# Is bypassing LEO more efficient when launching to Venus or Mars?

According to the table in the Wikipedia article on Hohmann transfer orbits, when traveling to Venus or Mars, the Δv to enter a Hohmann transfer orbit from Earth's orbit is less than the Δv from LEO:

• Venus
• Δv to enter Hohmann orbit from Earth's orbit = 2.5 km/s
• Δv from LEO = 3.5 km/s
• Mars
• Δv to enter Hohmann orbit from Earth's orbit = 2.9 km/s
• Δv from LEO = 3.6 km/s

Would launching a craft "straight up"--directly from the surface into Hohmann transfer (bypassing LEO)--be more efficient than first launching into LEO and then performing a transfer burn?

Have any spacecraft launched to either planet in this way? (If not, why not?)

• One reason might be to stop and smell the roses, or in this case exchange telemetry, get a few more position and velocity measurements from Earth stations and some GPS fixes, sanity checks, integrity checks, and give the ground crew a chance to use the rest room, grab another coffee, and otherwise commit to the next step. – uhoh Mar 31 '17 at 1:44
• By earth orbit, they mean a heliocentric orbit 1 A.U. from the sun. Imagine a 1 A.U. orbit with no earth. It'd take 2.9 km/s to enter a transfer orbit to a 1.52 A.U. orbit. The 2.9 km/s is the departure Vinfinity. – HopDavid Mar 31 '17 at 2:12
• Possible dup: space.stackexchange.com/questions/637/… – Antzi Mar 31 '17 at 5:36
• May or may not be a dup (jury's still out on that one as far as I am concerned), but definitely related. (That question appears to be primarily about Earth orbits, while this question is about heliocentric transfer orbits to other planets in the solar system.) – a CVn Mar 31 '17 at 8:48

No. It is not more efficient to bypass LEO. You can't bypass getting out of earth's gravity well.

2.9 km/s is what it'd take to enter a Mars transfer orbit if the ship were outside of earth's gravity well moving 30 km/s in a circular 1 A.U. orbit. (1 A.U. or one astronomical unit is earth's average distance from the sun.) So, again, that 2.9 km/s is what it takes to get from one sun centered orbit to the sun centered Mars transfer orbit. The planetary gravity wells aren't even considered in that number.

From earth's point of view, the exit orbit is a hyperbola and that 2.9 km/s is the hyperbola's Vinfinity. There's no getting around delta V to get out of earth's gravity well.

And a horizontal burn is the best way to achieve that Vinfinity. Every second you're thrusting vertically, you're losing delta V from gravity loss.

Atmospheric friction prevents a ship from achieving orbital velocity at earth's surface. So a rocket ascends above the thick atmosphere before turning sideways and doing the major horizontal burn.

Once above the atmosphere, it would be possible to do a single major horizontal burn that achieves TMI (Trans Mars Insertion). But at some period during that burn, the ship will have a velocity of 7.7 km/s with close to a zero flight path angle. At that point the ship could be said to be in LEO.

Virtually every spacecraft passes through LEO during some point of it's flight path. From Launch Vehicle Performance Estimation by John Schilling:

The Townsend technique begins by assuming that all space launches consist of a direct ascent to a low circular parking orbit, followed by a series of on-orbit maneuvers to the final destination orbit. In fact, many launch vehicles fly only a direct-ascent trajectory, even to a high or non- circular orbit. However, an observation of these trajectories almost invariably finds the launch vehicle, at an altitude of a few hundred kilometers, accelerating almost horizontally through the local circular orbit velocity. One may simplify the problem by treating this as an instantaneous "parking orbit", reached by direct ascent, and with all subsequent powered flight treated as an "on-orbit maneuver".

Antzi and uhoh are correct that spending awhile in LEO gives time to do checks and preparation. Also departing from LEO is the best way to do it when considering delta V.

• Your links led me to a thought experiment. Launch two identical vehicles up to a stationary platform 300 km above earth. There is no difference in energy cost to get to that platform. The platform is positioned prograde to the earth's orbit around the sun. One vehicle then thrusts horizontally to LEO (7.73 km/s), and later thrusts to a Hohmann transfer to Mars (3.6 km/s) = 11.33 km/s total. The other vehicle thrusts vertically to Vhyp (sqrt of ΔvEsc^2 + ΔvInf^2) = 11.31 km/s (ΔvEsc = 10.93 and ΔvInf = 2.9). – Brandon Bonds Mar 31 '17 at 20:28
• So the difference is 0.02 km/s and that could be rounding error. Regardless, I would expect a gravity turn to be much more efficient, which would make a significant difference in favor of a launch via LEO. – Brandon Bonds Mar 31 '17 at 20:31
• (Sorry for the deletes, needed to correct errors.) – Brandon Bonds Mar 31 '17 at 20:32
• @MarkAdler speed of a hyperbola is sqrt(Vesc^2 + Vinf^2). Vinf to Mars orbit is 2.9 km/s. Vesc is about 11 km/s. So speed of a hyperbola is 11.4 km/s. You were telling Antzi that the .7 km/s savings Brandon imagined is correct. I don't believe a steeper flight path angle will give TMI for 10.7 km/s. I don't see how a steeper flight path angle confers any savings. – HopDavid Apr 1 '17 at 4:22
• See my answer for the cost. – Mark Adler Apr 1 '17 at 8:26

