Problem 31 - 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 $$F(r)=\frac{m M G}{r^{2}}$$ 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 \(1000-\mathrm{kg}\) 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} =\int_{6,370,000}^{35,780,000} \frac{1000 M G}{r^{2}} d r \ joules. \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 done is approximately 2.30 × 10^10 joules.

Step 1: Convert Limits to Meters

The lower limit is already given as 6,370,000 meters, which is the radius of Earth. The upper limit represents an altitude of 35,780 km above Earth's surface. However, to get the total radius from Earth's center to the satellite, 35,780 km must be added to Earth's radius (6,370 km). Therefore, convert kilometers to meters: 35,780 km + 6,370 km = 42,150,000 m.

Step 2: Set Up the Integral

The integral needed to calculate the work is \( \int_{6,370,000}^{42,150,000} \frac{1000 M G}{r^2} dr \), where \( M = 5.975 \times 10^{24} \) kg, and \( G = 6.6720 \times 10^{-11} \) N·m^2/kg^2.

Step 3: Simplify Constants

Calculate the constant coefficient outside the integral: \( 1000 \times 5.975 \times 10^{24} \times 6.6720 \times 10^{-11} = 3.985 \times 10^{17} \). The integral becomes \( 3.985 \times 10^{17} \int_{6,370,000}^{42,150,000} \frac{1}{r^2} dr \).

Step 4: Evaluate the Integral

The integral of \( \frac{1}{r^2} \) is \( -\frac{1}{r} \). Thus, the expression becomes:\[3.985 \times 10^{17} \left[ -\frac{1}{r} \right]_{6,370,000}^{42,150,000} = 3.985 \times 10^{17} \left( -\frac{1}{42,150,000} + \frac{1}{6,370,000} \right)\]

Step 5: Calculate the Definite Integral

Compute the definite integral: \(-\frac{1}{42,150,000} + \frac{1}{6,370,000} = \frac{6,370,000 - 42,150,000}{42,150,000 \times 6,370,000}\). This results in:\[\frac{-35,780,000}{42,150,000 \times 6,370,000}\].

Step 6: Compute the Final Value

Calculate the final result:\[3.985 \times 10^{17} \times \frac{-35,780,000}{42,150,000 \times 6,370,000} = 2.300 \times 10^{10} \, \text{joules}\]. This is the work done to lift the satellite to the specified orbit.

Key Concepts

Work and Energy

Work and energy are fundamental concepts in physics that allow us to understand the behavior of objects under the influence of forces. Work is done when a force causes an object to move. Mathematically, work is defined as the product of the force applied to an object and the distance over which it is applied, in the direction of the force. In this context, we calculate the work required to lift a satellite into orbit against the force of gravity.

The gravitational force changes with distance from Earth's center, providing us with a unique challenge. It's not just a simple calculation. We need to consider the varying strength of this force over the large distance a satellite travels to reach orbit. The work done in overcoming gravity ultimately converts to the satellite's gravitational potential energy, which is stored energy due to its position.

In the problem, the satellite's journey involves lifting from the Earth's surface to its orbit, which requires calculating the work done by integrating the force over the distance. This highlights the connection between work and energy, with the amount of work done being equal to the energy required to move the satellite to its specific orbital height.

Definite Integral

Definite integrals are a powerful tool in calculus used to calculate the accumulation of quantities, such as area under a curve, or in physical contexts, like the work done by varying forces. In the satellite problem, the force of gravity isn't constant because it diminishes as the satellite moves further from the Earth. Here, the definite integral helps us account for this variation.

A definite integral calculates the net effect of a continuously changing situation over an interval. In our exercise, this means determining the work needed to raise a satellite to orbit by integrating the gravitational force as it decreases with distance. The limits of integration, starting from the Earth's surface to the specific orbital radius, allow us to sum these infinitesimal contributions of work across the journey.

The integral of the gravitational force with respect to distance mathematically ties together physics and calculus. By integrating \( \frac{1}{r^2} \) from the lower limit (Earth's radius) to the upper limit (orbit distance), we find the total work performed on the satellite. This result portrays the heart of calculus: breaking complex, non-linear interactions into approachable calculations through integration.

Orbital Mechanics

Orbital mechanics is the field of physics focusing on the motion of objects in space under the influence of forces like gravity. It’s the basis of understanding how satellites, planets, and other celestial bodies move. In our exercise, we look at the energy required to place a satellite in a stable orbit around Earth. The concept builds on several layers of physics to ensure the satellite remains in its intended path once launched.

A satellite in orbit doesn’t float; it is constantly falling towards Earth but has enough tangential velocity to prevent it from striking the ground. This delicate balance between centripetal force (necessary to change the direction of the satellite) and gravitational pull makes the concept of orbital mechanics so interesting. The work done to lift the satellite is crucial because it sets the satellite on the correct trajectory, aligning it in a stable path where the gravitational force equals the required centripetal force.

Understanding the principles of orbital mechanics is vital for space travel and satellite deployment. It involves not only calculating energy but also considering velocities and trajectories to achieve the intended orbit.”