what is the period of the space shuttle traveling at 250km above the earth and what is its velocity

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, ~1.407974x10^16 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 = 1.407974x10^16/[(3963+250)x5280] = 1.407974x10^16/22,244,640 = 25,155 fps. (17,147 mph.) Since velocity is inversely proportional to r, the higher you go, the smaller the required orbital velocity.

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 period from a simple expression. You might like to try it 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.

To calculate the period of an object in circular motion, we need the radius of the circular path and the gravitational constant. In this case, the space shuttle is traveling at a height of 250 km above the Earth's surface.

Step 1: Convert the height above the Earth's surface to meters.
Given that 1 km = 1000 m, we can convert the height as follows:
250 km = 250,000 m

Step 2: Add the Earth's radius to the height to get the radius of the circular path.
The average radius of Earth is about 6,371 km. Converting it to meters:
6,371 km = 6,371,000 m

The radius of the circular path would be:
Radius = height + Earth's radius = 250,000 m + 6,371,000 m = 6,621,000 m

Step 3: Calculate the period using the formula:
Period = 2π * √(radius^3 / gravitational constant)

The gravitational constant (G) is approximately 6.674 × 10^-11 m^3 kg^-1 s^-2.

Plugging in the values, we have:
Period = 2π * √((6,621,000)^3 / (6.674 × 10^-11))

Calculating this expression will give us the period of the space shuttle's orbit above the Earth.

Regarding the velocity of the space shuttle, it depends on which type of velocity you are referring to. There are two common types: orbital velocity and ground-relative velocity. Orbital velocity is the speed required to maintain a stable orbit around the Earth, while ground-relative velocity is the speed with respect to the Earth's surface.

If you want to calculate the orbital velocity, it can be determined using the formula:
Orbital velocity = √(gravitational constant * mass of Earth / radius).

Substituting the given values, we can find the orbital velocity for the space shuttle.

If you are looking for the ground-relative velocity, it would depend on the direction of the shuttle's motion (e.g., moving eastward, westward, or another direction).

To calculate the period and velocity of the space shuttle, we'll need some information. The period refers to the time it takes for the space shuttle to complete one orbit around the Earth, while the velocity refers to its speed.

1. Find the altitude above the Earth's surface: The space shuttle is traveling at an altitude of 250 km above the Earth.

2. Find the radius of the orbit: Subtract the radius of the Earth from the altitude. The average radius of the Earth is approximately 6,371 km.

Radius of orbit = Altitude + Radius of Earth
Radius of orbit = 250 km + 6,371 km

3. Calculate the circumference of the orbit: The formula for the circumference of a circle is given by 2πr, where "r" is the radius of the orbit.

Circumference of orbit = 2π × Radius of orbit

4. Calculate the period: The period is the time taken to complete one orbit. It can be calculated using the formula:

Period = Circumference of orbit / Velocity of the space shuttle

However, since we don't have the velocity yet, we need to calculate it first.

5. Find the velocity: The velocity of the space shuttle in orbit can be determined using the formula:

Velocity = 2π × Radius of orbit / Period

Now, let's calculate the velocity and period of the space shuttle:

Radius of orbit = 250 km + 6,371 km = 6,621 km

Circumference of orbit = 2π × 6,621 km

Next, if you have the velocity of the space shuttle, you can calculate the period using the formula mentioned earlier. Conversely, if you have the period, you can calculate the velocity using the second formula.

Please provide the velocity of the space shuttle so that I can calculate the period accordingly.