Problem 49 - Algebra 2

Question

AEROSPACE The radius \(r\) of the orbit of a satellite is given by \(r=\sqrt[3]{\frac{G M t^{2}}{4 \pi^{2}}}\) where \(G\) is the universal gravitational constant, \(M\) is the mass of the central object, and \(t\) is the time it takes the satellite to complete one orbit. Find the radius of the orbit if \(G\) is \(6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}, M\) is \(5.98 \times 10^{24} \mathrm{kg}\) , and \(t\) is \(2.6 \times 10^{6}\) seconds.

Step-by-Step Solution

Answer

The radius of the orbit is \( 4.034 \times 10^8 \) meters.

Step 1: Understand the formula

The formula given is for calculating the radius of the orbit: \( r = \sqrt[3]{\frac{G M t^2}{4 \pi^2}} \). Here, \( G \) is the gravitational constant, \( M \) is the mass of the central object, and \( t \) is the period of orbit.

Step 2: Insert known values into the formula

Substitute the given values into the formula: \( G = 6.67 \times 10^{-11} \), \( M = 5.98 \times 10^{24} \), and \( t = 2.6 \times 10^6 \). So the equation becomes \( r = \sqrt[3]{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})(2.6 \times 10^6)^2}{4 \pi^2}} \).

Step 3: Calculate inside the cube root

First, calculate \( t^2 = (2.6 \times 10^6)^2 = 6.76 \times 10^{12} \). Multiply it by \( M \) and then by \( G \): \( (6.67 \times 10^{-11})(5.98 \times 10^{24})(6.76 \times 10^{12}) \). This gives \( 2.7022 \times 10^{27} \).

Step 4: Divide by constant terms

Divide by \( 4 \pi^2 \). Since \( \pi \approx 3.1416 \), \( 4\pi^2 = 4(3.1416)^2 \approx 39.4784 \). So divide \( 2.7022 \times 10^{27} \) by \( 39.4784 \) to get \( 6.8416 \times 10^{25} \).

Step 5: Take the cube root

Find the cube root of \( 6.8416 \times 10^{25} \). The cube root of this value is \( 4.034 \times 10^8 \).

Step 6: Conclusion

The calculated radius of the orbit is \( 4.034 \times 10^8 \) meters.

Key Concepts

Gravitational Constant

The gravitational constant, often denoted by the symbol \( G \), is a key figure in physics known for its role in Newton's law of universal gravitation. This constant helps us understand the force of gravity between two objects.
Essentially, it quantifies the attractive force that a massive object like a planet exerts on other objects around it.
For our calculations, we use the value \( G = 6.67 \times 10^{-11} \ \mathrm{N \cdot m^2/kg^2} \). Despite its small size, the gravitational constant plays a huge role in holding planets, stars, and galaxies together.

Cube Root Calculation

When faced with an equation that involves a cube root, we are essentially looking for a number that can be multiplied by itself twice to give the original value inside the cube root. In this exercise, you may wonder why we cube root, not square root or any other root.
In orbital mechanics, the cube root helps to simplify calculations involving the motion of satellites and celestial bodies by reducing exponential values.

• To compute \( \sqrt[3]{value} \), use either a scientific calculator or the cube root function in math software.

This offers an efficient way to manage large numbers, particularly when dealing with physical constants or astronomical figures.

Orbital Mechanics

Orbital mechanics is the study of the motion of objects in space under the influence of gravitational fields. It is essential for understanding how satellites, planets, and other bodies travel through the cosmos.
The satellite orbit equation used in the exercise is derived from principles that govern these motions. Key elements like the gravitational constant \( G \), the mass of the central body \( M \), and the orbital period \( t \) help determine the orbit's radius through this formula.
Grasping orbital mechanics involves understanding how these variables interplay to maintain the satellite's trajectory around the planet without veering off into space or crashing back into the planet.

Algebraic Substitution

Algebraic substitution is a technique used to solve equations by replacing variables with known values. In our case, it simplifies the computation of the satellite's orbit radius.

• We start with the given formula: \( r = \sqrt[3]{\frac{G M t^2}{4 \pi^2}} \).
• Next, insert the values for \(G\), \(M\), and \(t\) as provided in the problem.
• This transforms the complex equation into a more manageable numeric expression.

Algebraic substitution breaks down the formula so that each part of the equation is easy to handle, enabling us to compute complicated expressions by working through simpler steps. This technique is invaluable in fields requiring precise calculations, like physics and engineering.