(a) Based on Kepler’s laws and information on the orbital characteristics of the Moon, calculate the orbital radius for an Earth satellite having a period of 1.00 h. (b) What is unreasonable about this result? (c) What is unreasonable or inconsistent about the premise of a 1.00 h orbit?
Solution 49 PE
The orbital radius for an Earth satellite can be calculated using the Kepler’s third law which relates the time period of the satellite to the radius of the orbit.
Step 1 of 4
The orbital characteristics of the Moon are
Time period of the moon Tm is Tm= 27.3 days
Radius of the orbit rm is rm= 3.84 m
Step 2 of 4</p>
Using Kepler’s third law the time period (TS) and radius of orbit (rS) for the Earth satellite can be related to the time period (Tm ) and radius of orbit (rm) for the Moon.
Given : = 1 hr , rm= 3.84 m,Tm= 27.3 days
Tm= 27.3 days
= 27.3 days
Tm = 655.2 hrs
Substitute 655.2 for Tm , 1 for and 3.84 for
rs= (3.84 )
= (0.0312 ) (3.84 )
= 5068800 m
rs= 5068.8 km
Therefore ,the orbital radius for an Earth satellite having a period of 1.00 h is 5068.8 km