To derive a function for the distance $r(t)$ between two objects initially at rest and separated by a distance $r_0$ in space, we assume that gravitational attraction is the only force acting between the objects. We'll proceed based on Newton's law of gravitation and his second law of motion.

Problem Assumptions:

• Two masses $m_1$ and $m_2$ are initially at rest in space.
• They are separated by an initial distance $r_0$.
• The only force acting between the two objects is gravity.

Newton's Law of Gravitation:

The gravitational force between two masses is given by:

$F = \frac{G m_1 m_2}{r^2}$

where:

• $G$ is the gravitational constant,
• $m_1$ and $m_2$ are the masses of the two objects,
• $r$ is the distance between them at time $t$.

Newton's Second Law of Motion:

The acceleration $a$ experienced by mass $m_1$ due to the gravitational force is:

$F = m_1 a_1 = m_1 \frac{d^2 r}{dt^2}$

So, equating the force from gravity to the acceleration, we get:

$m_1 \frac{d^2 r}{dt^2} = \frac{G m_1 m_2}{r^2}$

Dividing both sides by $m_1$, we obtain:

$\frac{d^2 r}{dt^2} = \frac{G m_2}{r^2}$

This is a second-order nonlinear differential equation for $r(t)$, which describes the motion of the two objects.

Energy Consideration (Alternative Approach):

We can also use the conservation of energy to simplify this problem. Initially, the objects are at rest, so the total mechanical energy $E$ is purely potential:

$E = U = - \frac{G m_1 m_2}{r_0}$

At any later time, the total mechanical energy $E$ is the sum of the kinetic and potential energy:

$E = \frac{1}{2} m_1 \left( \frac{dr}{dt} \right)^2 - \frac{G m_1 m_2}{r}$

Since total energy is conserved, we can set the initial energy equal to the energy at any time $t$:

$- \frac{G m_1 m_2}{r_0} = \frac{1}{2} m_1 \left( \frac{dr}{dt} \right)^2 - \frac{G m_1 m_2}{r}$

Dividing by $m_1$, we get:

$- \frac{G m_2}{r_0} = \frac{1}{2} \left( \frac{dr}{dt} \right)^2 - \frac{G m_2}{r}$

Rearranging this, we get:

$\frac{1}{2} \left( \frac{dr}{dt} \right)^2 = \frac{G m_2}{r} - \frac{G m_2}{r_0}$

Multiplying by 2:

$\left( \frac{dr}{dt} \right)^2 = 2 G m_2 \left( \frac{1}{r} - \frac{1}{r_0} \right)$

Taking the square root:

$\frac{dr}{dt} = \sqrt{2 G m_2 \left( \frac{1}{r} - \frac{1}{r_0} \right)}$

This is a separable differential equation, and we can integrate it to find $r(t)$. Rearranging, we have:

$\frac{dr}{\sqrt{\frac{1}{r} - \frac{1}{r_0}}} = \sqrt{2 G m_2} \, dt$

This integral will give you the exact form of $r(t)$, though solving it involves elliptical integrals.

In summary:

The distance function $r(t)$ is governed by the equation

$\frac{dr}{dt} = \sqrt{2 G m_2 \left( \frac{1}{r} - \frac{1}{r_0} \right)}$

and requires integration to express $r$ explicitly as a function of $t$.

Would you like me to go further into solving this integral? Also, here are five questions to deepen your understanding:

1. How does this equation change if the masses are not equal?
2. What happens to $r(t)$ as $t \to \infty$?
3. How does the inclusion of additional forces (e.g., a repulsive force) modify the equation?
4. Can this system reach an equilibrium state?
5. How would relativistic effects alter the outcome?

Tip: In complex systems, energy conservation often simplifies the process of solving motion equations.