Question

Imagine a large hollow sphere with mass \(M\) and outer radius \(R\) located in outer space. The hollow sphere has a thickness \(t\), where \(t \ll R\). What is the gravitational force on an object with mass \(m\) on the outer and inner surfaces of the hollow sphere, respectively? a) zero, zero b) \(m M G / R^{2}\), zero c) zero, \(m M G /(R-t)^{2}\) d) \(m M G / R^{2}, m M G /(R-t)^{2}\) e) zero, \(m M G / R^{2}\)

Short answer

Answer: The gravitational forces on the object are \(m M G / R^{2}\), zero for the outer and inner surface, respectively.

Step-by-step solution

Identify information given and needed

We are given the mass of the hollow sphere (M), its outer radius (R), the thickness of the sphere (t), and the mass of an object (m). We need to find the gravitational force on the object at the outer and inner surfaces of the hollow sphere.

Apply the shell theorem

According to the shell theorem, the gravitational force inside a uniformly distributed hollow sphere is zero. Therefore, the gravitational force on the object when placed on the inner surface of the hollow sphere will be zero.

Apply Newton's law of universal gravitation for the outer surface

For the object placed on the outer surface of the hollow sphere, we can consider all the mass of the sphere to be concentrated at the center. Then, we can apply Newton's law of universal gravitation: \(F = G \frac{mM}{r^2}\), where F is the gravitational force, G is the gravitational constant, m and M are the masses of the object and the sphere, and r is the distance between their centers. Since the object is on the outer surface, r = R. Therefore, \(F = G \frac{mM}{R^2}\).

Combine the results

The gravitational force on the object at the inner surface is zero, and the force at the outer surface is given by \(G \frac{mM}{R^2}\). The answer is, therefore, option b) \(m M G / R^{2}, \) zero.

Key concepts

Shell Theorem

Understanding the behavior of gravitational force around a hollow sphere might seem complex, but the shell theorem makes it a lot clearer. This principle, crucial in gravitational theory, states that a uniform spherical shell of matter exerts no net gravitational force on an object placed anywhere inside the hollow region. This means that if you're positioned within the empty space of a large spherical shell, like our hollow sphere with mass M, you wouldn't feel gravity pulling you in any direction, regardless of your location inside.

What's behind this phenomenon? Imagine the sphere as a collection of rings. For each ring pulling you towards it, there's an opposite ring pulling you the other way. These pulls cancel each other out completely, resulting in zero net gravitational force within the shell. However, this unique property of cancellation doesn't apply outside the sphere. Once an object is on the outer surface or beyond, the shell can be thought of as a point mass at its center, influencing other objects with its gravity.

Newton's Law of Universal Gravitation

The universal law that governs the gravitational attraction between any two bodies is known as Newton's law of universal gravitation. This fundamental physical law states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The formula to express this force is as simple as it is profound:

\( F = G \frac{m M}{r^2} \)

where \( F \) symbolizes the gravitational force, \( G \) is the gravitational constant, \( m \) and \( M \) are the respective masses of the two objects, and \( r \) is the distance between their centers. The beauty of this formula is in its directness and universality—it applies to objects as small as apples and as massive as planets. The law tells us that gravitational force is directly proportional to the masses and inversely proportional to the square of the distance between their centers, which is why gravity gets weaker as objects move further apart.

Gravitational Constant

At this point, you might be wondering: what exactly is the gravitational constant, often represented by the letter G, and why is it so important? In the universal equation of gravitation, the gravitational constant is the glue that holds the formula together, providing the proper scaling to calculate the force of gravity between any two masses. Its value is incredibly small, \( G = 6.674 \times 10^{-11} \) Newton meter squared per kilogram squared (N m²/kg²), illustrating just how feeble the force of gravity is compared to other fundamental forces.

Despite its minute magnitude, this constant is significant. Without it, you wouldn't be able to translate the abstract concept of gravitation into actual force values that can predict the orbits of planets or the trajectories of spacecraft. It's like the conversion rate between mass and gravitational force, ensuring the law of gravitation can be applied practically in scientific calculations across the universe.

Gravitational Force Calculation

To calculate the gravitational force according to Newton's law, we use the formula mentioned earlier. However, since we are dealing with a hollow sphere, we apply this formula in a slightly different way. When calculating the force on the outer surface, we can use the entire radius of the sphere as our \( r \), and the equation simplifies to:

\( F = G \frac{m M}{R^2} \)

This accounts for the gravitational force at the outer surface only. Conversely, when calculating the gravitational force on the inside, the shell theorem comes into play. As explained before, the force inside the hollow sphere is zero due to the cancellation of gravitational influences from every point on the sphere. Therefore, the object inside doesn't experience any gravitational pull from the sphere, simplifying the force calculation to \( F = 0 \), regardless of the object's position inside the hollow sphere.