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

Answer:

The weight of Earth's atmosphere exert is $516.6\times10^{17}\ N$

Explanation:

Given that,

Average pressure $P=1.01\times10^{5}\ Pa$

Radius of earth $R_{E}=6.38\times10^{6}\ m$

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

$P=\dfrac{F}{A}$  

$F=PA$

$F=P\times 4\pi\times R_{E}^2$

Where, P = pressure

A = area

Put the value into the formula

$F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2$

$F=516.6\times10^{17}\ N$

Hence, The weight of Earth's atmosphere exert is $516.6\times10^{17}\ N$

Answer 2

Answer:

5.164 x 10^19 N

Explanation:

P = 1.01 x 10^5 Pa

R = 6.38 x 10^6 m

Area = 4 π R²

$A= 4\times 3.14\times6.38\times10^{6}\times6.38\times10^{6}$

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