This question involves the concepts of the time period, orbital radius, and gravitational constant.
The distance of the satellite above the Earth's Surface is "10400 km ".
The theoretical time period of the satellite around the earth can be found using the following formula:
[tex]\frac{T^2}{R^3}=\frac{4\pi^2}{GM}\\\\[/tex]
where,
T = Time Period of Satellite = [tex]\frac{1}{frequency} = \frac{1}{4\ orbits/day}\frac{3600*24\ s}{1\ day} = 21600\ s[/tex]
R = Orbital Radius = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
Therefore,
[tex]\frac{(21600\ s)^2}{R^3}=\frac{4\pi^2}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}\\\\R = \sqrt[3]{\frac{4.66\ x\ 10^8\ s^2}{9.91\ x\ 10^{-14}\ s^2/m^3}} \\R = 1.675\ x\ 10^7\ m = 1.68\ x\ 10^4\ km[/tex]
Now, this orbital radius is the sum of the radius of the earth (r) and the distance of satellite above earth's (h) surface:
R = r + h
1.68 x 10⁴ km = 0.64 x 10⁴ km + h
h = 1.68 x 10⁴ km - 0.64 x 10⁴ km
h = 1.04 x 10⁴ km = 10400 km
Learn more about the orbital time period here:
brainly.com/question/14494804?referrer=searchResults
The attached picture shows the derivation of the formula for orbital speed.
