Atmosphere for Kepler 62e

This chemical reaction was used to produce oxygen of the following amount:

H₂O₂ -MnO₂-> O₂ + H₂O

0.015 cm³ of H₂O₂

0.5 g of the catalyst MnO₂

0.100 cm³ of O₂ were produced, the  amount of water produced was approximately the amount H₂O₂ of used

In order to calculate how oxygen could create an atmosphere for Kepler 62e we make the following assumptions for the calculations below:

-the atmosphere will be made up entirely by oxygen

-we assume that the oxygen fills a volume of about 2km*Kepler62e’s surface area

– density distribution due to gravity-> (air pressure) will be neglected, we assume the volume produced is stated at a pressure of 1 atmosphere( = 100 kPa )

Volume needed=( (4/3)*π*(1.6*6371km)³ – (4/3)*π*(1.6*6373km)³ ) * 10³

= 4.259*10^12 m³ of oxygen is needed

Now divide this number by the amount of oxygen produced in the experiment to find out the factor

4.259*10^12 m³/0.00001m³ =4.259*10^16

Multiply the amount of H₂O₂ used by this factor

0.000015m³*4.259*10^16= 6.388*10^11 m³ of would be needed to produce a 2km high atmosphere around Kepler 62e

Multiplying the same factor by the amount of catalyst used

0.5*10^(-6)* 4.259*10^16= 2.130 tonnes of MnO₂