A hypothetical planet has density r, radius R and surface gravitational acceleration g. If the radius of the planet were doubled, but the planetary density stayed the same, the acceleration due to gravity at the planet’s surface would be:

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$\frac{g}{2}$

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Solution

The correct option is B
2g

$1^{s t} c a s e g = \frac{G (\frac{4}{3} π R^{3} p)}{R^{2}} = \frac{4}{3} π G R_{p} I I^{n d} c a s e : g^{'} = \frac{G (\frac{4}{3} π (2 R)^{3} p)}{(2 R)} = 2 [\frac{4}{3} π R G_{p}] ∴ g^{'} = 2 g$

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A hypothetical planet has density $ρ$ , radius R, and surface gravitational acceleration g. If the radius of the planet is doubled, but the planetary density stayed the same, the acceleration due to gravity at the planet’s surface would be

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A hypothetical planet has density r, radius R and surface gravitational acceleration g. If the radius of the planet were doubled, but the planetary density stayed the same, the acceleration due to gravity at the planet’s surface would be:

The correct option is B 2g 1stcaseg=G(43πR3p)R2=43πGRpIIndcase:g′=G(43π(2R)3p)(2R)=2[43πRGp]∴g′=2g

The correct option is B
2g

$1^{s t} c a s e g = \frac{G (\frac{4}{3} π R^{3} p)}{R^{2}} = \frac{4}{3} π G R_{p} I I^{n d} c a s e : g^{'} = \frac{G (\frac{4}{3} π (2 R)^{3} p)}{(2 R)} = 2 [\frac{4}{3} π R G_{p}] ∴ g^{'} = 2 g$