# The Martian

## The Movie

The fantastic “The Martian” film came out in 2015 and itself was based on the equally good book by Andy Weir published in 2011. Matt Damon plays an astronaut who is presumed dead in an accident and left behind on Mars. He has to use a lot of science and maths to make it through a range of hairy survival situations.

## Survival Problem 1: Food

Astronaut Mark Watney is marooned and it looks like he’s going to be stuck on the planet for a long time. How long? Well, he estimates it’s 1400 *sols* (Martian days which are 24.5 earth hours long) until he will be rescued.

NASA provided the original team of 6 crew with enough food for about 50 sols.

Since there’s only one person left, that can be multiplied by 6:

*Sols supply = 6 × 50*

*Sols supply = 300*

Now since times are desperate, Watney can also ration out the food to last a bit longer – about 1.33 times longer. So that gets him a bit more time:

*Rationed supply time = ration factor × normal supply duration*

*Rationed supply time = 1.33 × 300*

*Rationed supply time = 400 sols*

That leaves Watney with another 1000 sols without food. His only solution – to grow potatoes on Mars using his own faeces (poo) as fertilizer! Yuck!

Mark calculates he needs a minimum of 1500 calories per sol. For the remaining 1000 sols:

*Total extra calories required = sols left × calories required per day*

*Total extra calories required = 1000 × 1500*

*Total extra calories required = 1500000*

He needs to grow a whopping 1.5 million calories of potatoes!

Mark estimates a single potato has about 150 calories. That means he needs to grow a total of:

*Number of grown potatoes required = total calories required / calories per potato*

*Number of grown potatoes required = 1500000 / 150*

*Number of grown potatoes required = 10000*

**A note about calories:** when people talk about calories in food, they’re actually talking about “kilocalories” (1 thousand calories). For example, “that burger has 250 calories” actually means that burger has 250 *kilocalories*, or 250,000 calories.

1 kilocalorie is equivalent to 4.184 kJ, which is another unit of energy in food that is used in many places around the world.

## Survival Problem 2: Disaster Strikes

At one point, a catastrophic explosion exposes Mark’s entire potato farm to the Martian atmosphere and kills his remaining crop. All he has are the rations, and the crops he’s already grown, which he estimates to be about 400 plants, with an average of 5 potatoes per plant.

We can calculate how long he can survive with just this limited crop and his rations:

Revised survival estimate = ration time + crop time

The rations we already know will last him about 400 sols. The crops will last him:

*Crop time = number of plants saved × number of potatoes per plant × calories per potato / calories required per day*

*Crop time = 400 × 5 × 150 / 1500*

*Crop time = 200 sols*

So:

*Revised survival estimate = 400 + 200*

*Revised survival estimate = 600 sols*

Mark now has a **major** problem, because the emergency resupply probe isn’t expected to reach him until Sol 856 – which is more than 250 days after his food is expected to run out.

## Survival Problem 3: Farming on Mars

Watney set up the roughly circular habitat to become a farm. If the circular farm area has a radius of 5.41 metres, we can calculate the total farming area:

*Total farming area = π × r ^{2}*

*Total farming area = 92 m ^{2}*

To successfully grow potatoes, Mark has to cover the floor to a depth of 10 centimetres:

*Total soil required = farming area × soil depth*

*Total soil required = 92 m ^{2} × 0.1 m*

*Total soil required = 9.2 m ^{3}*

Planting the potatoes won’t be enough – Mark will need to water them as well. He calculates that he needs about 40 litres of water per cubic metre of soil. So in total he’ll need:

*Total farming water required = number cubic metres soil × water required per cubic metre*

*Total farming water required = 9.2 × 40*

*Total farming water required = 368 litres*

## Survival Problem 4: Water Creation

Watney needs water to drink, but also to grow his crops (before the explosion squashes that plan).

From the various equipment and supplies on the base, he can get his hands on both oxygen and hydrogen. The figure above shows how two hydrogen molecules combine with one oxygen molecule to create two water (H_{2}0) molecules.

Using the above figure and some information about the molecular weights of the various molecules involved, we can calculate that 1 kilogram of oxygen combines with 0.125 kilograms of hydrogen to form 1.125 kilograms of water.

If we assume that 1 litre of water is the same as 1 kilogram of water, then we can calculate how much oxygen and hydrogen Watney will need to source to generate the required 368 litres of water:

*Oxygen required = water required × water to oxygen ratio*

*Oxygen required = 368 × 1 / 1.125*

*Oxygen required = 327.1 kg*

*Hydrogen required = water required × water to hydrogen ratio*

*Oxygen required = 368 × 0.125 / 1.125*

*Oxygen required = 40.9 kg *

## Survival Problem 5: Rover Journey to Schiaparelli Crater

Mark has to take the rover and make a 3200 km journey to Schiaparelli Crater to blast off and hopefully intercept the rescue mission as it passes overhead.

Watney works out a way to carry enough solar panels with him to recharge the rover on a daily basis.

