A rocket and its gas exhaust velocity

Homework Statement:
Along the present problem you may assume that there is no air friction, that the ignition processes of the different rocket phases are instantaneous, and that fuel capsules have negligible mass.

a) Consider a 1-phase rocket such that its gas exhaust velocity is $u$, its initial mass (payload plus fuel) is $m_0$, and its final mass (payload) is $m_f$. Assuming that the rocket is vertically launched from the Earth's surface, and that the Earth is at rest, determine the amount of fuel required for the rocket to reach a maximum height $h$.
Relevant Equations:
Energy conservation
This isn't right, is it?
$$-\dfrac{GM}{R}+\dfrac12 v^2=-\dfrac{GM}{R+h}$$
$$v=\sqrt{\dfrac{GM}{R}}\left( 1-\sqrt{\dfrac{R}{R+h}}\right)$$
He's doing energy conservation. The mechanical energy at the Earth's surface is equal to the energy when the speed is $0$.

anuttarasammyak
Gold Member
Then how do you determine the amount of fuel required to v ?

This isn't right, is it?
$$-\dfrac{GM}{R}+\dfrac12 v^2=-\dfrac{GM}{R+h}$$
$$v=\sqrt{\dfrac{GM}{R}}\left( 1-\sqrt{\dfrac{R}{R+h}}\right)$$
Don't worry about the rocket part, it's just this step

Then how do you determine the amount of fuel required to v ?
You can find the speed with
$$\Delta V=u\cdot \ln \left( \dfrac{m_0}{m_f}\right)$$
It's Tsiolkovski's equation.

jbriggs444
Homework Helper
He's doing energy conservation. The mechanical energy at the Earth's surface is equal to the energy when the speed is $0$.
Yes. In order for this step to become meaningful, one has to reason a bit about the problem. What pattern of burn will give you the best performance?

Do you burn steadily from surface to final altitude?
Do you hover for half an hour on your rockets and then burn hard and fast?
Do you burn halfway up, then turn off the rockets and coast?
Do you burn as hard and as fast as you can immediately and then coast the rest of the way?

Two heuristics help make that decision easy. First, the longer your flight takes, the more time the Earth's gravity has to decrease your upward velocity. Second, any excess velocity you have when you get to the goal is wasted.

anuttarasammyak
Gold Member
Tsiolkovski's equation is derived from the law of conservation of momentum in case of no external forces. I wonder if it stands with gravity force working.

I agree with v of OP as the initial speed of projectile popped up with no self propulsion. So the case is the rocket gets speed v by instantaneous burning of all the fuel at launching, like a cannon or a bullet.

Last edited:
haruspex
Homework Helper
Gold Member
2020 Award
I believe the question being asked by @Guillem_dlc is purely in regard to this algebraic step:
$$-\dfrac{GM}{R}+\dfrac12 v^2=-\dfrac{GM}{R+h}$$
$$v=\sqrt{\dfrac{GM}{R}}\left( 1-\sqrt{\dfrac{R}{R+h}}\right)$$
And I agree, that is quite wrong. It should yield
##v=\sqrt{\frac{2GMh}{R(R+h)}}##.

I did wonder if it is a defensible approximation, but it does not appear to be.

Delta2
I agree with v of OP as the initial speed of projectile popped up with no self propulsion. So the case is the rocket gets speed v by instantaneous burning of all the fuel at launching, like a cannon or a bullet.
Yes, in this problem it is considered to burn all the fuel instantly at first and continue until gravity cancels out all the initial speed.

I did wonder if it is a defensible approximation, but it does not appear to be.
I also think that, but I think it might indicate the speed variation.

Otherwise, there's no meaning in the minus sign.

If you do the velocity variation considering circular trajectories I think that's correct. Since the speed: $$V=\sqrt{\dfrac{GM}{R}}$$

Although not on top it shouldn't have speed

anuttarasammyak
Gold Member
Yes, in this problem it is considered to burn all the fuel instantly at first and continue until gravity cancels out all the initial speed.
OK. So you should consider how much fuel of relative speed u should go out instantly so that the rocket body gets velocity v, not by the post #4.

Delta2
Homework Helper
Gold Member
I think the algebra is wrong there (or there is a typo), it should have been
$$v=\sqrt\frac{2GM}{R}\sqrt{1-\frac{R}{R+h}}$$ which yields the same expression as that of post #7

And i dont know if we can use the approximation $$\sqrt{1-\frac{R}{R+h}}\approx \frac{1}{\sqrt{2}}\left (1-\sqrt\frac{R}{R+h}\right )$$ but thats the only way to justify the book's formula.

haruspex
OK. So you should consider how much fuel of relative speed u should go out instantly so that the rocket body gets velocity v, not by the post #4.
Yes

which yields the same expression as that of post #7
And here you substitute the velocity for the expression of $u\cdot \ln \left( \dfrac{m_o}{m_f}\right)$ and isolate the mass

"I) Transfer from initial (circular) to transfer (elliptical) orbit.
- The initial trajectory of $m$ is a circumference centred in the massive object $M$ of radius $r=r_p$. The velocity of $m$ is constant, and can be easily obtained as follows:
$$e=-\dfrac{GM}{2r_p}=\dfrac12 v_p^2-\dfrac{GM}{r_p}$$
$$v_p=\sqrt{\dfrac{GM}{r_p}}"$$
Perhaps it is related to this as the $2$ seems to disappear, and the rocket always makes an elliptical orbit when it leaves the Earth.

Delta2
jbriggs444
- The initial trajectory of $m$ is a circumference centred in the massive object $M$ of radius $r=r_p$. The velocity of $m$ is constant, and can be easily obtained as follows:
$$e=-\dfrac{GM}{2r_p}=\dfrac12 v_p^2-\dfrac{GM}{r_p}$$
$$v_p=\sqrt{\dfrac{GM}{r_p}}"$$
Perhaps it is related to this as the $2$ seems to disappear, and the rocket always makes an elliptical orbit when it leaves the Earth.