Question

Let $r$ be the distance of a particle from a fixed point to which it is attracted by an inverse square law force given by $F=k / r^{2} (k=$ constant). Let $m$ be the mass of the particle and $L$ be its angular momentum with respect to the fixed point. Which of the following formulae is correct about the total energy of the system?

• A. $\frac{1}{2} m\left(\frac{d r}{d t}\right)^{2}-\frac{k}{r}+\frac{L}{2 m r^{2}}=$ Constant
• B. $\frac{1}{2} m\left(\frac{d r}{d t}\right)^{2}-\frac{k}{r}=$ Constant
• C. $\frac{1}{2} m\left(\frac{d r}{d t}\right)^{2}+\frac{k}{r}+\frac{L^{2}}{2 m r^{2}}=$ Constant
• D. None

Answer 1

Answer: (A)

$F=\frac{k}{r^{2}}$
$U=-\int \vec{F} \cdot d \vec{r}=-\int \vec{F} d \vec{r} \cos \pi$
$\int F d r=\frac{k}{r^{2}} d r=-\frac{k}{r} ;$
$V_{2}=$ radial velocity $=\frac{d r}{d t}$
Hence, KE due to this velocity $=\frac{1}{2}\left(\frac{d r}{d t}\right)^{2} m$
$V_{1}=$ tangential velocity and $m V_{1} r=L$
$\therefore V_{1}=\frac{L}{m r}$
$KE$ due to tangential velocity $=\frac{1}{2} m v_{1}^{2}$
$\left(\frac{L}{m r}\right)^{2} m V_{1}^{2}=\frac{1}{2} m \frac{L^{2}}{m^{2} r^{2}}$
$=\frac{1}{2} \frac{L^{2}}{m r^{2}}$
Total energy $= KE _{\text {total }}+ PE$
$=\frac{1}{2} m\left(\frac{d r}{d t}\right)^{2}+\frac{1}{2} \frac{L^{2}}{m r^{2}}-\frac{k}{r}$