Question

The weight of Earth's atmosphere exerts an average pressure of 1.01 ✕ 105 Pa on the ground at sea level. Use the definition of pressure to estimate the weight of Earth's atmosphere (in N) by approximating Earth as a sphere of radius RE = 6.38 ✕ 106 m and surface area A = 4πRE2. HINT N

Answer 1

5.164 x 10^19 N

Step-by-step explanation:

P = 1.01 x 10^5 Pa

R = 6.38 x 10^6 m

Area = 4 π R²

`A= 4* 3.14*6.38*10^(6)*6.38*10^(6)`

A = 5.112 x 10^14 m^2

Pressure is defined as the force exerted per unit area.

The formula for the pressure is

P = F / A

Where, F is the force, A be the area

here force is the weight of atmosphere.

F = P x A

F = 1.01 x 10^5 x 5.112 x 10^14

F = 5.164 x 10^19 N

Answer 2

The weight of Earth's atmosphere exert is
`516.6*10^(17)\ N`

Step-by-step explanation:

Given that,

Average pressure
`P=1.01*10^(5)\ Pa`

Radius of earth
`R_(E)=6.38*10^(6)\ m`

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

`P=(F)/(A)`

`F=PA`

`F=P* 4\pi* R_(E)^2`

Where, P = pressure

A = area

Put the value into the formula

`F=1.01*10^(5)*4*\pi*(6.38*10^(6))^2`

`F=516.6*10^(17)\ N`

Hence, The weight of Earth's atmosphere exert is
`516.6*10^(17)\ N`