Comet and the Earth

one small typo r0 and v0 are the things when the comet is closest to the sun

Can you go ahead and finish it off?

arre bhaaii koi toh karooooooo......................

actually the basic equation is wat rohan has given
since its a parobolic trajectory we have binding energy 0
so we have

\frac{GM}{r}=\frac{v^2}{2}
everwhere !

r₁is earth's orbitol radius , we have its realtion with the time period T
now at any instant r , let the velocity be v
we have
v_{radial}=\sqrt{v^2-v_t^2}
also we know that angular momentum is conserved
so
v(t)=v(0)r(0)/r
putting that in eauation
also jsut now we showed relation b/w v and r
put that i9n equation too
also
v(r)=dr/dt
hence we get this differential equation !
\int_{r_1}^{r_0}{\frac{dr}{\sqrt{\frac{2GM}{r}-\frac{v_0^2r_0^2}{r^2}}}}=\int_{0}^{\frac{t}{2}}{dt}

r_1=\frac{TR\sqrt{g}}{2\pi}
we also have
\frac{GM}{r_0}=\frac{v_0^2}{2} \\ \text{hence we can obtain T purely}\\\text{ as a function of r(0) and then }\\\text{diffrentiate w.r.t }r_0 \text{to get the answer }

More or less this problem has been solved. However, just for the sake of completeness, I shall solve it.

The total energy of the comet + sun system is zero. So we get, at any r,
\dfrac{1}{2}m(v_r^2+v_t^2) -\dfrac{GM_\odot m}{r}=0 ------ (1)
where M_\odot is the mass of the sun, v_r is the radial velocity and v_t is the tangential velocity of the comet.

This is a two dimensional problem. However, we could reduce it to one dimensional problem by introducing the angular momentum which is a conserved quantity in any central force problem.
Since, L=mrv_t, equation (1) transforms to
\dfrac{1}{2}mv_r^2 + \dfrac{L^2}{2mr^2}-\dfrac{GM_\odot m}{r}=0
Further, if we denote the distance of closest approach as p, then using the fact that at this point v_r=0, we get from (1) that
\dfrac{L^2}{2mp^2}=\dfrac{GM_\odot m}{p}
i.e. \dfrac{L^2}{2m}=GM_\odot mp
So after some manipulations, equation (1) ultimately becomes,
v_r^2=2GM_\odot \left(\dfrac{r-p}{r^2}\right)
If we denote the total time as \tau, then in time \tau/2, the comet moves from r=p to r=R, the orbital radius of the Earth. Using the fact that v_r = dr/dt, we get
\sqrt{2GM_\odot}\int_0^{\tau/2}\mathrm dt=\int_p^R \dfrac{r}{\sqrt{r-p}}\,\mathrm dr =\dfrac{2}{3}R^{3/2}\left(1+\dfrac{2p}{R}\right)\sqrt{1-\dfrac{p}{R}}
i.e.
\tau =\dfrac{4}{3\sqrt{2}}\dfrac{1}{2\pi}\sqrt{\dfrac{4\pi^2}{GM_\odot}R^3}\left(1+\dfrac{2p}{R}\right)\sqrt{1-\dfrac{p}{R}}
Expressed in terms of Earth's time period of revolution T = 1 yr, we get
\tau =\dfrac{2}{3\pi\sqrt{2}}\cdot T \cdot \left(1+\dfrac{2p}{R}\right)\sqrt{1-\dfrac{p}{R}} -------- (2)

So basically we need to maximize the quantity
\left(1+\dfrac{2p}{R}\right)\sqrt{1-\dfrac{p}{R}}
Calling p/R = x, we need to essentially maximize
f(x)=(1+2x)\sqrt{1-x}\qquad \forall \,x\in [0,1].
Now according to AM-GM inequality
4f^2(x)=(1+2x)^2(4-4x)\le \left(\dfrac{1+2x+1+2x+4-4x}{3}\right)^3=8
So f(x)\le \sqrt{2},
the equality being attained when x=1/2 i.e. when p=R/2
And hence from (2), we obtain that
\tau \le \dfrac{2}{3\pi}T
So the maximum time that comet could spend withing the Earth's orbit as shown is
\boxed{\tau_\mathrm{max}= \dfrac{2}{3\pi}T\approx 77.5 \ \text{days}}