The basic rover is equipped to go 35 km on a full charge:

*Basic journey time = 3200 / 35*

*Basic journey time = 91.4 days*

To make the journey faster, he cannibalizes the second rover and doubles the daily travel distance to 70 km.

Now part of the power usage of the rover is based on heating. If Watney turns off all the heating, he can use *all* of the rover’s 18000 watt hours of power to drive. He calculates it takes about 200 watt hours to drive 1 km:

*New driving endurance = total battery capacity / power usage per km driving*

*New driving endurance = 18000 / 200*

*New driving endurance = 90 km*

## Survival Problem 6: Solar Power for the Rover

Every day, Mark has to stop the rover and recharge the batteries using solar panels. He needs to calculate how many solar panels to bring along.

He has 2 square metre solar panels that are 10.2% efficient, which means that they absorb 10.2% of the approximately 500 watts of sunlight hitting each square metre.

He also calculates that he has about 13 hours of good sunlight charging time each day.

*Area of solar panel required = battery charge required / (sunlight energy per square metre per hour × solar panel efficiency × charging hours)*

*Area of solar panel required = 18000 / (500 × 0.102 × 13)*

*Area of solar panel required = 27.1 m ^{2}*

That means he’ll need to bring along 14 solar panels (since each panel is 2 square metres) with him on the journey.

## Real Life Example – Survivalists

Survivalists are people who deliberately prepare themselves for possible disasters. One of their key tasks is stockpiling enough long life food to keep their family supplied for long periods of time where other food isn’t available.

Well prepared survivalist might stockpile enough food to feed their family for 2 years.

**The following calculations use American units of pounds (0.45 kg), feet (30.48 cm), inches (2.54 cm).**

Let’s say the family has 2 adults:

- 40 year old male weighing 180 pounds, 6 feet tall,
- 40 year old female weighing 135 pounds, 5 feet 6 inches tall,
- boy aged 8 weighing 70 pounds, 5 feet 6 inches tall,
- girl aged 6 weighing 50 pounds, 5 feet tall).

Let’s say they’re not doing much, so living a sedentary lifestyle.

We can use the Basal Metabolic Rate (BMR) to estimate caloric needs.

**Women:**BMR = 655 + ( 4.35 x weight in pounds ) + ( 4.7 x height in inches ) – ( 4.7 x age in years )**Men:**BMR = 66 + ( 6.23 x weight in pounds ) + ( 12.7 x height in inches ) – ( 6.8 x age in year )

These calculations are then modified based on type of lifestyle:

- If you are
**sedentary**(little or no exercise) : Calorie-Calculation = BMR x 1.2 - If you are
**lightly active**(light exercise/sports 1-3 days/week) : Calorie-Calculation = BMR x 1.375 - If you are
**moderately active**(moderate exercise/sports 3-5 days/week) : Calorie-Calculation = BMR x 1.55 - If you are
**very active**(hard exercise/sports 6-7 days a week) : Calorie-Calculation = BMR x 1.725 - If you are
**extra active**(very hard exercise/sports & physical job or 2x training) : Calorie-Calculation = BMR x 1.9

### For the adult male:

*BMR = 66 + (6.23 x 180 ) + ( 12.7 x 6 x 12 ) – ( 6.8 x 40 )*

*BMR = 1830 calories*

*Calorie-Calculation = BMR x 1.2*

*Calorie-Calculation = 2196 calories *

### For the adult female:

*BMR = 655 + (4.35 x 135 ) + ( 4.7 x (5 × 12 + 6) ) – ( 4.7 x 40 )*

*BMR = 1364 calories*

*Calorie-Calculation = BMR x 1.2*

*Calorie-Calculation = 1637 calories*

### For the boy:

*BMR = 66 + (6.23 x 70 ) + ( 12.7 x (5 × 12 + 6) ) – ( 6.8 x 8 )*

*BMR = 1286 calories*

*Calorie-Calculation = BMR x 1.2*

*Calorie-Calculation = 1543 calories*

### For the girl:

*BMR = 655 + (4.35 x 50 ) + ( 4.7 x (5 × 12) ) – ( 4.7 x 6 )*

*BMR = 1126 calories*

*Calorie-Calculation = BMR x 1.2*

*Calorie-Calculation = 1351 calories *

Now we can calculate the total family calories required for the 2 year period:

*Calories required for whole family = time period × calories required per day*

*Calories required for whole family = 2 × 365 × (2196 + 1637 + 1543 + 1351)*

*Calories required for whole family = 2 × 365 × 6727*

*Calories required for whole family = 4910710*

That’s about 5 million calories!

A common survival food combination is rice and beans. A large 30 pound bucket of white rice or beans is about 50000 calories. Let’s say the family met most of their calorie intake using a 50-50 mix of white rice and beans. How much would they need to stockpile?

*Amount of white rice = 0.5 × total calories required × bucket weight / calories per bucket*

*Amount of white rice = 0.5 × 4910710 × 30 / 50000*

*Amount of white rice = 1473 pounds*

And they’d need that same amount of beans again, for a total of almost 3000 pounds of food, or about one and a half tonnes.