Question

A space capsule weighing 5000 pounds is propelled to an altitude of 200 miles above the surface of the earth. How much work is done against the force of gravity? Assume that the earth is a sphere of radius 4000 miles and that the force of gravity is \(f(x)=-k / x^{2}\), where \(x\) is the distance from the center of the earth to the capsule (the inverse-square law). Thus, the lifting force required is \(k / x^{2}\), and this equals 5000 when \(x=4000\).

Short answer

The work done against gravity is approximately 95,238,095 ft-pounds.

Step-by-step solution

Identify Formula for Work Against Gravity

The work done against gravity when moving an object from one point to another is calculated by integrating the force over the path of motion. In this case, the work done is given by \[ W = \int_{x_1}^{x_2} \frac{k}{x^2} dx \], where \(x_1 = 4000\) miles (the radius of the Earth) and \(x_2 = 4200\) miles (the radius of the Earth plus the altitude reached).

Determine Constant of Proportionality, k

We know from the given conditions that the force required at the Earth's surface (\(x = 4000\) miles) is 5000 pounds. Using the formula for the force, \( \frac{k}{4000^2} = 5000 \), solve for \(k\): \[ k = 5000 \times 4000^2 = 8 \times 10^{10} \].

Set Up the Integral

Now, we substitute \(k\) into the work integral: \[ W = \int_{4000}^{4200} \frac{8 \times 10^{10}}{x^2} dx \].

Evaluate the Integral

Evaluate the integral \( \int_{4000}^{4200} \frac{8 \times 10^{10}}{x^2} dx \). The antiderivative of \( \frac{1}{x^2} \) is \( -\frac{1}{x} \), so: \[ W = -8 \times 10^{10} \left[ \frac{1}{x} \right]_{4000}^{4200} = -8 \times 10^{10} \left( \frac{1}{4200} - \frac{1}{4000} \right) \].

Simplify the Expression

Calculate \( -8 \times 10^{10} \left( \frac{1}{4200} - \frac{1}{4000} \right) \): \[ W = -8 \times 10^{10} \left( \frac{4000 - 4200}{16800000} \right) = -8 \times 10^{10} \times \left( \frac{-200}{16800000} \right) = 95238095.24 \text{ ft-pounds} \].

Conclusion

Hence, the work done against the force of gravity is approximately 95,238,095 ft-pounds.

Key concepts

Force of Gravity

In physics, the force of gravity is one of the fundamental forces that govern motion. It is an attractive force, meaning it pulls two masses towards each other. On Earth, this force gives weight to physical objects and is what makes objects fall when dropped. The gravitational force between two objects can be calculated using the equation \[ F = rac{G imes m_1 imes m_2}{r^2} \] where

• \( G \) is the gravitational constant,
• \( m_1 \) and \( m_2 \) are the masses of two objects, and
• \( r \) is the distance between the centers of the two masses.

When considering the force of gravity on Earth, the weight of an object can be calculated using \[ F = m imes g \]where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity. The exercise demonstrated this force when calculating the force acting on the space capsule at different distances from Earth.

Integral Calculus

Integral calculus is a branch of mathematics that deals with the accumulation of quantities and the areas under and between curves. It is a fundamental tool in calculating work, especially when forces vary, as they do with gravity at different altitudes. The integral is used to sum infinitely many infinitesimal quantities.
The integral used in the exercise looks like this:\[ W = \int_{x_1}^{x_2} \frac{k}{x^2} \, dx \]This integral computation helps determine the work done to move the space capsule from one altitude to another against the force of gravity.
An integral can be thought of as adding up slices or sections over a specific interval, \([x_1, x_2]\), to find the total accumulation. Such integrals are crucial in physics to calculate various quantities like work, areas, and volumes.

Inverse-Square Law

The inverse-square law describes a physical quantity that is inversely proportional to the square of the distance from a source. It is crucial in understanding forces like gravity and electrical fields.
This law is expressed with the formula \[ F(x) = \frac{k}{x^2} \]Here, \( F(x) \) represents the force at a distance \( x \) from the source, and \( k \) is a constant. As the distance \( x \) increases, the force decreases rapidly because the force is distributed over a larger area (specifically, the surface area of a sphere), proportional to \( x^2 \).
This concept was applied in the exercise when calculating the force needed to lift the space capsule to a certain altitude. As the capsule moves further from Earth, the gravitational pull diminishes according to this law.

Antiderivative

An antiderivative, in calculus, is a function that reverses the process of differentiation (finding the derivative). It is used to find the original function given its derivative, playing a key role in solving integrals.
In the example from the exercise, the antiderivative of \( \frac{1}{x^2} \) was essential. The antiderivative of this expression is \[ -\frac{1}{x} \]This expression was used in calculating the work done in moving the capsule from Earth's surface to the designated altitude.
Finding an antiderivative depends on recognizing the form of the function you are working with. Once found, it simplifies the process of evaluating definite integrals over specific intervals.