Question

If the diameter of the planet is 4 times that of the earth and it's density is equal to that of earth. The value of g on that planet is?

Answer 1

Density=mass/volume, mass is inversely proportional to the gravity so the gravity would be the same as density is equal.

(I hope this is correct)

Answer 2

Answer:

• Acceleration due to gravity at planet is 4 times that of the Earth.

Given:

1. Diameter of the planet is 4 times that of the earth

Explanation:

$\rule{300}{1.5}$

From the Formula we know.

$\large\bigstar \:{\boxed{\tt g = \dfrac{4}{3}G \pi R \rho}}$

$\bold{Here}\begin{cases}\text{G Denotes Universal Gravitational Constant} \\ \text{R Denotes Radius of Planet} \\ \rho \text{ Denotes Density }\end{cases}$

$\large{\boxed{\tt g = \dfrac{4}{3}G \pi R \rho}}$

Now,

$\large{\tt \longmapsto g \propto R}$

• As 4/3 , π , ρ & G are Constants.

Now,

$\large{\tt \longmapsto \dfrac{g_{(Earth)}}{g_{(Planet)}} = \dfrac{R_e}{R_p}}$

∵ [R = D/2]

Substituting,

$\large{\tt \longmapsto \dfrac{g_{(Earth)}}{g_{(Planet)}} = \dfrac{2 \times D_e}{2 \times D_p}}$

$\large{\tt \longmapsto \dfrac{g_{(Earth)}}{g_{(Planet)}} = \cancel{\dfrac{2 \times D_e}{2 \times D_p}}}$

$\large{\tt \longmapsto \dfrac{g_{(Earth)}}{g_{(Planet)}} = \dfrac{D_e}{D_p}}$

From question we Know,

Diameter of the planet is 4 times that of the earth I.e. $\sf{D_p = 4 D_e}$

Substituting,

$\large{\tt \longmapsto \dfrac{g_{(Earth)}}{g_{(Planet)}} = \dfrac{D_e}{4 D_e}}$

$\large{\tt \longmapsto \dfrac{g_{(Earth)}}{g_{(Planet)}} = \cancel{\dfrac{D_e}{4 D_e}}}$

$\large{\tt \longmapsto \dfrac{g_{(Earth)}}{g_{(Planet)}} = \dfrac{1}{4}}$

Cross - Multiplying,

$\large{\tt \longmapsto 1 \times g_{(Planet)} = 4 \times g_{(Earth)}}$

$\large\longmapsto{\underline{\boxed{\red{\tt g_{(Planet)} = 4 g_{(Earth)}}}}}$

∴ Acceleration due to gravity at planet is 4 times that of the Earth.

$\rule{300}{1.5}$