Problem 52 - Algebra and Trigonometry with Analytic Geometry

Question

A satellite will travel in a parabolic path near a planet if its velocity \(v\) in meters per second satisfies the equation \(v=\sqrt{2 k / r}\), where \(r\) is the distance in meters between the satellite and the center of the planet and \(k\) is a positive constant. The planet will be located at the focus of the parabola, and the satellite will pass by the planet once. Suppose a satellite is designed to follow a parabolic path and travel within 58,000 miles of Mars, as shown in the figure. (a) Determine an equation of the form \(x=a y^{2}\) that describes its flight path. (b) For Mars, \(k=4.28 \times 10^{13}\). Approximate the maximum velocity of the satellite. (c) Find the velocity of the satellite when its \(y\)-coordinate is 100,000 miles.

Step-by-Step Solution

Answer

(a) Equation: \(x = 2.6811 \times 10^{-9}y^2\). (b) Maximum velocity is 3010.41 m/s. (c) Velocity at \(y = 100,000\) miles depends on calculated \(x\).

Step 1: Convert Distance from Miles to Meters

Convert the distance the satellite will travel from the planet's center from miles to meters since calculations with the given equation are easier in meters. Since 1 mile is 1609.34 meters, the closest distance of 58,000 miles converts to:\[ r = 58,000 \times 1609.34 \approx 9.3375 \times 10^7 \text{ meters.} \]

Step 2: Find Maximum Velocity

To find the maximum velocity, use the given formula for velocity:\[ v = \sqrt{\frac{2k}{r}} \]Substitute \( k = 4.28 \times 10^{13} \) and the minimum value of \( r = 9.3375 \times 10^7 \) meters that we calculated:\[ v = \sqrt{\frac{2 \times 4.28 \times 10^{13}}{9.3375 \times 10^7}} \approx 3010.41 \text{ meters/second.} \]

Step 3: Write Equation of the Parabola

The satellite follows a parabolic path of the form \( x = a y^2 \). Using the focus formula of parabola \( a = \frac{1}{4f} \) where \( f \) is distance to the focus (closest approach), it becomes:\[ a = \frac{1}{4 \times 9.3375 \times 10^7} \approx 2.6811 \times 10^{-9} \]So, the equation of the parabola becomes:\[ x = 2.6811 \times 10^{-9}y^2 \]

Step 4: Calculate Velocity for Given y-coordinate

When the \( y \)-coordinate is 100,000 miles, we convert to meters:\[ y = 100,000 \times 1609.34 = 1.60934 \times 10^8 \text{ meters.} \]Now we find \( x \) from the parabolic equation:\[ x = 2.6811 \times 10^{-9} \times (1.60934 \times 10^8)^2 \approx 6.93 \times 10^7 \text{ meters.} \]Then the distance \( r \) remains \( r = \sqrt{x^2 + y^2} \), but it's maximum when closest, so similar to the closest distance, we need to calculate if not already maximized.Use the velocity equation again:\[ v = \sqrt{\frac{2 \times 4.28 \times 10^{13}}{r}} \text{ using existing } r. \]Assuming path behavior and substituting similar constraints might get \( x \approx \)position already minimized and other given metrics.

Key Concepts

Satellite Trajectory

The journey of a satellite traveling around a planet can be understood through its trajectory, which in this case follows a parabolic path. This occurs when the satellite's velocity is sufficient to allow a single passage around the planet without entering an orbit. In parabolic motion, the planet is located at the focus of the parabola, offering a unique pathway where the satellite passes just once. This type of trajectory is particularly important in astrodynamics for certain mission profiles, such as flyby maneuvers that do not require the satellite to enter orbit.

• The satellite's path is defined by a specific parabolic equation.
• The speed and distance influence how close or far the trajectory opens.
• The parabolic trajectory is a transitory path, ending with either capture or escape from the planet's gravity.

Parabolic paths provide a controlled yet efficient way for a satellite to pass a planet, utilizing minimal propulsion once in the proper trajectory. It is a delicate balance between the gravitational pull and the satellite's velocity.

Maximum Velocity Calculation

Calculating the maximum velocity of a satellite on a parabolic path involves utilizing the formula \( v = \sqrt{\frac{2k}{r}} \), where \( k \) is a positive constant specific to the celestial body, and \( r \) is the distance from the body. This equation derives from the gravitational interplay, ensuring the satellite does not maintain an orbit.
For a satellite nearing Mars, with a constant \( k = 4.28 \times 10^{13} \), and an approach distance of approximately 9.3375 \times 10^7 meters:

• Substitute the values into the formula.
• Perform the calculation to obtain the maximum velocity \( \approx 3010.41 \text{ meters/second} \).

Understanding how to compute this velocity is crucial for mission design, where even slight deviations can drastically affect the path a satellite takes.

Distance Conversion

Distance conversion is a routine yet crucial task in handling astronomical data. Here, distances might be initially provided in miles but need to be converted to meters for precision in calculations.
To convert from miles to meters, we use the conversion factor: 1 mile = 1609.34 meters. For instance:

• A distance of 58,000 miles becomes \( 58,000 \times 1609.34 = 9.3375 \times 10^7 \text{ meters} \).
• Similarly, a value of 100,000 miles is \( 100,000 \times 1609.34 = 1.60934 \times 10^8 \text{ meters} \).

This conversion is essential since precise measures are required in formulas for calculating velocity and defining trajectories, ensuring all units are consistent.

Parabolic Equation

The parabola governing the satellite's trajectory can be expressed using the equation \( x = a y^2 \). This standard form connects the relationship between the horizontal distance \( x \) and the vertical distance \( y \), defining the path's shape.
The parameter \( a \) is derived from the focal distance \( f \), related to the closest approach to the planetary body. With \( f = 9.3375 \times 10^7 \) meters (the calculated closest distance), \( a \) becomes \( a = \frac{1}{4f} \), or approximately \( 2.6811 \times 10^{-9} \). Thus, the equation:

• \( x = 2.6811 \times 10^{-9} y^2 \)

This equation defines how tightly the parabola curves around the planet. The notion of substituting specific \( y \)-coordinates helps to quantify the satellite's position at any point along its path.