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Newton's Law of Gravitation

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

A

\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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B

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

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C

\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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D

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

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Solution

The correct option is A $\frac{F_{1}}{F_{2}} = \frac{r_{1}}{r_{2}} i f r_{1} < R a n d r_{2} < R$
$g = \frac{4}{3} π ρ G r ∴ g \propto r i f r < R g = \frac{G M}{r^{2}} ∴ g \propto \frac{1}{r^{2}} i f r > R I f r_{1} < R a n d r_{2} < R t h e n \frac{F_{1}}{F_{2}} = \frac{g_{1}}{g_{2}} = \frac{r_{1}}{r_{2}} I f r_{1} > R a n d r_{2} > R t h e n \frac{F_{1}}{F_{2}} = \frac{g_{1}}{g_{2}} = {(\frac{r_{2}}{r_{1}})}^{2}$

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