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Gravitational Field Due to a Solid Sphere

The magnitude...

Question

The magnitudes of the gravitational force 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}}

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

r_{1} > R

and

r_{2} > R

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

r_{1} < R

and

r_{2} < R

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Solution

The correct option is A $\frac{F_{1}}{F_{2}} = \frac{r_{1}}{r_{2}}$ if $r_{1} < R$ and $r_{2} < R$
For $r < R$
$g = \frac{4}{3} π ρ G r ∴ g \propto r$

For $r > R$
$g = \frac{G M}{r^{2}} ∴ g \propto \frac{1}{r^{2}}$

If $r_{1} < R$ and $r_{2} < R$ then $\frac{F_{1}}{F_{2}} = \frac{g_{1}}{g_{2}} = \frac{r_{1}}{r_{2}}$

If $r_{1} > R and r_{2} > R$ then $\frac{F_{1}}{F_{2}} = \frac{g_{1}}{g_{2}} = {(\frac{r_{2}}{r_{1}})}^{2}$

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