The planet Mars has a diameter of 6800 km.  A satellite is in orbit 5000 km above the planet's surface travelling at a speed of 7100 m/s. 

(a)    How long does it take to orbit? 

(b)   What is the centripetal acceleration at this speed?

(c)    What is the acceleration due to gravity at this distance?

(d)   Will it remain in that orbit?  Mass of Mars = 6.42 × 10²³ kg                             

(a) Radius of orbit = (3.400 + 5.000) × 10⁶ m = 8.4 × 10⁶ m (P)

v = 2pfr Þ f = v/2pr = 7100 m/s ÷ (2 × p × 8.4 × 10⁶ m) = 1.35 × 10^-4 Hz (P)

Period = 1/1.35 × 10^-4 Hz = 7430 s (= 2.06 h)  (P)

(b) a = v² / r =  (7100 m/s)² ÷  8.4 × 10⁶ m  (P) = 6.0 m/s² (P)

(c) g = -GM/r² = - 6.67 ´ 10^-11 Nm² kg^-2 × 6.42 × 10²³ kg (P)

                                            (8.4 × 10⁶ m)²

g  = -0.61 m/s² (P)

(d) To remain in a given orbit the centripetal force per unit mass (P) would need to be in

equilibrium  (P) with the gravitational force per unit mass (P).  Since the centripetal force

is greater, the satellite will move to a higher orbit (P).