Problem 33 - Thomas Calculus

Question

Putting a satellite in orbit 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 \begin{equation}F(r)=\frac{m M G}{r^{2}}.\end{equation} 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 \(1000-\mathrm{kg}\) satellite from Earth's surface to a circular orbit \(35,780 \mathrm{km}\) above Earth's center is therefore given by the integral \begin{equation}\begin{array}{c}{\text { Work }=\int_{6,370,000}^{35,780,000} \frac{1000 M G}{r^{2}} d r \text { joules. }}\end{array}\end{equation} 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 calculated to lift the satellite is approximately \(1.365 \times 10^{10}\) joules.

Step 1: Formula Substitution

Substitute the given values for mass \(m = 1000 \text{ kg}\), Earth's mass \(M = 5.975 \times 10^{24} \text{ kg}\), and the gravitational constant \(G = 6.6720 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2\) into the integral.

Step 2: Simplify Constant Terms

Calculate the multiplication of the constant terms \(1000 \times 5.975 \times 10^{24} \times 6.6720 \times 10^{-11}\), and assign it to a single constant \(C\).

Step 3: Evaluate the Integral

Integrate the function \(\frac{C}{r^2}\) from \(r = 6,370,000\) to \(r = 35,780,000\). The integral of \(\frac{1}{r^2}\) with respect to \(r\) is \(-\frac{1}{r}\), so evaluate \(-C \cdot \left[\frac{1}{r}\right]_{6,370,000}^{35,780,000}\).

Step 4: Substitute the Limits

Substitute the upper and lower limits into the evaluated integral to find \[-C \cdot \left( \frac{1}{35,780,000} - \frac{1}{6,370,000} \right)\].

Step 5: Calculate the Result

Compute the expression from the previous step to find the total work done. This involves evaluating the fraction and multiplying it by the constant \(C\) found earlier.

Key Concepts

Satellite Orbit

When we talk about a satellite orbiting Earth, it means the satellite is moving around the Earth due to the gravitational pull. Earth's gravity helps keep satellites in orbit. This concept can be explained using Newton's law of universal gravitation, which tells us that everything with mass attracts everything else with mass. In the case of satellites,
Earth's gravity provides the centripetal force needed to keep a satellite in its orbit. The balance between the satellite's speed and Earth's gravitational pull prevents it from flying off into space or crashing back to Earth.

• This orbit depends on the satellite's velocity and altitude.
• Higher altitudes require less velocity to stay in stable orbit.
• The circular orbit mentioned is a common and stable type of satellite movement.

This understanding allows engineers to calculate the requirements for launching a satellite properly, ensuring it stays in orbit as desired.

Integral Calculus

Integral calculus is a branch of mathematics that helps us determine areas, volumes, and other quantities under curves, and it's crucial for calculating work in physics. Here, it helps to find how much work is done on the satellite.
The integral \[ \int_{6,370,000}^{35,780,000} \frac{1000 M G}{r^{2}} dr \]
represents the work needed to move the satellite from Earth's surface to its orbit. By integrating, we calculate the total change in gravitational force as the satellite changes position.

• Integration helps add up infinitely tiny changes over a range of distance.
• The limits of the integral (radius of Earth and orbital radius) define the start and end points of the satellite's journey.
• Performing this integration gives a measure of total energy exerted to move the satellite.

Integral calculus hence simplifies the complex problem of gravitational work into a solvable equation.

Gravitational Field

A gravitational field is an invisible force field that exists around any mass, like a planet. This field exerts a force on another mass, like a satellite, pulling it towards the larger mass. In our exercise, we use the gravitational field strength equation,
\[ F(r) = \frac{m M G}{r^2} \]
to understand how strong Earth's pull is at different distances.

• The gravitational field strength decreases with the square of the distance from the Earth's center, indicated by the \( \frac{1}{r^2} \) term.
• Mass \( m \) of the satellite and \( M \) of the Earth are central to determining the gravitational force.
• This field concept simplifies understanding how and why satellites remain in orbit.

By using this equation, we can calculate the necessary conditions for a satellite to achieve and maintain its orbit.

Work Done

Work done in physics refers to the energy required to move an object against a force. Lifting a satellite from Earth's surface requires a significant amount of work to overcome the gravitational pull. The work done is calculated using the concept of integration over a path.
In this exercise, calculating work involves integrating the gravitational force from the Earth's surface to a given altitude.

• The formula for work done, \( \text{Work} = \int F(r) \, dr \), calculates the energy needed to change the satellite's position.
• This considers the variable gravitational force as the satellite moves from one point to another.
• The final calculation reveals the total energy needed to lift and maintain the satellite in its particular orbit.

Understanding work done aids in comprehending the energy investment required in satellite launches and the mechanics of satellite movement in space.