Question

A planet is revolving in an elliptical orbit around the sun. Its closest distance from the sun is $\mathrm{r}$ and the farthest distance is $\mathrm{R}$. If ${\mathrm{k}}_1$ & ${\mathrm{k}}_2$ are the maximum and minimum kinetic energy of the planet then find $\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}$.

• A. $\frac{\mathrm{R}}{\mathrm{r}}$
• B. $\sqrt{\frac{\mathrm{R}}{\mathrm{r}}}$
• C. $\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2}$
• D. $1$

Answer 1

The correct option is C

$\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2}$

Step 1: Given data

It is given that, the closest distance from the sun is $\mathrm{r}$ and the farthest distance is $\mathrm{R}$.

Maximum kinetic energy is given as ${\mathrm{k}}_1$.

Minimum kinetic energy is given as ${\mathrm{k}}_2$.

We have to find the ratio of the maximum and minimum kinetic energy of the planet.

Step 2: Concept used

Conservation of angular momentum is a physical characteristic of a spinning system such that its spin remains constant unless it is acted upon by an external torque.

The angular momentum of the planet near the sun is equal to the angular momentum of the planet farthest from the sun.

Let, the velocity at closest and further distance be $v_1$ and $v_2$.

Now, we can see that net external torque is zero.

So, angular momentum is conserved.

Conserving Angular momentum is, $m{v}_1\mathrm{r}=m{v}_2\mathrm{R}$

So, $v_1\mathrm{r}=v_2\mathrm{R}$

$v_1=\frac{v_2\mathrm{R}}{\mathrm{r}}$

Step 3: Determine the maximum kinetic energy of the planet.

According to Kepler's law, the distance in the velocity of the planet when closer to the Sun is used for the maximum kinetic energy whereas the velocity of the planet when it is far away from the Sun is used for the minimum kinetic energy.

According to Kepler's law, the speed changes based on the distance far away from the sun.

Thus, the maximum kinetic energy of the planet is,

${\mathrm{k}}_1=\frac{1}{2}{\mathrm{mv}}_1^2$

$=\frac{1}{2}\mathrm{m}{\left(\frac{{\mathrm{v}}_2\mathrm{R}}{\mathrm{r}}\right)}^2$

$=\frac{1}{2}\frac{{\mathrm{mv}}_2^2{\mathrm{R}}^2}{{\mathrm{r}}^2}$

Step 4: Determine the minimum kinetic energy of the planet ratio to the maximum and minimum kinetic energy of the planet.

The minimum kinetic energy of the planet is,

${\mathrm{k}}_2=\frac{1}{2}{\mathrm{mv}}_2^2$

Therefore,

$\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{\frac{\mathit{1}}{\mathit{2}}\frac{m{v}_{\mathit{2}}^{\mathit{2}}{R}^{\mathit{2}}}{r^{\mathit{2}}}}{\frac{\mathit{1}}{\mathit{2}}m{v}_{\mathit{2}}^{\mathit{2}}}$

$={\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^2$

Thus, option C is the correct answer.