Question

Figure 6.77 is a graph of the energy of a system of a planet interacting with a star. The gravitational potential energy $\mathit{U}\mathit{g}\mathbf{}$is shown as the thick curve, and plotted along the vertical axis are various values of $\mathit{K}\mathbf{+}\mathit{U}\mathit{g}\mathbf{.}$

Suppose that $\mathit{K}\mathbf{+}\mathit{U}\mathit{g}$of the system is A. Which of the following statements are true? (a) The potential energy of the system decreases as the planet moves from ${\mathit{r}}_{\mathbf{1}}$to${\mathit{r}}_{\mathbf{2}}$. (b) When the separation between the two bodies is ${\mathit{r}}_{\mathbf{2}}$, the kinetic energy of the system is (A − B). (c) The system is a bound system; the planet can never escape. (d) The planet will escape. (e) When the separation between the two bodies is${\mathit{r}}_{\mathbf{2}}$, the kinetic energy of the system is (B − C). (f) The kinetic energy of the system is greater when the distance between the star and planet is ${\mathit{r}}_{\mathbf{1}}$than when the distance between the two bodies is${\mathit{r}}_{\mathbf{2}}$.

Suppose instead that $\mathit{K}\mathbf{+}\mathit{U}\mathit{g}$of the system is B. Which of the following statements are true? (a) When the separation between the planet and star is${\mathit{r}}_{\mathbf{2}}$, the kinetic energy of the system is zero. (b) The planet and star cannot get farther apart than${\mathit{r}}_{\mathbf{2}}$. (c) This is not a bound system; the planet can escape. (d) When the separation between the planet and star is${\mathit{r}}_{\mathbf{2}}$, the potential energy of the system is zero.

Short answer

For Part 1, statements B, C and F are correct.

For Part 2, it is inferred that option A and B is correct.

Step-by-step solution

Identification of given data

The given data can be listed below,

The gravitational energy if the system is, $U_g$.

The kinetic energy of the system when separation between 2 bodies is $r_2$is,$K.E=\left(A-B\right)$

When the separation between the two bodies is $r_2$, the kinetic energy of the system is $K.E=\left(B-C\right)$.

Concept/Significance of kinetic and potential energy.

Kinetic work happens only due to the potential energy stored in some component. Potential energy is not just about rising something up; it is also about the object's configuration.

Determination of the true statement for the kinetic energy of the system is (A − B).

The energy at point A is given by,

$E=U_g+k$

Here, $U_g$is the potential energy and k is the kinetic energy of the system

According to option A the energy of the planet is decreasing but from equation it is observed that the energy is increasing. So, it is incorrect.

The potential energy at B is given by,

$P.E=U_g$

From option B, $\left(A-B\right)$is kinetic energy at $r_2$is shown below,

$$\begin{gathered} \left(A-B\right)=K+U_g-U_g \\ =K \end{gathered}$$

Thus, option B is correct.

For option C, a state is said to be bound state when the total energy of the system is negative.

Hence, option C is correct.

The total energy of the system being negative, the planet cannot escape because negative energy automatically refers to stability of the system. So the planet will stay and this proves that option D is correct as it is inclusive effect of option C.

The energy at point B is given by,

$B=U_g$

Here, $Ug$is the potential energy of the system.

The potential energy at C is given by,

$C=U_g+K$

Where K is the kinetic energy at C.

From option E, (B-C) is kinetic energy at $r2$is shown below,

$$\begin{gathered} \left(B-C\right)=+U_g-U_g-K \\ =-K \end{gathered}$$

Thus, option E is incorrect.

From the data given, it is concluded that at $r_1$, there is less potential energy and so there is more kinetic energy.

Option F says the same thing and hence it is correct.

Hence, for Part 1, statements B, C and F are correct.

Determination of the true statement for the kinetic energy of the system is B

At point B, where the system is essentially at $r_2$, the total energy of the system is purely potential energy. This means kinetic energy is zero.

$B=U_g$

And,$K=0$

Hence, Option A is correct.

Because Kinetic energy is zero at $r_2$, the planet will be bound at $r_2$, which is the extreme allowed point of its trajectory. It cannot travel further and thus option B is also correct.

Both A and B is directly contradicting the statements C and D. And as it is proved that A and B are correct.

Thus, it can be inferred that option C and D is Incorrect and A, B is correct.