Given: G=6.67259*10^ -11 Nm^ 2 /kg^ 2 The acceleration of gravity on the surface of a planet of radius R = 7040l is 11.1 m/s ^ 2 What is the period of a satellite in circu lar orbit h = 15488kn above the surface? Answer in units of s
To find the period of a satellite in a circular orbit, we can use the formula:
T = 2π √(r^3 / (G*M))
Where:
T is the period of the satellite
π is a mathematical constant approximately equal to 3.14159
r is the total distance from the center of the planet to the satellite (radius of the planet plus the height of the satellite)
G is the gravitational constant
M is the mass of the planet
Let's calculate the period step by step:
1. Convert the given radius of the planet from kilometers to meters:
R = 7040 km = 7040 * 1000 m = 7,040,000 m
2. Convert the given height of the satellite from kilometers to meters:
h = 15488 km = 15488 * 1000 m = 15,488,000 m
3. Calculate the total distance from the center of the planet to the satellite:
r = R + h
4. Substitute the values into the formula:
T = 2π √(r^3 / (G*M))
Before proceeding, we need to know the mass (M) of the planet. If the mass is not given, we cannot determine the period accurately. Please provide the mass of the planet.