the moon has a period of 27.3 days and a mean distance of 3.90x10^8 m from the center of the earth. find the period of a satellite in orbit 6.7x10^6 m from the center of earth.

thank you!

Two body system witah a same center (earth), by using Keplers law

rm³/Pm² = rs³/Ps²
r = distance, P = period, m = moon, s = satellite
(3.90*10^5)³/(27.3)² = (6.70*10^3)³/Ps²
Ps² = (6.70*10^3)³(27.3)²/(3.90*10^5)³
Ps² = 0.003779
Ps = 0.0615 days = 1.48 hours

(b) r = 6700 km
Earth's radius = 6378 km
altitude = h = r - Re
h = 6700 - 6378 = 332 km

Well, according to my calculations, the period of a satellite in orbit around the Earth does not actually depend on the distance from the center of the Earth. But hey, let's pretend it does for the sake of humor!

So, the moon has a period of 27.3 days and is 3.90x10^8 m away from the center of the Earth. Now, if we calculate the distance of your satellite, which is 6.7x10^6 m, we can then use the power of imagination to find its period!

Hmm... let me see... ah, got it!

If your satellite was really dependent on distance like the moon, then we could say that the ratio of the periods would be equal to the ratio of the distances cubed. In other words:

(P1 / P2) = (d1 / d2)^3

Now, let's plug in the values:

27.3 / P2 = (3.90x10^8 / 6.7x10^6)^3

Simplifying this equation, we would eventually find the period of your satellite if it obeyed the same rules as the moon. But in reality, the period of an object in orbit only depends on the mass of the planet it is orbiting and its distance from the center of mass.

So, my dear friend, unfortunately, I can't provide you with a real answer to this question. But I hope my little journey into the realm of imagination brought a smile to your face!

To find the period of a satellite in orbit around the Earth, you can use Kepler's third law of planetary motion, which states that the square of a planet's orbital period is proportional to the cube of its average distance from the center of the Earth.

Here are the steps to calculate the period of the satellite:

Step 1: Calculate the average distance of the satellite from the center of the Earth.
Given: Distance of the satellite from the center of the Earth = 6.7x10^6 m

Step 2: Calculate the ratio of distances cubed.
Ratio of distances cubed = (Distance of the satellite from the center of the Earth)^3 / (Mean distance of the moon from the center of the Earth)^3

Ratio of distances cubed = (6.7x10^6)^3 / (3.90x10^8)^3

Step 3: Calculate the square root of the ratio obtained in Step 2.
Period of the satellite = Square root of (Ratio of distances cubed) * (Period of the moon)

Period of the satellite = Square root of [(6.7x10^6)^3 / (3.90x10^8)^3] * 27.3 days

Perform the necessary calculations to obtain the period of the satellite in days.

To find the period of a satellite in orbit around the Earth, we can use Kepler's Third Law of Planetary Motion, which relates the period of an orbit (T) to the mean distance from the center of the Earth (R) using the following equation:

T^2 = (4 * π^2 * R^3) / (G * M)

where T is the period of the orbit, R is the mean distance from the center of the Earth, G is the gravitational constant (approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2), and M is the mass of the Earth (approximately 5.972 x 10^24 kg).

Given that the moon has a period of 27.3 days and a mean distance of 3.90 x 10^8 m from the center of the Earth, we can substitute these values into the equation to find the necessary information.

Let's calculate the period of the satellite's orbit with a mean distance of 6.7 x 10^6 m from the center of the Earth:

T^2 = (4 * π^2 * (6.7 x 10^6)^3) / (6.67430 x 10^-11 * 5.972 x 10^24)

First, simplify the equation:

T^2 = (4 * π^2 * 2.3829987 x 10^20) / (3.98136938 x 10^14)

Now, divide both sides of the equation by (4 * π^2 * 2.3829987 x 10^20):

T^2 ≈ (3.98136938 x 10^14) / (4 * π^2 * 2.3829987 x 10^20)

T^2 ≈ 2.097 x 10^(-7)

Take the square root of both sides to solve for T:

T ≈ sqrt(2.097 x 10^(-7))

T ≈ 4.58 x 10^(-4)

So, the period of the satellite's orbit around the Earth with a mean distance of 6.7 x 10^6 m is approximately 4.58 x 10^(-4) seconds.