# Would a frog on the side of a rocket be enough to steer it substantially offcourse?

Recently, someone posted this picture in The Pod Bay, our chatroom:

It shows a frog being blown off of the launch of LADEE, a lunar probe.

It got me thinking: Would a frog, successfully holding on to the side of a rocket, be enough to steer the rocket substantially off course if the rocket didn't compensate for it? By how much?

• I could answer this, but I'd need some ballpark values for the mass and length of LADEE. Commented Sep 14, 2013 at 16:03
• @Manishearth - The rocket that launched LADEE to TLI was actually the Minotaur V (PDF), and Wikipedia quotes its height at 24.56 m, and it's mass at 89,373 kg. Other info that you might need should be in one of these two pages linked. ;) Commented Sep 14, 2013 at 16:16
• @TildalWave Thanks. The average speed is around 5000 m/s, right? Commented Sep 14, 2013 at 16:18
• @Manishearth - Didn't have the time to calculate, but its thrust is 1,607 kN and a 83 second burn. You'll have to calculate for the course deviation through the atmosphere tho. I also don't have any numbers on the average weight of a frog. Or their aerodynamic properties. :D Commented Sep 14, 2013 at 16:24
• Assume a spherical frog with density 1.0 gram/cc and mass 10,000 kg.... Commented Sep 15, 2013 at 20:22

There won't be much of an effect, if at all.

Note that this is an order of magnitude calculation. I'll be making approximations/assumptions left and right.

The drag force will be $\frac12\rho v^2c_D A$, and by taking some approximations (as there are too many variables) we get $\frac12 1.2 v^2 0.1 (\pi 0.05)^2 =0.00047v^2$. This can produce a torque up to $F\frac{L}2=0.0056v^2$, which in turn produces an angular acceleration of $\alpha=\frac\tau{m}=6\times 10^{-8}v^2 \:\mathrm{s^{-2}}$

Let's say that the rocket accelerates constantly till it leaves the bulk of the atmosphere (after which point the drag doesn't apply). I'm taking the acceleration as 17 from the thrust values.

Then,$\alpha(t)=6\times 10^{-8}v^2*17*17t^2\implies \theta(t)=6\times 10^{-8}17*17 frac{t^4}{12}$. For the first 10 seconds (when this constant-acceleration scheme holds --after this its velocity stabilizes and the atmosphere is thinner, so the net deviation should be of the same order of magnitude or less), this gives us a deviation of 0.01 radians. The total deviation will be of a similar order of magnitude.

This isn't much. Winds can generate a much larger deviation. Of course, the deviation of a couple hundredths of a radian is a lot when talking about the moon, but again, the deviation due to other effects is a lot more, so the onboard stabilizers can take care of this pretty well.

Values used in answer: Mass of rocket: $90,000\:\mathrm{kg}$, Acceleration: $17.8 \:\mathrm{m/s^2}$, Length: 24 m, Drag coefficient of frog: To the order 0.1, Radius of frog: $0.05 \:\mathrm{m}$. h/t @TildalWave for providing this info :)

• This is... awesome - especially the 'Drag coefficient of frog' part. Pure gold.
– user12
Commented Sep 14, 2013 at 18:28
• Also taking in to affect that the frog was almost certainly below the rocket, and... Commented Sep 15, 2013 at 1:07