at what altitude should a satellite be placed into circular orbit so that its orbital period is 48 hours? the mass of the earth is 5.976x10^24 kg and the radius of the eatrh is 6.378x10^6 m.
The velocity required to keep a satellite in a circular orbit derives from
Vc = sqrt(µ/r) where r = the radius of the orbit in feet and µ = the Earth's gravitational constant = 1.407974x10^16 yielding r = 57,064 miles.
Ignore the previous response as it did not answer your question.
The required altitude to maintain a period of 48 hours derives from
T = 2(Pi)sqrt[r^3/µ] where T = the period in seconds, r = the orbital radius in feet and µ = the Earth's gravitational constant = 1.407974x10^16.
From this, r = 41,668 miles making the altitude 37,705 miles.
The velocity at this altitude is 8000 fps.
To determine the altitude at which a satellite should be placed in a circular orbit with a desired orbital period of 48 hours, we can use Kepler's third law of planetary motion.
Kepler's third law states that the square of the orbital period (T) of a satellite is proportional to the cube of its semi-major axis (a), which in this case is related to the altitude. Mathematically, it can be represented as:
T^2 = (4π^2 / GM) * a^3
Where:
T = Orbital period in seconds (48 hours = 48 * 60 * 60 seconds)
G = Universal gravitational constant (6.67430 × 10^-11 m^3/(kg s^2))
M = Mass of the Earth in kilograms (5.976 × 10^24 kg)
a = Semi-major axis (radius + altitude)
Let's plug in the values and solve for a (semi-major axis):
(48 * 60 * 60)^2 = (4π^2 / (6.67430 × 10^-11 * 5.976 × 10^24)) * a^3
Simplifying:
(48 * 60 * 60)^2 = (4π^2 / (6.67430 × 5.976)) * a^3
a^3 = ((48 * 60 * 60)^2 * (6.67430 × 5.976)) / (4π^2)
a^3 = (2.176 × 10^15) * (6.67430 × 5.976) / (4π^2)
a^3 ≈ 5.3437 × 10^22
Taking the cube root of both sides:
a ≈ 7.042 × 10^7 meters
Now, subtracting the Earth's radius from the semi-major axis calculated above will give us the altitude at which the satellite should be placed:
Altitude = a - Radius of the Earth
Altitude ≈ 7.042 × 10^7 - 6.378 × 10^6
Altitude ≈ 6.404 × 10^7 meters
Therefore, a satellite should be placed at an altitude of approximately 6.404 × 10^7 meters to achieve an orbital period of 48 hours.
To determine the altitude at which a satellite should be placed into a circular orbit with a desired orbital period, we can use Kepler's Third Law of Planetary Motion.
Kepler's Third Law states that the square of the orbital period (T) of a satellite is directly proportional to the cube of its semi-major axis (a). Mathematically, it can be expressed as:
T^2 = (4π^2 / GM) * a^3
Where:
T is the orbital period
G is the gravitational constant (approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2)
M is the mass of the Earth
a is the semi-major axis of the satellite's orbit
In this case, we want the orbital period to be 48 hours. Let's substitute the values into the equation:
(48^2) = (4π^2 / GM) * a^3
Now, we can rearrange the equation to solve for a, the semi-major axis:
a^3 = [(48^2) * GM * 1 / (4π^2)]
a^3 = [(48^2) * (6.67430 x 10^-11) * (5.976 x 10^24) / (4π^2)]
a^3 ≈ 5.249 x 10^21
Taking the cube root of both sides, we find:
a ≈ 8303 km
The altitude of the satellite is equal to the semi-major axis minus the radius of the Earth. Therefore:
Altitude = a - radius of the Earth
≈ 8303 km - 6378 km
≈ 1925 km
Hence, the satellite should be placed at an altitude of approximately 1925 kilometers in order to have an orbital period of 48 hours.