To achieve a stable orbit around Earth, an object needs to reach a certain velocity known as the orbital velocity. The orbital velocity depends on the altitude above the Earth's surface. For an object approximately 1 meter off the ground, the altitude is negligible compared to the radius of the Earth, so we can assume the altitude to be effectively zero.
The orbital velocity at a specific altitude can be calculated using the following formula:
v = sqrt(G * M / r)
Where:
- v is the orbital velocity
- G is the gravitational constant (approximately 6.67430 × 10^(-11) m^3/(kg·s^2))
- M is the mass of the Earth (approximately 5.972 × 10^24 kg)
- r is the distance from the object's center of mass to the center of the Earth (approximately 6,371,000 meters for the Earth's radius)
Plugging in the values, we have:
v = sqrt((6.67430 × 10^(-11) m^3/(kg·s^2)) * (5.972 × 10^24 kg) / (6,371,000 meters))
Calculating this, the orbital velocity required to achieve orbit at an altitude of approximately 1 meter off the ground is approximately 7,905 meters per second (m/s).
As for the time it takes for an object to complete one orbit, it depends on the altitude and the size of the orbit. At such a low altitude, the orbit's radius would be very small, resulting in a very short orbital period. However, due to the extremely low altitude mentioned (1 meter off the ground), the object would encounter significant atmospheric drag, which would lead to a rapid decay in its orbit. In practice, maintaining a stable orbit at such a low altitude is not feasible due to atmospheric effects.
For objects in stable orbits at higher altitudes, the orbital period can be calculated using the following formula:
T = 2π * sqrt(r^3 / (G * M))
Where:
- T is the orbital period
- G is the gravitational constant
- M is the mass of the Earth
- r is the radius of the orbit
Given the extremely low altitude of 1 meter mentioned, it is not practical to calculate a meaningful orbital period using this formula.