# How much would cost to take a kg to the same distance of the planet than LEO in Mars?

In Mars it's needed a much lower velocity to attain orbit. How would this impact the costs of taking 1 kg to the orbit of the planet, at the same distance of the planet than Low Earth Orbit but in Mars?

The cost in money terms depends rather dramatically on your assumptions. If your rocket for launching from Mars has to be transported from Earth, then it's going to be pretty pricey, since a very much larger rocket will have been needed to launch it from Earth. On the other hand if you are imagining that Mars has been colonised and has a thriving local rocket-manufacturing industry, it could be quite cheap.

What we can give though is the velocity needed to reach that orbit, and from that, and some assumptions about the launcher we can make a guess at how big it will need to be, and how much fuel it will need.

The velocity is given on wikipedia as 4.1 km/s. This tells us, via the rocket equation

$${\displaystyle \Delta v=v_{\text{e}}\ln {\frac {m_{0}}{m_{f}}}=I_{\text{sp}}g_{0}\ln {\frac {m_{0}}{m_{f}}}}$$

the mass ratio (the ratio of starting mass $$m_0$$ on the surface of Mars to mass $$m_f$$ put in orbit (including empty fuel tanks, rocket engines, etc. as well as your 1kg payload). We can try a few plausible fuel combinations which will give us different values for $$I_{sp}$$:

• A solid rocket motor, such as a space shuttle solid rocket booster, with an $$I_{sp}$$ of 250, gives us a mass ratio of 5.3
• Standard storable hypergolic liquid propellants (widely used for long space missions), $$I_{sp}$$ about 340, mass ratio 3.3
• Liquid methane and liquid oxygen (SpaceX proposal) $$I_{sp}$$ about 360, mass ratio 3.1
• Liquid hydrogen and liquid oxygen (highest $$I_{sp}$$ commonly used) $$I_{sp}$$ about 450, mass ratio 2.5.

So, assuming an optimistic 1kg of rockets and tanks for your 1kg payload, you'd need somewhere between 5 and 11kg of rocket on the surface of Mars.

This ignores atmospheric resistance, among other things, which would be a big problem for a rocket this small, even, I suspect, on Mars.

Let's check the orbital velocities of low orbits around Earth and Mars.

Mars is 1/2 the diameter and 1/8 the mass. We can use the vis-viva equation to calculate the velocity of a circular orbit.

$$v = \sqrt{\frac{GM}{a}}$$

GM (m^3/s^2)      R (m)       a = R + 400 km   v (m/s)
Earth     3.986E+14      6.378E+06       6.778E+06       7905
Mars      4.283E+13      3.396E+06       3.796E+06       3551

It's a little harder to estimate the difference in the delta-v penalty to lift out of each planet's gravity well by 400 km because it depends in part on how fast you do it which means it depends on the specific rocket and how much you limit acceleration, but it's probably safe to estimate that for the small, uncrewed payload it will be roughy half on Mars than it is on Earth as well.

So for the same payload and exhaust velocity, using the rocket equation in the same way as in @SteveLinton's answer we can estimate that the mass of the rocket will be lower by the factor:

$$\frac{m_M}{m_E} = \exp\left( \frac{v_M - v_E}{v_{ex}} \right)$$

Assuming the exhaust velocity is moderate at 3000 m/s the Mars rocket's propellant is only 13% as massive as the Earth's rocket. Of course this is only a rough approximation, but we can say that it's going to be a much smaller rocket. (for 4000 m/s the ratio is about 21%).

Now the cost of putting that rocket on Mars to begin with, or building and fueling it there is going to be astronomical!