Question

A person stands on the surface of the Earth. 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) \(m g\) b) \(M g\). c) \(M^{2} g / m\). d) \(m^{2} g / M\). e) zero.

Short answer

In this exercise, you have to find the force exerted by a person on the Earth when they are at their maximum height during a jump. After deriving the gravitational force equation and substituting the appropriate values for radius (R) and height (h), the force exerted by the person at height h is found to be: \(F = m g \frac{1}{(1 + \frac{h}{R})^2}\). Comparing this expression with the given options, it can be concluded that the force at the highest point of the jump still has the same magnitude as the person's weight on the Earth's surface, which is a) \(mg\).

Step-by-step solution

(Step 1: Write the equation for gravitational force)

(We will start by writing Newton's law of universal gravitation formula which is \(F = G \frac{M m}{r^2}\), where \(F\) is the force of gravitation, \(G\) is the gravitational constant, \(M\) is the mass of the Earth, \(m\) is the mass of the person, and \(r\) is the distance between the centers of mass of the Earth and the person.)

(Step 2: Substitute \(r\) and \(h\) appropriately)

(Now we have to take into account the height reached by the person. When the person is at the maximum height \(h\), the distance between the centers of mass of the Earth and the person becomes \((R+h)\), where \(R\) is the Earth's radius. So, we now have the equation \(F = G\frac{Mm}{(R+h)^2}\). )

(Step 3: Relate gravitational force with weight of the person)

(The weight of a person on the Earth's surface, \(m g\), is the force acting on them due to gravity. We can relate this with the gravitational force between the Earth and the person. At the surface of the Earth, the distance between their centers of mass is simply \(R\). So at the surface, the gravitational force becomes \(m g = G \frac{Mm}{R^2}\). From this equation, we can see that \(g = G\frac{M}{R^2}\).)

(Step 4: Substitute value of 'g' in the equation)

(Now, using the derived relation between \(g\) and \(M, R, G\) i.e., \(g = G\frac{M}{R^2}\), we can substitute for \(g\) in the equation we found in Step 2, to get the force exerted by the person at height \(h\) as: \(F = m g \frac{1}{(1 + \frac{h}{R})^2}\).)

(Step 5: Compare obtained expression with given options to find the correct answer)

(Comparing the obtained expression for the force exerted by the person on the Earth at height \(h\) with the given options in the exercise, we can see that it does not directly match any of the options. However, in the equation for \(F\), we see that the force cannot be zero. Therefore, the only option that matches the obtained force expression is \(a) mg\), which means that the magnitude of the force at the highest point of the jump is still the same as the person's weight on the Earth's surface.) The answer is a) \(mg\).

Key concepts

Newton's law of universal gravitation

Understanding the gravitational force between two masses is essential when studying physics, and this is where Sir Isaac Newton's law of universal gravitation comes into play. This fundamental law states that every pair of particles in the universe attracts each other 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 formula for calculating this force is given by \( F = G \frac{Mm}{r^2} \), where \(F\) is the gravitational force, \(G\) is the gravitational constant, \(M\) is the mass of one object (e.g., Earth), \(m\) is the mass of the other object (e.g., a person), and \(r\) is the distance between the centers of mass of the two objects.

This law can be applied to objects ranging from apples falling from trees to planets orbiting the sun. When we consider a person jumping on Earth in our exercise, even at the height \(h\) above the Earth's surface, according to Newton's law, both the Earth and the person exert a gravitational force on each other, ensuring that the person will eventually come back down. The beauty of this law is that it is universal, meaning it applies to all objects with mass, regardless of where they are in the universe.

Gravitational acceleration

Gravitational acceleration, commonly denoted as \(g\), is a measure of the acceleration due to gravity that a freely falling object experiences when it is close to the surface of a massive body, such as a planet or a moon. Near the surface of the Earth, this acceleration has an average value of approximately \(9.81 \text{ m/s}^2\), but it can vary slightly depending on altitude and latitude.

The concept of gravitational acceleration is crucial in understanding motion in a gravitational field. For example, it is what gives us weight and determines the speed at which objects fall to the ground. In our exercise, we used the relationship between the gravitational force and weight to find that \(g = G\frac{M}{R^2}\), where \(R\) is the Earth's radius. This expression demonstrates that the gravitational acceleration at the surface of the Earth is indeed a result of the gravitational force exerted by the Earth's mass. As one moves away from the Earth's surface, this acceleration decreases, which affects the force exerted by an object at a height \(h\).

Gravitational acceleration is a critical factor in the calculation of gravitational force at different altitudes. It is the reason why astronauts experience weightlessness in space; as they move farther from Earth, \(g\) diminishes significantly, reducing the gravitational force and, subsequently, the sensation of weight.

Gravitational constant

The gravitational constant, denoted as \(G\), is a key component in the universal law of gravitation. It is the proportionality factor that helps to quantify the strength of the gravitational force between two masses. The value of \(G\) has been determined experimentally to be approximately \(6.674 \times 10^{-11} \text{ N}·\text{m}^2/\text{kg}^2\).

The constancy of \(G\) is essential because it ensures that the gravitational force can be calculated consistently anywhere in the universe. It is thanks to this constant that we can predict the gravitational interaction between celestial bodies or objects on Earth, as shown in our exercise.

Although the gravitational constant is incredibly small, the mere existence of large masses like planets or stars magnifies its effect, leading to significant gravitational forces. Furthermore, as \(G\) is a universally constant value, its inclusion in calculations provides us with a critical understanding of the gravitational interactions not only on our planet but also throughout the cosmos. The accurate measurement of \(G\) is so vital to physics that ongoing research aims to refine its value for even more precise scientific modeling.