Question

show that if a body is taken to a height h above the Earth's surface acceleration due to gravity is decreased by the factor R²/(R+H)², where R is the radius of the earth

Note :- I am of Class 9 and also, Explain, in simple way and Please don't copy Answers!


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Answer 1

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Acceleration due to gravity of earth is $g = \frac{GM}{ {R}^{2} } --- eq. (i)$

Now at height 'H' above surface of earth, acceleration due to gravity must be $g' = \frac{GM}{{(R+H)}^2} --- eq. (ii)$

On dividing both equations, we get,

$\frac{g'}{g} = \frac{R^2}{(R+H)^ 2} \\g' = \frac{g(R)^2}{(R+H)^2}$

Answer 2

Answer:

here is ur ans

we know that acceleration due to gravity of earth is

g = GM/ R^2

now at height H above surface of earth acceleration due to gravity must be

g' = GM/(R+H)^2

now divide both

g'/g = R^2/(R+H)^ 2

g' = g(R)^2 / (R+H)^2

Explanation: