Respuesta :
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 [tex]\sqrt{3} v_{e}[/tex]
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 , [tex]v_{i}=2 v_{e}[/tex]. Here the radius is R, so find the escape velocity as follows,
[tex]\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0[/tex]
[tex]\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}[/tex]
[tex]v_{e}^{2}=\frac{2 G M}{R}[/tex]
[tex]v_{e}=\sqrt{\frac{2 G M}{R}}[/tex]
Where, M = Planet’s mass and G = constant.
From given conditions,
Surface potential energy can be expressed as, [tex]U_{i}=-\frac{G M m}{R}[/tex]
R tend to infinity when far away from the planet, so [tex]v_{f}=0[/tex]
Then, kinetic energy at initial would be,
[tex]k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}[/tex]
Similarly, kinetic energy at final would be,
[tex]k_{f}=\frac{1}{2} m v_{f}^{2}[/tex]
Here, [tex]v_{f}=\text { final velocity }[/tex]
Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as
[tex]U_{i}+k_{i}=k_{f}+v_{f}[/tex]
[tex]\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0[/tex]
[tex]\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}[/tex]
'm' and [tex]\frac{1}{2}[/tex] as common on both sides, so gets cancelled, we get as
[tex]4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}[/tex]
We know, [tex]v_{e}=\sqrt{\frac{2 G M}{R}}[/tex], it can be wriiten as [tex]\left(v_{e}\right)^{2}=\frac{2 G M}{R}[/tex], we get
[tex]4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}[/tex]
[tex]v_{f}^{2}=3\left(v_{e}\right)^{2}[/tex]
Taking squares out, we get,
[tex]v_{f}=\sqrt{3} v_{e}[/tex]
