Question

If R is the radius of the earth and g is the acceleration due to gravity on the earth's surface, the mean density of the earth is ______.

Options

• `4/3 "G"/"R"`

• `(3pi"R")/(4"gG")`

• `(3"g")/(4pi"RG")`

• `(pi"Rg")/"u"`

Answer 1

If R is the radius of the earth and g is the acceleration due to gravity on the earth's surface, the mean density of the earth is `bb((3"g")/(4pi"RG"))`.

Explanation:

g_e = `"Gm"_"e"/"R"^2`

(Given R → Radius of earth, ρ → mean density of earth)

g_e = `("G"(4/3pi"R"^3rho))/"R"^2` ; g_e = `4/3`πR ρG

ρ = `3/4.g_e/(pi"RG")`