The value of g on the Earth's surface is 980 | vedclass

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(A) The acceleration due to gravity at a height $h$ above the Earth's surface is given by the formula: $g' = g \left( \frac{R}{R+h} \right)^2$. Given:

Vedclass · Accepted Answer

$960.40$

Vedclass · Answer

(A) The acceleration due to gravity at a height $h$ above the Earth's surface is given by the formula: $g' = g \left( \frac{R}{R+h} ight)^2$. Given: $g = 980 \, cm/s^2$,$R = 6400 \, km$,and $h = 64 \, km$. Substituting the values: $g' = 980 	imes \left( \frac{6400}{6400 + 64} ight)^2$ $g' = 980 	imes \left( \frac{6400}{6464} ight)^2$ $g' = 980 	imes \left( \frac{100}{101} ight)^2$ $g' = 980 	imes \left( 0.990099 ight)^2$ $g' \approx 980 	imes 0.980296$ $g' \approx 960.69 \, cm/s^2$. Using the approximation $g' \approx g(1 - 2h/R)$: $g' = 980 	imes (1 - 2 	imes 64 / 6400) = 980 	imes (1 - 128/6400) = 980 	imes (1 - 0.02) = 980 	imes 0.98 = 960.40 \, cm/s^2$. Thus,the correct option is $A$.

The value of $g$ on the Earth's surface is $980 \, cm/s^2$. Its value at a height of $64 \, km$ from the Earth's surface is ........ $cm/s^2$ (Radius of the Earth $R = 6400 \, km$).

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