A 782-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satellite's orbital speed. 9.82278e7 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s (b) Find the period of its revolution. h (c) Find the gravitational force acting on it.

a) v = 5.59x10³ m/s

b) T = 4 h

c) F = 1.92x10³ N

Explanation:

a) We can find the satellite's orbital speed by equating the centripetal force and the gravitation force as follows:

Where:

g is the gravity = 9.81 m/s²        

r: is the Earth's radius = 6371 km

h: is the satellite's height = r = 6371 km      

b) The period of its revolution is:

c) The gravitational force acting on it is given by:

Where:

M is the Earth's mass =  5.97x10²⁴ kg    

m is the satellite's mass = 782 kg

G is the gravitational constant = 6.67x10⁻¹¹ Nm²kg⁻²

I hope it helps you!

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2