Earth-Mars Hohmann transfer

Problem

Earth-Mars Hohmann transfer. Estimate the total $\Delta v$ needed for a Hohmann transfer orbit from Earth to Mars, knowing that the Earth’s semi-major axis is $a_T = 1{,}5\cdot 10^{11}$ m and Mars’s is $a_M = 2{,}28\cdot 10^{11}$ m. Use $v = \sqrt{GM_S(2/r - 1/a)}$ and remember that the semi-major axis of the transfer orbit is $(a_T + a_M)/2$.

The transfer ellipse (dashed) is tangent to the Earth’s orbit at perihelion and to Mars’s orbit at aphelion.

Solution

Let $GM_S = 6{,}67\cdot 10^{-11} \cdot 2{,}0\cdot 10^{30} = 1{,}33\cdot 10^{20}$ (SI). The semi-major axis of the transfer ellipse is $a_t = \frac{a_T + a_M}{2} = \frac{(1{,}5 + 2{,}28)\cdot 10^{11}}{2} = 1{,}89\cdot 10^{11}\ \text{m}$

Manoeuvre 1 (departure from Earth). Earth’s circular velocity: $v_T = \sqrt{GM_S/a_T} = \sqrt{1{,}33\cdot 10^{20}/1{,}5\cdot 10^{11}} \approx 2{,}98\cdot 10^4$ m/s. Velocity at the perihelion of the ellipse ($r = a_T$): $v_{p} = \sqrt{GM_S\!\left(\frac{2}{a_T} - \frac{1}{a_t}\right)} = \sqrt{1{,}33\cdot 10^{20}(1{,}33\cdot 10^{-11} - 5{,}29\cdot 10^{-12})} \approx 3{,}28\cdot 10^4\ \text{m/s}$ $\Delta v_1 = v_p - v_T \approx 32{,}8 - 29{,}8 \approx 2{,}9\ \text{km/s}$

Manoeuvre 2 (arrival at Mars). Velocity at the aphelion of the ellipse ($r = a_M$): $v_{a} = \sqrt{GM_S\!\left(\frac{2}{a_M} - \frac{1}{a_t}\right)} = \sqrt{1{,}33\cdot 10^{20}(8{,}77\cdot 10^{-12} - 5{,}29\cdot 10^{-12})} \approx 2{,}16\cdot 10^4\ \text{m/s}$ Mars’s circular velocity: $v_M = \sqrt{GM_S/a_M} \approx 2{,}42\cdot 10^4$ m/s. $\Delta v_2 = v_M - v_a \approx 24{,}2 - 21{,}6 \approx 2{,}6\ \text{km/s}$

Total. $\Delta v_\text{tot} = \Delta v_1 + \Delta v_2 \approx 2{,}9 + 2{,}6 \approx 5{,}6\ \text{km/s}$

$\ev{\Delta v_\text{tot} \approx 5{,}6\ \text{km/s}\quad(\Delta v_1 \approx 2{,}9\ \text{km/s at departure})}$

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Topics: Gravitation Concepts: Orbits · Kepler’s laws · Kinetic energy Skills: Symbolic setup Objects: Satellite