solve for Initial Velocity \(v_i\)

Solve the equation for \(v_i\):\[ \frac{1}{2} v_i^2 = \frac{4G M}{5R_E} \]Multiply both sides by 2:\[ v_i^2 = \frac{8G M}{5R_E} \]Take the square root of both sides:\[ v_i = \sqrt{\frac{8G M}{5R_E}} \]

Identify Known Quantities

Recognize that \(G = 6.674 \times 10^{-11} \, \mathrm{m^3/kg/s^2}, \ M = 5.972 \times 10^{24} \, \mathrm{kg}, \) and \(R_E = 6.371 \times 10^6 \, \mathrm{m}\). Substitute these into the equation for \(v_i\):\[ v_i = \sqrt{\frac{8 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{5 \times 6.371 \times 10^6}} \]

Calculate the Initial Speed \(v_i\)

Perform the calculation to find the numerical value:\[ v_i \approx 11083 \, \text{m/s} \]

Key concepts

Projectile Motion

Projectile motion describes the path taken by an object propelled into the air. In this context, the projectile is launched vertically from the Earth's surface. Unlike typical projectile motion, which accounts for horizontal displacement and various angles of launch, this example involves a trajectory that solely considers vertical movement. Here, the force propelling the projectile is designed to overcome Earth's gravity and allow it to reach a certain height. This situation simplifies the calculations by focusing only on vertical motion, as the effects of air resistance are ignored. This example is unique because it calculates the speed needed to reach a height defined as a multiple of Earth's radius, specifically five times. Understanding this calculation is crucial because it connects to real-world applications like spacecraft launches, where vertical motion and gravitational pull play significant roles.

• Vertical launch: Motion is straight up rather than at an angle.
• Ignore air resistance: Simplifies calculations by using only gravity.
• Reach specific height: Height is five times Earth’s radius.

Conservation of Energy

The conservation of energy principle is critical in solving this projectile problem. Total mechanical energy in an isolated system remains constant if only conservative forces, like gravity, are present. The energy in question is the sum of kinetic and gravitational potential energies. Initially, when the projectile is launched, it possesses kinetic energy due to its speed and gravitational potential energy because of its position relative to Earth. As the projectile ascends, it trades this kinetic energy for gravitational potential energy. At the peak of its height, the projectile’s speed becomes zero, converting all initial kinetic energy into gravitational potential energy. This transformation is beautifully captured by the equation representing the conservation of mechanical energy, highlighting how energy shifts from one form to another but remains unchanged in total:

• Kinetic energy converts to gravitational potential energy.
• Mechanical energy stays constant when only gravity acts.

Kinetic Energy

Kinetic energy is a fundamental energy form linked to an object's speed and mass. It is given by the equation \( \frac{1}{2} mv^2 \). In the context of this problem, the initial kinetic energy balances against the gravitational pull of the Earth. When the projectile begins its journey, it houses kinetic energy derived from its launch speed, weakening as it climbs higher. The calculation of the initial velocity \( v_i \) involves ensuring that this kinetic energy is sufficient for the projectile to reach a height of 5 times Earth's radius before its motion ceases momentarily at the top.Understanding how kinetic energy influences the motion and how it transitions into potential energy as height increases is crucial to grasping projectile energy exchanges.

• Depends on mass and speed: More speed or mass means more energy.
• Conversion: Changes to potential energy as the projectile rises.

Newton's Law of Universal Gravitation

Newton's law of universal gravitation explains the gravitational attraction between two masses. In this scenario, it defines the gravitational pull the Earth exerts on the launched projectile. This force is determined by the equation \( F = \frac{G M m}{r^2} \), where \( G \) is the gravitational constant, \( M \) is Earth's mass, \( m \) is the projectile's mass, and \( r \) is the distance from Earth's center.At the Earth's surface, the gravitational potential energy is calculated as \( -\frac{G M m}{R_E} \), and it changes as the projectile ascends. The guidance this law provides helps precisely predict how much energy is necessary for overcoming gravitational forces.Grasping these concepts lets you understand the relationship between the initial speed and the gravitational force, ensuring that the projectile reaches the designated height multiple of the Earth's radius.

• Explains: The attractive force between Earth and projectile.
• Guides energy conversion: Helps determine needed initial energy.