Calculating Fuel-Payload Ratio for Martian Rocket Lift-off

[Flisp]

Watching "Mars" on National Geographics I try to calculate what the cost would be to bring anything usefull back to Earth. More specifically I wonder what fuel-payload ratio a rocket could have when lifting off from Mars. I want to use Tsiolkovsky's rocket equation

Δv = v_e ln(m₀/m_f)

or

Δv = I_sp g₀ ln(m₀/m_f)

That contains g (9,8) and I thought I could replace it with g for Mars (3.7) but that gets me nowhere since that a lower g results in a lower Δ v, when it should not.

The fule-payoad ratio of a Saturn V for high orbit is about 60:1 (m₀/m_f). Cost for one launch about 1 billion $. I want to calculate how many launches are needed to bring the parts and fuel to Mars needed to assemble and tank a rocket, that can lift some payload and landing vehicle from Mars and bring it back to Earth. For that I need a Tsiolkovsky equation with a differnt ”g”, so to say. Isp I assume to be the same for both Saturn V and return rocket.
Any suggstions?

 

[jbriggs444]

Δv = ve ln(m0/mf)

or

Δv = Isp g0 ln(m0/mf)

That contains g (9,8) and I thought I could replace it with g for Mars (3.7) but that gets me nowhere since that a lower g results in a lower Δ v, when it should not.

The "g" in the latter equation is there in order to convert specific impulse into exhaust velocity. It is there because specific impulse has Earth gravity built into the unit -- it is the number of seconds that a quantity of fuel can support its own Earth weight.

Launch from Mars is going to take enough delta v to achieve Mars escape velocity. [The rest of the trip back to Earth is going to take additional delta-v]. You can calculate Mars escape energy by multiplying Mars surface gravity times Mars radius. From there it is easy to get escape velocity.

I pulled that formula for escape energy right out of my back side. Let us give it a sanity check using Earth as a test case.

Earth g is 9.8 meters per second. Earth radius is 6371 km. Multiplying the two we get 62435800 meters squared per second squared. Or, equivalently, 62 megajoules per kilogram. $E=\frac{1}{2}mv^2$. Solving for v, that's $v=\sqrt{2\frac{E}{m}}$. We've calculated $\frac{E}{m}$ as 62 megajoules per kilogram. Evaluating gives v=11,174 m/sec. Googling for the escape velocity of Earth, we see that it is 11.2 km/sec. Looks like the formula passes the sanity check.

Edit: I did not find that sanity check compelling enough so...

More formally, gravitational potential is the path integral of the dot product of gravitational acceleration a times incremental path length ds on a path leading from a test point out to the reference point at infinity. Any path will do. We can use a path leading straight up for convenience.

Gravitational force F is given by $F=G\frac{Mm}{r^2}$. For a small test mass m, gravitational acceleration, a is given by $a=G\frac{M}{r^2}$. Our path is directly upward and the acceleration is directly downward, so there is a minus sign on the resulting dot product. We want to evaluate $\int_r^\infty -G\frac{M}{s^2}\ ds$. That's an easy integral -- the integral of $\frac{-GM}{s^2}$ is $\frac{GM}{s}$. So we evaluate $\frac{GM}{s}$ at s=infinity (yielding zero) and at s=r (yielding $\frac{GM}{s}$) The difference is gravitational potential per unit mass = $-\frac{GM}{r}$

The result comes out negative as it should. Potential at the surface is negative relative to a reference at infinity. But that uses Mars mass M and Newton's universal gravitational constant G. We want to get Mars surface gravity in there instead.

Mars surface gravity $g_m$ is given by $g_m=\frac{GM}{r^2}$. Or, equivalently, $GM=r^2g_m$. Substituting that into $-\frac{GM}{r}$ we obtain $-g_m r$

So we have the desired result. Gravitational potential (per unit mass) at the surface is negative and is given by surface gravity times the radius of the gravitating body.

 

[Flisp]

The "g" in the latter equation is there in order to convert specific impulse into exhaust velocity. It is there because specific impulse has Earth gravity built into the unit -- it is the number of seconds that a quantity of fuel can support its own Earth weight.

Thank you! I can more or less follow the math. To put it into use I want to do this: Leaving everything constant I calculate a Mars-standard $I_{sp}$ by multiplying with $g_{earth}/g_{mars}$.
I don't know what final velocity a rocket really has to Mars. But whatever it is, I can use the same $v$ for the trip back and I know that this velocity is smaller by the difference between the escape velocity from Earth and the one from Mars. And then I solve all that for my fuel-payload ratio. Does that sound right?

