Question

Scientists want to place a 4200.0 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 1.6 times the radius of Mars above the surface of the planet. Here is some information that will help solve this problem:mmars = 6.4191 x 1023 kgrmars = 3.397 x 106 mG = 6.67428 x 10-11 N-m2/kg2

Answer 1

The period of the orbit is 12163.649 seconds.

Step-by-step explanation:

What quantity are you looking for? I'll assume it's the orbital period.

The period, P, of an orbit in seconds is found from

P = 2π √[a³/(GM)]

where

a = the semimajor axis of an elliptical orbit, or the radius of a circular orbit

G = 6.67428e-11 m³ kg⁻¹ sec⁻²

M = the sum of the masses of Mars and the satellite

The mass of the satellite is presumably negligible.

M = 6.4191e+23 kg

a = 1.6 Rmars = 1.6 (3.397e+6 meters) = 5435200 m

Therefore,

P = 12163.649 sec