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

Moderate

A

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

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

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

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

Solution

The uniform solid sphere
$\begin{array}{l} F = (G M / R^{3}) \times r \\ F_{1} = \frac{G M}{R^{3}} . r_{1} & F_{2} = \frac{G M}{R^{3}} . r_{2} \\ \frac{F_{1}}{F_{2}} = \frac{r_{1}}{r_{2}} \end{array}$
(or)
Outside the solid sphere, magnitude of gravitational field
$\begin{array}{l} F = \frac{G M}{r^{2}} \\ F_{1} = \frac{G M}{r_{1}^{2}}; F_{2} = \frac{G M}{r_{2}^{2}} \\ \frac{F_{1}}{F_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}} \end{array}$

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