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?

Options

A. (a) `1/2m((dr)/(dt))^2-k/r+(L)/(2mr^2)=Constant`
B. (b) `1/2m((dr)/(dt))^2-k/r=Const ant`
C. (c) `1/2m((dr)/(dt))^2+k/r+(L^2)/(2mr^2)=Const ant`
D. (d) None

Answer

A

Explanation

`F=k/r^2`
`U=-intvecF.dvecr=-intvecFdvecrcospi`
`intFdr=k/r^2dr=-k/r`, `V_2=radial velocity=(dr)/(dt)`
Hence, KE due to this velocity `=1/2((dr)/(dt))^2m`
`V_1=` tangential velocity and `mV_1r=L`
`:. V_1=(L)/(mr)`
KE due to tangential velocity `=1/2mv_1^2`
`((L)/(mr))^2mV_1^2=1/2m(L^2)/(m^2r^2)=1/2(L^2)/(mr^2)`
Total energy `=KE_(t otal)+PE`
`=1/2m((dr)/(dt))^2+1/2(L^2)/(mr^2)-k/r`