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Past exam of the physics 1 paper of the University of Oxford

2022

{c}

A <circular orbit> has constant $r=r_0>0$. Both its initial radial <velocity> and radial <acceleration> must vanish, giving
$$
\boxed{r(0)=r_0,\quad\dot r(0)=0,\quad
V_{\mathrm{eff}}'(r_0)=V'(r_0)-\frac{J^2}{mr_0^3}=0.}
$$
Equivalently, the initial tangential <speed> and <angular velocity> obey
$$
v_{\theta,0}^2=\frac{r_0f(r_0)}m,\qquad
\dot\theta(0)=\frac{J}{mr_0^2},\qquad
J^2=mr_0^3f(r_0).
$$
The initial <polar angle> is arbitrary, and the <energy> is fixed at $E=V_{\mathrm{eff}}(r_0)$. Either orientation is possible. These conditions give an exact <circular orbit> even if it is unstable; stability requires an additional positive second <derivative> of the <effective potential>.

Solution

Past exam of the physics 1 paper of the University of Oxford / 2022 / 8 / b / Solution

Past exam of the physics 1 paper of the University of Oxford / 2022 / 8 / b / Solution

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Past exam of the physics 1 paper of the University of Oxford / 2022 / 8 / b / Solution

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