Question #89554

For a special mission to Mars you need to know the smallest distance between Earth and Mars.
However, you have lost your astronomy book and you could only find these values:
Distance Earth to Sun: 149.6 million km
Orbital period Earth: 1.00 years
Orbital period Mars: 1.88 years
By using these values and assuming that Mars and Earth move an circular orbits, calculate the
smallest possible distance between Earth and Mars.

Expert's answer

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:


ΔR=RMRE\Delta R = R_M - R_E

Applying the 2nd Newton's law to the planetary motion, we obtain:


GMmERE2=mEaE,aE=4π2RETE2GM=4π2RE3TE2G \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:


RE3TE2=RM3TM2RM=RE(TMTE)2/3\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,


ΔR=RE[(TMTE)2/31]\Delta R = R_E \left[ \left(\frac{T_M}{T_E}\right)^{2/3} - 1\right]

Substituting the numerical values, we obtain:


ΔR=149.6[(1.881.00)2/31]78.3millionkm\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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