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.

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.

This should give you some help.

The length of time it takes for a satellite to orbit the earth, its orbital period, varies with the altitude of the satellite above the earth's surface. The lower the altitude, the shorter the period. The higher the altitude, the longer the period. For example, the orbital period for a 100 mile high satellite is ~88 minutes; 500 miles ~101 minutes; 1000 miles ~118 minutes; 10,000 miles 9hr-18min; 22,238 miles 23hr-56min-4.09sec. A satellite in an equatorial orbit of 22,238 miles altitude remains stationary over a point on the Earth's equator and the orbit is called a geostationary orbit. A satellite at the same 22,238 miles altitude, but with its orbit inclined to the equator, has the same orbital period and is referred to as a geosynchronous orbit as it is in sync with the earth's rotation.
Not surprisingly, the velocity of a satellite reduces as the altitude increases. The velocities at the same altitudes described above are 25,616 fps. (17,426 mph) for 100 miles, 24,441 fps. (16,660 mph.) for 500 miles, 23,177 fps. (15,800 mph.) for 1000 miles, 13,818 fps. (9419 mph) for 10,000 miles, and 10,088 fps. (6877 mph.) for 22,238 miles.

Depending on your math knowledge, you can calculate the orbital velocity and orbital period from two simple expressions. You might like to try them out if you have a calculator.
The time it takes a satellite to orbit the earth, its orbital period, can be calculated from

T = 2(Pi)sqrt[a^3/µ]

where T is the orbital period in seconds, Pi = 3.1416, a = the semi-major axis of an elliptical orbit = (rp+ra)/2 where rp = the perigee (closest) radius and ra = the apogee (farthest) radius from the center of the earth, µ = the earth's gravitational constant = 1.407974x10^16 ft.^3/sec.^2. In the case of a circular orbit, a = r, the radius of the orbit. Thus, for a 250 miles high circular orbit, a = r = (3963 + 250)5280 ft. and T = 2(3.1416)sqrt[[[(3963+250)5280]^3]/1.407974x10^16] = ~5555 seconds = ~92.6 minutes.

The velocity required to maintain a circular orbit around the Earth may be computed from the following:

Vc = sqrt(µ/r)

where Vc is the circular orbital velocity in feet per second, µ (pronounced meuw as opposed to meow) is the gravitational constant of the earth, ~1.40766x10^16 ft.^3/sec.^2, and r is the distance from the center of the earth to the altitude in question in feet. Using 3963 miles for the radius of the earth, the orbital velocity required for a 250 miles high circular orbit would be Vc = 1.40766x10^16/[(3963+250)x5280] = 1.40766x10^16/22,244,640 = 25,155 fps. (17,147 mph.) Since velocity is inversely proportional to r, the higher you go, the smaller the required orbital velocity.

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

Angular velocity (ω) = Velocity (v) / Radius (r)

Given:
Velocity (v) = 17,000 miles per hour
Radius (r) = Earth's radius + Height of satellite

Using the given values:
Velocity (v) = 17,000 miles per hour
Radius (r) = 3960 miles + 200 miles

Angular velocity (ω) = 17,000 miles per hour / (3960 miles + 200 miles)

Simplifying the expression:
ω = 17,000 miles per hour / 4160 miles

Therefore, the angular velocity needed to maintain a LEO satellite at 200 miles above Earth is approximately 4.09 radians per hour.

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 from the previous question:

Angular velocity (ω) = Velocity (v) / Radius (r)

Given:
Angular velocity (ω) = 4 radians per hour
Velocity (v) = 17,000 miles per hour
Radius (r) = Earth's radius + Height of satellite

Using the given values:
4 radians per hour = 17,000 miles per hour / (3960 miles + Height of satellite)

Simplifying the expression:
4 radians per hour * (3960 miles + Height of satellite) = 17,000 miles per hour

We can solve for the height of the satellite by rearranging the equation:
3960 miles + Height of satellite = (17,000 miles per hour / 4 radians per hour)
3960 miles + Height of satellite = 4250 miles

Subtracting 3960 miles from both sides of the equation:
Height of satellite = 4250 miles - 3960 miles

Therefore, the LEO satellite with an angular velocity of 4 radians per hour is approximately 290 miles above Earth.

c. The angular velocity of any LEO satellite is dependent on the speed required to maintain a constant distance from Earth. As stated in the question, LEO satellites need to achieve a speed of 17,000 miles per hour to remain in orbit between 200 and 500 miles above Earth. The angular velocity will be specific to the given height above Earth. In general, the angular velocity will decrease as the height of the satellite increases, since a larger orbital radius requires a slower angular velocity to maintain the same constant distance from Earth.

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

Angular velocity (ω) = Orbital speed (v) / Radius of orbit (r)

The orbital speed is given as 17,000 miles per hour, and the radius of orbit is the sum of the radius of the Earth (3960 miles) and the altitude of the satellite (200 miles).

So, the angular velocity is:

ω = 17000 miles/hour / (3960 miles + 200 miles)

Simplifying this calculation gives:

ω = 17000 miles/hour / 4160 miles

Therefore, the angular velocity needed to maintain a LEO satellite at 200 miles above Earth is approximately 4.09 radians per hour.

b. To determine how far above Earth a LEO satellite is with an angular velocity of 4 radians per hour, we need to rearrange the formula from part a to solve for the radius of the orbit.

Rearranging the formula gives:

Radius of orbit (r) = Orbital speed (v) / Angular velocity (ω)

Substituting the given values, we have:

r = 17000 miles/hour / 4 radians per hour

Simplifying this calculation gives:

r = 4250 miles

Therefore, the LEO satellite with an angular velocity of 4 radians per hour is approximately 4250 miles above Earth.

c. The angular velocity of any LEO satellite is determined by the speed of the satellite and the distance it is from Earth. As explained earlier, the angular velocity can be calculated using the formula:

Angular velocity (ω) = Orbital speed (v) / Radius of orbit (r)

In general, LEO satellites are designed to orbit at a constant distance from Earth, which means their angular velocity remains relatively constant. However, the specific angular velocity will depend on the altitude of the LEO satellite above the Earth's surface. The higher the altitude, the larger the radius of the orbit, and hence, the smaller the angular velocity needed to maintain that orbit. Conversely, if the altitude is decreased, the radius of the orbit decreases, and the angular velocity needed to maintain the orbit increases.