Unreasonable Results

(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.

a.)

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

Now, rs=

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

b.)