I want to make a rocket that will travel from Earth then fly from the moon then reach Mars Most of the things are done I just need the formula

  • $\begingroup$ First build something which really works and you can do it! It is enough if it can go a hundred meter high. Then use what you've learned building it, to build some bigger. But, the first step is, you need to show something what others can see! And then, step by step. $\endgroup$ – peterh - Reinstate Monica Sep 9 '19 at 19:52
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    $\begingroup$ I think he's making a simplified simulator with Python, Java, JS (Idk), not literally building a rocket. Even then, there's more to that than just knowing how to calculate thrust. $\endgroup$ – Star Man Sep 9 '19 at 20:07
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    $\begingroup$ @Devarsh Newar F=ma can be used to find the force (thrust) required to move a mass at a certain rate of acceleration. In reality, you want to use the Tsiolkovsky Rocket Equation to find the delta-v as it takes into account the decreasing mass from the burning propellant. $\endgroup$ – Star Man Sep 9 '19 at 20:30

What is the formula to calculate thrust needed to launch from the Moon to Mars?

I think this is a reasonable question and it shouldn't be so down-voted.

Here's how to calculate roughly how much thrust you need to launch from the Moon and reach Mars' orbit. Landing safely on Mars once you approach it is a separate and difficult challenge because of the atmosphere.

To escape the Moon's gravity you'll first need escape velocity of about 2400 km/s. You can calculate that from

$$v_{esc} = \sqrt{\frac{2GM_M}{R_M}}$$

where $G$ is the gravitational constant and $M_M$ is the mass of the Moon. You can find a table of the product $GM$ for the Moon and many other bodies, called the standard gravitational parameter. Note that the units there are meters, not kilometers.

Now you are free of the Moon, moving around the Sun at the Earth's orbital velocity around the Sun plus the Moon's orbital velocity around the Earth.

You can calculate those using the vis-viva equation.

$$v = \sqrt{GM \left( \frac{2}{r}-\frac{1}{a} \right) }$$

where $r$ is the radius from the orbiting body at the moment, and $a$ is the semimajor axis of the orbit. Lets call those orbits circular, so we can set $r=a$ and for circular orbits only write the simlified

$$v_{circ} = \sqrt{\frac{GM}{a}}$$

So using that table of standard gravitational parameters $GM$ for the Sun and the Earth, the Earth's velocity around the Sun is about 29,700 m/s and the Moon's velocity around the Earth is about another 1000 m/s.

Remember to make sure all of your units are meters and seconds! It's really easy to accidentally use some numbers in km and mess up the calculation.

Let's assume you are far enough from Earth that you can ignore the additional bump you'll need to completely escape Earth's pull, being so far away from it already.

Now you are going around the Sun with a velocity of 29,700 + 1,000 = 30,700 m/s, at a distance of 1 AU. What is the semimajor axis of your new orbit?

We can flip the vis-viva equation around to get:

$$\frac{1}{a} = \frac{2}{r} - \frac{v^2}{GM_{Sun}}$$

That turns out to be 160 million km. You are currently at your perihelion of 150 million km, which means that your farthest distance from the Sun (on the other side) will be 170 million km. But Mars' closest approach is out at about 208 million kilometers, so you'll need another bump in velocity.

If you need an orbit with a perihelion of 150 million km and aphelion of 208 million km, then your semimajor axis will have to be (150 + 208)/2 = 179 million km. The velocity of that orbit at $r=$ 150 million km will be 32,100 m/s, but you are only going 30,700 m/s, so in addition to your initial 2,400 m/s lunar escape velocity you'll need another 1,400 m/s, or a total of 3,800 m/s delta-vin order to reach Mars' orbit at its perihelion of 208 million km, doing it my way.

However, there's a better way. As pointed out in this answer to * Why is Wikipedia's delta-v from the Moon to Mars transfer so much lower than mine?* you can do it for less delta-v by using the Oberth effect. Instead of launching from the Moon directly into a transfer orbit, you should launch towards the Earth. Not directly at it, but in a hyperbolic orbit that swings past it. When you are at a very low altitude (say 200 km) you can do a 600 m/s burn instead of the 1,400 m/s burn away from the Earth.

That will put you in line with the numbers shown in the second plot below.

from the Moon to Mars transfer

Delta-Vs for inner Solar System

Source: Delta-Vs for inner Solar System

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    $\begingroup$ I think the downvotes stem more from a lack of showing personal work, but great answer. I like reading these high level explanations you do. Helps to cement the basic concepts in my head even more. $\endgroup$ – Magic Octopus Urn Sep 10 '19 at 14:57

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