Question

A gallon of gasoline contains about \(130 \mathrm{MJ}\) of chemical energy at a mass of around \(3 \mathrm{~kg}\). How high would you have to lift the gallon of gasoline to get the same amount of gravitational potential energy? Compare the result to the radius of the earth.

Short answer

Answer: The height required to achieve the same amount of gravitational potential energy as the given chemical energy is approximately 4.42 x 10^6 meters. This height is approximately 69.4% of the Earth's radius.

Step-by-step solution

Gravitational Potential Energy Formula

Recall the formula for gravitational potential energy (GPE) is given by: GPE = m * g * h Where m is the mass, g is the acceleration due to gravity (approximately \(9.8 \mathrm{m/s^2}\)), and h is the height. We will use this formula to solve for the height h.

Convert Chemical Energy to Joules

We are given the chemical energy of the gasoline as \(130 \mathrm{MJ}\). To use it in our calculations, we have to convert it to Joules (J). This can be done using the conversion factor: \(1 \mathrm{MJ} = 1 \times 10^6 \mathrm{J}\) So, the chemical energy in Joules is: \(130 \mathrm{MJ} \times \dfrac{1 \times 10^6 \mathrm{J}}{1 \mathrm{MJ}} = 130\times10^6 \mathrm{J}\)

Set Up and Solve the Equation

Equate the gravitational potential energy to the chemical energy, and then solve for the height (h): \(m * g * h = 130\times10^6 \mathrm{J}\) Substitute the given values for mass (m) and gravitational acceleration (g): \(3 \mathrm{~kg} \times 9.8 \mathrm{m/s^2} \times h = 130\times10^6 \mathrm{J}\) Now, solve for the height: \(h = \dfrac{130\times10^6 \mathrm{J}}{3 \mathrm{~kg} \times 9.8 \mathrm{m/s^2}}\)

Calculate Height

Perform the calculations to find the height: \(h = \dfrac{130\times10^6 \mathrm{J}}{3 \mathrm{~kg} \times 9.8 \mathrm{m/s^2}} = 4.42 \times 10^6 \mathrm{m}\)

Compare the Height to the Earth's Radius

The Earth's average radius is about \(6.37\times10^6 \mathrm{m}\). Let's calculate the ratio of the height to the Earth's radius: \(\dfrac{4.42 \times 10^6 \mathrm{m}}{6.37\times10^6 \mathrm{m}}\) Perform the calculation to obtain the ratio: \(\dfrac{4.42 \times 10^6 \mathrm{m}}{6.37\times10^6 \mathrm{m}} \approx 0.694\) This implies that the height required to achieve the same amount of gravitational potential energy as the given chemical energy is approximately 69.4% of the Earth's radius.

Key concepts

Chemical Energy Conversion

In our daily lives, we often see energy transform from one form to another. For instance, when you fill up your car with gasoline, the chemical energy stored in the gasoline is converted into kinetic energy, propelling your vehicle forward. This process is known as chemical energy conversion. The energy in gasoline comes from the chemical bonds between atoms in the fuel. These bonds are formed through chemical reactions that occurred millions of years ago and stored energy from sunlight.

When the gasoline is burnt in the engine of a car, these chemical bonds break. The released energy is then used to generate motion. The energy within a gallon of gasoline, or any other fuel, can also be equated to other forms of energy using certain calculations. In our example, we have equated the chemical energy to gravitational potential energy by determining how high the gasoline would need to be lifted to represent the same energy in a different form. This conceptual bridge between different types of energy is pivotal in understanding energy conservation and conversion.

Physical Science Calculations

Physical science calculations are fundamental tools used in understanding the quantitative relationships in various phenomena. They allow us to solve problems such as converting one unit of energy to another, as was done in our exercise. To convert chemical energy in megajoules (MJ) to joules (J), we employ simple multiplication, noting that 1 MJ equals one million joules.Once we have our energy units consistent, we can use the physical science formulas, such as the gravitational potential energy (GPE) equation \(GPE = m \times g \times h\), to find unknown variables. In this equation, the mass \(m\) and acceleration due to gravity \(g\) are often known quantities. The challenge typically lies in solving for the height \(h\) which represents the potential energy equivalence we are seeking to find.

Energy Conservation

Energy conservation is a principle stating that energy cannot be created or destroyed in an isolated system. It can only be transformed from one form to another. This principle underpins nearly every physical science calculation related to energy, as seen in our exercise. Here, we have implicitly used the principle of energy conservation to determine the height that a given mass of gasoline would need to be raised to have gravitational potential energy equivalent to its chemical energy content.

In the context of our problem, the chemical energy does not disappear when it is converted to gravitational potential energy. Instead, it is stored in a new form. If the gasoline were lifted to the calculated height, the energy would manifest as potential energy which could do work if the gasoline were to fall back to Earth's surface. Just as in nature or a machine, no energy is lost in the conversion; it simply changes form, adhering to the law of energy conservation.

Earth's Radius Comparison

When we describe quantities in physics, it's often helpful to compare them to known references for better understanding. The Earth's radius comparison is one such example. Giving context to the height required to match the gravitational potential and chemical energy by comparing it to Earth's radius allows us to comprehend the sheer scale involved.

The Earth's average radius is about \(6.37\times10^6 \:m\). In our solution, the height of \(4.42 \times 10^6 \:m\) is a substantial distance - approximately 69.4% of the Earth's radius. Such comparisons are not only intriguing but they enhance our understanding of the issue at hand. It is vital to note that lifting an object close to the Earth's radius is not practical due to numerous factors, including the decreasing gravitational force with altitude, but as a theoretical exercise, it provides a powerful visualization of energy equivalence.