Question

at what depth from the earths surface does the accelerartion due to gravity become 1/4 times the value at the surface

Answer 1

Answer:

Since

$g= \frac{GM }{ {r}^{2} }$

Now assume that distance is d . we will only take mass which is inside this . So

let p be density of earth

$M = \rho( \frac{4}{3} \pi {r}^{3} )$

$m = \rho( \frac{4}{3} \pi {d}^{3} ) = \frac{M {d}^{3} }{ {r}^{3} }$

Now

$\frac{g}{4} = \frac{G \frac{M {d}^{3} }{ {r}^{3} } }{ {d}^{2} } = ( \frac{GM }{ {r}^{2} } ) \frac{d}{r} = g. \frac{d}{r} \\ \\ d = \frac{r}{4}$