Problem 66 - Calculus with Concepts in Calculus

Question

According to Newton's Law of Gravitation, if two bodies are a distance \(r\) apart, then the gravitational force \(F(r)\) exerted by one body on the other is given by $$ F(r)=\frac{k}{r^{2}} \quad \text { for } r>0 $$ where \(k\) is a positive constant. Suppose that as a function of time \(t\), the distance between the two bodies is given by $$ r(t)=4000\left(\frac{1+t}{1+t^{2}}\right) \text { for } t \geq 0 $$ Find a formula for the force in terms of time.

Step-by-Step Solution

Answer

\( F(t) = \frac{k}{16000000} \cdot \frac{(1 + t^2)^2}{(1+t)^2} \)

Step 1: Substitute the Expression for Distance

Given the formula for gravitational force \[ F(r) = \frac{k}{r^2} \] and the expression for distance as a function of time \[ r(t) = 4000\left(\frac{1+t}{1+t^2}\right), \] we substitute \( r(t) \) into the force equation to express the force in terms of time \( t \).

Step 2: Calculate [r(t)]^2

First, we need to calculate \( [r(t)]^2 \) to substitute into the force equation: \[ [r(t)]^2 = \left(4000\left(\frac{1+t}{1+t^2}\right)\right)^2 = 4000^2 \cdot \left(\frac{(1+t)^2}{(1+t^2)^2}\right). \] Simplifying this expression: \[ [r(t)]^2 = 16000000 \cdot \frac{(1+t)^2}{(1+t^2)^2}. \]

Step 3: Substitute [r(t)]^2 into F(r)

Now substitute \( [r(t)]^2 \) into the force equation to express force in terms of time: \[ F(t) = \frac{k}{[r(t)]^2} = \frac{k}{16000000} \cdot \frac{(1+t^2)^2}{(1+t)^2}. \] This gives the expression for the force \( F(t) \) as a function of time.

Step 4: Final Expression for Force in Terms of Time

After simplifying and rearranging we arrive at the final expression: \[ F(t) = \frac{k}{16000000} \cdot \frac{(1 + t^2)^2}{(1+t)^2}, \] which describes the gravitational force between the two bodies as a function of time \( t \).

Key Concepts

Gravitational Force

Gravitational force is a fundamental interaction that attracts two bodies toward each other. According to Newton's Law of Gravitation, the force (\( F(r) \)) between two bodies is given by the equation:

• \( F(r) = \frac{k}{r^2} \)

where \( r \) is the distance between the bodies, and \( k \) is a constant that depends on the masses of the bodies and the gravitational constant. This equation highlights two key elements:

• The force is inversely proportional to the square of the distance \( r \). This means that the gravitational force decreases rapidly as the distance between two objects increases.
• \( k \) includes values that make the units and calculations consistent, providing a measure of how strongly two objects pull towards each other based on their mass and proximity.

Gravitational force is crucial in understanding the movement of celestial bodies and satellites, and it also explains many everyday phenomena we observe on Earth.

Distance as Function of Time

Distance as a function of time describes how the separation between two objects changes over time. In our problem, the distance \( r(t) \) is given by the function:

• \( r(t) = 4000\left(\frac{1+t}{1+t^2}\right) \)

This equation indicates how the distance changes as time \( t \) evolves from zero onwards. Breaking it down:

• At \( t = 0 \), the distance is \( 4000 \) units.
• As \( t \) increases, the formula suggests that the distance changes dynamically, influenced by the ratio of \( 1+t \) to \( 1+t^2 \).

The function smoothly adjusts the distance based on the growth of \( t \), illustrating how gravitational interactions can transform over time due to motion or other factors affecting \( t \). Observing \( r(t) \) lets us predict how forces like gravity modify with temporal shifts.

Substitute into Equation

The process of substituting into equations is fundamental in solving problems involving dynamic systems. Given the context:

• A gravitational force expression \( F(r) = \frac{k}{r^2} \).
• A distance-time relation function \( r(t) = 4000\left(\frac{1+t}{1+t^2}\right) \).

You combine these by substituting \( r(t) \) into the gravitational force equation to find the force as a function of time. Begin by calculating \([r(t)]^2\) as follows:

• \( [r(t)]^2 = 16000000 \cdot \frac{(1+t)^2}{(1+t^2)^2} \)

This squared form of \( r(t) \) is critical because the gravitational formula requires the inverse of \( r^2 \). By substituting \([r(t)]^2\) back, you express the force in terms of time:

• \( F(t) = \frac{k}{16000000} \cdot \frac{(1 + t^2)^2}{(1+t)^2} \)

This process demonstrates the powerful technique of substitution, allowing complex systems to be understood in simpler, time-dependent terms. As time \( t \) changes, the effect on the gravitational force can now be directly analyzed.