Question

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice the escape speed ve of the rock from the planet. If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of _____ .

Answer 1

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$

Explanation:

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , $v_{i}=2 v_{e}$. Here the radius is R, so find the escape velocity as follows,

            $\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

            $\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

            $v_{e}^{2}=\frac{2 G M}{R}$

            $v_{e}=\sqrt{\frac{2 G M}{R}}$

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as,  $U_{i}=-\frac{G M m}{R}$

R tend to infinity when far away from the planet, so $v_{f}=0$

Then, kinetic energy at initial would be,

                  $k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Similarly, kinetic energy at final would be,

                $k_{f}=\frac{1}{2} m v_{f}^{2}$

Here, $v_{f}=\text { final velocity }$

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

                 $U_{i}+k_{i}=k_{f}+v_{f}$

                 $\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0$

                  $\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

'm' and $\frac{1}{2}$ as common on both sides, so gets cancelled, we get as

                   $4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

We know, $v_{e}=\sqrt{\frac{2 G M}{R}}$, it can be wriiten as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$, we get

                $4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

                $v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking squares out, we get,

                $v_{f}=\sqrt{3} v_{e}$