Two satellites, one in geosynchronous orbit (T = 24 hrs) and one with a period of 12 hrs, are orbiting Earth. How many times larger than the radius of Earth is the distance between the orbits of the two satellites.

Johannes Kepler was the first to propose three laws for the planetary motion. According to him, the orbits in which planets are rotating are elliptical in nature and Sun is at the focus of the ellipse. Also the area of sweeping is same.

So based on these three assumptions, Kepler postulated three laws. One among them is Kepler's third law of planetary motion. According to the third law, the square of the time taken by a planet to cover a specified region is directly proportional to the cube of the major elliptical axis or the radius of the ellipse.

So, [tex]T^{2} = r^{3}[/tex]

Thus, for the geosynchornous satellite, as the time taken is 24 hours, then the radius or the major axis of this satellite is

[tex](24)^{2}= r^{3} \\(2*2*2*3)^{2} = r^{3}\\r = \sqrt[\frac{2}{3} ]{2*2*2*3} =(2)^{2} * (6)^{\frac{2}{3} } =13.21 m[/tex]

Similarly, for the another satellite orbiting in time period of 12 hours, the major axis of this satellite is

[tex](12)^{2}= r^{3} \\(2*2*3)^{2} = r^{3}\\r = \sqrt[\frac{2}{3} ]{12} =5.24 m[/tex]

So, the difference between the two radius will give the distance between the two orbits, 13.21-5.24 = 7.97 m.

So the distance between the orbits of two satellites is 7.97 m.