Enceladus is a moon of Saturn - Scottish Highers Physics - Question 5 - 2018

To find the orbital radius (r) of Enceladus around Saturn, we can use the formula derived from Newton's law of universal gravitation:

$F = G \frac{m_1 m_2}{r^2}$

Where:

• $F$ is the gravitational force ($7.24 \times 10^{19}$ N)
• $G$ is the gravitational constant, approximately $6.674 \times 10^{-11} \text{ m}^3 \text{kg}^{-1} \text{s}^{-2}$
• $m_1$ is the mass of Saturn ($5.68 \times 10^{26}$ kg)
• $m_2$ is the mass of Enceladus ($1.08 \times 10^{20}$ kg)

Rearranging the formula for $r$, we get:

$r = \sqrt{G \frac{m_1 m_2}{F}}$

Substituting the known values:

$r = \sqrt{(6.674 \times 10^{-11}) \frac{(5.68 \times 10^{26})(1.08 \times 10^{20})}{7.24 \times 10^{19}}}$

Calculating this gives:

$r \approx 2.38 \times 10^{10} \text{ m}$

Thus, the correct answer is option A: 2.38 × 10¹⁰ m.