Question

If g is the acceleration due to gravity on the surface of earth, its value at a height equal to double the radius of earth is

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Answer 1

Answer:

g/9

this is the correct answer

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Answer 2

$\color{blue}\mathfrak{\underline{Acceleration \: due \: to \: gravity \: at \: the \: surface \: of \:the \: earth}:} \\ g=G \frac{m}{ {R}^{2} }-(i) \\ \color{blue}\mathfrak{\underline{Acceleration \: due \: to \: gravity \: at \:height \:equal \: to \: double \: the \: radius \: of \:the \: earth}:}\\ \rightarrow Radius \: of \: the \: earth = R \\ \rightarrow Acceleration \: at \: some\: height = 2R \\ \rightarrow Total \: distance \: from \: centre \: of \: the \: earth \\ to \:some \: point(say \: P) = R + 2R = 3R \\ \\ g' = G \frac{m}{ {(3R)}^{2} } \\ g' = G \frac{m}{ {9R}^{2} }-(ii) \\ \color{blue}\mathfrak{\underline{On \: divding \: {eq}^{n} \: (ii) \: by \: (i)}:} \\ \frac{g'}{g} = \frac{G \frac{m}{ {9R}^{2} }}{ G \frac{m}{ {R}^{2} }} \\ \\ \frac{g'}{g} = \frac{1}{ 9} \\ \\ \color{green} \rightarrow \boxed{ g' = \frac{g}{9} }.Answer \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$