Calculate where the top of the atmosphere would be.

  • Thread starter Littlemac
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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.)
 

Answers and Replies

  • #2
BvU
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Hi Mac, :welcome:

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

However, you do mention under '3. Attempt at solution' (:nb)) 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 :wink:)

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.
 

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