Generalized projectile motion [closed]

Question

Toss an object from the surface of earth at speed $V_1=\sqrt{MG/R}$. Here $R$ is the radius of the earth and $M$ is the mass of the earth and $G$ is universal gravitational constant. How long will it take for it to fall back?

My approach is to calculate the maximum height of the toss. Let the maximum height be $h$ above the ground and $m$ be the mass of the object. Using the relationship between kinetic energy and gravitational energy we get $$\Delta E_k=-\Delta E_p\Rightarrow\frac{GMm}R-\frac{GMm}{R+h}=\frac12m\sqrt{MG/R}^2$$ . Thus $h=R$ but how to get the time? I suppose the $\text{Theorem of Momentum}$ is the most suitable only the gravitational force changes with the object's position.

Answer 1

Extending your method slightly, using energy conservation $$ \frac 1 2 m\dot{y}^2 - \frac 1 2 mv_0^2 = \frac{GMm}{R+y} - \frac{GMm}{R} $$ Here, $y$ is the height from the earth's surface, $v_0$ is the initial speed, and $\dot{y} \equiv \frac{dy}{dt}$ is the speed of the particle. Substituting, $$ \dot{y}^2 = \frac{2GM}{R + y} - \frac{GM}{R} \implies \dot{y} = \pm \sqrt{\frac{2GM}{R + y} - \frac{GM}{R}} $$ We want $$ T = 2\int_0^R \frac{1}{|\dot{y}|} dy $$ Complicated integrals aren't my speciality – I stuck it into WolframAlpha and it gave me $$ T = (\pi+2)\sqrt{\frac{R^3}{GM}} $$