Hohmann Transfer Calculator

A 1-tonne satellite has to be moved from a 300 km parking orbit to geostationary altitude. Find both burns, and then find what fraction of the vehicle leaving the parking orbit has to be propellant. Take r₁ = 6671 km, r₂ = 42164 km, and Earth’s gravitational parameter μ = 398600 km³/s².

1. A Hohmann transfer is half of an ellipse that just touches both circles — perigee on the inner one, apogee on the outer. That fixes its semi-major axis before any speed is calculated, because the major axis simply spans from one orbit to the other.

   $$a_t=\frac{r_1+r_2}{2}=\frac{6671+42164}{2}=24417.5\mathrm{km}$$

2. Everything else comes from one equation, vis-viva, which gives the speed at any radius on any orbit: v² = μ(2/r − 1/a). On a circle a = r, and it collapses to the familiar circular speed. That is how fast the satellite is already going.

   $$v_{c1}=\sqrt{\frac{\mu }{r_1}}=\sqrt{\frac{398600}{6671}}=7.730km/s$$

3. Now apply the same equation at the same radius but on the transfer ellipse, where a = a_t. The satellite is at the lowest point of a much larger orbit, so it needs to be moving faster. The first burn buys exactly that difference — the engine changes the orbit, not the position.

   $$v_p=\sqrt{\mu \left(\frac{2}{r_1}-\frac{1}{a_t}\right)}=10.158km/s⟹\Delta v_1=v_p-v_{c1}=2.428km/s$$

4. Coast to apogee and the ellipse has slowed right down: angular momentum is conserved, so climbing costs speed. But a circular orbit at that height needs more than the ellipse has left, which is why the second burn is a speed-up too, not a brake. Without it the satellite falls straight back to perigee.

   $$v_a=\sqrt{\mu \left(\frac{2}{r_2}-\frac{1}{a_t}\right)}=1.607km/s,v_{c2}=\sqrt{\frac{\mu }{r_2}}=3.075km/s⟹\Delta v_2=1.468km/s$$

5. A rocket cannot spend Δv directly; it spends mass, at an exchange rate set by the exhaust velocity. For a storable upper stage with a specific impulse of 320 s, v_e = 320 × 9.80665 = 3.138 km/s, and Tsiolkovsky converts the total.

   $$\frac{m_0}{m_f}=e^{\Delta v/v_e}=e^{3.895/3.138}=3.46⟹\text{propellant}=1-\frac{1}{3.46}=0.711$$

You are scheduling a Mars mission. Work out how long the crossing takes, and then the thing that actually decides the launch date: where Mars has to be when you leave, and how often that happens. Take Earth’s orbit as 1.000 AU, Mars’ as 1.523 AU, and Mars’ year as 687 days.

1. Same construction as before, one scale up: the transfer ellipse touches Earth’s orbit at perihelion and Mars’ at aphelion, so its semi-major axis is the mean of the two.

   $$a_t=\frac{r_1+r_2}{2}=\frac{1.000+1.523}{2}=1.262\mathrm{AU}$$

2. Kepler’s third law then gives the period outright — in astronomical units and years the constants vanish and it is just T = a^3/2. You only fly half the ellipse, so halve it. Check that against the calculator above: it reports 258.8 days for this transfer, from the same geometry.

   $$T=a_t^{3/2}=1.262^{3/2}=1.417\mathrm{yr}⟹t_{\text{transfer}}=\frac{T}{2}=0.709\mathrm{yr}=258.8\mathrm{d}$$

3. Here is what the transfer time is really for. Mars does not wait at the arrival point — it has to be delivered there too, by its own orbit, in exactly the time you spend crossing. So work out how far round Mars gets while you are in flight.

   $${\theta }_{\text{Mars}}=360^{\circ }\times \frac{258.8}{687}=135.6^{\circ }$$

4. You arrive 180° round from where you left. Mars covers 135.6° in the same interval, so at the instant of departure it must still be short of the arrival point by the difference. Aim at where Mars is and you arrive months behind it.

   $$\varphi =180^{\circ }-{\theta }_{\text{Mars}}=180^{\circ }-135.6^{\circ }=44.4^{\circ }$$

5. That alignment is not a date, it is a beat frequency: the two planets return to the same relative geometry at the synodic period, which is the difference of their angular rates.

   $$\frac{1}{S}=\frac{1}{365.25}-\frac{1}{687}⟹S=780\mathrm{d}$$