For some perspective, imagine you were able to drive your car up a ramp to an altitude characteristic of low-earth orbit (about \(320 \mathrm{~km}\), or 200 miles). It takes about \(5 \times 10^{10} \mathrm{~J}\) of energy \(^{33}\) to win the fight against gravity. Meanwhile, each gallon of gasoline can do about \(25 \times 10^{6} \mathrm{~J}\) of useful work. How many gallons would it take to climb to orbital height in a car? Roughly how many miles per gallon is this (just counting vertical miles)?

The concept of energy efficiency is pivotal when discussing any system that converts energy from one form to another. It is defined as the ratio of useful work performed by a machine or in a process to the total energy expended or heat taken in.

In the case of our exercise, we're considering the conversion of chemical energy in gasoline to mechanical work to lift a car to a great height against the gravitational field of the Earth. The energy efficiency can be characterized by the amount of fuel required to attain a certain altitude and is inversely proportional to the number of gallons consumed for the climb. We intuitively recognize that fewer gallons per unit of height reached signifies higher energy efficiency. However, because the car is not designed to climb vertically, the sheer amount of energy required per gallon indicates an extremely low energy efficiency for this hypothetical scenario.

Solving physics problems requires a step-by-step approach that often involves identifying known and unknown variables, selecting appropriate formulas, and doing careful unit conversions.

In our exercise, the problem-solving process began with a clear understanding of the related physics concepts, namely energy conversion and gravitational potential energy. By dividing the total energy required for the lift by the energy yield per gallon of gasoline, we determined the number of gallons needed.

Further, converting kilometers to miles as required for the 'miles per gallon' calculation demonstrates the importance of unit consistency in physics problems. These steps collectively form the structured approach essential for successful physics problem solving. By following this methodology, students can not only find the solution but also gain a deeper understanding of the underlying physical principles.

Gravitational potential energy (GPE) is the energy held by an object because of its position relative to a gravitational source, usually Earth. The formula for GPE is given by \(GPE = mgh\), where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity, and \(h\) is the height above the reference point.

In this problem, the car gains GPE as it ascends to the altitude of low-earth orbit. The total energy required to win the fight against gravity mentioned in the exercise aligns with the concept of GPE. Although the exercise does not explicitly involve computing GPE (since the total energy required was given), understanding this concept is fundamental in explaining why energy consumption increases as the car ascends. It's the increase in GPE with height that necessitates such a large energy input for seemingly small elevation gains, thereby making the vertical miles per gallon efficiency incredibly low.