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A. Is least on the equator.

B. Is least on poles.

C. Is the same on equator and poles.

D. Increases from pole to equator.

Answer
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Hint: Using concept of variation in gravity with shape of the earth. As from Newton’s law of gravitation we can say acceleration due to gravity is inversely proportional to square of the radius of the planet.

Formula used:\[\dfrac{{{g}_{e}}}{{{g}_{p}}}=\dfrac{r_{p}^{2}}{r_{e}^{2}}\]and $F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$

Complete step by step answer:

We know by newton law of gravitation that gravitational force (F) can be written as

$F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$ ---equation (1).

And gravitational force (F) can also be written as

$F=mg$-- equation (2).

After equating equation (1) and (2)

$mg=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$

That can be written as

$g\alpha \dfrac{1}{{{r}^{2}}}$-- equation (3).

As G, ${{m}_{1}}$, ${{m}_{2}}$ and m are constant. So we get the result that acceleration due to gravity is inversely proportional to square of the radius of the planet.

Earth has the shape of an oblate spheroid that means it’s radius near the equator is more than it’s radius near the poles.

Let

${{g}_{e}}$ be gravity near equator,

\[{{g}_{p}}\] be gravity near poles,

\[{{r}_{e}}\] be radius of earth at equation

and ${{r}_{p}}$ be the radius of earth at poles.

Using equation (3) we compare the gravity at poles and equator.

We get

$\dfrac{{{g}_{e}}}{{{g}_{p}}}=\dfrac{r_{p}^{2}}{r_{e}^{2}}$

This is equation (4).

Now as we know because Earth is an oblate spheroid so

${{r}_{e}}>{{r}_{p}}$

Using this information in equation (4)

We get that

${{g}_{p}}>{{g}_{e}}$

So, the correct answer is option (a). is least at the equator.

Note: Students can also use the concept of variation of gravity with the rotation of the earth. As the Earth rotates, the body on the surface rotates in a circular path and hence, experiencing the centrifugal force. In which we can relate the angular velocity of rotation of earth with gravity and the way it varies from equator to poles.

Formula used:\[\dfrac{{{g}_{e}}}{{{g}_{p}}}=\dfrac{r_{p}^{2}}{r_{e}^{2}}\]and $F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$

Complete step by step answer:

We know by newton law of gravitation that gravitational force (F) can be written as

$F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$ ---equation (1).

And gravitational force (F) can also be written as

$F=mg$-- equation (2).

After equating equation (1) and (2)

$mg=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$

That can be written as

$g\alpha \dfrac{1}{{{r}^{2}}}$-- equation (3).

As G, ${{m}_{1}}$, ${{m}_{2}}$ and m are constant. So we get the result that acceleration due to gravity is inversely proportional to square of the radius of the planet.

Earth has the shape of an oblate spheroid that means it’s radius near the equator is more than it’s radius near the poles.

Let

${{g}_{e}}$ be gravity near equator,

\[{{g}_{p}}\] be gravity near poles,

\[{{r}_{e}}\] be radius of earth at equation

and ${{r}_{p}}$ be the radius of earth at poles.

Using equation (3) we compare the gravity at poles and equator.

We get

$\dfrac{{{g}_{e}}}{{{g}_{p}}}=\dfrac{r_{p}^{2}}{r_{e}^{2}}$

This is equation (4).

Now as we know because Earth is an oblate spheroid so

${{r}_{e}}>{{r}_{p}}$

Using this information in equation (4)

We get that

${{g}_{p}}>{{g}_{e}}$

So, the correct answer is option (a). is least at the equator.

Note: Students can also use the concept of variation of gravity with the rotation of the earth. As the Earth rotates, the body on the surface rotates in a circular path and hence, experiencing the centrifugal force. In which we can relate the angular velocity of rotation of earth with gravity and the way it varies from equator to poles.

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