Would the jerk of this rocket be constant?

Question

I was working on this interesting problem I came up with for a couple hours and got stuck - decided to ask here.

A rocketship with is accelerating upward with constant thrust. Ignoring air resistance and the change in mass due to loss of propellant, what would the graph of $j(t)$ (the jerk of the rocket at a given time) look like? Constant? Linear? Something else?

Here is what I have considered thus far:

• While the thrust is constant, the force of gravity is not. As the rocket accelerates further and further away from the surface, the force of gravity becomes ever so slightly weaker. It is obviously a negligible effect at first, but it is non-zero.
• Therefore, the net force on the rocket is increasing, so by Newton's Second Law, the rocket's acceleration is increasing. So the jerk is not equal to zero.
• Newton's Universal Law of Gravitation states that the force of gravity is inversely proportional to the square of the distance between the rocket and the center of the Earth. Therefore, transitively, $a \propto \frac{1}{d^2}$
• As displacement from the earth increases, the acceleration increases, but the acceleration causes displacement to increase at faster and faster rates. Is there a mathematical way to account for this dual relationship?

How would all of this work?

Answer 1

Jerk is the change in acceleration, and in your scenario acceleration is constant and therefore jerk is zero.

But no, because now we are considering the change in gravity due to height. So at any point, the acceleration is a function of height $h$ only.

$$ a= \frac{F}{m} - \frac{GM}{(R+h)^2} \tag{1}$$

Considering the velocity as $v = \tfrac{\rm d}{{\rm d}t} h$, jerk is defined by the chain rule

$$ j = \tfrac{\rm d}{{\rm d}t} a = \tfrac{{\rm d} a}{{\rm d}h} \tfrac{{\rm d}h}{{\rm d}t} = v \left( \frac{2 G M}{(R+h)^3} \right) \tag{2}$$

But since velocity $v$ changes with height, this is still not an answer. But we can integrate (1) to get the velocity at each height $h$.

$$ \tfrac{1}{2} v^2 - \tfrac{1}{2} v_0^2 = \int_0^h a\,{\rm d}h $$

$$ v = \sqrt{ v_0^2 + \frac{2 F h}{m} - \frac{2 G M h}{R(R+h)} } \tag{3} $$

So now we can find jerk as a function of height

$$ j = \sqrt{ v_0^2 + \frac{2 F h}{m} - \frac{2 G M h}{R(R+h)} } \left( \frac{2 G M}{(R+h)^3} \right) \tag{4} $$

If initially, the rocket is at rest $v_0=0$ and considering different forces which are more than the weight at the surface, then the shape of jerk vs. height looks something like this

Now the peak of the curve (maximum jerk) occurs at the following height (with $\gamma$ the ratio of Force/Weight)

$$ h_\text{max jerk} = R \frac{\sqrt{9 \gamma^2 -17 \gamma + 9} -2 \gamma + 3}{5 \gamma} \tag{5}$$

In the limiting case where $\gamma=1$, the force equals the weight, then $ h_\text{max jerk} = \tfrac{2}{5} R$, which is a rather large distance, considering $R$ is the radius of the earth.

The higher the force, the sooner peak jerk occurs. The closest it can be though is $\tfrac{1}{5}R$ when $F \rightarrow \infty$