 

[Janus]

Thank you! I can more or less follow the math. To put it into use I want to do this: Leaving everythin constant I calculate a Mars-standard $I_{sp}$ by multiplying with $g_{earth}/g_{mars}$.

No. the ISP is a rating for the rocket engine. It remains the same no matter what planet you are launching from (for example, a Liquid hydorgen-LOX engine has a ISP of ~450). What you need to do is work out what type of delta V you would require to lift your lander into orbit, and then use these number to to get the mass ratio and then figure out the mass ratio needed for a rocket engine with a given ISP.
So, for example, if we assume our return craft is in orbit around Mars at an altitude of 100 km, then we can use
$$ V = \sqrt { GM \left ( \frac{2}{R_m}- \frac{1}{a} \right )}$$
where
M is the Mass of Venus
R_m is the radius of Mars
a is R_m+ 50km
To work out how much delta v you would need to get your lander up to orbital height .
Using the 450 ISP given above, this works out to ~3.58 km/sec.
Once reaching this altitude, the lander will be at the apapis of an elliptical orbit and moving little bit slower than the return craft. You can use the formula below to work
In this case it will work out to be ~.05 km/sec for a total dv of 3.63 km/sec. For comparison, escape velocity from the surface of Mars is ~5 km/sec
Using an isp of 450, and solving for the mass ratio, we get an MR of ~2.28, meaning you need a minimum* of 1.28 kg of fuel per Kg of payload to get your lander to the the return craft.

How this effects the fuel needed to get to Mars in the first place depends on the ISP of the rocket you use doing so (a chemical rocket will require much more fuel than the more efficient ion rocket)* in reality you'll need more, as this assumes an ideal trajectory which is not practical for an actual Mars lift off. For example, if we were to use the same method to calculate the mass ratio for the Saturn V, you would get a smaller value for the same orbital insertion that you gave a value of 60 for.

I don't know what final velocity a rocket really has to Mars. But whatever it is, I can use the same $v$ for the trip back and I know that this velocity can smaller by the difference between the escape velocity from Earth and the one from Mars. And then I solve all that for my fuel-payload ratio. Does that sound right?

Going either way uses the pretty much the same total delta v.
Leaving LEO to LMO:
1.You change velocity to leave LEO
2.You change velocity to enter the transfer trajectory
3. You change velocity to match speed with Mars
4 You change velocity to enter LMO

Coming back you just reverse the order, with the magnitudes of each velocity change being the same (other than that caused by any shift in the relative positions of Earth and Mars for the transfer trajectory.)
The total delta v would be the sum of all these velocity changes.
The real difference comes in during the return to Earth from LEO, where you can make use of Earth's thick atmosphere to do a lot of the braking.

 

[Flisp]

Thank you, Janus. It will take me some time to work through all that. I came back when I have som eresults or questions...

 

[Flisp]

I am sorry, but this is for me simultaniously to simple and to complicated.
If it have $\frac {Δv} {v_e} = ln \frac {m_0} {m_f}$ then the $\frac {Δv} {v_e}$ -part somehow expresses the performance of the ship. Now I know that $\frac {m_0} {m_f} = 83$. This value I can find for the SpaceX Heavy for shipping cargo to Mars. Whatever $Δv$ may be (including all the unfavorable trajecories, course corrections, air-resistance, deccelerations upon arrival and other factors and values that I have no information about) and whatever $v_e$ may be, (given that there are different stages with different fuels and $I_{sp}$, that I also have no information about), it is still reasonable to assume that the return rocket has (about) the same $\frac {Δv} {v_e}$ for the travel back. However, due to Mars lower gravity $\frac {m_0} {m_f}$ ought to be lower. How much lower? If I go through all through all these steps

1.You change velocity to leave LEO
2.You change velocity to enter the transfer trajectory
3. You change velocity to match speed with Mars
4 You change velocity to enter LMO

without the correct values for $Δv, v_e$ and $I_{sp}$ and all the other factors the sum of all faulty assumptions will make my results useless.
So, how do I plug the lower gravity on Mars into Tsiolkovsky's equation so I can solve it for $\frac {m_0} {m_f}$?

 

[jbriggs444]

I am sorry, but this is for me simultaniously to simple and to complicated.
If it have $\frac {Δv} {v_e} = ln \frac {m_0} {m_f}$ then the $\frac {Δv} {v_e}$ -part somehow expresses the performance of the ship.

