A particle is moving in an elliptical orbit as shown in the figure. If $p$, $L$, and $r$ denote the linear momentum, angular momentum, and position vector of the particle (from focus $O$) respectively at a point $A$, then the direction of $\alpha = p \times L$ is along
We are given that a particle is moving in an elliptical orbit, and we need to determine the direction of \( \alpha = p \times L \), where \( p \) is the linear momentum, \( L \) is the angular momentum, and \( r \) is the position vector of the particle from the focus \( O \) at a point \( A \).
Step 1: Understanding the quantities - The linear momentum \( p \) is given by: \[ p = m \cdot v \] where \( m \) is the mass of the particle and \( v \) is its velocity. - The angular momentum \( L \) is given by: \[ L = r \times p \] where \( r \) is the position vector of the particle from the focus \( O \), and \( p \) is the linear momentum vector.
Step 2: Direction of \( p \times L \) We are asked to determine the direction of \( \alpha = p \times L \). The cross product of two vectors \( p \) and \( L \) gives a vector that is perpendicular to the plane formed by \( p \) and \( L \). The direction of this vector follows the right-hand rule.
Step 3: Applying the right-hand rule - The velocity vector \( v \) is tangential to the elliptical orbit, and the position vector \( r \) points from the focus \( O \) to the position of the particle at point \( A \). - The angular momentum \( L \) is directed along the axis perpendicular to the plane of the orbit. - The cross product \( p \times L \) will be directed along the axis that is perpendicular to both the velocity and angular momentum vectors.
Step 4: Conclusion From the figure and the application of the right-hand rule, we find that the direction of \( \alpha = p \times L \) is along the **+ve x axis**.
