Problem 104 - Algebra and Trigonometry Real Mathematics, Real People

Question

A satellite in a 100 -mile-high circular orbit around Earth has a velocity of approximately 17,500 miles per hour. When this velocity is multiplied by \(\sqrt{2},\) the satellite has the minimum velocity necessary to escape Earth's gravity and follows a parabolic path with the center of Earth as the focus. (See figure.) (a) Find the escape velocity of the satellite. (b) Find an equation of its path (assume the radius of Earth is 4000 miles).

Step-by-Step Solution

Answer

The escape velocity of satellite is 24748.78 miles per hour. The equation of its path around the Earth is \(x^2 = 16400y\).

Step 1: Calculate Escape Velocity

The escape velocity is given as the initial velocity of the satellite multiplied by \(\sqrt{2}\). Therefore, the escape velocity is \(17500 \times \sqrt{2}\) = 24748.78 miles per hour.

Step 2: Formulating the equation for the path of the satellite

For a parabola with focus at the origin and directrix \(y=-p\), the standard equation is \(x^2 = 4py\). As given, the focus of the parabola (the center of the Earth) is at the origin since we are assuming the orbit to start from there. The satellite is at a height of 100 miles from the surface of the earth, therefore making the distance from the focus (earth's center) to the directrix (satellite's starting position), equal to Earth's radius + 100 which is 4100 miles. Thus, the value of \(p\) in the standard equation corresponds to this distance i.e. \(p = 4100\). Substituting this into the standard equation gives \(x^2 = 4 \times 4100y\).

Step 3: Simplify the equation

Simplifying the equation from step 2 gives \(x^2 = 16400y\)

Key Concepts

Escape Velocity

Escape velocity is the minimum speed a body must reach to break free from a celestial object's gravitational pull without further propulsion. For Earth, we calculate this by multiplying the object’s current orbital velocity by \(\sqrt{2}\). This arises because the kinetic energy required for escape must equal the gravitational energy binding the object. Imagine a satellite orbiting Earth at 17,500 miles per hour. To escape, it must travel at \(17500 \times \sqrt{2}\), equaling approximately 24,748.78 miles per hour.

This speed frees the object from Earth’s gravity, allowing it to soar into space without falling back. Earth's gravitational pull decreases with distance, but initially achieving this threshold velocity near the surface is critical for successful escape.

• Escape velocity is independent of the mass of the escaping object.
• It depends only on the mass and radius of the celestial body being escaped from.
• Higher escape velocities are needed for more massive or smaller celestial bodies.

Parabolic Trajectory

When an object reaches escape velocity, it can follow a parabolic trajectory. Unlike circular or elliptical orbits, a parabolic path indicates the boundary between the bound and unbound states in space dynamics. These orbits are mathematically represented as conic sections and can be understood visually as an open curve.

In a parabolic trajectory, the satellite has just enough energy to overcome Earth's gravitational attraction but not enough for a hyperbolic path. The path is an open cone section, meaning the satellite will not return. It is not bound to any repetitive motion and will continue away from Earth indefinitely.

When visualized, a parabolic path creates a U-shape with:

• The vertex representing the closest approach to Earth.
• The arms extending infinitely as gravity’s pull gradually diminishes.

Equation of Path

The equation of the path of the satellite describes mathematically how it moves once it reaches escape velocity. For a parabolic trajectory with Earth as the focal point, the equation of the path can be modeled as \(x^2 = 4py\). The vertex of the parabola is at the closest point to Earth.

Understanding the specifics:

• The focus is at the center of Earth, since the orbital path is centered there.
• The directrix, a line perpendicular to the axis of symmetry, lies at the satellite’s starting position.

Given the Earth's radius is 4000 miles and the satellite is 100 miles above the surface, the distance from Earth's center to the directrix equals 4100 miles.

This results in \(p = 4100\) miles. Thus, substituting \(p\) in the path equation gives \(x^2 = 16400y\), which describes the parabola the satellite follows after achieving escape velocity.