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Universal Law of Gravitation

The magnitude...

Question

The magnitude of the gravitational field at distance $r_{1}$ and $r_{2}$ from the centre of a uniform sphere of radius $R$ and mass $M$ are $F_{1}$ and $F_{2}$ respectively. Then

\frac{F_{1}}{F_{2}} = \frac{r_{1}}{r_{2}}

r_{1} < R

and

r_{2} < R

and

\frac{F_{1}}{F_{2}} = \frac{r_{1}^{2}}{r_{2}^{2}}

r_{1} > R

and

r_{2} > R

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\frac{F_{1}}{F_{2}} = \frac{r_{1}}{r_{2}}

r_{1} > R

and

r_{2} > R

and

\frac{F_{1}}{F_{2}} = \frac{r_{2}}{r_{1}}

r_{1} < R

and

r_{2} < R

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\frac{F_{1}}{F_{2}} = \frac{r_{1}^{2}}{r_{2}^{2}}

r_{1} < R

and

r_{2} < R

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none of these

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Solution

The correct option is D none of these
We know, $F = \frac{G M}{r^{2}}, i f r \geq R$

Force is inversely proportional to $r^{2}$ . When $r\ge R$.

So,

$\begin{matrix} \frac{F_{1}}{F_{2}} = \frac{{r_{2}}^{2}}{{r_{1}}^{2}} W h e n, r_{1} \geq R, r_{2} \geq R a g a i n, F = \frac{G M r}{R^{3}}, W h e n r \geq R \end{matrix}$

Force is directly proportional to r. When
$\begin{matrix} r < R h e n c e, \frac{F_{1}}{F_{2}} = \frac{r_{1}}{r_{2}} w h e n, r_{1} < R, r_{2} < R \end{matrix}$

Therefore, Option $D$ none of these is the correct answer.

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The magnitude of the gravitational field at distance r1 and r2 from the centre of a uniform sphere of radius R and mass M are F1 and F2 respectively. Then

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