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 $E_{1}$ and $E_{2}$ respectively. Then:

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

, if

r_{1} < R

and

r_{2} < R

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

, if

r_{1} > R

and

r_{2} > R

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\frac{E_{1}}{E_{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}

, if

r_{1} < R

and

r_{2} < R

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

, if

r_{1} < R

and

r_{2} < R

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Solution

The correct options are
A $\frac{E_{1}}{E_{2}} = \frac{r_{1}}{r_{2}}$ , if $r_{1} < R$ and $r_{2} < R$
B $\frac{E_{1}}{E_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}}$ , if $r_{1} > R$ and $r_{2} > R$

If $r \leq R$ , then $E = \frac{G M}{R^{3}} (r) ⟹ E \propto r$

$⟹ \frac{E_{1}}{E_{2}} = \frac{r_{1}}{r_{2}}$ if $r_{1} < R$ and $r_{2} < R$

If $r \geq R$ , then $E = \frac{G M}{r^{2}} ⟹ E \propto \frac{1}{r^{2}}$

$⟹ \frac{E_{1}}{E_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}}$ if $r_{1} > R$ and $r_{2} > R$

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