Question

The weight of a body at a height equal to the radius of the earth is 'N'. What will be its weight at a height equal to three times the radius of the earth?​

Answer 1

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\pink{Answer:-}$

⛄ We know that,

$g = G \frac{M}{d^{2} }$

Therefore, when d= 3d, then,

$g = G \times \frac{M}{(3d)^{2} }$

$= ) \: g' = \: \frac{GM}{9d^{2} }$

or

$g' = \frac{g}{9}$

Therefore,

★ The original weight = N = mg

and

★ Final weight= N' = m × g/9 or N' = N/9.

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Answer 2

Hello lovely mate!!♥️♥️

your answer....

3 times the radius of the earth means 3R from the center of the earth. So you go from 2R to 3R so 1/r^2 goes from a factor of 1/4 to 1/9. Further away from the center the weight decreases giving a smaller number so use (1/9)/(1/4) = 4/9.

So the 1 N

becomes (4/9) N at 3R.

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