Find the period of a satellite with a circular orbit 350 above Earth's surface.
To find the period of a satellite with a circular orbit 350 km above Earth's surface, we need to use the formula for the period of a satellite:
T = 2π√(r^3/GM)
Where:
T is the period of the satellite,
r is the distance of the satellite from the center of the Earth,
G is the gravitational constant, and
M is the mass of the Earth.
First, we need to determine the distance from the center of the Earth to the satellite. The altitude of the satellite above the Earth's surface is given as 350 km. However, the radius of the Earth is about 6,371 km.
Therefore, the distance from the center of the Earth to the satellite (r) can be calculated by adding the radius of the Earth to the altitude of the satellite:
r = radius of the Earth + altitude of the satellite
r = 6,371 km + 350 km
r = 6,721 km
Next, we can substitute the values into the formula to find the period of the satellite:
T = 2π√(r^3/GM)
T = 2π√((6,721 km)^3 / (6.67430 × 10^-11 N(m/kg)^2 × (5.972 × 10^24 kg))
Calculating the expression within the square root:
(6,721 km)^3 ≈ 3.6087 × 10^11 km^3
Substituting the values and simplifying:
T = 2π√(3.6087 × 10^11 km^3 / (6.67430 × 10^-11 N(m/kg)^2 × (5.972 × 10^24 kg))
T = 2π√(5.3997 × 10^34 km^3/N(m/kg)^2)
Converting km^3 to m^3:
(5.3997 × 10^34 km^3) * (1000 m/km)^3 ≈ 5.3997 × 10^25 m^3
Substituting the value and simplifying:
T ≈ 2π√(5.3997 × 10^25 m^3/ (6.67430 × 10^-11 N(m/kg)^2 × (5.972 × 10^24 kg))
T ≈ 2π√(8.0588 × 10^11 s^2)
T ≈ 2π * 8.9876 × 10^5 s
T ≈ 56,545 s
Therefore, the period of the satellite with a circular orbit 350 km above Earth's surface is approximately 56,545 seconds.