How Much Does Earth's Atmosphere Weigh?

[nikita33]

Homework Statement

above each square inch of Earth's surface is 14.7 lbs of air. the density of air at sea level is about 1.27g/L. the Earth is 7930 miles in diameter and has a mass of 5.98 x 10^24kg.
a.) use the data above to determine the mass in kg of the Earth's atmosphere.

b.) determine the average density of Earth in grams per milliliter.

Homework Equations

i think the only relevant equation is d=m/v. everything else seems to be unit conversions. this problem really isn't about the atmosphere, but rather how well i can convert units, and apparently not well.

The Attempt at a Solution

im looking for the mass of the atmosphere, so i use m = dv
d is 1.27g/mL
i derive v from 4/3pir^3
but i have the radius in miles. does this matter?
i would get (1.27g/mL)(2.55x10^11) since i need it in kg
(1.27 x 10^-3kg)(2.55x10^11) = 3.24 x 10^8 kg/mL?

im sure I am hopelessly wrong. i don't really want to do part b until i get part a.

 

[Redbelly98]

(a) We're given that there are 14.7 pounds of air per square inch of Earth's surface area. If you can figure out how many square inches of surface area the Earth has, can you take it from there?

(b) The equation you had, d=m/v, applies here.

 

[Moetasim]

Your approach is in the right direction, but there are a few errors in your calculations.

a.) To find the mass of the Earth's atmosphere, we can use the given information that 1 square inch of Earth's surface has 14.7 lbs of air. We can convert this to grams by using the conversion factor 1 lb = 453.592 g. So, 14.7 lbs of air is equal to 14.7 x 453.592 g = 6662.66 g of air per square inch of Earth's surface.

Next, we need to find the total surface area of the Earth. The radius of the Earth is given in miles, but we need it in inches to match the units of the density of air. 1 mile = 63360 inches. So, the radius of the Earth in inches is 7930 x 63360 inches = 502848000 inches.

Now, we can use the formula for the surface area of a sphere, A = 4πr^2, to find the total surface area of the Earth. Plugging in the radius in inches, we get A = 4 x 3.14 x (502848000 inches)^2 = 3.16 x 10^18 inches^2.

Finally, we can find the mass of the Earth's atmosphere by multiplying the mass of air per square inch with the total surface area of the Earth. So, the mass of the Earth's atmosphere is 6662.66 g/in^2 x 3.16 x 10^18 in^2 = 2.11 x 10^22 g.

b.) To find the average density of the Earth, we can use the formula d = m/v, where m is the mass of the Earth and v is the volume of the Earth. We already know the mass of the Earth from the given information, which is 5.98 x 10^24 kg.

To find the volume of the Earth, we can use the formula v = 4/3πr^3, where r is the radius of the Earth. We can convert the radius from miles to meters by using the conversion factor 1 mile = 1609.34 meters. So, the radius of the Earth in meters is 7930 x 1609.34 meters = 1.27 x 10^7 meters.

Plugging in the values,