Answer:
Step-by-step explanation:
Assuming the satellite is in a circular orbit, we can use the following equations to calculate its time period, velocity, and acceleration:
Time period (T): T = 2π√(a^3/GM), where a is the radius of the orbit, G is the gravitational constant, and M is the mass of the Earth.
Velocity (v): v = √(GM/a), where G is the gravitational constant, M is the mass of the Earth, and a is the radius of the orbit.
Acceleration (a): a = v^2/a, where v is the velocity and a is the radius of the orbit.
Using the given radius of the orbit (a = 49870 km = 4.987 x 10^7 m) and the known values of G and M, we can calculate the time period, velocity, and acceleration as follows:
Time period: T = 2π√(a^3/GM) = 2π√((4.987 x 10^7)^3/(6.674 x 10^-11 x 5.972 x 10^24)) = 85514 seconds.
Therefore, the time period of the satellite is approximately 23.75 hours (85514 seconds / 3600 seconds per hour).
Velocity: v = √(GM/a) = √((6.674 x 10^-11 x 5.972 x 10^24)/(4.987 x 10^7)) = 3074.4 m/s.
Therefore, the velocity of the satellite is approximately 3074.4 m/s.
Acceleration: a = v^2/a = (3074.4 m/s)^2/4.987 x 10^7 m = 0.190 m/s^2.
Therefore, the acceleration of the satellite is approximately 0.190 m/s^2.