Question

The acceleration due to gravity at the surface of planet having mass M and density p is proportional to​

Answer 1

it is proportional to density and also to radius

i

Answer 2

The density is  proportional to the R⁻¹.

Explanation:

Given that,

Mass = M

Density =ρ

We know that,

The acceleration due to gravity

$g=\dfrac{GM}{R^2}$

The acceleration due to gravity at the surface of planet  

We need to calculate the Mass

Using formula of density

$\rho=\dfrac{M}{V}$

$M=\rho\times V$

Put the value into the formula

$M = \rho\times\dfrac{4}{3}\pi R^3$

Put the value of M in to the formula of gravity

$g=\dfrac{G\times \rho \times\dfrac{4}{3}\pi R^3}{R^2}$

$g=\dfrac{4}{3}\times G\times \rho\times R$

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

$\rho\propto\dfrac{1}{R}$

$\rho\propto R^{-1}$

Hence, The density is  proportional to the R⁻¹.

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Topic : acceleration due to gravity

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