Question

A satellite circles the earth in an orbit whose radius is 4.67 times the earth's radius. The earth's mass is 5.98 times 10^24 kg, and its radius is 6.38 times 10^6 m. What is the period of the satellite? Number Units the tolerance is +/-2%

Answer 1

The velocity of the rotation v = sqrt (G M / r )
Where G is the gravitational constant ,M is the mass of the earth , r is the radius of the orbit
r = 4.67 x R = 4.67 x 6.38 x10⁶ = 29.7946 x 10⁶ m
v = sqrt (6.67 x 10^-11 N m² / kg² x 5.98 x 10²⁴ kg /   29.7946 x 10⁶ m)
v = 3.659 x10³ m /s
The angular velocity is
$\omega$ = v / r
$\omega$   = 3.659 x 10³ / 29.7946 x 10⁶
$\omega$   = 0.1228 x 10^-3 s^-1
T = 2 $\pi$ / $\omega$    
T = 2 x 3.14 /  0.1228 x 10^-3 s^-1
T = 51140.06 s