Question

The ratio g/gh , where g and gh are the accelerations due to gravity at the surface of the earth and at a height h above the earth’s surface respectively, is:

Answer 1

Answer:

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Answer 2

The ratio of acceleration due to gravity at surface of earth and at height h above the surface of the earth is $1+\frac{2h}{R}$.

Given:

Acceleration due to gravity at surface of earth $=g$

Acceleration due to gravity at height h above earths surface $=g_{h}$

To find:

The ratio of acceleration due to gravity at earths surface and at height h above the earths surface.

Explanation:

We know, according to Newton's Law of gravitation, the attractive force between two earth of mass $M$ and any object of mass $m$ lying on its surface separated by distance $R$ which is the radius of earth is given by

$F=G\frac{Mm}{R^{2} }$

Also, according second law of motion, the force exerted by the object on the surface of earth subject to its mass which is nothing else but the  weight of an object lying on the surface of earth is given by

$F=mg$

Equating both the force equations, we get

$mg=G\frac{Mm}{R^{2} }$

$g=\frac{GM}{R^{2} }$                              $(i)$

Similarly,

When this object is taken at a height (lets say) $h$ above the surface of earth, the separation between the centers of the earth and the object becomes $(R+h)$

Therefore, the acceleration due to gravity at this height will be

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

Dividing the equation $(i)$ by $(ii)$, we get

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

$\frac{g}{g_{h} }= (\frac{R+h }{R})^{2}$

$\frac{g}{g_{h} }= (1+\frac{h }{R})^{2}$

$\frac{g}{g_{h} }= 1+\frac{2h}{R}+(\frac{h}{R} )^{2}$

Since, $R>>>h$, hence, the quantity $\frac{h^{2} }{R^{2} }$ ≅ $0$

$\frac{g}{g_{h} }= 1+\frac{2h}{R}$

Final answer:

Hence, the ratio of acceleration due to gravity at surface of earth and at height h above the surface of the earth is $1+\frac{2h}{R}$.