Question

Acceleration due to gravity at height h = R/2 is equal to the acceleration due to gravity at depth d from the surface of earth then find d in terms of R. when R is the radius of the earth.

Answer 1

Answer:

option 3 is the answer

Explanation:

explanation is in picture

Answer 2

ANSWER.

D in terms of R. when R is radius of earth

=> 5/9 R

option [ 3 ] is correct answer.

EXPLANATION.

$\sf \to \: acceleration \: due \: to \: gravity \: at \: height \: h \: = \dfrac{R}{2} \\ \\ \sf \to \: it \: is \: equal \: to \: the \: acceleration \: due \: to \: gravity \: at \: depth \: \: d \: \\ \sf \to \: from \: the \: surface \: of \: earth$

$\sf \to \: as \: we \: know \: that \: \\ \\ \sf \to \: F \: = \frac{ G M_{1} M_{2} }{ {d}^{2} } \\ \\ \sf \to \: g_{1} = g_{2} \\ \\ \sf \to \: \frac{G M}{(r + h) {}^{2} } = \frac{G M}{ {r}^{3} } (r - d) \\ \\ \sf \to \: \frac{G M}{(r + \frac{r}{2}) {}^{2} } = \frac{G M}{ {r}^{2} } (1 - \frac{d}{r} )$

$\sf \to \: \dfrac{4G M}{9 {r}^{2} } = \dfrac{G M}{ {r}^{2} } (1 - \dfrac{d}{r} ) \\ \\ \sf \to \: \frac{4}{9} = 1 - \frac{d}{r} \\ \\ \sf \to \: \frac{4}{9} = \frac{r - d}{r} \\ \\ \sf \to \: 4r = 9r - 9d \\ \\ \sf \to \: 5r \: = 9d \\ \\ \sf \to \: d \: = \frac{5r}{9} = answer$