The moon is 384,000 km from the earth and makes 1 revolution in 27.3 days. Find its angular and linear speeds in its orbit.

angular speed = 2 pi radians/27.3 days

linear speed = (orbit circumference)/(27.3 days)

They may want you to convert to radians and km per second. To do that, divide by the number of seconds in 27.3 days

Moon is 384000 km from earth and makes one revolution in 27.3 days find its angular and linear velocity in its orbit

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To find the angular and linear speeds of the moon in its orbit, we'll need some basic formulas related to circular motion.

Angular speed (ω) is defined as the angle (θ) covered per unit of time (t).
Angular speed (ω) = Δθ / Δt

Linear speed (v) is the distance (s) traveled in a certain time (t).
Linear speed (v) = Δs / Δt

Let's start by finding the angular speed of the moon.

The moon completes one revolution around the Earth in 27.3 days. To find the angular speed, we need to convert this time into seconds, as angular speed is commonly measured in radians per second.

27.3 days = 27.3 × 24 hours × 60 minutes × 60 seconds ≈ 2,360,320 seconds

The angle covered in one revolution is 2π radians (or 360 degrees).

Therefore, the angular speed (ω) of the moon can be calculated as follows:

ω = (2π radians) / (2,360,320 seconds) ≈ 2.66 × 10^(-6) radians/second

Next, let's determine the linear speed of the moon.

The distance from the Earth to the Moon is given as 384,000 km.

To find the linear speed, we need to convert this distance into meters and then divide it by the time taken for one revolution.

384,000 km = 384,000 × 1000 meters = 384,000,000 meters

Using the formula for linear speed:

v = (384,000,000 meters) / (2,360,320 seconds) ≈ 1627.99 meters/second

Therefore, the angular speed of the moon in its orbit is approximately 2.66 × 10^(-6) radians/second, and the linear speed is roughly 1627.99 meters/second.