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If a particle of mass \(m\) moves in a plane under the influence of gravitational force of magnitude \(\dfrac{m}{r^{2}}\) directed towards origin. Then Lagrangian of the system is :

  • A.\(\dfrac{m}{2}(\dot{r}^{2} + r^{2}\dot{\theta}^{2}) - \dfrac{m}{r}\)
  • B.\(\dfrac{m}{2}(\dot{r}^{2} + r^{2}\dot{\theta}^{2}) + \dfrac{m}{r}\)
  • C.\(\dfrac{m}{2}(\dot{r}^{2} + r^{2}\dot{\theta}^{2}) - \dfrac{m}{r^{2}}\)
  • D.\(\dfrac{m}{2}(\dot{r}^{2} + r\dot{\theta}^{2}) + \dfrac{m}{r}\)

Answer: B

The Lagrangian is \(\dfrac{m}{2}(\dot{r}^{2} + r^{2}\dot{\theta}^{2}) + \dfrac{m}{r}\).

In plane polar coordinates the kinetic energy is \(T = \tfrac{m}{2}(\dot{r}^{2} + r^{2}\dot{\theta}^{2})\). An attractive force of magnitude \(m/r^{2}\) towards the origin comes from the potential \(V = -m/r\), since \(-dV/dr = -m/r^{2}\), a pull inwards. So

\[ \begin{aligned} L &= T - V \\ &= \frac{m}{2}(\dot{r}^{2} + r^{2}\dot{\theta}^{2}) + \frac{m}{r} \end{aligned} \]

Option A has the sign of \(V\) wrong, which would make the force repulsive. Option C uses \(m/r^{2}\), the force, in place of the potential, and option D has \(r\dot{\theta}^{2}\) for \(r^{2}\dot{\theta}^{2}\).

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