Problem 238 - Calculus Volume 1

Question

Newton's law of gravity states that the gravitational force exerted by an object of mass \(M\) and one of mass \(m\) with centers that are separated by a distance \(r\) is \(F=G \frac{m M}{r^{2}}, \quad\) with \(G\) an empirical constant \(G=6.67 x 10^{-11} \mathrm{m}^{3} /\left(\mathrm{kg} \cdot \mathrm{s}^{2}\right) .\) The work done by a variable force over an interval \([a, b]\) is defined as \(W=\int_{a}^{b} F(x) d x\) . If Earth has mass \(5.97219 \times 10^{24}\) and radius 6371 \(\mathrm{km}\) , compute the amount of work to elevate a polar weather satellite of mass 1400 \(\mathrm{kg}\) to its orbiting altitude of 850 \(\mathrm{km}\) above Earth.

Step-by-Step Solution

Answer

The work required is approximately \(7.088 \times 10^{9}\) Joules.

Step 1: Analyze the Problem

We need to compute the work done to move a satellite from the Earth's surface to an orbiting altitude. The work done by a variable force can be found by integrating the force over the given path.

Step 2: Identify Variables

From the problem, we have these values:- Mass of Earth, \(M = 5.97219 \times 10^{24} \text{ kg}\)- Mass of the satellite, \(m = 1400 \text{ kg}\)- Radius of Earth, \(r = 6371 \text{ km} = 6371000 \text{ m}\)- Altitude above Earth, \(h = 850 \text{ km} = 850000 \text{ m}\)- Gravitational constant, \(G = 6.67 \times 10^{-11} \text{ m}^3/\text{kg} \cdot \text{s}^2\)

Step 3: Determine Limits of Integration

The integration limits are determined by the initial and final distances from the Earth's center. The initial distance is the Earth's radius, \(r = 6371000 \text{ m}\), and the final distance is \(r + h = 7221000 \text{ m}\), the Earth's radius plus the orbiting altitude.

Step 4: Set Up the Integral

The work done, \(W\), is calculated using the integral of the gravitational force:\[ W = \int_{r}^{r+h} F(x) \; dx = \int_{6371000}^{7221000} G \frac{m M}{x^2} \; dx \] where \(x\) represents the variable distance from the center of the Earth.

Step 5: Evaluate the Integral

We know \( F(x) = G \frac{m M}{x^2} \), thus the integral to solve is:\[ W = G m M \int_{6371000}^{7221000} x^{-2} \; dx \]This integral evaluates to:\[ W = G m M \left[ -\frac{1}{x} \right]_{6371000}^{7221000} \]Substituting the limits, \(W\) becomes:\[ W = G m M \left(-\frac{1}{7221000} + \frac{1}{6371000}\right) \]

Step 6: Compute the Numerical Value

Substitute known values into the expression:\[ W = (6.67 \times 10^{-11} \text{ m}^3/\text{kg} \cdot \text{s}^2)(1400 \text{ kg})(5.97219 \times 10^{24} \text{ kg}) \times \left(\frac{1}{6371000} - \frac{1}{7221000}\right) \]Calculating gives:\[ W \approx 7.088 \times 10^{9} \text{ Joules} \]

Step 7: Conclusion

The amount of work required to elevate the satellite to its orbit is approximately \(7.088 \times 10^{9}\) Joules.

Key Concepts

Newton's Law of Gravity

Newton's Law of Gravity is a fundamental principle that explains the gravitational attraction between two masses, such as the Earth and a satellite. The law is given by the equation:

• \( F = G \frac{m M}{r^2} \)

Here, \(F\) is the gravitational force, \(m\) and \(M\) are the masses of the two objects, and \(r\) is the distance between their centers. The constant \(G\), known as the gravitational constant, has a value of \(6.67 \times 10^{-11} \text{ m}^3/\text{kg} \cdot \text{s}^2\).

This force is mutual and attractive, meaning each mass pulls on the other equally but in opposite directions. By understanding this relationship, we can calculate how much work is needed to move an object within a gravitational field, such as elevating a satellite into orbit.

In practical scenarios, this law helps us understand various phenomena in physics, like falling objects or orbits of planets.

Integration in Calculus

Calculus allows us to calculate complex quantities that are predictable under changing conditions. Integration, in particular, is a technique used to find areas under curves, total accumulated quantities, and in physics, it helps evaluate work done by variable forces.

In the context of gravitational work, we use integration to sum up infinite contributions of infinitesimally small bits of force over a distance. The integration of the gravitational force over a path gives us the work performed:

• \( W = \int_{a}^{b} F(x) \; dx \)

This expression represents the work \(W\) done by a force \(F\) as it acts over a path from point \(a\) to point \(b\).

When computing the work required to elevate a satellite, we integrate the gravitational force equation \(G \frac{m M}{x^2}\) over the initial and final distances from Earth's center. The integral becomes a precise sum of all minute forces acting along the variable path of the satellite, providing the total gravitational work required.

Orbital Mechanics

Orbital mechanics is the study of objects in motion due to gravitational forces. It tells us how satellites and other celestial bodies move in space.

To place a satellite into orbit, we need to understand the concept of work in a gravitational context. The work done to elevate a satellite considers the change in gravitational potential energy as the satellite moves from Earth's surface to a higher altitude. This involves balancing the satellite's velocity and the gravitational pull of Earth, allowing it to enter a stable orbit.

When launching a satellite, the calculation of the work performed ensures the satellite reaches the necessary altitude and speed to commence orbit. This is a crucial step for maintaining functions like communication, weather monitoring, and scientific exploration.

The principles of orbital mechanics rely heavily on Newton's laws and calculus, making them essential for anyone interested in understanding how objects like satellites move through space. By mastering these concepts, we can design missions that explore our solar system and beyond.