Payload to LEO of the BFR 2017 compared to the 2016 version

Question

The 2016 Mars vehicle design had a payload to LEO of 500 tons in expendable mode.

Its replacement, the 2017 "BFR" design, is planned to have a payload to LEO of 150 tons with reuse.

What would the payload of the 2016 version have been be if reused? And what would the payload of the 2017 version be in expendable mode?

Then we would know how much weight penalty the downsizing cost.

Answer 1

We can make a base assumption first about how much $\Delta V$ we need to get to LEO and if you wanna update that the rest of the calculations are super easy to redo. Let's say we need $10km/s$ for now.

A stage has a $\Delta V$ given by Tsiolkovsky's Rocket Equation:

$\Delta V = g_0 \cdot I_{sp} \cdot ln \Bigl( \frac{m_{dry} + m_{prop} + m_{payload}}{m_{dry} + m_{payload}} \Bigr) $

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In 2016, Elon Musk's presentation in Guadalajara mentioned that the booster uses 7% of its propellant to perform the boostback and landing maneouvres, and this gives us almost all of the information we need. The one quantity which we are unsure of is the value of $I_{sp}$ to use for the first stage, since it is constantly changing as the ambient atmospheric pressure changes around it during ascent. A quick and dirty (but surprisingly accurate) approximation is to take the mid-point of the sea-level and vacuum values, and round up a bit. In this case we get

$I_{sp} = (334 + 361)/2 = 347.5s$

So let's just take 350s. Now, our $\Delta V$ is given by

$\Delta V = 9.81 \cdot 350 \cdot ln \Bigl( \frac{275 + 6700 + 2100 + m_{payload}}{275 + 0.7\cdot6700 + 2100 + m_{payload}} \Bigr) + 9.81 \cdot 382 \cdot ln \Bigl( \frac{2100 + m_{payload}}{150 + m_{payload}} \Bigr) $

Wolfram Alpha can solve this for us, for $\Delta V = 10km/s$. We get a payload mass of ~303t.

All figures are taken from the IAC presentation slides here:

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This year, Musk didn't go into detail about the downsized booster so we don't know its dry mass or propellant mass, which is a problem. But to get some idea, we could assume that the booster has the same propellant mass fraction. (It's a bit hand-wavy, but you can make different assumptions and follow the same methodology.) Then we can derive the dry mass and propellant mass, modify our expression to use 100% of propellant, and then solve for our new payload mass.

Firstly, the 2016 booster had a propellant mass fraction of

$pmf_{2016} = \frac{6700}{6700+275} = 0.961 $

Then our 2017 booster gives us

$4400 - 1335 = 3065 = \frac{m_{prop}}{0.961}$

$ m_{prop} = 2944.2t$

$ m_{dry} = 3065 - 2944.2 = 120.8t$

Plugging these numbers in actually gives us $ \Delta V = 9.685km/s$ but we can live with that. In reality, the $pmf_{2017}$ will be higher since the stage has a lower dry mass than in 2016.

Now for an expendable version, we get

$\Delta V = 9.81 \cdot 345 \cdot ln \Bigl( \frac{2944.2 + 120.8 + 1185 + m_{payload}}{120.8 + 1185 + m_{payload}} \Bigr) + 9.81 \cdot 375 \cdot ln \Bigl( \frac{1185 + m_{payload}}{85 + m_{payload}} \Bigr) = 9685 $

Solving this using Wolfram Alpha gives us an expendable payload mass of 183t.

All figures are taken from the IAC presentation slides here: