## Solution to 1986 Problem 75

 We use Kepler's Third Law, which states that\begin{align*}\frac{s_1^3}{T_1^2} = \frac{s_2^3}{T_2^2}\end{align*}for any two bodies orbiting the same mass. Here, we let $s_1 = R_e$ be the radius of the earth, $s_2$ be the radius of the synchronous satellite orbit, $T_1 = 80 \mbox{ min}$ be the period of the hypothetical satellite orbiting at sea level, and $T_2 = 24 \cdot 60 \mbox{ min}$ be the period of the synchronous satellite. Then \begin{align*}s_2 = \left(\frac{24 \cdot 60}{80}\right)^{2/3} R_e = 6.87 R_e\end{align*}Thus, answer (B) is correct.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}