A tunnel is dug along a diameter of the earth | vedclass

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Question

$\mathrm{F}=\frac{\mathrm{GM}^{1} \mathrm{m}}{\mathrm{r}^{2}}$

$\mathrm{M}^{1}=\frac{\mathrm{GMe}}{\mathrm{Re}^{3}} 	imes \mathrm{r}^{3}$

$\mat

Vedclass · Accepted Answer

$\frac{{G{M_e}m}}{{R_e^3}}.r$

Vedclass · Answer

$\mathrm{F}=\frac{\mathrm{GM}^{1} \mathrm{m}}{\mathrm{r}^{2}}$

$\mathrm{M}^{1}=\frac{\mathrm{GMe}}{\mathrm{Re}^{3}} 	imes \mathrm{r}^{3}$

$\mathrm{F}_{	ext {attaction }}=\frac{	ext { GMeM }}{\mathrm{Re}^{3}} 	imes \mathrm{r}$

A tunnel is dug along a diameter of the earth. If $M_e$ and $R_e$ are the mass and radius of the earth respectively. Then the force on a particle of mass $'m'$ placed in the tunnel at a distance $r$ from the centre is

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