Question

The gravitational force exerted by the earth on an object having mass \(m\) that is a distance \(r\) from the center of the earth is $$ g(r)=\left\\{\begin{array}{ll} \frac{G M m r}{R^{3}}, & \text { if } r

Short answer

Yes, the function \( g(r) \) is continuous for all \( r \).

Step-by-step solution

Identify if `g` is continuous for `r < R` and `r > R`

Evaluate the given function piecewise. For \( r < R \) and \( r \geq R \), each piece of the function \( g(r) \) is smooth and defined without division by zero or other discontinuities. Thus, \( g(r) \) is continuous individually in the intervals \( r < R \) and \( r \geq R \).

Check continuity at the boundary `r = R`

To determine continuity at \( r = R \), ensure the left-hand limit (\( \lim_{r \to R^{-}} g(r) \)) equals the right-hand limit (\( \lim_{r \to R^{+}} g(r) \)) and \( g(R) \). Calculate the left-hand limit: \[ \lim_{r \to R^{-}} g(r) = \lim_{r \to R^{-}} \frac{G M m r}{R^{3}} = \frac{G M m R}{R^{3}} = \frac{G M m}{R^{2}}. \] Calculate the right-hand limit: \[ \lim_{r \to R^{+}} g(r) = \lim_{r \to R^{+}} \frac{G M m}{r^{2}} = \frac{G M m}{R^{2}}. \]

Verify if the function's value at `r=R` equals the limits

The value of \( g(R) \) is determined by the expression in the case for \( r \geq R \): \[ g(R) = \frac{G M m}{R^{2}}. \]Both limits and \( g(R) \) are equal, continuously transitioning from the mass-point gravitational model to the volume-integrated one as \( r \) passes through \( R \).

Conclude the continuity of `g`

Since \( \lim_{r \to R^{-}} g(r) = \lim_{r \to R^{+}} g(r) = g(R) = \frac{G M m}{R^{2}} \), the function \( g(r) \) is continuous at \( r = R \). Coupled with the individual continuity on the intervals, it confirms \( g(r) \) is continuous everywhere.

Key concepts

Piecewise Functions

Piecewise functions are mathematical functions defined by different expressions based on the input value. Each "piece" of the function applies to a certain interval of the independent variable. In the context of the gravitational force problem, the function \( g(r) \) is expressed in two distinct pieces depending on the distance \( r \) from the Earth's center.
The piece for \( r < R \) is different from the piece for \( r \geq R \). These expressions represent two models of gravitational force:

• For \( r < R \), the function accounts for a model assuming a non-uniform gravitational force inside the Earth.
• For \( r \geq R \), it uses the inverse square law model, which is typical for points outside a spherically symmetric mass like Earth.

Understanding this function requires analyzing each piece separately and ensuring they fit together smoothly, particularly at the boundaries where they meet. This ensures the function captures the physical phenomena accurately and transitions correctly between these zones.

Gravitational Force

Gravitational force is the attractive force exerted by one mass on another due to their masses and the distance between them. In classical physics, this force is defined by Newton's law of universal gravitation. For an object of mass \( m \) near or within the Earth, the force experienced depends on its distance from the Earth's center.
In our example, for \( r < R \), the expression \( \frac{G M m r}{R^{3}} \) models the force, reflecting a different force distribution within Earth's volume. As \( r \) becomes equal to or more significant than \( R \), the expression changes to \( \frac{G M m}{r^{2}} \), reflecting the standard model where the force diminishes with the square of the distance from Earth's surface.
These models are essential in predicting how objects behave under Earth's gravitational influence, whether inside the planet, on the surface, or even in orbit.

Limits

The concept of limits helps us determine the behavior of a function as it nears a particular point. For piecewise functions, it's crucial to evaluate limits, especially at boundary points where one piece transitions into another. In this exercise, limits are used to check the continuity of \( g(r) \) at \( r = R \).
The left-hand limit, \( \lim_{r \to R^{-}} g(r) \), checks the function's behavior as \( r \) approaches \( R \) from values less than \( R \). The right-hand limit, \( \lim_{r \to R^{+}} g(r) \), checks it as \( r \) approaches from values greater than \( R \).
If both limits yield the same result and equal the function's value at \( R \), it indicates the function is continuous at this point. Limits thus confirm the smooth transition between different expressions of \( g(r) \) at \( r = R \).

Continuity at a Point

A function is continuous at a point if, at that point, the following three conditions are met:

• The function is defined at the point.
• The limit of the function exists as we approach the point from both sides.
• The function's value at the point equals this common limit.

In this gravitational force problem, continuity at \( r = R \) was examined in detail. The steps shown in the exercise ensured that \( g(r) \) met all three conditions at \( R \). Both the left and right limits at \( R \) were calculated and found to be \( \frac{G M m}{R^{2}} \), which also matches the function value \( g(R) \).
This finding indicates a seamless continuity, demonstrating that there's no sudden change in the model of gravitational force at Earth's surface boundary, which anchors our understanding of gravitational physics through a continuous model of behavior.