How do you calculate the pressure that is exerted on a surface (say from a rocket engine) at an arbitrary altitude? For example, pretend I know the exit velocity and pressure of a rocket engine at an altitude of 100 m, how would I calculate the pressure exerted on the surface by the rocket exhaust? I assume you would need to know the ambient pressure as that affects the pressure exerted on a surface. I know how to calculate pressure exerted on a surface which is calculated with $P = {{F}\over{A}}$. However that only works if the object is touching the surface and thus, does not apply for rocket engines and its exhaust.

  • 1
    $\begingroup$ What do you mean by surface? Surface of what? $\endgroup$ Nov 8 '19 at 0:47
  • $\begingroup$ @AdrianHoward The ground. For example, a rocket that's 100 m off the ground, what would be the pressure on the ground exerted by the rocket? $\endgroup$
    – Star Man
    Nov 8 '19 at 1:49
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    $\begingroup$ Are you talking about the force exerted by the exhaust gases? $\endgroup$
    – Bob D
    Nov 8 '19 at 9:18
  • $\begingroup$ @BobD Yes.That is correct, $\endgroup$
    – Star Man
    Nov 8 '19 at 20:36
  • $\begingroup$ @StarMan in that case, It is a common misconception that a rocket's exhaust pushes on the ground. See the following: courses.lumenlearning.com/boundless-physics/chapter/… $\endgroup$
    – Bob D
    Nov 8 '19 at 20:53

As people in the comments pointed out, the pressure exerted at the ground will go to 0(due to rocket thrust) very quickly if the atmosphere is present. To solve that problem exactly, you will need to solve Navier's stokes equation, which as you would probably know, is difficult because of the non-linearity and has to be done numerically. I can give you some idea about what will happen in a vacuum. Let's assume the rocket is using ion thrusters and ejecting gas it at a speed $v_{rel}$ with mass rate $dm/dt$ relative to the rocket. As the gas is not hindered by particles in the atmosphere and neglecting the collisions due to it's own fuel bouncing back from the ground, it will reach the ground with speed $v_{fin} = \sqrt{({v_{rel}+v_{rocket}})^2+2*g*h}$, assuming g to be constant at small heights. Now, the pressure caused due to the gas is $F/A$ which is $\frac{2(\cos\phi) v_{fin}dm}{A_{fin}dt}$, assuming elastic collisions and $\phi$ is the angle at the point you are measuring. In this equation, the $A_{fin}$ will be the effective area that the fuel lands in. Assuming the area of the rocket exhaust circular with radius $r_{init}$ and it shoots the fuel with a divergence of angle $\theta$, you will get $A_{fin}=\pi (r_{init}+h\tan \theta)^2$. You can make the formula more accurate by relaxing more and more assumptions.

  • $\begingroup$ This simplification does not give the right answer, because it neglects the fluid dynamics of the exhaust stream itself, which can be quite complicated. Even though the rocket is in a vacuum, the exhaust plume is certainly not vacuum. $\endgroup$
    – Al Nejati
    Nov 12 '19 at 19:41
  • $\begingroup$ @AlNejati In an ion thruster rocket, the rocket gain momentum by accelerating particles through an electromagnetic field. Their ejection velocity is much higher than their random velocity and therefore, the assumption that the particles do not collide midway. Moreover, I have taken the exhaust to be launched not directly vertically down but in a cone of angle $\theta$ . $\endgroup$ Nov 13 '19 at 9:28
  • $\begingroup$ @RishabhJain That is only for a ion thruster rocket. What about for a conventional combustion rocket engine? I assume that will be more complicated and requires a vast knowledge in fluid dynamics. $\endgroup$
    – Star Man
    Nov 13 '19 at 20:42
  • $\begingroup$ @StarMan In a rocket where the thrust relies more on the amount of mass ejected rather than the velocity of the ejection, you are right that you would need to fall back on Navier stokes only. $\endgroup$ Nov 14 '19 at 9:53
  • $\begingroup$ Also, you can see from youtube videos like youtube.com/watch?v=49eIoaY8pVM that the landings even on the earth have a significant impact on the ground at few 100 meters. $\endgroup$ Nov 14 '19 at 9:58


If we take the aggressive assumptions that the exhaust stream maintains its shape and faces zero drag going back to the surface, then we can just use $F=ma$ to decelerate the jet, or even conservation of energy. Or the similarly large assumption that we know it’s speed, $v_f$, when reaching the ground.

This just calculates what level of force streams put on the ground when hitting. In a vacuum, or very low altitude, or with particle exhaust, the results would give estimates worth considering.

F= ma

Assume we know the speed when it hThe force to stop a column of exhaust (from $v=v_f$ to $v=0$) over some time $\Delta t$ on a mass $M$. At $t=0$ the amount of column being accelerated is $m=0$, and by $t=\Delta t$, the whole column is finishing up being accelerated.

$$F= (m) (\frac{dv}{dt})=(m_{average})(\frac{\Delta v}{\Delta t}) =(\frac{M}{2})(\frac{\Delta v}{\Delta t}) $$ $$= \tfrac{1}{2}(\frac{M}{\Delta t})(v_f-0) = \tfrac{1}{2} \dot{m}v_f$$ $$P=\frac{F}{A}=\frac{\dot{m}v_f}{2A} $$

Of course $M$ and $\Delta t$ were arbitrary and dropped out.

Conservation of energy

Alternatively, we can do work on the exhaust column of mass $M$ and length $L$ in a frame where the column is still, and use that work to impart kinetic energy (decelerate it to $v=0$ in a still frame, a frame where the jet is doing work on the ground in impact and damage and hence jet is losing energy).

$$\tfrac{1}{2}M(v_f^2-0) = (F)(L)= (F)(v_f~dt)$$

$$\implies F= \tfrac{1}{2}\dot{m}v_f$$

$$P= \frac{\dot{m} \sqrt{v_i^2+ 2gh}}{2A} $$

The two methods agree.


The good thing about the energy derivation is that it implies we could adjust if the velocity profile is not a constant velocity across the exhaust jet.

Here is prof Skakya discussing a jet hitting a wall. He does only the first method ($F=ma$ to decelerate the stream) and gets a similar answer but does not break down the column as I did to verify, nor do conservation of energy: https://youtu.be/kNL5w73owzU

If we have particles:

Assuming they fall to actually gain speed, let $v_i$ be the exhaust’s velocity relative to the earth when it leaves the rocket (That might be the normal exhaust stream speed, minus the current velocity of the rocket.. if those are what we know.) $v$ is the exhaust speed when it hits the earth. Then the speed at earth is:

$$\tfrac{1}{2} m~(v_f^2-v_i^2) = mgh \implies v_f= \sqrt{v_i^2+ 2gh}$$

If we know the area of impact, it’s not necessary to assume that it maintains its shape, but there is still the strong assumptions that it stays evenly dispersed, all of it reaches the impact area, and most of all that it didn’t lose speed due to drag (actually gained speed from falling). If the speeds are high, we can estimate loss, if we know particle sizes, from $$F_D = \tfrac{1}{2} \rho v_f^2 C_D a$$

Where $\rho$ is for air.


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