Calculate where the top of the atmosphere would be.

[Littlemac]

1. The weight of the atmosphere above 1 square meter of Earth's surface is about 100,000 N. Density, of course, decreases with altitude, but suppose the density of air were a constant 1.2 kg/m^3. Calculate where the top of the atmosphere would be.
- weight = 100,000 N (or 10197.16 kg)
- density = 1.2 kg/m^3

2. Convert the weight from N to kg then divide that answer (10197.16 kg) by the density (1.2 kg) to achieve the answer. Which comes out to 8497.63 m? (this is where I am having problems. I do not know which equation to use.)

 

[BvU]

Hi Mac,

No wonder you don't know which equation to use: you deleted '2. Homework Equations ' from the template !

However, you do mention under '3. Attempt at solution' () something that can be translated into $$ m = \rho V$$ with m = mass, $\rho$ = density and V is volume. So you're doing fine, physically. (But PF rules are a bit unhappy )

One further comment about number of digits: if your given data are 'about 100000 N' and 'íffy' (the 1.2 kg/m³), then you don't answer in six digits that pretend an incredible accuracy. Two digits is sufficient.