Assuming circular orbits for both Mars and Earth, one can deduce that minimal distance can be observed when both planets are aligned within the common radial line:
"\\Delta R = R_M - R_E" Applying the 2nd Newton's law to the planetary motion, we obtain:
"G \\frac{M m_E}{R_E^2} = m_E a_E, \\quad a_E = \\frac{4 \\pi^2 R_E}{T_E^2} \\, \\Rightarrow \\, GM = \\frac{4 \\pi^2 R_E^3}{T_E^2}"The same expressions are valid for Mars (one has only to replace index "E" by "M"). Hence, we obtain:
"\\frac{R_E^3}{T_E^2} = \\frac{R_M^3}{T_M^2} \\, \\Rightarrow \\, R_M = R_E \\left(\\frac{T_M}{T_E}\\right)^{2\/3}" As a result,
"\\Delta R = R_E \\left[ \\left(\\frac{T_M}{T_E}\\right)^{2\/3} - 1\\right]" Substituting the numerical values, we obtain:
"\\Delta R = 149.6 \\cdot \\left[ \\left(\\frac{1.88}{1.00}\\right)^{2\/3} - 1 \\right] \\approx 78.3 \\, million \\, km" Answer: 78.3 million km.
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