Problem 16 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Work against the force of gravity. Suppose a 3 kilogram mass instrument is lifted 20,000 kilometers above the surface of the Earth, the force acting on it is not constant throughout the motion. See Figure 9.14. The acceleration of gravity may be computed for a distance \(x\) above the Earth as Acceleration of gravity at altitude \(x=9.8 \times \frac{R^{2}}{(R+x)^{2}}\) where \(R \doteq 6,370\) kilometers is the radius of the Earth. We partition the interval [0,20,000] into 4 equal subintervals and assume the acceleration of gravity to be constant on each of the subintervals. The work to lift the instrument the first 5000 kilometers is approximately $$ 5.05 \times 3 \times 5000=75814 \quad \text { Newton-meters } $$ a. Approximate the total work to lift the instrument to \(20,000 \mathrm{~km}\). b. Approximate the total work done to lift the \(3 \mathrm{~kg}\) instrument \(40,000 \mathrm{~km}\). (You already know the work required to lift it \(20,000 \mathrm{~km} .)\) c. Approximate the total work done to lift the \(3 \mathrm{~kg}\) instrument \(100,000 \mathrm{~km}\). It is a very interesting question as to whether the instrument can 'escape from the Earth's gravity field' with a finite amount of work. We will return to the question in Section 11.5.1, Escape Velocity.

Step-by-Step Solution

Answer

Total work is summed over subintervals using average gravity, calculated at midpoints for 0-20,000 km. Repeat for higher distances.

Step 1: Understanding the Equation for Gravity

First, we need to understand how to compute the acceleration due to gravity at a given altitude \( x \), which is given by the formula: \[ g(x) = 9.8 \times \frac{R^2}{(R + x)^2} \] where \( R = 6,370 \) km is the Earth's radius. This formula adjusts the gravitational force to reflect changes at different altitudes.

Step 2: Partitioning the Interval

We partition the interval from 0 to 20,000 km into 4 subintervals, each representing 5,000 km. For each subinterval, we'll assume the gravitational force is constant, and compute the work done using average gravitational acceleration within that subinterval.

Step 3: Calculating Gravitational Force Across Subintervals

For each subinterval: - Calculate the gravitational force at the midpoint to estimate the average force.- For example, for the first interval [0, 5,000], calculate \( g(2,500) \) as \[ g(2,500) = 9.8 \times \frac{R^2}{(R + 2,500)^2} \].- Then calculate the work as \( W = g(2,500) \times 3 \times 5,000 \). Repeat for additional subintervals.

Step 4: Calculate Work for Each Interval

Perform similar calculations for the intervals [5,000, 10,000], [10,000, 15,000], and [15,000, 20,000]. Sum up the work amounts for each interval to find the total work.

Step 5: Approximating Total Work to 20,000 km

Add the work values from all four intervals to approximate the total work done to lift the instrument to an altitude of 20,000 km. This can be represented as \( W_{total} = W_1 + W_2 + W_3 + W_4 \).

Step 6: Work for 40,000 km

To compute the work for 40,000 km, we continue the partitioning process with two additional intervals from 20,000 to 40,000 km just like above. Add the previously calculated work for 20,000 km to the new calculations to find this total.

Step 7: Work for 100,000 km

For 100,000 km, perform similar partitioning and calculations up to the desired altitude. Follow the same method: compute gravitational force for midpoints, calculate work for each interval, and sum them. This will give you the total work required to reach 100,000 km.

Key Concepts

Gravitational Force

Gravitational force is a fundamental concept in physics that describes the attraction between two masses. The gravitational force that acts on an object near Earth's surface is usually approximated by a constant, thanks to the nearly uniform gravitational field. However, when an object is moved far above the Earth, this force changes.
The force depends on the mass of the Earth, the mass of the object, and the distance to the center of the Earth. The formula for gravitational force is given by Newton's law as:

• Gravitational Force: \[ F = G \cdot \frac{m_1 \cdot m_2}{r^2} \]
• Where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between the centers of the two masses.

For objects close to Earth's surface, a simplification occurs with the constant average value of gravitational acceleration. However, as the distance from Earth increases, the radius \( r \) changes, altering the gravitational force. The formula becomes essential in calculating how much work is needed to move an object to a substantial height.

Acceleration due to Gravity

Acceleration due to gravity is the rate at which an object will speed up as it falls, due to the gravitational pull of Earth. At the surface of the Earth, this is approximately \( 9.8 \, \text{m/s}^2 \). However, as you move further away from Earth’s surface, this acceleration decreases.
It can be calculated as follows at a height \( x \):

• Acceleration due to gravity: \[ g(x) = 9.8 \times \frac{R^2}{(R + x)^2} \]
• Here, \( R \) is Earth's radius (\( 6,370 \) km) and \( x \) is the altitude above the Earth's surface.

At greater altitudes, the factor \( \frac{R^2}{(R+x)^2} \) becomes smaller, thus reducing the value of \( g(x) \). This variation needs to be considered when calculating work done against gravity for varying altitudes, making accurate computations necessary for non-constant acceleration in space.

Physics of Motion

The physics of motion involves understanding objects' movement under various forces, including gravity. When lifting an object vertically, we work against the gravitational force acting on it, requiring energy expenditure.
To calculate the work done, it’s necessary to understand:

• Work is defined as the force applied over a distance: \[ W = F \cdot d \]
• For lifting an object, \[ W = m \cdot g \cdot d \]

Where \( m \) is the mass, \( g \) is the gravitational acceleration, and \( d \) is the distance over which the object is moved. The work varies with different altitudes due to changing gravitational force. To tackle this, physicists often simplify scenarios by dividing the journey into intervals and assuming forces are constant within each, allowing for manageable calculations.

Calculus Applications

In physics, calculus is a powerful tool to solve complex problems involving continuous change. In the context of motion, it helps with calculating work and forces over changing distances and non-uniform fields.
When dealing with varying gravitational force:

• The work done can be found through integration, by considering the changes in force over small intervals.
• Breaking the entire movement into sections, allows using calculus to sum work done over these tiny distances leading to the expression:
\[ \text{Total Work} = \int_{0}^{d} F(x) \, dx \]

This approach, where the distance \( d \) is broken down into smaller measurable units, allows for the calculation of work in segments with varying gravitational pull, giving a precise total work done. Essentially, calculus allows for control over complexities of changing variables in physics.