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
$F_{1} / F_{2} = r_{1} / r_{2}$ , if $r_{1} < R$ and $r_{2} < R$
$F_{1} / F_{2} = r_{2}^{2} / r_{1}^{2}$ , if $r_{1} > R$ and $r_{2} > R$
$F_{1} / F_{2} = r_{1} / r_{2}$ , if $r_{1} > R$ and $r_{2} < R$
$F_{1} / F_{2} = r_{1}^{2} / r_{2}^{2}$ , if $r_{1} < R$ and $r_{2} < R$

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
$F_{1} / F_{2} = r_{1} / r_{2}$ , if $r_{1} < R$ and $r_{2} < R$
$F_{1} / F_{2} = r_{2}^{2} / r_{1}^{2}$ , if $r_{1} > R$ and $r_{2} > R$
$F_{1} / F_{2} = r_{1} / r_{2}$ , if $r_{1} > R$ and $r_{2} < R$
$F_{1} / F_{2} = r_{1}^{2} / r_{2}^{2}$ , if $r_{1} < R$ and $r_{2} < R$

Toppr Admin · Accepted Answer

-Gravitational field-inside a sphere of radius R at a distance r is F-GMR3r-And- outside the-sphere- it is F-GMr2-Therefore- for r-lt-R-xA0-F1F2-r1r2-And for r-gt-R-xA0-F1F2-r22r12-

The magnitudes of the gravitational field at distances r1 and r2 from the centre of a uniform sphere of radius R and mass M are F1 and F2 respectively. ThenF1/F2=r1/r2, if r1<R and r2<RF1/F2=r22/r21, if r1>R and r2>RF1/F2=r1/r2, if r1>R and r2<RF1/F2=r21/r22, if r1<R and r2<R

Gravitational field inside a sphere of radius R at a distance r is F=GMR3r And, outside the sphere, it is F=GMr2 Therefore, for r<R F1F2=r1r2 And for r>R F1F2=r22r12

Gravitational field inside a sphere of radius $R$ at a distance $r$ is $F = \frac{G M}{R^{3}} r$

And, outside the sphere, it is $F = \frac{G M}{r^{2}}$

Therefore, for $r < R$
$\frac{F_{1}}{F_{2}} = \frac{r_{1}}{r_{2}}$

And for $r > R$
$\frac{F_{1}}{F_{2}} = \frac{{r_{2}}^{2}}{{r_{1}}^{2}}$