Distance from ground to start suicide burn with initial height [closed]

Question

In the figure below, the rocket is dropped with no initial velocity at a height of $h$. $d_f$ (free fall distance) is the distance in which the rocket is in free fall. $v_b$ is the velocity due to free fall when the burn is started. $d_b$ (burn distance) is the distance in which the thruster is then active with constant acceleration. Assume the acceleration vector is straight down. Assume the mass of the rocket is constant. The rocket should then provide just enough thrust so that its velocity is exactly zero exactly as it reaches the ground. This is also known as a suicide burn.

I want to find an expression that calculates $d_b$, or the last possible point at which the rocket must fire its thruster to be able to achieve zero velocity when it reaches the ground. I did some calculations but I am unsure if this is the correct approach.

This is not homework, I am just curious the relationship of height and distance required to burn for different gravitational force and thrusts.

Below is my attempt, can anyone verify if this is correct, or tell me a better way to calculate this?

I tried using these kinematic equations:

$$(v_b)^2 = (v_i)^2 + 2gd_f$$

$$h = d_b + d_f$$

$$(v_f)^2 = (v_b)^2 + 2a_bd_b$$

Setting the initial velocity:

$$(v_b)^2 = (0)^2 + 2gd_f$$

$$(v_b)^2 = 2gd_f$$

Substituting for $d_f$:

$$d_f = h - d_b$$

Setting the final velocity:

$$(0)^2 = (v_b)^2 + 2a_bd_b$$

$$(v_b)^2 = -2a_bd_b$$

Therefore:

$$2gd_f = -2a_bd_b$$

$$2g(h - d_b) = -2a_bd_b$$

$$2gh - 2gd_b = -2a_bd_b$$

$$2gh = -2a_bd_b + 2gd_b$$

$$2gh = (2g - 2a_b)d_b$$

$$\frac{gh}{g - a_b} = d_b$$

Where:

$v_i$ is the initial velocity, which is zero

$v_b$ is the velocity when the burn is started

$v_f$ is the final landed velocity which is zero

$d_b$ is the distance from the ground that the burn is started

$d_f$ is the distance of free fall from the initial position

$a_b$ is the net acceleration upwards when the thrusters are active, with acceleration due to gravity accounted for

$g$ is the acceleration due to gravity (is negative downwards)

Answer 1

Almost got it right. Looks like there may be one sign error (or a definition issue)

I'm a big fan of abusing symmetry to solve problems like these. I hate actually running equations, since I'll miss something. Since you start and end at zero velocity, it's rather nice. I'm also a big fan of approaching problems in multiple different ways to get a result. If you get the same result, you're in good shape. You went down a very strong "write down the equations and turn the crank" approach with a great deal of rigor. I'll go down the more intuitive approach, trying to find related equations which are much simpler, but require more connect-the-dots thinking. If they both agree, that's a really good sign!

You spend some time speeding up, and then some time slowing down. Since $\Delta V=at$, and $\Delta V_f = -\Delta V_b$ (to ensure the net change in velocity is zero), we see that $gt_f=a_bt_b$. Thus $\frac{t_f}{t_b}=\frac{a_b}{g}$.

Now we can look at the distance traveled. The distance traveled falling is pretty straight forward, but we'd need the velocity when we start the burn to use the simple equations to find the distance traveled while burning. Fortunately, physics is symmetric. The distance traveled while decelerating at $a_b$ to 0 velocity is the same distance traveled in the opposite direction: accelerating at $a_b$ from 0 velocity. This happens to be independent of the falling phase, makign the equations easy. Now we have two simple equations:

$$d_f = \frac{1}{2}gt_f^2$$ $$d_b = \frac{1}{2}a_b t_b^2$$

Dividing these, we get

$$\frac{d_f}{d_b}=\frac{gt_f^2}{a_b t_b^2}$$

Now we have a ratio for $\frac{t_f}{t_b}$, so we can use it

$$\frac{d_f}{d_b}=\frac{g}{a_b}(\frac{a_b}{g})^2 = \frac{a_b}{g}$$

Is't that nice and sweet. The ratios of distances are equal to the reciprocal of the ratios of the accelerations.

all that's left is to add $h=d_f+d_b$ into the puzzle. If we rewrite that as $d_f=h-d_b$, we can substitute

$$\frac{h-d_b}{d_b} = \frac{a_b}{g}$$ $$(h-d_b)g = a_b d_b$$ $$gh-gd_b = a_b d_b$$ $$gh = (a_b+g)d_b$$ $$d_b = \frac{gh}{a_b+g}$$

So we almost agree. You have a minus sign, and I have a plus sign. How did that happen?

Looking at your equations, I think I see the line where we differ, right near the top: $$(v_b)^2=(v_i)^2 + 2gd_f$$ $$(v_f)^2=(v_b)^2 + 2a_bd_b$$

Note that the signs of the $2ad$ terms in these two are the same. That means you are defining your accelerations in the same direction, so $g$ and $a_b$ have opposite signs. However, you define these to be:

$a_b$ is the net acceleration upwards when the thrusters are active, with acceleration due to gravity accounted for

$g$ is the acceleration due to gravity

Typically $g$ is defined to be a positive number (such as 9.8m/s on the Earth's surface). The direction of the force of gravity is down. You define $a_b$ to be a "net acceleration upwards," which is the opposite direction of gravity. By that definition $a_b$ is positive as well.

So if you intended to use the definition you gave, my answer is correct. However, if instead you intended to have all accelerations be in a consistent direction (which is a very reasonable thing to do when crunching numbers), then what I call $a_b$, you would call $-a_b$ because you would want a negative number. If you carry this minus sign all the way through my logic, you will indeed get the result you got. If you change your 2nd equation to $(v_f)^2=(v_b)^2 - 2a_bd_b$, you will get the result I got.