Problem 64 - Precalculus

Question

A satellite in a 100 -mile-high circular orbit around Earth has a velocity of approximately 17,500 miles per hour. If this velocity is multiplied by \(\sqrt{2},\) then the satellite will have the minimum velocity necessary to escape Earth's gravity and will follow a parabolic path with the center of Earth as the focus (see figure). (a) Find a polar equation of the parabolic path of the satellite (assume the radius of Earth is 4000 miles). (b) Use a graphing utility to graph the equation you found in part (a). (c) Find the distance between the surface of the Earth and the satellite when \(\theta=30^{\circ}\). (d) Find the distance between the surface of Earth and the satellite when \(\theta=60^{\circ}\).

Step-by-Step Solution

Answer

The polar equation for the satellite path is \(r = \frac{100\sqrt{2}}{1+\cos(\theta)} - 4000\). The distances must be calculated by substituting the values for \(\theta\).

Step 1: Finding Polar Equation for Satellite's Path

A polar form of a conic section with eccentricity intact as 1 can be represented by \(r = \frac{p}{1 + \cos(\theta)}\), where p corresponds to the semi-latus rectum. Given, the velocity to be multiplied by \(\sqrt{2}\), it implies that the semi-latus rectum also increases by this factor. So, 100 miles is multiplied by \(\sqrt{2}\). Also, as the radius of the earth is subtracted from the r, the equation becomes \(r = \sqrt{2}*100 - 4000 = \frac{100\sqrt{2}}{1+\cos(\theta)} - 4000\)

Step 2: Graphing the Polar Equation

A graphing utility is used to plot the polar equation. Since the output in this format can't include a graph, it’s not possible to display a graphical interpretation here.

Step 3: Finding Distance at \(\theta=30^{\circ}\)

To find the distance between the satellite and Earth, insert \(\theta = 30^{\circ}\) into the polar equation: \(r(\theta=30) = \frac{100\sqrt{2}}{1+\cos(30^{\circ})} - 4000\), and calculate the result. Convert the degrees into radians before inserting into cosine function.

Step 4: Finding Distance at \(\theta=60^{\circ}\)

To find the distance between the satellite and Earth, insert \(\theta = 60^{\circ}\) into the polar equation: \(r(\theta=60) = \frac{100\sqrt{2}}{1+\cos(60^{\circ})} - 4000\), and calculate the result. Convert the degrees into radians before inserting into cosine function.

Key Concepts

Conic Sections in Polar Coordinates

Understanding the representation of conic sections in polar coordinates is fundamental for analyzing paths followed by celestial bodies like satellites. Polar coordinates are a two-dimensional coordinate system where each point on a plane is determined by a distance from a reference point and an angle from a reference direction.

In the context of conic sections, such as parabolas, ellipses, and hyperbolas, polar coordinates provide a unique way to describe these curves. A key parameter is the eccentricity of the conic section, which determines its shape. For a parabola, the eccentricity is exactly 1. In polar form, the equation of a conic section with eccentricity 'e' and semi-latus rectum 'p' is given by:
\r

• \r\(r = \frac{p}{1 + e\cos(\theta)}\) for an ellipse or a parabola
• \r\(r = \frac{p}{1 - e\cos(\theta)}\) for a hyperbola

Here, \(r\) is the radius from the focus to any point on the curve, and \(\theta\) is the angle from the horizontal axis.

By tweaking these parameters, we can model the paths of objects in space. In the exercise provided, the satellite's path is a parabola, which can be described using the above equation with the eccentricity set to 1, modifying the equation to illustrate the path trajectory.

Polar Equation of a Conic

The polar equation for a conic section is instrumental in determining the trajectory of a satellite orbiting a planet like Earth. The general polar equation can be expressed in terms of the semi-latus rectum (p) and the eccentricity (e). For the specific case of a parabola, where the eccentricity is 1, the equation simplifies to \(r = \frac{p}{1 + \cos(\theta)}\).

The semi-latus rectum relates to the object's physical characteristics and velocity. As the escape velocity is reached, which is the speed necessary to break free from Earth's gravitational pull, the semi-latus rectum of the path changes. For the satellite in the exercise, upon reaching escape velocity, effectively the semi-latus rectum is extended by a factor of \(\sqrt{2}\).

By applying this change to the equation, one can derive the precise polar equation that models the new trajectory of the satellite. This equation reflects the balance of gravitational force and the satellite's kinetic energy, which shapes the parabolic escape path. The exercise further demonstrates how the computed polar equation can yield the satellite's distance from Earth's surface at various angles, highlighting the utility of this form in practical applications.

Escape Velocity and Orbital Mechanics

Escape velocity is a fundamental concept in orbital mechanics, which is the field of study concerned with the motion of objects in space, such as satellites and planets. Escape velocity is defined as the minimum velocity an object must have to overcome the gravitational pull of a celestial body without further propulsion.

In our exercise, the satellite must reach a speed of 17,500 miles per hour multiplied by \(\sqrt{2}\) to escape Earth's gravity. This multiplication reflects the acquirement of escape velocity, wherein the kinetic energy of the satellite becomes sufficient to escape the gravitational potential energy of the Earth.

Once the object reaches escape velocity, it follows a parabolic path, which is a specific type of conic section trajectory with an eccentricity of 1. By using the polar equation derived from the modified semi-latus rectum, we can observe the dynamics of how an object escapes from Earth's gravitational field. This velocity is crucial for space exploration as it informs the required energy for a spacecraft to leave Earth’s orbit and venture into deep space. The understanding of escape velocity paired with the application of polar equations for conic sections is vital in predicting the paths of satellites and planning space missions.