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I am trying to create an equation which allows for me to change the aspects of the rocket so i can calculate the distance traveled vertically. My idea is for a rocket that only moves vertically; with this i can calculate the amount of time it would take to make it past the first Lagrange point. So far I have made an equation which calculates the acceleration of the rocket, this being: $$ \frac{32T}{W_0+F-Bt}-\frac{Gm_e(W_0+F-Bt)}{r^2} $$

$T$=Thrust

$W_0$=Initial weight

$F$=initial weight of fuel

$B$=Burn rate(lbs of fuel per second)

$t$=time in seconds

$G$=gravity constant

$m_e$=mass of earth

$r$=radius from earths center of gravity

Thank You for taking the time to read this.

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closed as off-topic by ACuriousMind, CuriousOne, Kyle Oman, Qmechanic Sep 4 '15 at 22:23

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  • $\begingroup$ What about the velocity of the ejected gas? $\endgroup$ – Oussama Boussif Sep 4 '15 at 15:48
  • $\begingroup$ There doesn't seem to be a question here. Also, your "equation" lacks an equals sign? $\endgroup$ – Kyle Oman Sep 4 '15 at 16:55
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You've almost got it!

The constant thrust comes from a mass rate $\mu$ of fuel being expelled at a velocity $v_f$ as opposed to the speed of the rocket itself $v$. Therefore the equation is instead:

$$ (m_0 - \mu t) \frac{dv}{dt} = \mu v_f - \alpha \frac {(m_0 - \mu t)}{r^2},$$

where $\alpha = G M.$ Hence the gravitational term you wrote as $G m_e (W_0 + F - Bt) / r^2$ is deeply incorrect as an acceleration (it is a force!) and your equation has a type error. The corrected equation is simply:

$$ a = \frac{dv}{dt} = \frac{\mu v_f}{m_0 - \mu t} - \frac{\alpha}{r^2}.$$

Of course if $\alpha = 0$, the solution to this is famously $$\int dv = v - v_0 = \int dt~\frac{\mu ~v_f}{m_0 - \mu t},$$ and defining $m = m_0 - \mu t;\;dm = -\mu~dt,$ one gets$$v - v_0 = -\int dm ~ \frac {v_f}{m} = - v_f~\ln\left(\frac{m}{m_0}\right).$$In other words, starting from rest, you have $v(t) = -v_f \ln\left( 1 - \frac{\mu t}{m_0}\right).$

A crude heuristic then equates $v_1$ at the end of rocket acceleration with the escape velocity of the planet you're exiting, $v_e = \sqrt{2\alpha/r_0},$ hence you need $m_0 / m_1 = \exp(v_f / v_e).$ Earth's escape velocity is 11.2 km/s, some rocket exhaust speeds can be in the 3-4 km/s range, so that suggests that you need $m_1/m_0$ to be something like 10-50ish, or in other words your payload tends to be somewhere between 2% to 10% of the needed weight.

If you're going to model this equation in more depth than that, you should probably add a friction force $a_f = -\gamma v^2 / (m_0 - \mu t)$ and simply solve the equations numerically. It's about a day's coding, maybe less, to get some reasonably fast explorations of the parameter space in NumPy.

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You are assuming constant thrust $T$ during flight, presumably until the rocket runs out of propellant.

You are also assuming (or neglecting) any air drag to be zero. The resulting drag force could be very significant at high speeds and assuming your rocket is launched from the Earth's surface.

With those limitations in mind the balance of forces of the rocket becomes:

$T=m(t)a+F_g$.

$a=\frac{T}{m(t)}-\frac{F_g}{m(t)}$.

Here $m(t)=W_0+F-Bt$ and $F_g=G\frac{m_em(t)}{r^2}$.

Substituting we obtain:

$a=\frac{T}{W_0+F-Bt}-G\frac{m_e}{r^2}$ (I'm assuming that by 'weights' $W_0$ and $F$ you mean masses).

So I don't know where your factor $32$ came from.

The total burn time is $t_b=\frac{F}{B}$.

Since as $a=\frac{d^2r}{dt^2}$, in principle we have a second order Differential Equation in $(r,t)$ which when integrated between $t=0$ and $t=t_b$ would yield the total distance traveled during thrust. Slightly rearranged it becomes:

$\frac{d^2r}{dt^2} + \frac{Gm_e}{r^2}= \frac{T}{W_0+F-Bt}$

But this is a non-linear Differential Equation and I don't believe it has analytical solutions, only numerical (iterative) ones.

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  • $\begingroup$ But normally the speed of the ejected gas should be taken in consideration right? $\endgroup$ – Oussama Boussif Sep 4 '15 at 16:20
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    $\begingroup$ @Oussama Boussif: To calculate thrust, yes. But here the OP specifies thrust $T$ to be constant. It would be better to start from the Rocket Equation: en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation $\endgroup$ – Gert Sep 4 '15 at 16:32
  • $\begingroup$ ah okay thanks. I guess I just misunderstood the term thrust(I didnt know it had something to do with the ejected gas). $\endgroup$ – Oussama Boussif Sep 4 '15 at 16:36

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