Question

the ratio of acceleration due to gravity at a height 3R above earth surface to the acceleration due to gravity on the surface of the Earth ?​

Answer 1

Explanation:

As we know that

the value of acceleration due to gravity changes with the change in height.

Let g' be the acceleration due to gravity at a point which is h distance above the surface of earth , then

$\boxed{\underline{\underline{\bold{\mathsf{g' = \frac{GM}{{(R +h)}}^{2}}}}}}$.....(i)

"h" distance height will be added to r

but as we know

$g = \frac{GM}{ {(R + h)}^{2} }$....(ii)

now divide (i) by (ii)

$\frac{g'}{g} = \frac{GM}{( {(R + h)}^{2} } \times \frac{ {R}^{2} }{GM} \\ \\ = \frac{ {R}^{2} }{ {(R+ h)}^{2} }$

Final g'/g we get =

$= \frac{gm}{ {(r + h)}^{2} }$

Now put value of h

$= \frac{GM}{{(R + 3R)}^{2} } \\ \\ = \frac{GM}{ {(4R)}^{2} } \\ \\ = \frac{GM}{16 {R}^{2} }$

As we can observe that in g'/g the constant number we get is 1/16

°•° Ratio will be 1:16