# How hard would I hit the ground on Mars?

How do you calculate with how much force a mass would hit the ground when falling from a certain height?

In particular, I'm curious to know how the force of impact on Earth compares with Mars, so surface gravity needs to be accounted for.

According to wikipedia, Mars has a surface gravity of $3.711 m/s^2$ or $0.33895G$ and Earth has a surface gravity of $9.8067 m/s^2$ or $1G$. So, assuming a 100 meter fall, how hard would 100kg hit the ground?

• That's why I'm asking @Floris. I don't understand the subtleties of the question. It's been a long time since I took a physics class. Jan 14 '15 at 20:06
• @Floris If I remember my physics classes correctly then $v = \sqrt{2gh}$ Jan 14 '15 at 20:43
• @Vogel612 - of course you are right. $\frac12 m v^2 = m g h$ so $v = \sqrt{2 g h}$. Thanks for catching that. The formula in my answer is correct. Jan 14 '15 at 20:45
• Downvoter, I'm new here. If there's something wrong with my question, please let me know so I can improve it. Jan 14 '15 at 21:27
• By force do you mean impulse? Jan 14 '15 at 23:42

The answer to this is relatively simple. Assuming an inexisting atmosphere (and thus terminal velocity) on mars, the kinetic energy of the weight equals its potential energy.

Potential energy is calculated relatively simple:

$$E_p = m \cdot g \cdot h$$

now if you wanted to find out how fast the mass would impact on mars, you can set this equal to the kinetic energy you get after completely using up that potential (aka. when you fall 100 m):

$$\frac{1}{2} \cdot m \cdot v^2 = m \cdot g \cdot h$$

now with a little standard mathematics, you arrive at a formula for your velocity, depending on height of where you start from:

$$v = \sqrt{2 \cdot g \cdot h}$$

Interestingly (and as mentioned in the already existing answer) the velocity can be removed from the equation. For the sake of the demonstration let's calculate it either way:

$$v = \sqrt{2 \cdot 3.711 \frac{m}{s^2} \cdot 100 m} = \sqrt {742.2 \frac {m^2}{s^2}} = 27.24... \frac {m}{s}$$

This is a roundabout $100 \frac{km}{h}$, give or take. (or a little more than 60 mph).

Anyways since you wanted to know about the "Force" of the collision...

How much force does it take to stop $100kg$ with a velocity of $100 \frac{km}{h}$?

The question here is... How fast do you want to stop? The slower you stop, the less continuously applied force is required.

When we multiply our velocity and the mass we get an Impulse (which is $F \cdot\Delta t$ or $m \cdot v$). We now know the impulse of our mass: $2700 Ns$

To get this stopped in a tenth of a second (which is an extremely optimistic guess, assuming falling on the hard mars), you need a whopping: $27 kN$

If you absorb the fall from a height $h$ by bending your knees (so you decelerate over a distance $d$) then the force is simply the weight multiplied by the ratio $\frac{h}{d}$ (because the work done by gravity over distance $h$ must be absorbed over distance $d$). The general formula (for any mass m, gravity g, drop height h, absorption distance d) is

$$m\cdot g \cdot h = F \cdot d\\ F = \frac{h}{d} m g$$

So if you can absorb the shock over 1 m, you drop from 100 m, and gravity is $3.7 m/s^2$, then

$$F = \frac{100}{1}\cdot 3.7 \cdot 100 = 37 kN$$

The above explains why falling onto a trampoline or air bag doesn't hurt (much) - and why a hard floor does. It also explains why parachutists try to "fall" as they land - they again try to increase the distance over which they absorb the impact of landing.

If you are interested instead in decelerating in a certain time, then the formula you want is the impulse equation:

$$F\Delta t = m \Delta v$$

From conservation of energy, we get

$$v = \sqrt{2gh}$$

and thence you find

$$F = \frac{m\sqrt{2gh}}{\Delta t}$$

Note that this ignores the force of gravity during the deceleration.

As for the atmosphere: Mars atmosphere has a density of about $\mathrm{0.02 kg/m^3}$. This is <2% of earth's density at sea level so it is pretty safe to ignore it in these calculations.

Also note that the mean surface acceleration on Mars is $\mathrm{3.68 m/s^2}$ - the rotation slows you down about 1% so it matters where you fall (more than or earth because Mars is smaller).

Nasa website with data on Mars

• However air resistance is less on Mars than on Earth. Jan 15 '15 at 8:28
• There is so little atmospheric resistance that the terminal velocity on Mars is more than five times the one on Earth. So if you fall from high enough (around 600m), you will hit the martian ground harder. Jan 15 '15 at 10:23