Question

Consider a uniform distribution of total mass \(m^{\prime}\) over a spherical shell of radius The potential energy \(\phi\) of a mass \(m\) in the gravitational field of the spherical shell is. $$\phi=\left\\{\begin{array}{ll}\text { const. } & \text { if } m \text { is inside the spherical shell, } \\ -\frac{C m^{\prime}}{r} & \text { if } m \text { is outside the spherical shell, where } r \text { is the distance } \\\ & \text { from the center of the sphere to } m, \text { and } C \text { is a constant. }\end{array}\right.$$ Assuming that the earth is a spherical ball of radius \(R\) and constant density, find the potential and the force on a mass \(m\) outside and inside the earth. Evaluate the constants in terms of the acceleration of gravity \(g,\) to get \(\mathbf{F}=-\frac{m g R^{2}}{r^{2}} \mathbf{e}_{r}, \quad\) and \(\quad \phi=-\frac{m g R^{2}}{r}\) \(m\) outside the earth; \(\mathbf{F}=-\frac{m g r}{R} \mathbf{e}_{r}, \quad\) and \(\quad \phi=\frac{m g}{2 R}\left(r^{2}-3 R^{2}\right)\) \(m\) inside the earth.

Short answer

Outside the earth: \( \phi = -\frac{mg R^2}{r} \), \( \mathbf{F} = -\frac{mg R^2}{r^2} \mathbf{e}_r \). Inside the earth: \( \phi = \frac{m g}{2 R} ( r^2 - 3 R^2 ) \), \( \mathbf{F} = - \frac{m g r}{R} \mathbf{e}_r \).

Step-by-step solution

Identify the Conditions

Given a spherical shell with a uniform mass distribution, the potential energy \( \tilde{\rho} \) differs depending on whether the mass \( m \) is inside or outside the shell.

Potential Energy Inside the Earth

If the mass \( m \) is inside the spherical shell, the potential energy \( \phi \) is constant according to the given conditions.

Potential Energy Outside the Earth

If the mass \( m \) is outside the spherical shell, the potential energy \( \phi \) is given by: \[ \tilde{\rho} = -\frac{C m^{\prime}}{r} \]

Evaluate Constants with Gravitational Parameters

Consider the second part of the problem which specifies that the earth is a spherical ball of radius \( R \) and constant density. Use the acceleration due to gravity \( g \) to find the constants in the formula for potential energy and force.

Compute Potential Outside the Earth

Using Newton's law of gravitation for a mass \( m \) outside a spherical mass distribution, the gravitational potential energy \( \phi \) is given by \[ \phi = -\frac{G M}{r} \] where \( M \) is the mass of the spherical earth, \( G \) is the gravitational constant, and \( r \) is the distance from the center of the earth. The relationship between \( g \) and \( G \) for the earth is \( g = \frac{GM}{R^2} \). Substituting this in, we get: \[ \phi = -\frac{mg R^2}{r} \]

Force Outside the Earth

The gravitational force on the mass \( m \) outside the earth is the negative gradient of the potential energy: \[ \mathbf{F} = - abla \left( -\frac{mg R^2}{r} \right) = -\frac{mg R^2}{r^2} \mathbf{e}_r \]

Potential Inside the Earth

Now, consider the mass \( m \) inside the earth. Using similar methods, derive: \[ \phi = \frac{m g}{2 R} \left( r^2 - 3 R^2 \right) \]

Force Inside the Earth

The gravitational force on the mass \( m \) inside the spherical earth is then obtained by differentiating the potential: \[ \mathbf{F} = - abla \left( \frac{m g}{2 R} \left( r^2 - 3 R^2 \right) \right) = - \frac{m g r}{R} \mathbf{e}_r \]

Key concepts

Potential Energy

Potential energy is a measure of the energy an object has due to its position in a gravitational field. For a uniform spherical shell, the potential energy of a mass \(m\) placed at a distance \(r\) from the center varies:

• Inside the shell: The potential energy is a constant.
• Outside the shell: The potential energy \(\theta\) is given by \(\theta = -\frac{C m'}{r}\), where \(C\) is a constant and \(m'\) is the mass of the shell.

Understanding potential energy helps in determining how much work would be required to move the mass \(m\) in the gravitational field of the shell. Only the positions outside of the shell experience a varying potential energy, inversely proportional to \(r\).

Gravitational Force

When dealing with gravitational forces, we examine how a mass \(m\) experiences a pull towards another mass. For masses distributed uniformly in a spherical shell:

• The force outside the earth (a spherical mass distribution) is given by: \textbf{\(\boldsymbol{F}=-\frac{mgR^2}{r^2}\)}. Here, \(R\) is the radius of the earth and \(g\) is the acceleration due to gravity.
• Inside the earth, the force is found to be \(\boldsymbol{F} = -\frac{mgr}{R}\), showing that the force increases linearly with the distance from the center.

These expressions explain why gravitational force varies depending on whether we are inside or outside a mass distribution, crucial for understanding gravitational interactions.

Uniform Mass Distribution

A uniform mass distribution means that the mass is spread equally throughout the volume of an object. For a spherical shell or Earth with uniform mass distribution, we consider:

• Mass is evenly spread, leading to simpler calculations of gravitational potential and force.
• Results in symmetrical forces and potentials centered around the center of the object, simplifying the math.

This concept means we treat the Earth or a spherical shell as if all the mass were concentrated at the center when calculating forces and potential energy, an important simplification provided by Newton's Shell Theorem.

Spherical Shell

A spherical shell is a three-dimensional object where mass is distributed over a thin layer in the shape of a sphere. For such a shell, properties include:

• Potential energy within: Constant, due to the uniform distribution of mass.
• Potential energy outside: Varies inversely with distance from the center, \(-\frac{Cm'}{r}\).

This distribution implies that inside, no matter the position, the potential remains the same, aiding in understanding phenomena like hollow planetary bodies or gravitational fields created by ring-like structures.

Acceleration Due to Gravity

The acceleration due to gravity (\(g\)) is a constant measure of the gravitational force exerted by Earth per unit mass. For Earth:

• \(g = 9.8 \text{m/s}^2\) at the surface.
• Relates to the gravitational constant (\(G\)) and Earth's mass (\(M\)) by \(g = \frac{GM}{R^2}\).

This value helps in deriving other gravitational attributes like potential energy and force both inside and outside Earth, providing a standard measure for comparing gravitational effects in varied contexts.