If the value of g at the surface of the earth | vedclass

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(A) The acceleration due to gravity at the surface of the earth is given by $g = \frac{GM}{R^2} = 9.8 \, m/s^2$.At a height $h$ above the surface of t

Vedclass · Accepted Answer

$8.4$

Vedclass · Answer

(A) The acceleration due to gravity at the surface of the earth is given by $g = \frac{GM}{R^2} = 9.8 \, m/s^2$.At a height $h$ above the surface of the earth,the acceleration due to gravity $g'$ is given by $g' = g \left( \frac{R}{R+h} ight)^2$.Given: $g = 9.8 \, m/s^2$,$R = 6400 \, km$,and $h = 480 \, km$.Substituting these values into the formula:$g' = 9.8 	imes \left( \frac{6400}{6400 + 480} ight)^2$$g' = 9.8 	imes \left( \frac{6400}{6880} ight)^2$$g' = 9.8 	imes (0.9302)^2$$g' = 9.8 	imes 0.8653 \approx 8.48 \, m/s^2$.Rounding to the nearest provided option,we get $8.4 \, m/s^2$.

If the value of $g$ at the surface of the earth is $9.8 \, m/s^2$,then the value of $g$ at a place $480 \, km$ above the surface of the earth will be .......... $m/s^2$. (Radius of the earth is $6400 \, km$)

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