Venera 9 was a Soviet space mission to Venus that consisted of an orbiter and lander with a launch mass of 4936 kg and with an entry mass of 1560 kg. It was the first spacecraft to return an image from the surface of another planet.

Venus surface
(Click on the image for a closer view.)

The launch rocket was the Proton-K/D that was part of the program to land Soviet cosmonauts on the Moon, with 3 stages each having specific impulses varying from 316 to 327 sec.
Could Starship with its Raptor engines with specific impulses at sea level of 330 sec. accomplish such a mission to Venus with a (much) heavier lander to prolong the lifetime of its instruments on the surface ?
And could landing near a pole be more challenging because the descent would have to start at one pole (since its axis is perpendicular to the spacecrafts trajectory plane), while the spacecraft would still have to orbit half the planet to finally land at the other one ?

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    $\begingroup$ From the Wikipedia article: "The entry sphere weighed 1,560 kg (3,440 lb) and the surface payload was 660 kg (1,455 lb)." It shed the entry sphere much like how M2020 & MSL shed the aeroshell $\endgroup$ Commented Feb 13, 2022 at 16:05
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    $\begingroup$ ...if you're looking for polar ice on Venus, you're going to be disappointed. $\endgroup$ Commented Feb 13, 2022 at 16:09
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    $\begingroup$ @Cornelis I'm very confident that's the whole lander, but no proof. $\endgroup$ Commented Feb 13, 2022 at 16:54
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    $\begingroup$ Coming in from an interplanetary trajectory, with a Venusian atmosphere to aerobrake in, if you have any ability to do fine adjustments en-route, the delta-v requirements are negligible to aim for any landing zone on the side of the planet facing you on arrival. $\endgroup$
    – notovny
    Commented Feb 13, 2022 at 17:57
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    $\begingroup$ @BrendanLuke15 Even if you don't aerobrake, interplanetary distances are large, and planets are tiny in comparison. If you want to pick a point on the planet to smack into (or land propulsively, if you've got the thrust and fuel for it), if you have the control and time, the cost to hit any point on the facing surface is effectively the same. $\endgroup$
    – notovny
    Commented Feb 13, 2022 at 18:05

1 Answer 1


Orbital refilling (as Musk stresses) is critical to any deep space ambitions for Starship. In the latest (Feb. 2022) Starship update Musk said so (re: orbital refilling, emphasis added):

This is going to be an important thing to demonstrate [...] Won't be in the near term for Starlink launches but it will be necessary for Mars and the Moon.

This can be confirmed looking at approximate Starship specs. Starship's LEO payload is ~100+t and Musk raised the possibility of 150 t in the Feb. 2022 update. Even if this payload was all propellant there is not enough $\Delta V$ to even reach the Moon:

Thus an unrefilled Starship cannot escape Earth.

Most (if not all?) sole in-situ atmospheric & beyond missions (i.e., no orbiter component) directly enter the planet's atmosphere from their interplanetary trajectory (makes sense: why bring propellant to slow down where "air" will suffice?). This turns out to be rather difficult to do for a Venus polar mission, not entirely due to dynamics, but geometry.

A typical low energy transfer window to Venus from Earth looks like this in a porkchop plot:

Venus 2025-2026 window (Personal work, click to enlarge)

The spoiler parameter is the Spacecraft-Venus-Earth angle on the right. At low angles the arrival $v_{\infty}$ vector is basically parallel with the Venus-Earth line. It is important to not forget that Venus has significant mass and will deflect the spacecraft's hyperbolic trajectory prior to atmospheric interface and subsequent landing:

low angle example (Personal work)

The above figure is a 3D (unfortunately only for me) visualization of an example low energy (L: 7-Jun-2026, A: 29-Nov-2026, C3=15.9 km^2/s^2, $V_{entry}$=10.8 km/s), low angle (2.34°) trajectory. The black solid lines are 12 individual hyperbolic trajectories that represent constant hyperbolic impact parameter B-plane coordinates (a circle in a B-plane plot). The magenta dots represent the atmospheric interface point (250 km) and the green points represent 0 altitude (no atmospheric/entry effects considered, so a real trajectory would likely have a shorter in-atmosphere down track distance). The dotted black line is the $v_{\infty}$ vector. The red, green, and blue lines show the J2000 ecliptic XYZ (respectively) reference frame. The yellow line is pointing towards the Sun and the light blue line is pointing towards the Earth. Note that the angle between the dashed black line and the light blue Earth line is the Spacecraft-Venus-Earth angle.

There's a lot going on in this figure, and it's unfortunately only 2D for viewers, but understanding it is critical.

The above example uses an impact parameter value that puts the periapsis of the hyperbolic trajectory at Venus' surface ($||\vec{B}||=18,432 km$, entry flight path angle (EFPA) of -12.2°, FWIW). Also note that this visualization/sim ignores the orientation/pointing of Venus' pole, though its tilt and inclination are both low enough to not play that important of a role here (maybe ~5° error between +Z, blue, and Venus' north pole).

Note how not all latitudes are accessible for landing despite spanning all of the "theta" angles in the B-plane. The green dots begin to coalesce in a ring of accessible landing points. The spread of the accessible surface points depends on the arrival energy ($v_{\infty}$) and impact parameter. A lower arrival energy means more deflection and thus more latitude locking.

