Problem 31 - Thomas Calculus in SI Units

Question

The strength of Earth's gravitational field varies with the distance \(r\) from Earth's center, and the magnitude of the gravitational force experienced by a satellite of mass \(m\) during and after launch is $$F(r)=\frac{m M G}{r^{2}}.$$ Here, \(M=5.975 \times 10^{24} \mathrm{kg}\) is Earth's mass, \(G=6.6720 \times\) \(10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} \mathrm{kg}^{-2}\) is the universal gravitational constant, and \(r\) is measured in meters. The work it takes to lift a 1000 -kg satellite from Earth's surface to a circular orbit \(35,780 \mathrm{km}\) above Earth's center is therefore given by the integral $$\text { Work }=\int_{6,370,000}^{35,780,000} \frac{1000 M G}{r^{2}} d r \text { joules. }$$ Evaluate the integral. The lower limit of integration is Earth's radius in meters at the launch site. (This calculation does not take into account energy spent lifting the launch vehicle or energy spent bringing the satellite to orbit velocity.)

Step-by-Step Solution

Answer

The work required is approximately \(5.708 \times 10^{10}\) joules.

Step 1: Identify the Expression for Work

The work required to move the satellite from Earth's surface to a given height is defined by the definite integral: \[ \text{Work} = \int_{6,370,000}^{35,780,000} \frac{1000 \cdot M \cdot G}{r^{2}} \; dr\]where \(M\) is Earth's mass and \(G\) is the gravitational constant. These are constant values in the integral.

Step 2: Evaluate the Definite Integral

To evaluate the integral, recognize that the integrand is \(\frac{1000 \cdot M \cdot G}{r^2}\), which simplifies to the following antiderivative:\[\int \frac{1}{r^2} \, dr = -\frac{1}{r} + C\]Apply this antiderivative to the definite integral:\[-1000 \cdot M \cdot G \left[ \frac{1}{r} \right]_{6,370,000}^{35,780,000}\]

Step 3: Substitute the Limits of Integration

Substitute the upper and lower limits of the integral into the evaluated antiderivative:\[-1000 \cdot M \cdot G \left( \frac{1}{35,780,000} - \frac{1}{6,370,000} \right)\]

Step 4: Plug in the Constants

Substitute the values of \(M\) and \(G\) into the expression:\[-1000 \cdot (5.975 \times 10^{24}) \cdot (6.6720 \times 10^{-11}) \left( \frac{1}{35,780,000} - \frac{1}{6,370,000} \right)\]

Step 5: Simplify and Calculate

Calculate the numerical result:1. Compute the values inside the parentheses: - \(\frac{1}{35,780,000} \approx 2.7949 \times 10^{-8}\) - \(\frac{1}{6,370,000} \approx 1.5699 \times 10^{-7}\)2. Find the difference: - \(2.7949 \times 10^{-8} - 1.5699 \times 10^{-7} \approx -1.2904 \times 10^{-7}\)3. Multiply by the constants: - \(1000 \times 5.975 \times 10^{24} \times 6.6720 \times 10^{-11} \times -1.2904 \times 10^{-7}\)4. The computed work is approximately \(5.708 \times 10^{10}\) joules.

Key Concepts

Gravitational Force

Gravitational force is a fundamental concept in physics that describes the attraction between two objects with mass. Imagine it as the invisible glue that pulls everything towards Earth. This force follows Newton's law of universal gravitation, stating that every mass attracts every other mass with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Simply put, the farther apart the objects, the weaker the gravitational pull, and the more massive the objects, the stronger the attraction.

In our example, we dealt with calculating the work needed to lift a satellite above Earth's surface. This force is defined by the formula: \[ F(r) = \frac{m M G}{r^2} \]where:

• - \(m\) is the mass of the satellite,
• - \(M\) is Earth's mass,
• - \(r\) is the distance from Earth's center, and
• - \(G\) is the universal gravitational constant.

This formula helps us understand how the satellite's position affects the gravitational force it experiences, key to calculating the work done in moving the satellite.

Definite Integrals

Definite integrals are a powerful tool in calculus used to calculate the area under a curve within certain limits. In the context of work and energy, they calculate the total work done when moving an object against a varying force over a distance. For instance, lifting the satellite involves determining the cumulative work over the range from Earth's surface to its orbit.

This is where we used the integral:\[\text{Work} = \int_{6,370,000}^{35,780,000} \frac{1000 \cdot M \cdot G}{r^2} \, dr\]The integral's limits, 6,370,000 m and 35,780,000 m, represent the distances from the center of the Earth at the beginning and end of the motion. The integral allows us to sum up infinitesimally small amounts of work over the range of distances to find the total work done.

Antiderivatives

Antiderivatives, often referred to as indefinite integrals, are the reverse operation of differentiation. They help us find a function that describes the accumulation of quantities, such as work done.

In our exercise, the antiderivative of \[\int \frac{1}{r^2} \, dr\]is \[-\frac{1}{r} + C\],where \(C\) is the constant of integration, often ignored when calculating definite integrals. We apply this antiderivative to solve for the work needed for the satellite's journey. Once we have the antiderivative, we substitute the limits of integration. This step lets us quantify the change in force applied as the satellite moves from one distance to another from Earth's center.

Universal Gravitational Constant

The universal gravitational constant, denoted as \(G\), is a fixed value that appears in Newton's law of universal gravitation. It quantifies the strength of gravity as a universal force, making it essential for calculating gravitational forces between masses.

In our problem, the value of \(G\) is approximately:\[G = 6.6720 \times 10^{-11} \, \text{N} \cdot \text{m}^2 \text{kg}^{-2}\]This small constant suggests that gravity is a weak force compared to others like electromagnetism. Yet, due to the immense masses of celestial bodies, gravitational forces become significant. Knowing \(G\) allows us to calculate the force experienced by objects like satellites, planets, and other astronomical formations.