# Single-impulse coplanar orbit transfer

Calculate Hohmann circular-to-circular orbit transfers, circular to GTO, elliptical to circular, and elliptical-to-elliptical transfers with apse line rotation.

Press **⏎** to see the model. Within the model C **⏎** is the central body that is orbited, 0 **⏎** is the initial orbit some time before the Δv is applied, 0i **⏎** is the initial orbit the moment before the Δv is applied, 1i **⏎** is the new orbit the moment after the Δv is applied, and 1 **⏎** is the new orbit some time later.

Try your own inputs or the examples below. Refresh your browser to clear results.

## Circular orbit to GTO (geosynchronous transfer orbit)

- A satellite begins in a 300 km 0.h = 300
**⏎**circular orbit 0.isCircular = true**⏎**. This infers the initial orbit characteristics as shown in the orbit calculator examples.

- The ‘cheapest’ GTO passes through periapsis at initial altitude 1i.hp = 300
**⏎**and apoapsis at geosynchronous altitude of 35,786 km 1i.ha = 35786**⏎**. The model infers that this takes a ΔV of 2.4 km/s. - But how long does it take to reach geosynchronous altitude? As said in the introduction, orbit 1 is the new orbit some time after the ΔV. The time it takes to reach geosynchronous altitude is implied by first asserting that orbit 1 has reached apoapsis 1.h==ha = true
**⏎**. The model has already inferred that orbit 1’s intersection point is at its periapsis, so it remains to specify that orbit 1 begins after zero full rotations 1i.assume_fractional_orbit = true**⏎**and ends after zero full rotations 1.assume_fractional_orbit = true**⏎**(to specify a different number of full rotations, you can directly set the variable*θ.n*using 1.«ctrl+g»y.n = 3**⏎**, for example). The model infers that the journey from periapsis at point 1i to apoapsis at point 1 takes 5.3 hours dt1.h**⏎**, which is exactly half the period of this GTO 1.T.h**⏎**.

## Elliptical to circular: Rocket cutoff to circular orbit

- A microwave thermal rocket, let’s say, reaches 7.3 km/s 0.v = 7.3
**⏎**at 80 km altitude 0.h = 80**⏎**with a flight-path angle of 10° 0.g = 10**⏎**0i.assume_fractional_orbit = true**⏎**. Given this state at rocket cutoff, the model infers that the initial orbit 0**⏎**has an eccentricity of 0.2 and an apoapsis of 550 km but a periapsis of -1950 km, meaning that the rocket will not complete a full orbit and instead reenter. - What ΔV is needed to prevent the rocket from reentering and instead circularize the orbit at 500 km exactly? The previous calculation found that the rocket overshoots to 550 km, so the burn must occur a little before apoapsis 0i.isOutbound = true
**⏎**at 500 km h = 500**⏎**to circularize the orbit 1i.isCircular = true**⏎**and cannot simply point horizontally. The model infers that the rocket takes 8 minutes after cutoff to reach 500 km, and that it then takes a ΔV of 1 km/s with the nose of the rocket pointed 27° below horizontal (γ_{ΔV}= -27°) for the orbit to be made circular. The correct tilt exactly zeroes the radial component of the velocity and changes the tangential component to be that of a 500 km circular orbit, all in a single compound maneuver that takes less propellant than zeroing the tangential velocity and increasing the radial velocity separately.

## Transfer with apse line rotation

- Following the apse line rotation example of Curtis, a satellite orbits Earth at a radius of 8,000 km by 16,000 km 0.I.rp = 8000
**⏎**0.I.rp = 16000**⏎**, then transits to an orbit with a radius of 7,000 km by 21,000 km 1.I.rp = 7000**⏎**1.I.rp = 21000**⏎**whose apse line is rotated 25° counterclockwise dw = 25**⏎**. The orbits intersect at two points, to select the outbound one from orbit 0’s perspective 0i.isOutbound = true**⏎**. The model infers that this takes a ΔV of 1.5 km/s directed at an angle of γ_{ΔV }= 91.3° relative to horizontal when the true anomaly 0i.θ reaches 153°.

# Two-impulse coplanar orbit transfer

# Hohmann transfer

Hohmann transfers are two-impulse coplanar circular-to-circular transfers. The first impulse changes a circular orbit into an elliptical orbit that transfers between the initial and final orbits. The second impulse circularizes the orbit at the destination altitude. The two-impulse Hohmann transfer model is built by joining two single-impulse transfer models, which the Hohmann model regards as the ΔV_{0T} dV0T **⏎ **and ΔV_{T1} dVT1 **⏎ ** variables.

## Saving the International Space Station

NASA has decided to de-orbit the International Space Station (ISS). But what if your sovereign wealth fund / family foundation is interested in buying ISS and moving it to a higher orbit to preserve it as a museum for future space tourists? Is that realistic and affordable using the rockets of today? The Hohmann transfer calculator can generate some useful insights:

- The ISS is currently in circular orbit at 250 miles 0.h.mi = 250
**⏎**altitude and needs to be raised to 700 km 1.h.km = 700**⏎**to provide a century in space before the next boost (and if humanity is unable to provide this boost a century from now, it will likely have more to worry about than a reentering space station!). The model infers that an 82 m/s burn will shift the ISS into a 250 mi x 700 km transfer orbit and that a subsequent 81 m/s burn once it reaches 700 km will circularize the orbit at that altitude. The period of the transfer orbit T.T**⏎**is 99 minutes and the ISS will only take half that time to ascend to its destination altitude. - Is a Falcon Heavy rocket big enough to move the ISS through this orbit transfer? Let us assume that we’ll be using the Falcon’s upper stage to provide both burns and that its payload is propellant for raising the ISS’s orbit; the Falcon upper stage’s specific impulse is 348 seconds Stage.1.<Isp> = 348
**⏎**and thrust is 934 kN Stage.1.<T>.kN = 934**⏎**. The Hohmann transfer model incorporates a 2-stage rocket submodel wherein the initial mass of stage 2 is the payload mass of stage 1. To represent that the 4-ton Falcon stage provides both burns and no mass is ejected between burns, the stage 2 vehicle mass is set to 4 metric tons Stage.2.m.Mv.t = 4**⏎**and the stage 1 vehicle mass is set to zero Stage.1.m.Mv = 0**⏎**, whereas the overall payload (which is the same as the stage 2 payload) is set to the 430 metric tons that the ISS weighs: Stage.m.M2.t = 430**⏎**. The model infers that it takes 21 metric tons of propellant to perform both burns Stage.Mx.t**⏎**, which is good news because the Falcon Heavy rocket is able to carry up to 23 metric tons to LEO. At the level of thrust the second stage provides, it takes 78 seconds of burn time Stage.t**⏎**to produce the total required impulse, which is short compared to the orbit transfer duration of 50 minutes. Therefore it is feasible in principle to save the ISS using a $150M Falcon Heavy rocket.