Yes.

Consider a simplifying case of a direct escape vs. a parking orbit departure from an airless, non-rotating body using instantaneous maneuvers. The most efficient way to get to the parking orbit is an initial horizontal $\Delta V$ at the surface to get an orbit with a apoapsis at the parking orbit, followed by a circularization $\Delta V$ at apoapsis to raise the periapsis to that same radius. Then depart from the parking orbit to escape with a third $\Delta V$. Subtract from that an escape to the same velocity at infinity directly from the surface with a single $\Delta V$.

I get that difference to be:

$$\sqrt{\frac{\mu }{r}}\left(\sqrt{e-1+\frac{2}{q+1}}-\sqrt{e+1}+\frac{\sqrt{2} q}{\sqrt{(q+1) (q+2)}}\right)$$

where $\mu$ is the $GM$ of the body, $r$ is the surface radius, $e$ is the eccentricity of an escape hyperbola tangent to the surface, i.e. $e=1+{C_3 r\over\mu}$, and $q$ is the parking orbit altitude as a fraction of $r$.

For an escape from Earth to Mars with $C_3\approx 10\,\mathrm{{km}^2/s^2}$, $e\approx 1.16$. For a $100\,\mathrm{nmi}$ parking orbit, $q\approx 0.03$. I then get, using Earth's $\mu$ and $r$, that the cost of going to the parking orbit as opposed to a direct escape is about $74\,\mathrm{m/s}$.

About $58\,\mathrm{m/s}$ of that is the circularization burn, with the remaining $16\,\mathrm{m/s}$ being effectively a loss of Oberth effect due to an escape from a higher altitude.

A launch vehicle will get to a parking orbit more directly, but it can only cost more $\Delta V$. As for the atmosphere, you can imagine the launch vehicle getting out of the atmosphere in the same way to the same initial conditions at, say, $60\,\mathrm{nmi}$ altitude, and comparing direct escape vs. an intermediate parking orbit starting from there.

That $74\,\mathrm{m/s}$ is pretty small compared to the total of $\Delta V\approx 11,600\,\mathrm{m/s}$. The benefits of an intermediate parking orbit easily outweigh this small cost, both in terms of timing flexibility and in terms of lining up the outgoing asymptote with respect to the launch latitude.

The only time I am aware of a real consideration of a direct escape was for a mission to Mars that I was working on using an Ariane V before they had qualified upper stage restarts. If you can't restart the upper stage, then you can't use a parking orbit.

• The O.P. seems to be confusing the 3.6 km/s LEO TMI burn with the 2.9 km/s Vinf for the hyperbolic orbit. For a surface burn to hyperbolic orbit vs surface to a 100 km apogee, circularize and then a burn to hyperbolic orbit I get a .03 km/s savings. Nowhere near the .7 km/s savings the O.P. seems to imagine. Moreover, an 11.56 km/s burn from earth's surface is impractical. – HopDavid Apr 1 '17 at 15:39
• But for the sake of argument, let's say we're starting from an altitude of 60 nmi. At that altitude you still have earth's gravity of 9.5 m/s^2. Subtract that from the vertical component of thrust. After awhile you can subtract ω^2r from gravity but you're not going to get centrifugal force without a horizontal thrust. And without that ω^2r, the ship would fall back into the atmosphere. Maintaining altitude would incur gravity loss unless you have a horizontal speed of 7.9 km/s – HopDavid Apr 1 '17 at 15:47
• For the sake of argument, do the calculation. – Mark Adler Apr 1 '17 at 23:09
• Okay. I have a spreadsheet based on the accelerations of the Apollo rocket on the way to LEO. I did have the flight path ascending vertically at the outset but gradually going horizontal. But I changed it, keeping the flight path angle straight up, as suggested in the O.P.. Over 680 seconds of acceleration I get a gravity loss of 5.47 km/s. Final vertical velocity is 3.4 km/s at an altitude of 1524 kilometers. – HopDavid Apr 2 '17 at 22:32