Two objects initially at rest in space separated by a distance of r_0. Find a function for r(t)

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.

Two objects initially at rest in space separated by a distance of r_0. Find a function for r(t).

Explore how to derive the function r(t) for two objects initially at rest in space and separated by a distance r_0. This problem involves key physics concepts like Newton’s law of gravitation, differential equations, and energy conservation. Ideal for university-level physics or advanced math students.

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