Final answer:
The orbital period of a moon that orbits around a 1.0 x 10^26 kg planet with an orbit radius of 2.48 x 10^7 meters is approximately 2.34 x 10^2 seconds.
Step-by-step explanation:
Orbital Period of a Moon
To calculate the orbital period of a moon, we can use Kepler's third law, which states that the square of the period of an orbit is proportional to the cube of the orbit radius. The formula can be written as:
T^2 = (4π^2 / GM) * r^3
Where T is the orbital period, G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2), M is the mass of the planet (1.0 x 10^26 kg), and r is the orbit radius (2.48 x 10^7 meters).
Substituting the given values into the formula, we can solve for T:
T^2 = (4π^2 / (6.67 x 10^-11 Nm^2/kg^2 * 1.0 x 10^26 kg)) * (2.48 x 10^7 meters)^3
Simplifying the equation gives us:
T^2 = 3.6 x 10^3 seconds
Taking the square root of both sides, we find:
T = 2.34 x 10^2 seconds
Therefore, the orbital period of the moon is approximately 2.34 x 10^2 seconds, which corresponds to option b.