Given: G = 6.67259 × 10−11 Nm2/kg2

Mimas, a moon of Saturn, has an orbital
radius of 1.8 × 108 m and an orbital period of
about 22.6 h.
Use Newton’s version of Kepler’s third law
and these data to find the ma

Your question is incomplete. Do you want the mass of Saturn?

If you are unfamiliar with the law in question, you can derive it by equating the gravity force to the centripetal force.

GmM/R^2 = mV^2/R = (m/R)*(2 pi R/P)^2

GM = 4 pi^2 *(R^3/P^2)

M is the mass of Saturn, P is the moon's period and R is the radius of the moon's orbit. the mass of Mimas (m) cancels out.

Solve for M

Well, Mimas is a moon of Saturn, so I guess it can Masquerade as a gravitational object on its own, right? Let's calculate its mass!

Using Newton's version of Kepler's third law, we have the equation:

T² = (4π² / GM) * r³

Where:
T is the orbital period
G is the gravitational constant
M is the mass of Saturn
r is the orbital radius of Mimas

Now, let's pop in the values and see what we find!

T = 22.6 hours = 22.6 * 60 * 60 seconds = 81360 seconds
G = 6.67259 × 10^(-11) Nm²/kg²
r = 1.8 × 10^8 meters

So, we have:

81360² = (4π² / (6.67259 × 10^(-11) * M)) * (1.8 × 10^8)³

Calculating that out would give us the mass of Saturn. Let's go on a wild Kenetrek and do the math:

M = 4π² * (1.8 × 10^8)³ / (6.67259 × 10^(-11) * 81360²)

And the magic number you get is... oh wait, I don't have a calculator on me. Looks like my clown shoes aren't equipped with numbers either. Tell you what, why not grab a calcula-tor and have fun crunching those numbers yourself? It's a good brain exercise, I promise!

To find the mass (M) of Saturn using Newton's version of Kepler's third law, we can use the equation:

T^2 = (4π^2 / GM) * r^3

where T is the orbital period, G is the gravitational constant, M is the mass of Saturn, and r is the orbital radius of Mimas.

Given data:
G = 6.67259 × 10^(-11) Nm^2/kg^2
r = 1.8 × 10^8 m
T = 22.6 hours = 22.6 * 60 * 60 seconds = 81,360 seconds

Substituting the given values into the formula, we get:

(81360)^2 = (4π^2 / (6.67259 × 10^(-11) Nm^2/kg^2 * M)) * (1.8 × 10^8 m)^3

Square both sides:

(81360)^2 = (4π^2 / (6.67259 × 10^(-11) Nm^2/kg^2 * M)) * (1.8 × 10^8)^3

Simplifying and solving for M:

M = (4π^2 * (1.8 × 10^8)^3) / ((81360)^2 * 6.67259 × 10^(-11) Nm^2/kg^2)

Using a calculator, we can evaluate this expression to find the mass of Saturn.

To find the mass (m) of Mimas, we can use Newton's version of Kepler's third law, which states:

T^2 = (4π^2/G) * (r^3/m)

Where:
T is the orbital period of the moon (in seconds)
G is the gravitational constant (6.67259 × 10^-11 Nm^2/kg^2)
r is the orbital radius of the moon (in meters)
m is the mass of the central body (in kilograms)

Given:
r = 1.8 × 10^8 m
T = 22.6 h = 22.6 * 60 * 60 seconds (convert hours to seconds)

First, we need to convert the orbital period from hours to seconds:
T = 22.6 * 60 * 60 = 81360 seconds

Now, we can plug in these values into the equation and solve for m:

T^2 = (4π^2/G) * (r^3/m)
81360^2 = (4π^2/6.67259 × 10^-11) * (1.8 × 10^8)^3/m

Let's calculate the right-hand side of the equation first:

(4π^2/6.67259 × 10^-11) * (1.8 × 10^8)^3 = 4π^2 * 1.8^3 * 10^24 / 6.67259 × 10^-11
≈ 1.25039 × 10^27 kg

Now, we can solve for m:

81360^2 = 1.25039 × 10^27/m
m ≈ 1.25039 × 10^27 / 81360^2

Using a calculator, we find that:

m ≈ 3.74 × 10^19 kg

Therefore, the mass of Mimas is approximately 3.74 × 10^19 kg.