Answer:
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
