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Variation of G Due to Height and Depth

The magnitude...

Question

The magnitudes of the gravitational field at distances $r_{1}$ 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}} i f r_{1} < R a n d r_{2} < R

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\frac{{r_{2}}^{2}}{r_{2}} i f r_{1} > R a n d r_{2} < R

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\frac{F_{1}}{F_{2}} = \frac{r_{1}}{r_{2}} i f r_{1} > R a n d r_{2} > R

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\frac{F_{1}}{F_{2}} = \frac{{r_{1}}^{3}}{{r_{1}}^{2}} i f r_{1} < R a n d r_{2} < R

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Solution

The correct option is A $\frac{F_{1}}{F_{2}} = \frac{r_{1}}{r_{2}} i f r_{1} < R a n d r_{2} < R$
For $r > R$ , $F = \frac{G m}{r^{2}}$
For $r < R$ : Mass of the sphere of radius r, $m^{'} = ρ \times \frac{4 π}{3} r^{3} = \frac{m}{\frac{4 π}{3} R^{3}} \times \frac{4 π}{3} r^{3}$
$m^{'} = \frac{r^{3}}{R^{3}} m$
Thus, $F^{'} = \frac{G m \frac{r^{3}}{R^{3}}}{r^{2}} = \frac{G m r}{R^{3}}$
Now for $r_{1} < R ⟹ F_{1} = \frac{G m r_{1}}{R^{3}}$ and for $r_{2} < R ⟹ F_{2} = \frac{G m r_{2}}{R^{3}}$
Hence, $\frac{F_{1}}{F_{2}} = \frac{r_{1}}{r_{2}}$
For $r_{1} > R ⟹ F_{1} = \frac{G m}{r_{1}^{2}}$ and for $r_{2} < R ⟹ F_{2} = \frac{G m r_{2}}{R^{3}}$
Hence, $\frac{F_{1}}{F_{2}} = \frac{R^{3}}{r_{2} r_{1}^{2}}$
For $r_{1} > R ⟹ F_{1} = \frac{G m}{r_{1}^{2}}$ and for $r_{2} > R ⟹ F_{2} = \frac{G m}{r_{2}^{2}}$
Hence, $\frac{F_{1}}{F_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}}$

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