An object orbits Earth at an altitude of 200 kilometers above the planet’s surface. What is its speed and orbital period?

To begin with, we need to draw a graph with all of the forces acting on a body.

From that, we can see that the net force Fnet = Fg - Fc , where Fg is gravitational force and Fc is centripetal force acting on a body.

Due to the fact, that the object is in a stable orbit around the Earth, Fnet = 0 N and Fg = Fc -> GmM/(R+h)^2=mv^2/(R+h) -> GM/(R+h)=v^2 and v = (GM/(R+h))^0.5

Inserting values for Earth’s mass (M=5.972×10^24 kg) , and radius (R=6 371 km), we get that v=7785.9 m/s

To find the orbital period, we use the formula T = 2 * pi * (R+h) / v = 5302.8 s