Question

Saturn's mass is M= 5.69 x 10²⁶ kg and its radius R=60,300 km. If a moon orbits Saturn at a distance equal to 5 times its planetary radius, what is its period of orbit?

(Hint, use Newton's version of Kepler's 3rd law, and you can neglect the mass of the moon)
Express your answer in days to three significant figures.

Answer 1

The orbital period of a moon orbiting at five times the radius of Saturn can be determined using Newton's revision of Kepler's 3rd law, involving the mass and radius of Saturn as well as the gravitational constant.

Step-by-step explanation:

The question is asking for the orbital period of a moon orbiting Saturn at a distance of five times its planetary radius, using Newton's version of Kepler's 3rd law. Given Saturn's mass (M = 5.69 x 10²⁶ kg) and radius (R = 60,300 km), and the orbital distance being 5R, we can use the formula T² = ⅔³ / (GM), where G is the gravitational constant, to find the period T.

First, calculate the semimajor axis (a) which is the orbital distance:

• a = 5 * R = 5 * 60,300 km

Then, use the orbital equation to find T:

T² = (4π²a³) / (GM)

Finally, convert the period T from seconds to days and round to three significant figures.

To find the period of orbit for a moon orbiting Saturn at a distance equal to 5 times its planetary radius, we can use Newton's version of Kepler's 3rd law. According to the law, the square of the orbital period is proportional to the cube of the semi-major axis. In this case, the semi-major axis is equal to 5 times the radius of Saturn, which is 5 x 60,300 km = 301,500 km. Plugging this value into the equation, we have p^2 = a^3, where p is the period and a is the semi-major axis. Solving for p, we get p = square root of (a^3). Substituting the values, p = square root of (301,500 km)^3. Calculating this, the period of orbit is approximately 33.04 days to three significant figures.