Question

The Earth is not a uniform sphere, but has regions of varying density. Consider a simple model of the Earth divided into three regions-inner core, outer core, and mantle. Each region is taken to have a unique constant density (the average density of that region in the real Earth):
Region Radius (km) Density(kg/m3)
Inner Core 0-1220 13,000
Outer Core 1220-3480 11,100
Mantle 3480-6371 44,00
A) Use this model to predict the average density of the entire Earth.
B)The measured radius of the Earth is 6371km and its mass is 5.974*1024kg Use this data to determine the actual average density of the Earth.

Answer 1

Answer:

Explanation:

Volume of inner core = 4/3 π R₁³

where R₁ is radius of inner core so R₁ = 1220 km

Volume of inner core = 4/3 π (1220 km )³

= 1764.4 x 10¹⁷ m³ - 4/3 π (1220 km )³

= 76.02 x 10⁸ x 10⁹ m³

= 76.02 x 10¹⁷ m³

Density = 13000 kg / m³

mass = volume x density

= 76.02 x 10¹⁷ x 13000

= 988.26 x 10²⁰ kg

Volume of outer  core = 4/3 π (3480 km )³ -  4/3 π (1220 km )³

= 1764.4 x 10⁸ km³ - 4/3 π (1220 km )³

= 1764.4 x 10⁸ x 10⁹m³ - 4/3 π (1220 km )³

= 1764.4 x 10¹⁷ m³ - 4/3 π (1220 km )³

= 1764.4 x 10¹⁷ m³ - 76.02 x 10¹⁷ m³

= 1688.38 x 10¹⁷ m³

density = 11100 kg /m³

mass of outer  core= 1688.38 m³ x10¹⁷ x  11100 kg /m³

= 18741 x 10²⁰ kg

Volume of outer mantle  =  4/3 π (6371  km )³ - 4/3 π (3480 km )³

= 10826 x 10¹⁷ m³- 1764.4 x 10¹⁷ m³

= 9062 x 10¹⁷ m³

density = 4400 kg /m³

mass of outer  core= 9062 x 10¹⁷ m³ x  4400 kg /m³

= 39872.8 x 10²⁰ kg

Total mass of the earth = 988.26 x 10²⁰ kg + 18741 x 10²⁰ kg +39872.8 x 10²⁰ kg

= 59602.06 x 10²⁰ kg .

Total volume = 4/3 π (6371  km )³

= 10826 x 10¹⁷ m³

Density = mass / volume

= 59602.06 x 10²⁰ kg /  10826 x 10¹⁷ m³

= 5505.45 kg / m³

B )

Mass = 5.974 x 10²⁴ kg

volume = Total volume = 4/3 π (6371  km )³

= 10826 x 10¹⁷ m³

density = 5.974 x 10²⁴ kg / 10826 x 10¹⁷ m³

= 5518.2 kg / m³