a satellite orbits the earth at an altitude of 5000 m above the surface of the earth. Assuming that acceleration due to gravity is still 9.8 m/s/s, find the time for 1 revolution.
g=v^2/r where v=2PI r/Period
solve for period.
To find the time for one revolution of the satellite, we need to calculate the orbital period. The orbital period is the time taken for a satellite to complete one full revolution around the Earth.
The formula to calculate the orbital period, T, is:
T = 2π√(r³/g)
where:
T is the orbital period
π is a mathematical constant approximately equal to 3.14159
r is the radius of the orbit, which is the sum of the radius of the Earth (6,371 km) and the altitude of the satellite above the surface of the Earth
g is the acceleration due to gravity, which is approximately 9.8 m/s²
Let's substitute the values into the formula and solve the equation step by step:
1. Convert the altitude to meters: 5000 m.
2. Add the radius of the Earth to get the orbit's radius:
r = (6371 km + 5000 m) = 6371000 m + 5000 m = 6376000 m.
3. Substitute the values into the formula:
T = 2π√((6376000 m)³ / 9.8 m/s²).
4. Simplify the equation:
T = 2π√(258173760000000000 m³ / 9.8 m/s²).
T = 2π√(26306261224489795918.367346938775510204081632653 m³/s²).
5. Calculate the square root:
T = 2π * 5129192.973 m/s.
T ≈ 2 * 3.14159 * 5129192.973 m / 9.8 m/s².
T ≈ 32093062.395 s.
6. Round off the result to an appropriate number of decimal places:
T ≈ 32093062.4 s.
Therefore, the time for one revolution of the satellite is approximately 32,093,062.4 seconds.