The Density of the Earth Using the known radius of the Earth and that g = 9.80 m/s2 at the Earth's surface, find the average density of the Earth. SOLUTION Conceptualize Assume the Earth is a perfect sphere. The density of material in the Earth varies, but let's adopt a simplified model in which we assume the density to be uniform throughout the Earth. The resulting density is the average of the Earth. Categorize This example is a relatively simple problem. Using the equation mg = G MEm RE2 , solve for the mass of the Earth. (Use the following as necessary: g, G, m, and RE.) ME = Substitute this mass and the volume of a sphere into the definition of density. (Use the following as necessary: g, G, and RE. Enter the final calculated value in kg/m3.) rhoE = ME VE = gRE2 G 4 3 πRE3 = 3g = kg/m3 EXERCISE Imagine that we discovered a planet that has gravity at its surface 3.7 times that on Earth's surface, while its radius is 4.80 ✕ 107 m. Determine the average density of the planet (in kg/m

5502.0 kg m⁻³ for Earth, 2702 kg m⁻³ for the other planet

Explanation:

Here are the equations that you will need to answer questions of this kind.

The Gravitational Constant,

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

To find mass (in kilograms)

M(R,g) = gR²/G

M(R,v) = v²R/(2G)

M(R,ρ) = 4πρR³/3

M(g,v) = v⁴/(4gG)

M(g,ρ) = 9g³/(16π²ρ²G³)

M(v,ρ) = v³ √[3/(32πρG³)]

To find radius (in meters)

R(M,g) = √(GM/g)

R(M,v) = 2GM/v²

R(M,ρ) = ∛[3M/(4πρ)]

R(g,v) = v²/(2g)

R(g,ρ) = 3g/(4πρG)

R(v,ρ) = v √[3/(8πρG)]

To find surface gravity (in m sec⁻²)

g(M,R) = GM/R²

g(M,v) = v⁴/(4GM)

g(M,ρ) = G ∛(16π²ρ²M/9)

g(R,v) = v²/(2R)

g(R,ρ) = 4πGρR/3

g(v,ρ) = v √[(2πρG)/3]

To find escape speed from surface (in m sec⁻¹)

v(M,R) = √(2GM/R)

v(M,g) = ∜(4GMg)

v(M,ρ) = (32πρG³M²/3)^⅙

v(R,g) = √(2gR)

v(R,ρ) = R √(8πGρ/3)

v(g,ρ) = g √[3/(2πρG)]

To find average density (in kg m⁻³)

ρ(M,R) = 3M/(4πR³)

ρ(M,g) = [3/(4π)] √[g³/(G³M)]

ρ(M,v) = 3v⁶/(32πG³M²)

ρ(R,g) = 3g/(4πGR)

ρ(R,v) = 3v²/(8πGR²)

ρ(g,v) = 3g²/(2πGv²)

With your specific question, you want the equation

ρ(R,g) = 3g/(4πGR)

If R=6.371e+6 meters and g=9.80 m sec⁻², then ρ=5502 kg m⁻³

If R=4.8e+7 meters and g=36.26 m sec⁻², then ρ=2702 kg m⁻³

the answer is b

explanation:

the answer would be a