Problem 16 - The Calculus with Analytic Geometry

Question

A meteorite is \(a\) miles from the center of the earth and falls to the surface of the earth. The force of gravity is inversely proportional to the square of the distance of a body from the center of the earth. Find the work done by gravity if the weight of the meteorite is \(w \mathrm{lb}\) at the surface of the earth. Let \(R\) miles be the radius of the earth.

Step-by-Step Solution

Answer

The work done by gravity is \( w R \left( 1 - \frac{R}{a} \right) \).

Step 1: Define the gravitational force

The force of gravity is inversely proportional to the square of the distance from the center of the earth. Therefore, we can express the gravitational force as: \( F = \frac{k}{r^2} \), where \( k \) is a constant and \( r \) is the distance from the center of the earth.

Step 2: Determine the constant \( k \)

At the earth’s surface, the force of gravity is equal to the weight of the meteorite, \( w \) lbs. So, when \( r = R \), we have: \( F = w = \frac{k}{R^2} \). Solving for \( k \), we get: \( k = wR^2 \).

Step 3: Express the force as a function of distance

Now we can write the gravitational force as a function of the distance \( r \): \( F(r) = \frac{wR^2}{r^2} \).

Step 4: Set up the work integral

The work done by gravity as the meteorite falls from the distance \( a \) to the earth’s surface \( R \) is given by the integral of the force over that distance: \( W = \int_{R}^{a} F(r) \, dr = \int_{R}^{a} \frac{wR^2}{r^2} \, dr \).

Step 5: Evaluate the integral

We need to integrate \( F(r) \): \( W = wR^2 \int_{R}^{a} \frac{1}{r^2} \, dr \). The integral of \( r^{-2} \) is \( -r^{-1} \), so: \( W = wR^2 \left[ \frac{-1}{r} \right]_{R}^{a} \). Evaluating this, we get: \( W = wR^2 \left( -\frac{1}{a} + \frac{1}{R} \right) = wR^2 \left( \frac{R - a}{aR} \right) = w R \left( 1 - \frac{R}{a} \right) \).

Key Concepts

gravitational force

Gravitational force is the attractive force that pulls two masses toward each other. The formula for the gravitational force between two objects is given by Newton's Law of Universal Gravitation: \[ F = G \frac{m_1 m_2}{r^2} \]Here,

• F is the gravitational force
• G is the gravitational constant, approximately equal to \(6.674 \times 10^{-11} \text{ N(m/kg)}^2\)
• m_1 and m_2 are the masses of the two objects
• r is the distance between the centers of the two masses

This means that the gravitational force decreases as the distance between the two masses increases. It’s inversely proportional to the square of the distance, which is why if you double the distance, the gravitational force becomes one-fourth.

inverse square law

The inverse square law is a principle that states that a specified physical quantity or intensity is inversely proportional to the square of the distance from the source of that physical quantity. This law not only applies to gravity but also to other phenomena like light, sound, and radiation. The formula for the law in the context of gravity is: \[ F = \frac{k}{r^2} \]Here,

• F is the force of gravity
• k is a constant that includes the masses of the objects and the gravitational constant
• r is the distance from the center of the Earth to the object

As the distance (r) increases, the force decreases, and vice versa. This principle is crucial in understanding how various forces weaken with distance.

definite integrals

Definite integrals allow us to calculate the total accumulation of a quantity, such as area under a curve or work done by a force over a distance. In this exercise, we use a definite integral to find the work done by gravity as a meteorite falls from a height 'a' miles above the Earth to its surface. The integral used is: \[ W = \text{\int}_{R}^{a} \frac{k}{r^2} \text{ dr} \]Here,

• W represents the work done by gravity
• R is the radius of the Earth
• a is the initial distance from the center of the Earth to the meteorite
• r is a variable representing the distance from the center of the Earth

Evaluating this definite integral, we can solve for the total work done.

distance from the center of the earth

The distance from the center of the Earth is a key factor in determining the gravitational force experienced by an object. The Earth is nearly a sphere, so the distance from the center is simply the radius of the Earth (R) when the object is on its surface. As the object moves further away, the distance increases. For instance, when a meteorite at a distance 'a' from the center of the Earth falls to the surface, we have:

• Initial distance: a miles
• Final distance: R miles (where R is the radius of the Earth)

The change in distance affects the gravitational force as per the inverse square law.