Question 705046

Step 1: Identify the Force

The given potential is $V(r) = -\frac{k}{r}$.

The central force is the negative gradient of this potential:

$\vec{F} = -\nabla V(r) = -\frac{d}{dr} \left(-\frac{k}{r} \right) \hat{r} = -\frac{k}{r^2} \hat{r}$

According to Newton's Second Law, the force is equal to the rate of change of linear momentum ($\vec{p}$):

$\frac{d\vec{p}}{dt} = -\frac{k}{r^2} \hat{r}$

Step 2: Angular Momentum Property

For a central force, the torque is zero ($\vec{\tau} = \vec{r} \times \vec{F} = 0$). Therefore, the angular momentum $\vec{L}$ is conserved (constant in time):

$\frac{d\vec{L}}{dt} = 0$

Step 3: Set up the Constant of Motion Condition

We are given that $\vec{A}$ is a constant of motion, which means its time derivative must be zero:

$\frac{d\vec{A}}{dt} = \frac{d}{dt} (\vec{L} \times \vec{p} + \alpha m k \hat{r}) = 0$

First, let's evaluate the time derivative of $\vec{L} \times \vec{p}$:

$\frac{d}{dt} (\vec{L} \times \vec{p}) = \frac{d\vec{L}}{dt} \times \vec{p} + \vec{L} \times \frac{d\vec{p}}{dt}$

Since $\frac{d\vec{L}}{dt} = 0$, the first term vanishes:

$\frac{d}{dt} (\vec{L} \times \vec{p}) = \vec{L} \times \left( -\frac{k}{r^2} \hat{r} \right) = -\frac{k}{r^2} (\vec{L} \times \hat{r})$

Step 4: Solve the Vector Triple Product

We know that $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \dot{\vec{r}})$. Substituting this into the cross product:

$\vec{L} \times \hat{r} = m(\vec{r} \times \dot{\vec{r}}) \times \hat{r} = -m \left[ \hat{r} \times (\vec{r} \times \dot{\vec{r}}) \right]$

Now, apply the vector triple product identity, $\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B})$:

$= -m \left[ \vec{r}(\hat{r} \cdot \dot{\vec{r}}) - \dot{\vec{r}}(\hat{r} \cdot \vec{r}) \right]$

Using the standard properties $\hat{r} \cdot \vec{r} = r$ and $\hat{r} \cdot \dot{\vec{r}} = \dot{r}$:

$= -m [ \vec{r}\dot{r} - \dot{\vec{r}}r ] = m [ r\dot{\vec{r}} - \vec{r}\dot{r} ]$

Step 5: Re-substitute and Recognize the Derivative

Substitute this result back into the derivative equation from Step 3: $\frac{d}{dt} (\vec{L} \times \vec{p}) = -\frac{k}{r^2} \times m [ r\dot{\vec{r}} - \vec{r}\dot{r} ] = -mk \left[ \frac{r\dot{\vec{r}} - \vec{r}\dot{r}}{r^2} \right]$

The expression inside the bracket is exactly the quotient rule applied to the time derivative of the unit vector $\hat{r}$ (where $\hat{r} = \frac{\vec{r}}{r}$):

$\frac{d}{dt} \left( \frac{\vec{r}}{r} \right) = \frac{r\dot{\vec{r}} - \vec{r}\dot{r}}{r^2} = \frac{d\hat{r}}{dt}$

Thus, the equation simplifies to:

$\frac{d}{dt} (\vec{L} \times \vec{p}) = -mk \frac{d\hat{r}}{dt}$ Rearranging the terms:

$\frac{d}{dt} (\vec{L} \times \vec{p} + mk \hat{r}) = 0$

Final Step: Compare to Find $\alpha$ Our derived constant vector is: $\vec{L} \times \vec{p} + mk \hat{r}$

The vector given in the problem is: $\vec{A} = \vec{L} \times \vec{p} + \alpha m k \hat{r}$

By directly comparing the two expressions, we get:

$\alpha m k \hat{r} = m k \hat{r}$

$\alpha = 1$