The value of g at the surface of the earth is | vedclass

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(C) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula:$g_h = g \left[ \frac{R}{R+h} \right]^2$Giv

Vedclass · Accepted Answer

$8.5$

Vedclass · Answer

(C) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula:$g_h = g \left[ \frac{R}{R+h} ight]^2$Given:$g = 9.8 \, m/s^2$$R = 6400 \, km$$h = 480 \, km$Substituting the values into the formula:$g_h = 9.8 \left[ \frac{6400}{6400 + 480} ight]^2$$g_h = 9.8 \left[ \frac{6400}{6880} ight]^2$$g_h = 9.8 \left[ \frac{40}{43} ight]^2$$g_h = 9.8 	imes (0.9302)^2$$g_h = 9.8 	imes 0.8653$$g_h \approx 8.48 \, m/s^2$Rounding to the nearest value,we get $8.5 \, m/s^2$.

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 nearly .......... $m/s^2$ (radius of the earth is $6400 \, km$).

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