Subsidiary question: Imagine a sphere 10cm in diameter in low venusian orbit. Slow it down a little in order to deorbit it. What's the density of the sphere, in order to touch the ground at 0 vertical m/s, before climbing up again in the venusian sky because it's less dense than venusian atmosphere? (regarding pressure gradient, aerodynamic drag, high speed winds effects on trajectory, and other things i forget.(see images below))(rough approximations and thoughts are welcome)

Pressure on surface of Venus is 90 times greater than Earth's sea level pressure.

Are there studies about some sort of "low & variable density buoyancy braking lander" designed with removable onionlike heatshields -or a single deflatable heatshield- which would provide control over the density -and therefore the speed- of the whole lander during the descent?

The idea is about bringing multipurpose to parts, in order to minimise the number of parts. Aerobraking starts in high altitude, and stops on the floor, Buoyancybrake should start at a precise altitude, and stop at surface level, 0 m/s vertical speed, with the separation of the last low density, buoyant-heatshielding onionskin layer.

The less onionskins layers needed in the descent, (of low density high temperature resistant & thermical insulant, some sort of aerogel(?)) the better.

links towards atmospheric plots:



venus buoyant lander venus atmosphere graph pressure temperature altitude venus winds

  • 1
    $\begingroup$ The question is, is this even necessary on Venus? Venera 9 ditched its last parachute at an altitude of 50 km, and descended at low speed using a simple horizontal metal disk as an aerodynamic brake. $\endgroup$
    – Hobbes
    Apr 25, 2017 at 19:11

1 Answer 1


Surprisingly the answer is yes there was studies done on that subject.

A simple google search could yield this result:



In this study, it was assured that the large buoyant volune is deployed prior to atmospheric entry. The effect of buoyancy on the entry dynamics was investigated, using a first-order entry model. That is, a two- dimensional entry trajectory, a perfectly spherical planet, a constant gravity, and no wind were assumed. It was found that the effect of buoyancy on the velocity, maxiimm deceleration, and altitude of maximum deceleration of planetary entryvehiclesisinsignificant. Thisistrueforallentryangles, even if the entry velocity is decreased considerably by rocket braking, and even if the buoyant volune diameter is very large (greater than 500 feet). There is one case, however,for which the buoyant effect is not altogether insignificant, though still small. This is the case of equilibrium-gliding entry. For example, for constant lift-drag ratios of 0.1 and spherical buoyant volune diameters of 300 feet, the maxiirain deceleration is decreased by 2.6% for Mars and 1.8% for Venus from the value of maximum deceleration for non-buoyant entry vehicles. For constant lift-drag ratios of 1.0 and diameters of 300 feet, the maximum deceleration is decreased by 0.8% for Mars and 0.7% for Venus.

However, unsurprisingly, the result is that the buoyancy effect is insignificant.

  • $\begingroup$ Thank you for your answer, this study talks about Earth's atmospheric pressure, in which buoyancy effect is insignificant, parachutes work better. What about Venus, 90atm pressure and a light weight lander? $\endgroup$
    – user19132
    Apr 12, 2017 at 8:20
  • $\begingroup$ You are mixing two different steps in your question: The braking (where you need a heatshield; in the upper atmosphere, where the pressure is < 1 ATM, even on venus), and the descent where buoyancy can be significant. $\endgroup$
    – Antzi
    Apr 12, 2017 at 8:41
  • $\begingroup$ Edited question, @Antzi I'd like to mix the two, thinking of a solution where the lander brakes all the way until it touches the ground. $\endgroup$
    – user19132
    Apr 12, 2017 at 9:01
  • $\begingroup$ A more practical design would probably deploy a balloon once a safe speed is reached. $\endgroup$
    – Antzi
    Apr 12, 2017 at 9:59
  • 1
    $\begingroup$ @uhoh links added in the main question; I'll rewrite the subsidiary question to make a new one in relation to this one, thanks $\endgroup$
    – user19132
    Apr 25, 2017 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.