Classical Mechanics - Kepler's Law

The period of a hypothetical Earth satellite orbiting at sea level would be 80 minutes. In terms of the Earth’s radius R_e, the radius of a synchronous satellite orbit (period 24 hours) is most nearly

A. 3 R_e
B. 7 R_e
C. 18 R_e
D. 320 R_e
E. 5800 R_e

(GR8677 #75)

Solution:

Kepler’s 3rd Law:

R³ / T²  = constant    

Given:
T_earth = 80 minutes
T_satellite = 24 × 60 minutes

R_e³ / T_e²  = R_s³ / T_s²  

R_s³ = ( T_s² / T_e² ) R_e³ 

R_s³ = (24 × 60 / 80)² R_e³ = (18)² R_e³ 

R_s = (18)^2/3 R_e = (324)^1/3 R_e ≈ 7 R_e

Answer: B