Have you ever been outside on a particularely windy day? Not the sort of windy you get here in Oslo, but the proper sort of windy of the western part of the land. The sort where your hair is in your mouth and eyes and you can't even use an umbrella cause it would fly away and take you with it. That sort.
Then you have a pretty good idea of what air resistance is. Whenever something crashes into our atmosphere, the force of air resistance acts opposite to the direction of motion, affecting it. Air resistance is commonly described using the drag equation
\(F_d = \frac{1}{2}\rho C_d A v_{drag}^2\)
where \(F_d\) is the air resistance experienced by a body with cross-sectional area, \(A\) , travelling with speed \(v\) with respect to air with density \(\rho\). The constant \(C_d\) is a dimensional variable called the drag coefficient, which in general is not a constant, but we will assume this is equal to 1.
Before we can land, we need to understand the effects of air resistance on the spacecraft, and so we want to include the atmosphere's motion in the model we previously made for our planet's atmospheric composition.
If you're in a car and you throw a ball in the air, the ball doesn't fly away into the backseat, but rather moves up and then down into your hands again (from your point of view at least). This is because both you and the ball have the same velocity as the car, and so you both move with the car - even if the ball isn't touching the car while it's in the air. The atmosphere (and you!) also moves with the motion of the rotation of the earth. If the car suddenly stopped while the ball was in the air, it would seem to fly forwards, as it still has the velocity the car has when it was let go! So let's hope the earth does not suddenly stop rotating, as this would mean a big collision.
Why are we telling you this? You see, the planet's atmosphere follow the planet's rotation, traveling with the exact same angular velocity! This gives it a velocity \(\mathbf{w}=r\omega \mathbf{e}_{\theta}\) with respect to the planet, where \(\omega\) is the angular velocity of our planet and \(r\) is the distance from the planet to the atmosphere. We use this \(r\) to ensure that the atmosphere has higher velocities the further away from the planet surface it is (see here for why). As we are later going to integrate these vectors numerically, we want to go from polar coordinates to Cartesian coordinates, so let's do something about that pesky \( \mathbf{e}_{\theta}\), shall we? Unit vectors in polar coordinates are described here, but let's do a quick recap anyway. We have both \( \mathbf{e}_{\theta}\) and \( \mathbf{e}_{r}\) in polar coordinates. The latter is easily found, as it's the unit vector in the radial direction, or \(\frac{\mathbf{r}}{|\mathbf{r}|} = \hat{r} = [\cos{\theta},\sin{\theta}]\). Knowing that \( \mathbf{e}_{\theta}\) is perpendicular to \( \mathbf{e}_{r}\), we get \( \mathbf{e}_{\theta}= [-\sin{\theta}, \cos{\theta}]\). (You can take the dot product and see for yourself, it should be 0!) Using this, we can rewrite the velocity vector for the atmosphere like this: \(\mathbf{w}=r\omega [-\sin{\theta}, \cos{\theta}] = r\omega [-\hat{r}_y,\hat{r}_x]\). And voilà, no more polar coordinates!
If we set our spacecrafts velocity with respect to the planet as \(\mathbf{v}\), we can describe our velocity with respect to the atmosphere as \(\mathbf{v}_{drag}=(\mathbf{v}-\mathbf{w})\). (The air resistance works in the opposite direction to the velocity, meaning we write \(-(\mathbf{v}-\mathbf{w})\). This gives us our a new expression for the air resistance
\(F_d(r) = \frac{1}{2}\rho(r) C_d A v_{drag}^2(-\hat{v}_{drag})\)
where we have set density to depend on the distance, and we use the new expression for the \(\mathbf{v}_{drag}\)-vector to add the unit vector \(\hat{v}_{drag}\) so the drag force will work along the direction of the drag velocity.
For our landing to be successful we need to use both a parachute and thrusters. We deploy our parachute at a specific time, once we know that we've braked sufficiently so that the drag force does not exceed 250 000 Newtons, as this would cause the parachute to fail. We plan on launching a lander from the spacecraft at a suitable time, with a velocity we can choose ourselves. Once this lander unit is at a specific distance, the landing thrusters are ignited, hopefully ensuring a smooth landing.
#2: Just after we enter the atmosphere.
#3: After the tangential velocity with respect to the atmosphere is 0.
#4:Reaching terminal velocity.
Before we enter the atmosphere, only the force of gravity \(\mathbf{F_g}=G\frac{Mm}{r^2}\hat{r}\) does work, and then only in the radial direction.
When we enter the atmosphere our velocity will be very high, and so the air resistance will be higher than the force of gravity, actually giving us a negative acceleration and slowing us down (both in the tangential and the radial direction)!
As the gravitational force only does work in the radial direction, the air resistance in the tangential direction (always against the velocity) will eventually have slowed the rocket down enough for it to have 0 velocity in the tangential direction!
As we slow down, the air resistance will lower until the force of gravity is equal to the drag force, \(\sum \mathbf{F} = \mathbf{F}_g - \mathbf{F}_d = 0\), and at this point the velocity will be constant, what we call terminal velocity! If we assume that \(v_{\theta}=0\), and that we are at a point where \(\mathbf{F}_g=\mathbf{F}_d\), we can write
\(\frac{1}{2}\rho Av_r^2 = G\frac{Mm}{r^2}\)
and by rearranging these for \(v_r\) we have an expression for the terminal velocity:
\(v_r=\sqrt{\frac{2GMm}{r^2\rho_0 A}}\)
We can rearrange this for \(A\) in order to find the cross-sectional area of the lander
\(A=\frac{2GMm}{r^2\rho_0 v_r^2}\)
If we set \(v_r=3\) m/s (the maximum velocity required for a soft landing) this gives us \(A\) if we assume to know \(r\) at the point where the sum of the forces are equal, and \(\rho_0\), as we know the pressure at the surface of our planet.
Finally, we should know how to find the force of our thrusters in order to be able to determine the velocity we want when landing. We know that the force of our thrusters should be equal to the force of gravity and the drag force, \(F_L=F_g-F_d\), and if we assume we've reached terminal velocity we know that \(F_g=F_d\), meaning that we can write
\(\frac{1}{2}\rho_0Av^2_{t}= \frac{1}{2}\rho_0Av^2_{safe}\)
where \(F_d\) is dependent on the terminal velocity and \(F_g\) is dependent on the maximum velocity required for a soft landing, giving
\(F_L = \frac{1}{2}\rho_0A(v_t^2-v_{safe}^2)\)
As this force works in the radial direction of motion, we are going to add a unit vector \(\hat{r}\) to showcase the direction:
\(\mathbf{F}_L = \frac{1}{2}\rho_0A(v_t^2-v_{safe}^2)\hat{r}\)
And with that, we can simulate the landing using Euler-Cromer. Stay tuned for our next post, where we will both simulate the landing, and perform the actual landing!