This question involves the concepts of the time period, orbital radius, and gravitational constant.
The distance of Hailey's comet above the Earth's Surface is "3.19 x 10⁴ km".
The theoretical time period of the comet 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 = 75 years[tex](\frac{365\ days}{1\ year})(\frac{24\ h}{1\ day})(\frac{3600\ s}{1\ h})[/tex] = 2.36 x 10⁹ s
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{(2.36\ x\ 10^9\ 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{(6.67\ x\ 10{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)(2.36\ x\ 10^9\ s)^2}{4\pi^2}}[/tex]
R = 3.83 x 10⁷ m = 3.83 x 10⁴ km
Now, this orbital radius is the sum of the radius of the earth (r) and the distance of comet above earth's (h) surface:
R = r + h
3.83 x 10⁴ km = 0.64 x 10⁴ km + h
h = 3.83 x 10⁴ km - 0.64 x 10⁴ km
h = 3.19 x 10⁴ km = 31900 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.
