91Ó°ÊÓ

Mechanical energy (ME)is $-\text{2.0}×{\text{10}}^{\text{9}}\text{J}$.

Use the formula for mechanical energy to find kinetic energy at a particular point.Mechanical energy is the energy possessed by an object due to its motion or due to its position.

The formula is as follows:

$\mathit{M}\mathit{E}\mathbf{}\mathbf{=}\mathbf{}\mathit{K}\mathit{E}\mathbf{}\mathbf{+}\mathbf{}\mathit{U}$

whereMEis mechanical energy, KE is kinetic energy and U is potential energy.

Use the formula of mechanical energy to find kinetic energy as follows:

$\text{ME}=\text{KE}+\text{U}$

From graph,

U at $\text{r}=1.25R_s$ is $-\text{4.0}×{\text{10}}^{\text{9}}\text{J}$

So,

$$\begin{gathered} -2.0×{10}^{9}J=-4×{10}^{9}J+KE \\ KE=2.0×{10}^{9}J \end{gathered}$$

Hence, kinetic energy at radius $\text{r}=1.25R_s$is $\text{2}×{\text{10}}^{\text{9}} \text{J}$.

The turning point is the point where mechanical energy is equal to potential energy.

Now,$U=-4.0×{10}^9\text{J}$ at$\text{r}=1.25{\text{R}}_{\text{s}}$.So, reducing potential energy by a factor of 2, means the r value must be increased by a factor of 2.

Hence,the turning point in terms of${\text{R}}_{\text{s}}$is$\text{r}=1.25{\text{R}}_{\text{s}}$.

Gravitational potential energy is inversely proportional to the distance between the objects. If the distance is reduced, the potential energy increases. In this problem, thisconcept can be used to calculatethe change in gravitational potential energy.