The $\frac {Δv} {v_e}$ tells you the requirements placed upon the ship -- how many times greater than exhaust velocity is the required delta v for the trip. That, in turn, tells you the minimum required ratio of gross vehicle weight at takeoff to payload delivered on arrival.

Now I know that $\frac {m_0} {m_f} = 83$. This value I can find for the SpaceX Heavy for shipping cargo to Mars.

So instead of starting with the mission requirements, you are starting with an already-worked design to get there. That is exactly backward.

 

[Flisp]

Yes, I know it is backward. But still. If $Δv$ can be assumed to be the same both ways, and $v_e$ can be lower due to lower gravity on Mars, than I can solve for a new $\frac {m_0} {m_f}$. I can, so to say, use the same rocket, travel at the same speed, but repalce some of the fuel with payload. The $v_e$ I have now is a result of Earth gravity. There must be a simple way to determine how much lower $v_e$ can be based on the ratio between Earth gravity and Mars gravity.

 

[jbriggs444]

Yes, I know it is backward. But still. If $Δv$ can be assumed to be the same both ways, and $v_e$ can be lower due to lower gravity on Mars, than I can solve for a new $\frac {m_0} {m_f}$. I can, so to say, use the same rocket, travel at the same speed, but repalce some of the fuel with payload. The $v_e$ I have now is a result of Earth gravity. There must be a simple way to determine how much lower $v_e$ can be based on the ratio between Earth gravity and Mars gravity.

Surely by $v_e$ you mean exhaust velocity. If not, what are you talking about?

 

[Flisp]

Surely by vev_e you mean exhaust velocity. If not, what are you talking about?

Well, sure, $v_e$ must be replaced by something else. Like by $I_{sp}*g$ or something. Someting containing Earth and Mars gravity

 

[jbriggs444]

Well, sure, $v_e$ must be replaced by something else. Like by $I_{sp}*g$ or something. Someting containing Earth and Mars gravity

Exhaust velocity is not a parameter that you are free to modify based on your starting location. It depends on your choice of propellant, not your planet of origin.

All other things being equal, choosing the propellant that maximizes $v_e$ minimizes $\frac{\Delta v}{v_e}$ which, in turn minimizes $ln(\frac{m_{rocket}}{m_{payload}})$ which minimizes $m_{rocket}$ required for a given payload.

 

[Flisp]

OK, then I rephrase my question. I have the following known values for a rocket carrying payload to Mars: Mars gravity $g_m = 0.37$, Earth gravity $g_e = 9.8$ and a fuel-payload ratio $fpr = 83$ for the travel from Earth to Mars. What $fpr$ would be allowed for a trip back, using the same (now refilled) rocket, the same speed, the same whatever there might be relevant exept for the lower gravity. And in what equation do I put the values?

 

[Flisp]

That sounds to me like a very strange answer. The value of $\frac {Δv} {v_e}$ derives from the conditions caused by Earth gravity. Lower gravity must alter these conditions in a simple and predictable way.

 

[jbriggs444]

That sounds to me like a very strange answer. The value of $\frac {Δv} {v_e}$ derives from the conditions caused by Earth gravity.

No, it does not. But perhaps you should define for us what you mean by Δv.

 

[Flisp]

Then I am pretty much att a loss here... If I told you the $I_{sp}$ of a rocket and the $Δv$ of, let'say $16 km/sec$, could you then calculate my $fpr$ from that? Or what more would you need?

 

[jbriggs444]

Then I am pretty much att a loss here... If I told you the $I_{sp}$ of a rocket and the $Δv$ of, let'say $16 km/sec$, could you then calculate my $fpr$ from that? Or what more would you need?

Please define specific impulse for us.

My concern is that you believe that specific impulse on Mars is somehow different from specific impulse on Earth in a way that means that exhaust velocity is different for the two.

 

[Flisp]

I can copy that from Wikipedia: Specific impulse (usually abbreviated Isp) is a measure of how effectively a rocket uses propellant or a jet engine uses fuel. I understand what that means. But you said yourself that it is a value that tells the number of seconds that a quantity of fuel can support its own Earth weigh. So while a specific propellant surelly has a specific $I_{sp}$ calculated for or measured under the conditions here on Earth, if you did the same measurements for the same propellant under different conditions you would derive on a different value. The same propellant could support its own weight on Mars for a longer time. $I_{sp}$ as a standard value would be the same, of course, but the number of seconds would differ.

