Given: G = 6.67259 × 10−11 Nm2/kg2

The acceleration of gravity on the surface of a planet of radius R = 3430 km is 4.43 m/s2.What is the period T of a satellite in circular orbit h = 7683.2 km above the surface?
Answer in units of s

To find the period T of a satellite in a circular orbit above the surface of a planet, we can use Kepler's third law of planetary motion.

Kepler's third law states that the period T of a satellite in a circular orbit is related to the radius of the orbit, R, by the formula:

T^2 = (4 * π^2 * (R+h)^3) / (G * M)

Where:
T is the period of the satellite
π is the mathematical constant pi (approximately 3.14159)
R is the radius of the planet
h is the height of the satellite above the surface of the planet
G is the gravitational constant
M is the mass of the planet

Given:
G = 6.67259 × 10^(-11) Nm^2/kg^2
R = 3430 km = 3430 * 10^3 m
h = 7683.2 km = 7683.2 * 10^3 m

Now, we need the mass of the planet, which is not given in the given information. Without the mass, we cannot calculate the period T.