Recently, some information about a distant planet was learned. It has a radius of 6000000 meters, and the density of the atmosphere as a function of the height h (in meters) above the surface of the planet is given by δ(h)=3h+6000000 kilograms per cubic meter. Calculate the mass of the portion of the atmosphere from h=0 to h=57.

M = 9.8 ×10²²kg

Explanation:

This question involves the mass, density and volume relationship. The density of the atmosphere varies with the height above the surface of the planet. Given the density function δ(h)=3h+6000000. We can then calculate the value of the densities at the two given altitudes (height above the planet surface).

We will use the relationship between the mass, volume and density.

M = ρ × V

At h = 0m (the surface of the planet)

The radius of the planet = 6×10⁶m

ρ = 3h+6000000 = 3×0 + 6000000

= 6000000 = 6×10⁶ kg/m³

V = 4/3× r³ (volume of a sphere)

= 4/3× (6×10⁶)³ = 2.88 ×10²⁰ m³

M = 6×10⁶ × 2.88 ×10²⁰ = 1.728 ×10²⁷

At a height of h = 57m

r = 6000000 + 57 = 6000057m

V = 4/3 × (6000057)³ = 2.880082×10²⁰

ρ = 3h+6000000 = 3×57 + 6000000

= 6000171 kg/m³

M = 6000171 × 2.880082 × 10²⁰

= 1.728098 × 10²⁷

The mass of the atmosphere is the difference between the masses at the different altitudes.

So the mass of the atmosphere

= 1.728098 × 10²⁷ – 1.728000 × 10²⁷

= 9.8 ×10²² kg.

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation: