Last December, Elon Musk the billionaire founder of SpaceX said that the company is planning on catching their new Super Heavy Booster in mid-flight during landing. His aim is to reduce the vehicle downtime between flights in the same way that passenger jets are serviced on the tarmac between flights.
The current plan is for an arm from the launch tower to reach out and grab the grid fins on the rocket while it hovers over the pad in mid-flight. Once the booster is captured in flight, the engines are extinguished and the booster is lowered onto the launch pad. The booster can then be quickly checked, refueled, and relaunched.
But Elon didn’t explain the mechanics of a booster in flight.
What are the flight characteristics of the Super Heavy booster? At present the peculiar specifications of the Super Heavy are dark. However we can still investigate the physics of flight through the lens of the Starship.
Wind Drag
Wind drag is the force imparted on an object within a fluid. We normally consider a fluid to be a liquid like water but gases like the atmosphere also flow like a fluid.
When objects are in motion and at or below the speed of sound, the drag can be computed by the formula,
Where the drag force F is dependent on the density of fluid p, the velocity v, the drag coefficient C, and the projected area A.
The drag coefficient C is the Reynolds number. This is a dimensionless number that is dependent on the geometry of the object. Using a table .82 is a fair approximation for a cylinder.
The last term A is the area of the object. The Starship is a 50 meter tall structure from figures provided by Popular Mechanics. The nose can be approximated as a 8 meter tall cone with a radius of 4 meters. Using the area of a cone A=rland adding this to the area of a cylinder A=2r(h+r)The total area is 1,427.84 meters squared.
We still need the air density to compute the drag. Using tables the average air density at sea level is 1.27 kilograms per cubic meter. Plugging these figures into the first equation gives,
F(v)=743.47v2
We can now compute the drag force on a Starship rocket. Using the charts from Willyweather the average wind speed in Boca Chica, Texas is 13 miles per hour with gusts hitting 28.9 miles per hour. Plugging these values into the equation gives an average wind drag of F=25,096.64 Newtons with gusts reaching 124,104.77 Newtons. The gusts are roughly equivalent to the force of a single large diesel locomotive pressing against the rocket.
Flight Dynamics
The flight of the Starship is governed by the rocket equation. Assuming that the Starship is flying in a gravitational field the equation is,
In this equation c is the exhaust velocity of the rocket engine and is 3034 meters per second for a methane fueled rocket. m is the final mass of the rocket while m0 is the initial mass. The difference in mass is the reaction mass or the total propellant that the vehicle can carry. m dot is the mass flow rate of the propellant into the engines. g is 9.8 meters per second squared, the acceleration of gravity. v is the final speed of the rocket while v0 is the initial speed of the rocket. This is typically zero at liftoff but it could be any speed for a staging or landing rocket. t it is the time of the rocket burn.
These problems are normally complicated to solve and are best graphed, but Elon gave us a shortcut. His plan is for the booster to hover.
The rocket equation is normally solved for the burn of the booster from its liftoff to staging or MECO (Main Engine Cutoff). However if we compute the burn of the hover then it’s simple. Let’s assume that the booster is nearly exhausted of propellant. This seems reasonable for a booster arriving to land. The remaining propellant should then be only slightly higher than the original mass. In this condition the second term in the equation is about zero. The first term is slightly above 1 but we don’t know the exact value. If we set the ratio to 1 then the natural log is 0 and the initial and final velocities are equal. This makes sense for a hovering rocket. Then the velocity is 9.8 meters per second to counterbalance the acceleration of gravity.
We can also compute the forces of a Starship at landing from the equation
F=mawhere m is the final mass of the rocket and a is the acceleration of gravity. The mass of an empty Starship is 120,000 kilograms. Multiplying this with the gravity g produces a downward force of 1,176,000 Newtons. This force is comparable to the thrust of the three Space Shuttle Main Engines (SSME) at liftoff.
Last Thoughts
The great Douglas Adams foreshadowed the Starship with the quote “Flying is learning how to throw yourself at the ground and miss”.
When the rocket is successfully grappled in flight, it will be an amazing feat. A rocket as tall as a 16-story building, pressurized with explosive propellants, and pushed by engines as powerful as the Space Shuttles.
Plus the wind is blowing.
How would you capture the booster?