 

[jbriggs444]

I can copy that from Wikipedia: Specific impulse (usually abbreviated Isp) is a measure of how effectively a rocket uses propellant or a jet engine uses fuel. I understand what that means. But you said yourself that it is a value that tells the number of seconds that a quantity of fuel can support its own Earth weigh. So while a specific propellant surelly has a specific $I_{sp}$ calculated for or measured under the conditions here on Earth, if you did the same measurements for the same propellant under different conditions you would derive on a different value. The same propellant could support its own weight on Mars for a longer time. $I_{sp}$ as a standard value would be the same, of course, but the number of seconds would differ.

And exactly what effect does this have on exhaust velocity? or on a computation of fuel to payload ratio?

 

[jbriggs444]

A fuel with a specific impulse of 1 can support its own weight for one second. In one second, the object would have attained a velocity of 9.8 meters per second. By Newton's third, that means that the corresponding exhaust velocity is 9.8 meters per second. Two seconds = 19.6 meters per second. And so on.

We can compute exhaust velocity by multiplying specific impulse (in seconds) by 9.8 meters per second per second, yielding a velocity in meters per second. A quick trip to wikipedia gives agreement that this yields "effective exhaust velocity".

So yes, given a total delta V and a specific impulse, we can compute a fuel payload ratio.

$$g=9.8 m/sec^2$$
$$\Delta v=16 km/sec$$
$$v_e=g\ I_{sp}$$
$$ln\ (fpr) =\frac{\Delta v}{v_e}$$
$$fpr = e^{\frac{\Delta v}{v_e}}$$

 

[Flisp]

If would not effect $v_e$. But in order to reach the same $Δv$ when lifting off from Mars you need less fuel. Since the $Δv$ in Tsiolkovsky is constant (I say it is constant) and the fuel-payload ratio neccessarily must be lower when lifting off from a planet with lower gravity only the $x$ in $\frac {Δv} {x}$ can be changed. If $v_e$ where also constant it meant that gravity does not matter when you lift off from a planet. This is of course not true. In order to arrive at a different $fpr$ for the same $Δv$ under lower gravity you must change something. Ergo: $v_e$ can not be constant. So, you tell me that $v_e$ IS contant. But $v_e = I_{sp} * g_0$ and while also $I_{sp}$ is constant $g_0$ can not be. But since $g_0$ is definitely constant it must be $I_{sp}$. However way you turn it. In order to arrive at a differnt $fpr$ you must change something.

 

[jbriggs444]

If would not effect $v_e$. But in order to reach the same $Δv$ when lifting off from Mars you need less fuel. Since the $Δv$ in Tsiolkovsky is constant (I say it is constant) and the fuel-payload ratio neccessarily must be lower when lifting off from a planet with lower gravity only the $x$ in $\frac {Δv} {x}$ can be changed. If $v_e$ where also constant it meant that gravity does not matter when you lift off from a planet. This is of course not true. In order to arrive at a different $fpr$ for the same $Δv$ under lower gravity you must change something. Ergo: $v_e$ can not be constant. So, you tell me that $v_e$ IS contant. But $v_e = I_{sp} * g_0$ and while also $I_{sp}$ is constant $g_0$ can not be. But since $g_0$ is definitely constant it must be $I_sp$. However way you turn it. In order to arrive at a differnt $fpr$ you must change something.

Again, please define $\Delta v$ for us.

 

[Flisp]

So yes, given a total delta V and a specific impulse, we can compute a fuel payload ratio.
g=9.8 m/sec^2

So, if I put $g_{mars} = 3.7$ into that equaion I arrive at the value I seek? No, I do not.

 

[jbriggs444]

So, if I put $g_{mars} = 3.7$ into that equaion I arrive at the value I seek? No, I do not.

That is correct. You do not.

 

[Flisp]

$Δv$ is the sum of all speed changes of the rocket on its way home with 0 at start, including all course corrections, deccelerations etc that are needed and assumed to be equal both ways.

 

[jbriggs444]

Let me have a second try reading through your post below.

If would not effect $v_e$. But in order to reach the same $Δv$ when lifting off from Mars you need less fuel.

I do not understand this sentence at all. The delta v Earth to Mars is the same as the delta v Mars to Earth. At least until we consider atmospheric braking.

