Tethys, one of Saturn's moons, travels in a circular orbit at a speed of 1.1x104 m/s. The
mass of Saturn is 5.67x1026 kg. Calculate
a) the orbital radius in kilometres.( i know how to do this part, the answer is 3.1x10^8 m)
b) the orbital period in Earth days.
To calculate the orbital period of Tethys, we need to use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period is directly proportional to the cube of the semi-major axis of the orbit.
Let's go step by step:
Step 1: Calculate the orbital radius in kilometers.
You mentioned that you know how to do this part, and the answer is 3.1x10^8 m. To convert this value to kilometers, we divide by 1000:
Orbital radius = 3.1x10^8 m / 1000 = 3.1x10^5 km
Step 2: Calculate the gravitational constant.
The gravitational constant is represented by the letter G and has a value of 6.67x10^-11 N(m/kg)^2.
Step 3: Calculate the mass of Saturn in kilograms.
Given that the mass of Saturn is 5.67x10^26 kg, we can use this value in our calculations.
Step 4: Use Kepler's Third Law to calculate the orbital period.
The equation for Kepler's third law is:
T^2 = ((4π^2) / (G * M)) * r^3
Where:
T is the orbital period in seconds.
G is the gravitational constant.
M is the mass of the planet or central object.
r is the orbital radius in meters.
Now, let's plug in the values we have into this equation:
T^2 = ((4π^2) / (G * M)) * r^3
T^2 = ((4 * (3.1416)^2) / (6.67x10^-11 * 5.67x10^26)) * (3.1x10^8)^3
Calculate the right side of the equation:
T^2 = (0.59x10^16) * (0.9431x10^24)
Multiplying these values results in:
T^2 = 0.556x10^40
To find T, we take the square root of both sides:
T = sqrt(0.556x10^40)
Using a calculator or computer, we find that T ≈ 8.372 days.
Therefore, the orbital period of Tethys around Saturn is approximately 8.372 Earth days.