If you view the above figure along the Earth-Venus line (i.e., from Earth), you'll see this:

low angle from Earth (Personal work)

Notably you don't see any magenta or green dots meaning there is no Earth visibility of the critical EDL events. There are solutions for this (i.e., MarCO-A,B or a relay satellite), but it is another complexity to consider. Also consider that because Venus rotates so slowly those landing sites won't have direct line of site to Earth for ~months.

This is an extreme example, but it is very illustrative and noteworthy that low arrival energy trajectories are associated with low Spacecraft-Venus-Earth angles.

Higher arrival energies and oblique Spacecraft-Venus-Earth angles mitigate this, along with some declination angle for the arrival $v_{\infty}$ vector to "tilt" the ring of accessible surface points up or down towards a pole. Here is a promising looking example of this (using all 3 effects to its advantage):

promising Earth view promising oblique view

(Personal work, Earth view left, L:5-Sept-2026, A:18-Feb-2027, C3=11.6 km^2/s^2, $V_{entry}$=12.1 km/s, S/C-Venus-Earth angle of 54.8°, click to enlarge)

This solves the geometry problem but introduces extreme dynamics problems for Starship. Table 1 from Dutta et al. and this figure from Wercinski et al. show the extreme nature of a Venus direct entry (click to enlarge):

Table 1 Figure

The above plot is for a very different vehicle design (45° sphere cone), but the trends are representative and the absolute values are probably also representative within an order of magnitude. Also note that the above trajectory has a considerably higher entry velocity than any listed in the table.

Note that depending on Starship's loading/configuration its ballistic coefficient, $\beta$, can range from a minimum of a few hundred to a maximum of a few thousand $\frac{kg}{m^2}$. Higher $\beta$ values necessitate steeper entries (larger EFPA) which lead to higher inertial loading and higher peak heating. I will emphasize that the green contours represent intervals of 100's of g's and the blue contours 1000's of $\frac{W}{cm^2}$! This is almost certainly outside of Starship's planned capabilities, and by no means easy to address.

For completeness, work has been done to explore very low stress Venus direct entries (same source for above figure), but Starship as we know it is incapable of those low ballistic coefficients.

For these reasons, and because Starship has an enormous amount of $\Delta V$ (when refilled), I think it makes sense to first insert into orbit around Venus before entry.

A Starship entry from Venusian orbit is probably comfortably within the vehicle's capabilities. Summing the total $\Delta V$ from a 250 km LEO to a low, 300 km Venusian polar orbit (+ 100 m/s for deorbit) for that same 2025-2026 window looks like this:

Total dV 2025-26 window (Personal work)

Interestingly, this construct shifts the window a few weeks earlier than the C3 constrained representation at the beginning. The minimum $\Delta V$ for this window is 7335 m/s and occurs in the less than ideal Type III/IV trajectories in the upper "porkchop" region. It can be shown that the payload mass for a fully fueled Starship with 7335 m/s of $\Delta V$ is ~75 t (using available specs):

$$\Delta V = 7335 = I_{sp} \cdot g_0 \cdot \ln(\frac{120t+m_{pay}+1200t}{120t+m_{pay}}) \to m_{pay}=75t$$

Where 120t and 1200t are the dry and propellant masses of a full Starship, respectively.


  1. Dutta, Soumyo & Smith, Brandon & Prabhu, Dinesh & Venkatapathy, Ethiraj. (2012). Mission sizing and trade studies for low ballistic coefficient entry systems to Venus. IEEE Aerospace Conference Proceedings. 10.1109/AERO.2012.6187002. Free on ResearchGate

  2. Wercinski et al. "Enabling Venus In-Situ Science - Deployable Entry System Technology, Adaptive Deployable Entry and Placement Technology (ADEPT): A Technology Development Project funded by Game Changing Development Program of the Space Technology Program," (2012) NTRS ID: 20130001711

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    $\begingroup$ @Cornelis No, that figure assumes Starship does the entry, descent, and landing. $\endgroup$ Commented Feb 15, 2022 at 13:46
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    $\begingroup$ @Cornelis if you open the imgur link in a new tab and remove the "m" at the end of the url it will show full size, this goes for all of the smaller images. The table and plot next to it are also available in their respective sources $\endgroup$ Commented Feb 15, 2022 at 15:23
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    $\begingroup$ @Cornelis like I said: "I'm very confident that's the whole lander, but no proof". Also like I said: "that figure assumes Starship does the entry, descent, and landing" so the... $\endgroup$ Commented Feb 15, 2022 at 16:42
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    $\begingroup$ @Cornelis ...75t payload + Starship reaches the surface with a terminal velocity of ~30 km/h $\endgroup$ Commented Feb 15, 2022 at 16:43
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    $\begingroup$ Yes, I feel ashamed ! With the formula given by @Mark Adler I calculated a delta-v of 6812 m/s, for the normal "in the planets plane" case. That would be even better than the 6900 m/s that Elon Musk twittered. $\endgroup$
    – Cornelis
    Commented Feb 16, 2022 at 3:29

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