A Hohmann Transfer Swaps Orbits with Two Burns
Suppose a spacecraft in a low circular orbit needs to reach a higher circular orbit. Instead of burning fuel continuously and spiraling out, the thriftiest route uses a single ellipse that just kisses both circles. You speed up once on the inside to climb onto the ellipse, coast half a lap, and speed up once more the instant you touch the outer circle to settle into it. Two burns, one new orbit: this is the Hohmann transfer. First we pin down escape speed, then we build the transfer ellipse by hand.
Before swapping orbits, how fast must you go to leave entirely? The speed of a circular orbit of radius r is vcirc = √(GM/r). Nudge the speed up and the orbit stretches into a longer ellipse, until at exactly vesc = √(2GM/r) the ellipse opens into a parabola. That value is the escape speed, and it is precisely vesc = √2 · vcirc. Drag the slider up and watch the closed orbit open the instant you reach the √2 multiple.
Now build the transfer ellipse yourself. Draw a single ellipse whose perigee touches the inner circle (radius r₁) and whose apogee touches the outer circle (radius r₂). Its semi-major axis is a = (r₁ + r₂) / 2. The speed anywhere on it comes from the vis-viva equation v = √(GM(2/r − 1/a)). How much faster than the circular speed you must go at perigee is Δv₁; the amount needed to settle from apogee into the outer circle is Δv₂. Slide r₂ up and the ellipse redraws while both burns update together.
Between the two burns the spacecraft cuts its engine and simply coasts. One burn of Δv₁ at perigee puts it on the ellipse, and then it does nothing for half a lap, gliding on inertia. Following Kepler, it moves fast near perigee and slows near apogee. The moment it reaches apogee, a burn of Δv₂ settles it onto the outer circle. Press play and watch the two burns with the half-period coast between them.
Add up all the burns and you get the mission's Δv budget. The catch is that this budget translates straight into fuel. The bridge is the Tsiolkovsky rocket equation Δv = ve · ln(m₀ / mf), where ve is the speed the exhaust leaves at, m₀ is the starting mass, and mf is the mass once the propellant is spent. Because of the logarithm, a small rise in Δv makes the required mass ratio balloon. Slide ve down and see how steeply the propellant fraction climbs to pay for the same budget.
Is a Hohmann transfer always best? When the ratio of the two radii r₂ / r₁ is very large, another route can be thriftier. It is the bi-elliptic transfer: you fire out once to a point far above the target, then come back, spending three burns in all. Going so far out seems wasteful, but up high it costs very little speed change to shift the orbit, and that pays off. Past a radius ratio of about 11.94, an extreme bi-elliptic beats Hohmann. Toggle between the two methods and raise the ratio to find where the winner flips.