Mechanics

First we assume that the motion is under a central force. Applying logarithmic differentiation:$r=a\left( 1-\cos { \theta } \right) \\ \Rightarrow \frac { 1 }{ r } \frac { dr }{ d\theta } =\frac { a\sin { \theta } }{ a\left( 1-\cos { \theta } \right) } =\frac { 2\sin { \frac { \theta }{ 2 } } \cos { \frac { \theta }{ 2 } } }{ 2\sin ^{ 2 }{ \frac { \theta }{ 2 } } } =\cot { \frac { \theta }{ 2 } } =\cot { \phi } \\ \Rightarrow \phi =\frac { \theta }{ 2 }$ where $\phi$ is the polar-tangential angle in pedal coordinates. Now we have $p=r\sin { \phi } =r\sin { \frac { \theta }{ 2 } } =\frac { r }{ \sqrt { 2 } } \sqrt { 2\sin ^{ 2 }{ \frac { \theta }{ 2 } } } =\frac { r }{ \sqrt { 2 } } \sqrt { 1-\cos { \theta } } =\frac { r }{ \sqrt { 2 } } \sqrt { \frac { r }{ a } } =r\sqrt { \frac { r }{ 2a } } \\ \Rightarrow 2a{ p }^{ 2 }={ r }^{ 3 }$ Differentiating both sides w.r.t. r:$4ap\frac { dp }{ dr } =3{ r }^{ 2 }\\ \Rightarrow \frac { dp }{ dr } =\frac { 3{ r }^{ 2 } }{ 4ap } \\ \Rightarrow \frac { { h }^{ 2 } }{ { p }^{ 3 } } \frac { dp }{ dr } =\frac { { h }^{ 2 }.3{ r }^{ 2 } }{ { p }^{ 3 }.4ap } =3a\frac { { h }^{ 2 } }{ { r}^{ 4 } } =F$ Thus force is inversely proportional to fourth power of distance. Now, at an apse$\frac { dr }{ d\theta } =0\\ \Rightarrow \sin { \theta } =0\\ \Rightarrow \theta =0\quad or\quad \pi$ But $\theta =0\\ \Rightarrow r=0$ which is a cusp of the cardioid. Thus $\theta =\pi \\ \Rightarrow r=2a=p\\ \Rightarrow h=vp=2av\\ \Rightarrow F=3a{ \left( 2av \right) }^{ 2 }{ \left( \frac { 1 }{ 2a } \right) }^{ 2 }=\frac { 3{ v }^{ 2 } }{ 4a } \\ \Rightarrow 4aF=3{ v }^{ 2 }$[Q.E.D.]