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A planet has density, radius $R$ and acceleration due to gravity as $g$ . If the radius of the planet were doubled, keeping the density the same, the acceleration due to gravity at the surface will be which of the following?

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g / 2

g

2 g

4 g

answer is C.

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Detailed Solution

A planet has density

ρ

, radius

R

and acceleration due to gravity as

g

. If the radius of the planet were doubled, keeping the density the same, the acceleration due to gravity at the surface will be

2 g .

We know that,

Density = mass per unit volume = \frac{m}{v}

Hence,

ρ = \frac{m}{v}

,
i.e.;

m = ρ \times v

As we know,

g = \frac{GM}{r^{2}}

\Rightarrow g = \frac{G (ρ \times v)}{r^{2}}

volume,

v = \frac{4}{3} π r^{3}

Therefore,

g = (\frac{4}{3} Gπρ) r

Above equation show that,

g

is directly proportional to the radius of the planet. If

g_{1} and g_{2} (when radius is doubled)

are acceleration due to gravity for the planet. Then-

\frac{g_{1}}{g_{2}} = \frac{r_{1}}{r_{2}}

\Rightarrow \frac{g_{1}}{g_{2}} = \frac{R}{2 R}

, here

r_{1} = R

and

r_{2} =

2R.

\Rightarrow \frac{g_{1}}{g_{2}} = \frac{1}{2}

So,

g_{2} = 2 g_{1} = 2 g

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