The radii of two planets are respectively R1 and R2- and their densities are respectively - 1 and - 2- The ratio of the acceleration due to gravity at their surface is

Accelaration due to gravity on the surface of nbsp; a planet is given by, g=GMR^2 And we know that, nbsp; ρ=MV. so ⇒ g=GR^2×43π R^3 ρ=43π GRρ nbsp; He ...

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Standard XII

Physics

Principle of Superposition

The radii of ...

Question

The radii of two planets are respectively $R_{1}$ and $R_{2}$ , and their densities are respectively $ρ_{1}$ and $ρ_{2}$ . The ratio of the acceleration due to gravity at their surface is

\frac{ρ_{1}}{R_{1}^{2}} : \frac{ρ_{2}}{R_{2}^{2}}

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R_{1} R_{2} : ρ_{1} ρ_{2}

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R_{1} ρ_{2} : R_{2} ρ_{1}

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R_{1} ρ_{1} : R_{2} ρ_{2}

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Solution

The correct option is D $R_{1} ρ_{1} : R_{2} ρ_{2}$
Accelaration due to gravity on the surface of a planet is given by,

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

And we know that, $ρ = \frac{M}{V}$ . so

$\Rightarrow g = \frac{G}{R^{2}} \times \frac{4}{3} π R^{3} ρ = \frac{4}{3} π G R ρ$

Here, $π & G$ are constant

$∴ g \propto R ρ$

$\Rightarrow \frac{g_{1}}{g_{2}} = \frac{R_{1}}{R_{2}} \times \frac{ρ_{1}}{ρ_{2}}$

$\Rightarrow g_{1} : g_{2} = R_{1} ρ_{1} : R_{2} ρ_{2}$

Hence, option $(c)$ is correct.

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The radii of two planets are respectively R1 and R2, and their densities are respectively ρ1 and ρ2. The ratio of the acceleration due to gravity at their surface is

The correct option is D R1ρ1:R2ρ2Accelaration due to gravity on the surface of a planet is given by, g=GMR2 And we know that, ρ=MV. so ⇒g=GR2×43πR3ρ=43πGRρ Here, π & G are constant ∴g∝Rρ ⇒g1g2=R1R2×ρ1ρ2 ⇒g1:g2=R1ρ1:R2ρ2 Hence, option (c) is correct.

The radii of two planets are respectively $R_{1}$ and $R_{2}$ , and their densities are respectively $ρ_{1}$ and $ρ_{2}$ . The ratio of the acceleration due to gravity at their surface is