Question

The distances from the Sun at perihelion and aphelion for Pluto are \(4410 \cdot 10^{6} \mathrm{~km}\) and \(7360 \cdot 10^{6} \mathrm{~km}\), respectively. What is the ratio of Pluto's orbital speed around the Sun at perihelion to that at aphelion?

Short answer

Answer: The ratio is approximately 1.67.

Step-by-step solution

Write down the conservation of angular momentum

The conservation of angular momentum states that the initial angular momentum (\(L_i\)) is equal to the final angular momentum (\(L_f\)). So, we have \(L_i = L_f\).

Express the angular momentum at perihelion and aphelion

Since the mass of Pluto (\(m\)) remains constant, the angular momentum at perihelion (\(L_p\)) can be expressed as \(L_p = m \cdot v_p \cdot r_p\), where \(v_p\) is the orbital speed at perihelion and \(r_p\) is the distance at perihelion. Similarly, the angular momentum at aphelion (\(L_a\)) can be expressed as \(L_a = m \cdot v_a \cdot r_a\), where \(v_a\) is the orbital speed at aphelion and \(r_a\) is the distance at aphelion.

Set up the equation for the conservation of angular momentum

According to the conservation of angular momentum, \(L_p = L_a\). So we have \(m \cdot v_p \cdot r_p = m \cdot v_a \cdot r_a\). Since the mass of Pluto (\(m\)) remains constant, we can simplify this equation to \(v_p \cdot r_p = v_a \cdot r_a\).

Solve for the ratio of orbital speeds

We want to find the ratio \(\frac{v_p}{v_a}\). Divide both sides of the equation obtained in step 3 by \(v_a \cdot r_p\): \( \frac{v_p \cdot r_p}{v_a \cdot r_p} = \frac{v_a \cdot r_a}{v_a \cdot r_p}\) Now, simplify the equation to get: \(\frac{v_p}{v_a} = \frac{r_a}{r_p}\).

Substitute the given distances and find the ratio

Now that we have the equation for the ratio of orbital speeds, we can substitute the distances given in the problem. We have \(r_p = 4410 \cdot 10^{6} \mathrm{ ~km}\) and \(r_a = 7360 \cdot 10^{6} \mathrm{ ~km}\). Plug in these values into the equation: \(\frac{v_p}{v_a} = \frac{7360 \cdot 10^{6}}{4410 \cdot 10^{6}}\). Now, simplify the equation to get the ratio: \(\frac{v_p}{v_a} = \frac{7360}{4410} \approx 1.67\) So the ratio of Pluto's orbital speed at perihelion to aphelion is approximately 1.67.