Question

if the density of the earth is doubled keeping its radius constant then acceleration due to gravity is a) 20m/s^2 b) 10m/s^2 c) 5m/s^2 d) 2.5m/s^2​

Answer 1

Answer:-

20m/s² [Option.a]

Explanation:-

Let the density of earth be ρ.

Thus mass of earth:-

=> Mass = Volume × Density

=> M = 4/3πR³×ρ

=> M = 4/3πR³ρ

Acceleration due to gravity(g) = GM/R²

=> g = G×4/3πR³ρ/R²

=> g = 4/3πGRρ

Value of 'g' at the surface of earth is

10m/s², hence:-

=> 10 = 4/3πGRρ ------(1)

Now, according to given condition, density of the earth is doubled i.e 2ρ and the radius is constant , then the equation for g' will be:-

=> g' = 4/3πGR2ρ

=> g' = 8/3πGRρ ----(2)

On dividing eq.2 by eq.1, we get :-

=> g'/10 = [8/3πGRρ]/[4/3πGRρ]

=> g'/10 = 2

=> g' = 10(2)

=> g' = 20m/s²

Thus, acceleration due to gravity in given condition is 20m/s² .

Answer 2

Let initial density of earth be = $ρ$

According to the Question,

New density = $2ρ$

Formula:-

$\boxed{\tt{ M = ρv} }$

(Assume that density of Earth = ρ, and volume be = v, Mass = M)

∴ Final mass = (2ρ)(v) = 2ρv = 2M

We know,

$\boxed{ \tt{{g = G \frac{M}{R^2}}}}$

• (Can be obtained by finding relationship between g and G)

∴ Initially, $g_1 = G \frac{M}{R^2}$

(Let R be the radius)

But in this case,

Acceleration due to gravity = $G \frac{2M}{R^2}$

$\implies g_2 = 2(G \frac{M}{R^2})$

$\implies g_2 = 2g_1$

• (Value of g = 10 m/s² — can be accepted in case of solving equations)

$\implies g_2 = 20 m/s^2$ (a)