If the radius of a celestial body is k times that of the earth and its density is p times that of the earth, then the gravitational acceleration on the surface of the celestial body is () A. KP2 times B. KP & nbsp; times C. KP times D. p2k times

If the radius of a celestial body is k times that of the earth and its density is p times that of the earth, then the gravitational acceleration on the surface of the celestial body is () A. KP2 times B. KP & nbsp; times C. KP times D. p2k times

Let the radius of the earth be r and the density be ρ, then the mass of the earth is m = 43 π R3 ρ, and the gravity on the earth surface is equal to the universal gravitation: Mg = gmmr2, then the acceleration of gravity on the earth surface is g = gmr2 = G43 π R3 ρ R2 = 43g π R ρ, so the acceleration of gravity on the surface of the celestial body whose radius is k times of the earth and density is p times of the earth is g ′ = 43g π kr· P ρ = kp· G

If the radius of a celestial body is k times that of the earth and its density is p times that of the earth, then the gravitational acceleration on the surface of the celestial body is ()
A. KP2 times B. KP & nbsp; times C. KP times D. p2k times

Let the radius of the earth be r and the density be ρ, then the mass of the earth is m = 43 π R3 ρ, and the gravity on the earth surface is equal to the universal gravitation: Mg = gmmr2, then the acceleration of gravity on the earth surface is g = gmr2 = G43 π R3 ρ R2 = 43g π R ρ, so the acceleration of gravity on the surface of the celestial body whose radius is k times of the earth and density is p times of the earth is g ′ = 43g π kr· P ρ = kp· G

If the radius of a celestial body is known to be r and the gravitational acceleration g at its surface, then the density of the celestial body is?

GMm/R*R=mg
GM=g*R*R M=g*R*R/G
ρ=3M/4π*R*R*R
=3g*R*R/4π*R*R*R*G
=3g/4πRG