Rocketship problem- fuel burn and gravity

1. Nov 12, 2008

Savant13

Consider a spaceship, initially at rest on the surface of a planet that has no atmosphere. It takes off with constant force from thrust. The fuel mass decreases linearly with time, and the rocket accelerates directly away from the planet (no curving off orbit the planet). Use GMm/r^2 for gravitational force. Produce an equation describing the rocket's distance from the planet for all time from 0 when the rocket starts to burn fuel until it runs out of fuel.

I came up with this problem a while ago for fun, but it turned out to be much harder than I had imagined. The method that got me the farthest is energy density.

2. Nov 13, 2008

rcgldr

There was an earlier thread somewhat similar to this. The math got really nasty, and if I recall, the variables couldn't be seperated to form a proper equation. This could be one of those cases where differential equations and numerical intergration are needed.

3. Nov 13, 2008

Savant13

I was afraid it would be something like that. If someone could show how that would be done, I would appreciate it.

4. Nov 18, 2008

MrEnergy

E_{f}=∫_{R}^{R+h}[G⋅((M⋅m)/(r²))dr] ===> (integral from R to R+h and E_{f}=Fuel energy)

E_{f}=G⋅M⋅m⋅(h/(R(R+h)))

is it the solution?? By calculating the fuel energy in terms of joule and defining R, you can find the maximum height i guess...

5. Nov 18, 2008

rcgldr

No, because "m" decreases as fuel is burnt.

6. Nov 18, 2008

Savant13

That can't tell you where you are at a given time, and you didn't take loss of mass into account. Mass loss is accompanied by kinetic energy loss

7. Nov 19, 2008

MrEnergy

oh sorry i forgot!!! bad mistake:P

Anyways I have remade my answer and found a very long one. All the integrals might need to change into sum!Don't know if it's correct... but this is what i found:

i)The mass function depending on time is: M(t)=-(w/t0)

t+q+w ==> where q=mass of spaceship and w=initial mass

of fuel.(t0 is the time that fuel ends)

ii)let the thrust force be F, a be the acceleration of

spaceship;
F-M(t)g(t)=M(t)a then,(since mass and gravitional force changes over time)
(F/M(t))-g(t)=a(t) ===>if V is the velocity of spaceship then;

V=at so the V function depending on time is;
V(t)=[(F/M(t))-g(t)]t

iii)integral(from 0 to t0)[V(t)dt]+integral(from t0 to t1)[g(t)tdt]==>gives the maximum height reached.

to find t1 we use;

(1/2)g(t)(t1-t0)2=integral(from 0 to t0)[V(t)dt]

iv)We have to define g(t) if the missile gets too high because then the g would change of course, but if not we can use "g" and not bother rest.

g(t)=GMM(t)/(R+integral(from 0 to t)[V(t)dt])2 ===>V(t) explained later

v)Finally since F is constant(till fuel ends up) and we

know the dx(maximum height) we can now equlize the E_{f}

(fuel energy) to F*dx

F*dx=E_{f} i think that gives it. only t is variable.

others like q, w, t0 ... are constant.But the

integration might be changed into a sum if needed. It got a little messy but ill put some graphs to make it clearer.Hope its correct now.

Last edited: Nov 19, 2008