Since the $Δv$ in Tsiolkovsky is constant (I say it is constant) and the fuel-payload ratio neccessarily must be lower when lifting off from a planet with lower gravity only the $x$ in $\frac {Δv} {x}$ can be changed.

You are assuming a false conclusion here. The fuel-payload ratio must be the same (up until we consider the possibility of atmospheric braking).

If $v_e$ where also constant it meant that gravity does not matter when you lift off from a planet. This is of course not true.

It actually is true. It does not matter whether you are landing or taking off in a gravity well, the required delta v is the same in the two directions. If you contemplate a complete trip from takeoff to landing, it does not matter whether you take off from the low gravity end and land high or take off from the high gravity end and land low.

The rest of the post goes downhill from a false premise.

The win you get on the trip from Mars to Earth is that instead of burning a ton of fuel to slow down and land on Earth you can use atmospheric braking -- ablate some heat shield and carry some parachutes. You can model that as a $\Delta v$ budget that eliminates or reduces the expenses for slowing and landing on Earth.

 

[Flisp]

Let me have a second try reading through your post below.
You are assuming a false conclusion here. The fuel-payload ratio must be the same (up until we consider the possibility of atmospheric braking).

What? How can that be? The fuel-payload ratio was certainly not the same both ways when we flew to the moon!

 

[jbriggs444]

What? How can that be? The fuel-payload ratio was certainly not the same both ways when we flew to the moon!

Atmospheric braking. And the use of a separate command module/lunar module.

Have you looked at https://en.wikipedia.org/wiki/Delta-v_budget#/media/File:Delta-Vs_for_inner_Solar_System.svg

Edit to add an unrelated comment:

$\Delta v$ is what you need.
$v_e$ is what you have.
The fuel to payload ratio is what Tsiolkovski's rocket equation forces on you to get a craft that can use what you have to get what you need.

 

[Flisp]

So, what you basically say is, if we go from Earth to some place and burn fule at a $fpr$ of 60, we will need the same fpr to get back, regardless if our goal is a tiny asteroid with nearly no gravity, an Earth sized planet with the same gravity, Jupiter (if we could land on it), or for that matter a black hole. You can't be serious.

 

[jbriggs444]

So, what you basically say is, if we go from Earth to some place and burn fule at a $fpr$ of 60, we will need the same fpr to get back, regardless if our goal is a tiny asteroid with nearly no gravity, an Earth sized planet with the same gravity, Jupiter (if we could land on it), or for that matter a black hole. You can't be serious.

I am perfectly serious. In the absence of atmospheric braking, that is the truth. It takes just as much fpr to brake to a stop on the Earth's surface as it took to launch from Earth's surface.

Edit: I think that time reversal still works with trajectories that stay outside a black hole horizon, but I could easily be mistaken. General relativity is not easy on the intuition.

[Having played with lunar landing programs in the 70's, I can also say that the optimal landing burn looks just like the optimal launching burn in reverse. You wait until the last moment and go full thrust, hoping you got it exactly right so that you touch down just as you come to a stop.]

 

[Flisp]

Oh, that was a surprise! Does that also mean that if we assume atmospheric breaking on both ends of the travel, it will be the same fuel-payload ratios both ways? And that that is the very simple answer to my question?

 

[jbriggs444]

Oh, that was a surprise! Does that also mean that if we assume atmospheric breaking on both ends of the travel, it will be the same fuel-payload ratios both ways? And that that is the very simple answer to my question?

No. It is asymmetric. You gain a big advantage with atmospheric braking on Earth (high escape velocity/orbital velocity and good available atmosphere). Not so much on Mars (much lower escape velocity/orbital velocity and thinner atmosphere) and none at all on the moon (even lower escape velocity/orbital velocity and no atmosphere).

So there is some advantage to be had on the outbound trip but a much greater advantage to be had on the return.

 

[Flisp]

But if we would go from orbit to orbit (outside atmosphere) it would be symmetrical?

 

[jbriggs444]

But if we would go from orbit to orbit (outside atmosphere) it would be symmetrical?

Yes, indeed.

 

[Flisp]

Yes, indeed.

Great, thank you. And if I want a more accurate comparisson I have to piece the trips together the way you discribed it before. Earth to orbit, orbit to orbit, orbit to Mars surface and then back.

 

[jbriggs444]

Great, thank you. And if I want a more accurate comparisson I have to piece the trips together the way you discribed it before. Earth to orbit, orbit to orbit, orbit to Mars surface and then back.

Credit where credit is due, I think that was @Janus, not me.