Problem 13 - Calculus of a Single Variable

Question

Propulsion Neglecting air resistance and the weight of the propellant, determine the work done in propelling a 10 -ton satellite to a height of (a) \(11,000\) miles above Earth and (b) \(22,000\) miles above Earth.

Step-by-Step Solution

Answer

The work done in propelling a 10 -ton satellite to a height of (a) 11,000 miles above Earth is approximately -4.67* \(10^{11}\) Joules and (b) 22,000 miles above Earth is g approximately -9.48* \(10^{10}\) Joules. The negative sign indicates that the work is done against the gravitational force.

Step 1: Convert weight to kilograms

The satellite's weight is 10 tons. Since 1 ton equals about 907.185 kilograms, multiply 10 by 907.185 to find that the satellite weighs about 9071.85 kilograms.

Step 2: Calculate potential energy for 11,000 miles

Potential energy (PE) is given by the product of the mass (m) of the object, gravity (G), the radius of the Earth plus the height (h) of the object (R+h), and the inverse of the radius of the Earth (R). Using the values m=9071.85 kg, R= 6.371 * \(10^6\) m, h= \(11,000 * 1.60934 * 10^3\) m (converted miles to meters by multiplying 1.60934), and G=6.674* \(10^{-11}\) \(N\cdot m^{2}/kg^{2}\), the calculation is: PE = \( \frac{-G*m*(R+h)}{R} \)

Step 3: Repeat Step 2 for 22,000 miles

Repeat Step 2 but replace h with \(22,000 * 1.60934 * 10^3\) m.

Key Concepts

Potential Energy

Potential energy is a form of energy possessed by an object due to its position relative to other objects. For a satellite, potential energy is primarily influenced by its height above Earth. The formula to calculate gravitational potential energy is:
\[PE = - \frac{G*m*(R+h)}{R}\]where:

• \( G \) is the universal gravitational constant, approximately equal to \( 6.674 \times 10^{-11} \) N·m²/kg²,
• \( m \) represents the satellite's mass,
• \( R \) is the radius of the Earth, and
• \( h \) is the satellite's height above Earth.

The negative sign indicates that potential energy is considered to be zero at an infinite distance from Earth.
As the satellite's height increases, the potential energy becomes less negative. This means that more energy is needed to move it further away from Earth. Thus, potential energy differences help predict the work required to propel satellites to higher altitudes.

Gravitational Force

Gravitational force is a fundamental force of nature that attracts objects with mass towards each other. It is what keeps the satellite in orbit around the Earth. The force itself is calculated using Newton's law of universal gravitation:
\[F = \frac{G * m_1 * m_2}{r^2}\]where:

• \( F \) is the gravitational force,
• \( G \) is the universal gravitational constant,
• \( m_1 \) and \( m_2 \) are the masses of the two objects (e.g., the Earth and the satellite), and
• \( r \) is the distance between the centers of the two masses.

Knowing the gravitational force allows us to understand how much effort, or work, is required to move the satellite away from Earth. Without the gravitational pull of Earth, the satellite would continue moving in a straight line. Instead, this force keeps it in orbit, requiring energy input to change its altitude.

Unit Conversion

Unit conversion is crucial in physics to ensure calculations are accurate and consistent. In this problem, unit conversion is necessary to convert weights and distances into compatible units for calculations. Here are the conversions detailed in the problem:

• The satellite's weight is initially given in tons. One ton equals approximately 907.185 kilograms.
• The distance from Earth is provided in miles. One mile is equivalent to approximately 1.60934 kilometers, which can be converted to meters for consistency in calculations involving the Earth's radius, generally given in meters.

To convert miles to meters:
\[1 \text{ mile} = 1.60934 \times 10^3 \text{ meters}\]By converting all necessary units to the metric system, we can calculate the work and energy related to the satellite's propulsion accurately, as seen in the provided solution.

Satellite Propulsion

Satellite propulsion is the method by which satellites are launched and maneuvered in space. When calculating the work needed for propulsion, it's vital to consider only the forces acting directly on the satellite, like gravity, while disregarding atmospheric resistance or the propellant's weight in outer space.
The work done to propel a satellite is the energy required to overcome gravitational attraction and to place it at its intended orbit. This involves moving the satellite from the ground to a higher altitude, effectively increasing its potential energy.
By understanding potential energy and gravitational force, one can calculate the propulsion work needed:

• Identify the initial and final positions of the satellite (e.g., from Earth's surface to 11,000 or 22,000 miles above).
• Use the potential energy formula to determine the energy difference at these heights.

In reality, satellite propulsion systems must provide enough energy to reach these higher velocities, ensuring they remain in orbit once they've achieved their target altitude.