Question

A person stands on the surface of the Farth. The mass of the person is \(m\) and the mass of the Earth is \(M\). The person jumps upward, reaching a maximum height \(h\) above the Earth. When the person is at this height \(h,\) the magnitude of the force exerted on the Earth by the person is a) \(m \mathrm{~g}\). c) \(M^{2} g / m\) e) zero. b) \(\mathrm{Mg}\) d) \(m^{2} g / M\).

Short answer

Answer: The magnitude of the force exerted on the Earth by the person at the maximum height above the Earth can be calculated using the expression: $$F = m\mathrm{~g}\frac{(R+h)^2}{R^2}$$ where \(m\) is the mass of the person, \(g\) is the acceleration due to gravity, \(R\) is the radius of the Earth, and \(h\) is the maximum height the person reaches above the surface of the Earth.

Step-by-step solution

Identify the formula

We use Newton's law of universal gravitation to find the force between the person and the Earth. The formula for gravitational force is: $$F = G\frac{m_{1}m_{2}}{r^2}$$ where \(F\) is the gravitational force, \(G\) is the gravitational constant, \(m_{1}\) and \(m_{2}\) are the masses of the two objects, and \(r\) is the distance between their centers.

Substitute the given values

We have the following given values: Mass of the person: \(m\) Mass of the Earth: \(M\) Distance between the center of Earth and the person: \(r=R+h\) where \(R\) is the radius of Earth and \(h\) is the maximum height the person reaches above the surface of Earth. We substitute these values into the formula: $$F = G\frac{mM}{(R+h)^2}$$

Express force in terms of gravity

We know that \(F = m\mathrm{~g}\) at the Earth's surface, where \(r=R\). Therefore, the gravitational force at the Earth's surface can be expressed as: $$mg = G\frac{mM}{R^2}$$ We can now rewrite the formula in step 2 in terms of \(g\): $$F = \frac{m\mathrm{~g}(R+h)^2}{R^2}$$

Evaluate the expression and select the correct choice

We can simplify the expression in step 3 as: $$F = m\mathrm{~g}\frac{(R+h)^2}{R^2}$$ We cannot further simplify the expression to match any of the given options, so we will compare the expression with the given options. Comparing the expression to the given options, we find that none of the options match the derived expression. This means that the correct answer to the problem is not among the given choices.

Key concepts

Newton's Law of Universal Gravitation

One of the most influential discoveries in physics was articulated by Sir Isaac Newton in the 17th century, which is known as Newton's law of universal gravitation. This law reveals that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

The mathematical representation of this concept is beautifully simple yet profoundly deep: \begin{align*}F = G\frac{m_1m_2}{r^2}\end{align*}Here, \(F\) represents the gravitational force between two masses, \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses, and \(r\) is the distance between the centers of the two masses. Understanding this law is crucial for students as it helps to explain phenomena ranging from the apple falling from a tree to the orbiting of planets in the solar system.

By analyzing the gravitational interactions between objects, students can apply this fundamental principle to predict the behavior of physical systems. For instance, when a person jumps upward on Earth, their maximum height reached and the force exerted is an application of this law. This law also stresses the universal scope of gravity — it's not just an Earth-bound phenomenon but applies to the entirety of the cosmos.

Gravitational Constant

At the heart of Newton's law of universal gravitation is the gravitational constant, often denoted by \(G\). This constant is a key value that allows us to calculate the gravitational force between two objects. Its value is approximately \(6.674 \times 10^{-11} N(m/kg)^2\), which is surprisingly small, yet its impact on the universe is immense.

The universality of \(G\) means it is constant and does not change regardless of the location or the properties of the objects involved. This small number ensures that objects don't suddenly gravitate towards each other with unmanageable forces – it's a fundamental constant that ensures the stability of the cosmic order.

In practice, when students come across problems involving gravitational force, it is this constant that they'll plug into the equations to solve for the unknowns. For example, when looking at the force exerted by the Earth on a person, \(G\) plays a pivotal role in the calculations. Despite its constancy, the gravitational constant was incredibly difficult to measure accurately, and it wasn't until the late 1700s that Henry Cavendish was able to make a precise measurement, emphasizing the critical yet subtle influence of gravity on our lives.

Earth's Gravitational Field

The Earth's gravitational field is a fascinating concept that describes the influence exerted by the Earth's mass on objects in its proximity. Simply put, the gravitational field can be visualized as an invisible force field that surrounds the Earth, pulling objects towards its center.

When we talk about \(g\), the acceleration due to gravity (approximately \(9.8 m/s^2\) at the Earth's surface), we're referring to the strength of this gravitational field. The gravitational field strength, though mostly consistent near the Earth's surface, does vary slightly depending on altitude or depth and the density of the geological structures beneath. This means that when an individual jumps, and reaches a height \(h\) above the surface of the Earth, the force that the Earth exerts on the person changes with this varying altitude because the distance \(r = R + h\), where \(R\) is the Earth's radius, increases.

Gravitational fields are crucial for understanding how objects will move under the influence of gravity without making contact. For instance, satellite orbits can be calculated precisely because of our understanding of these fields. When solving problems, it's important for students to realize that the strength of the gravitational field affects the force experienced by objects, a principle that is elegantly factored into the mathematical expressions they use in physics.