Low Earth orbiting satellites orbit between 200 and 500 miles above Earth. In order to keep the satellites at a constant distance from Earth, they must maintain a speed of 17,000 miles per hour. Assume Earth's radius is 3960 miles.

a. Find the angular velocity needed to maintain a LEO satellite at 200 miles above earth.

b. How far above Earth is a LEO with a angular velocity of 4 radians per hour?

c. Describe the angular velocity of any LEO satellite

I posted another topic, but no one replied. THX <3!

The key relationship is

linear velocity = angular velocity x radius,

a) since the satellite is 200 miles above the earth, the radius would be 4160 miles.

lin. vel = ang vel. x radius
17000 = av(4160)
4.07 = av

so the angular velocity would be 4.07 radians/hour
(remember that 2pi, or 6.28 radians is one rotation, which makes this 234º per hour)

b) this seems to be the reverse question of a). Notice we got appr 4 radians per hour as the answer to the angular velocity, so the satellite would have to be about 200 miles above the earth.

Check my calculations, this seems rather odd.

a. To find the angular velocity needed to maintain a LEO satellite at 200 miles above Earth, we can use the formula:

Angular velocity = Linear velocity / Radius

The radius can be calculated by adding the altitude of the satellite to the Earth's radius.

Radius = Earth's radius + Altitude
= 3960 miles + 200 miles
= 4160 miles

Then, the linear velocity can be calculated using the given speed.

Linear velocity = 17,000 miles per hour

Now, we can plug in the values in the formula:

Angular velocity = Linear velocity / Radius
= 17,000 miles per hour / 4160 miles

Converting hours to seconds for consistency, we get:

Angular velocity ≈ 4.088 radians per second

Therefore, the angular velocity needed to maintain a LEO satellite at 200 miles above Earth is approximately 4.088 radians per second.

b. To find how far above Earth a LEO satellite is with an angular velocity of 4 radians per hour, we can rearrange the formula:

Angular velocity = Linear velocity / Radius

We need to convert angular velocity from hours to seconds by multiplying by 3600 (since there are 3600 seconds in an hour).

Angular velocity (in radians per second) = 4 radians per hour * (3600 seconds per hour)
= 14,400 radians per second

Now, we can rearrange the formula to solve for the radius:

Radius = Linear velocity / Angular velocity

However, in this case, we don't know the linear velocity. So, it is not possible to determine how far above Earth the LEO satellite is with an angular velocity of 4 radians per hour without knowing the linear velocity.

c. The angular velocity of any LEO satellite can vary depending on its altitude and linear velocity. However, to maintain a constant distance from Earth, the angular velocity must be adjusted accordingly. As the altitude increases, the radius of the orbit increases, hence the angular velocity needs to decrease to maintain a constant distance from Earth. Additionally, the linear velocity needs to also be adjusted to maintain the desired altitude. Therefore, the angular velocity of any LEO satellite will depend on the specific altitude and linear velocity chosen to maintain its orbit.

a. To find the angular velocity needed to maintain a LEO satellite at 200 miles above Earth, we can start by determining the circumference of the satellite's orbit.

The distance between the satellite and the Earth's surface is the sum of the satellite's altitude (200 miles) and the Earth's radius (3960 miles). So the total distance from the satellite to the center of the Earth is 200 + 3960 = 4160 miles.

The orbit's circumference is calculated using the formula C = 2πr, where r is the distance from the center of the Earth to the satellite.

C = 2π * 4160 miles

Now, we know that the satellite needs to complete this orbit in one hour, so the angular velocity (ω) can be found by dividing the circumference by the time taken for one complete revolution.

ω = (2π * 4160 miles) / 1 hour

Simplifying the equation gives us:

ω = 2π * 4160 miles/hour

Therefore, the angular velocity required to maintain a LEO satellite at 200 miles above Earth is approximately 26,198.94 miles/hour.

b. To determine how far above Earth a LEO satellite is with an angular velocity of 4 radians per hour, we can use the formula for angular velocity.

The angular velocity (ω) is given as 4 radians per hour. We can rearrange the formula to find the distance from the center of the Earth to the satellite (r), using the formula ω = v / r, where v is the linear velocity.

Since the given information states that the LEO satellites need to maintain a linear speed of 17,000 miles per hour, we can calculate the radius (r) using the equation:

r = v / ω

Substituting the known values:

r = 17000 miles per hour / 4 radians per hour

Simplifying the equation gives us:

r = 4250 miles

Thus, a LEO satellite with an angular velocity of 4 radians per hour is approximately 4250 miles above Earth's surface.

c. The angular velocity of any LEO satellite remains constant. It does not depend on the satellite's altitude above Earth. As long as the satellite maintains a constant circular orbit, its angular velocity will stay the same regardless of its distance from Earth. The specific angular velocity value depends on the linear velocity required to maintain the desired orbit. In the given scenario, the satellites should maintain a speed of 17,000 miles per hour, resulting in a specific angular velocity.