# General relationship between gravity and rocket design.

1. Jan 2, 2016

### mrspeedybob

I was pondering Mars missions and wondered about what kind of vehicle an astronaut on Mars would need in order to get home. My first thought was that mars gravity is .38 that of earth, so he would need a rocket about 38% as large to get off of mars as he needed to get off of earth, This of course neglects the fact that when he left earth he had to take his return vehicle with him, and also neglect the fact that the martian atmosphere is thinner, but at this point I'm not ready to think those factors yet.

So, in order to understand the relationship between the gravity of the planet and what size rocket that planet would require in order to leave it, I decided to try understand 4 simpler scenarios...

1. Rocket leaves earth and enters earth orbit.
2. Rocket leaves mars and enters mars orbit
3. Rocket leaves earth, travels to mars on a 1 way trip.
4. Rocket leaves mars, travels to earth on a 1 way trip.

If you neglect the atmosphere and only consider gravity, does #2 require a rocket 38% as large as #1, or are there compounding factors that make the relationship exponential or some other function of gravity?

Is the difference between #3 and #4 basically just the difference between 1 and 2 plus the ability to change orbital energy with respect to the sun by the same amount? In other words, if identical craft were already in orbit around earth and mars, could each make the transfer into orbit around the other planet? Or are the requirements different one way vs. the other?

2. Jan 2, 2016

### Bandersnatch

Yes.
It all comes down to the mass of propellant that you need the rocket to have to overcome gravity given its mass, and how more propellant on board means the rocket is heavier, therefore requires more propellant, which makes it heavier, and so on.
You can calculate the mass using the Tsiolkovsky equation:
https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation
It is tied to gravity via the escape velocity term, which is easy to calculate.

At least to the first approximation, they should be the same. Changing orbits means changing orbital energy, and the energy difference stays the same no matter which way you go.

3. Jan 2, 2016

### mrspeedybob

It looks to me like that would be a lower limit on the mass of the rocket, assuming instant acceleration to orbital velocity. Is that correct?
It seems to me like finite acceleration would require a larger rocket. The extreme case being a rocket with zero acceleration that just hovers and can use an infinite amount of fuel without going anywhere. So when calculating the delta V that I'd need to use for the Tsiolkovsky equation I should use orbital velocity + average gravitatioan acceleration x burn time. Does all this sound right?

4. Jan 2, 2016

### Bandersnatch

I had to look at the derivation, and good that I did, because it turns out I mislead you - the escape velocity is not the $V_e$ in the Tsiolkovsky equation. That's the effective exhaust velocity. Orbital/escape velocity is in the delta-v. Shows how well I remember this stuff.

Yes and no. It is a lower limit, because it doesn't include gravity and aerodynamic drag slowing the rocket down. It also doesn't account for changes in direction of thrust as the rocket climbs. But it doesn't assume instantaneous acceleration. The effective exhaust velocity accounts for the efficiency of the rocket engine in terms of how quickly it delivers the force.

I guess. But look at the wiki page for 'gravity drag' - there's more to consider, including the effect of direction of thrust.

Looking more into it - it would be the same, if you were only concerned with changing orbits w/r to the sun (and ignoring Oberth effect). But in this case you're also concerned with escaping the gravitational well of the planet you're at/orbiting, which are different.
They've got a nice breakdown of the delta-v budget for a Mars ('interplanetary' table) mission here:
https://en.wikipedia.org/wiki/Delta-v_budget

In any case, it's hard to sound authoritative with an egg on the face, so I'll maybe pass this on to somebody else, and play some Kerbal SP in the meantime. ;)

5. Jan 2, 2016

### Janus

Staff Emeritus
To escape Mars, you would need ~2kg of fuel for every kg you want to lift, compared to ~11 kg of fuel per kg to escape the Earth ( assuming a 4.5 km/sec Ve for your rocket)
However, for the same craft to land on Mars and then take off again would require 8.35 kg of fuel per kg. ( the rocket equation is logarithmic). Because of this, it would make sense to leave behind on Mars anything you are not going to need for the return trip. (in fact, you might even leave part of your lander behind, much like the LEM did).