JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Gravitation

R_{E}

be the radius of Earth, then the ratio between the acceleration due to gravity at a depth '

r

' below and a height '

r

' above the earth surface is:

(A) $1 + \frac{r}{R _{E}} + \frac{r ^{2}}{R _{E}^{2}} + \frac{r ^{3}}{R _{E}^{3}}$
(B) $1 - \frac{r}{R _{E}} - \frac{r ^{2}}{R _{E}^{2}} - \frac{r ^{3}}{R _{E}^{3}}$
(C) $1 + \frac{r}{R _{E}} - \frac{r ^{2}}{R _{E}^{2}} - \frac{r ^{3}}{R _{E}^{3}}$ ✓ Correct
(D) $1 + \frac{r}{R _{E}} - \frac{r ^{2}}{R _{E}^{2}} + \frac{r ^{3}}{R _{E}^{3}}$

Answer: (C)

Solution

At depth

r

the gravity is

g_{d} = g (1 - \frac{r}{R _{E}})

, and at height

r

it is

g_{h} = \frac{g}{( 1 + r / R _{E} ) ^{2}}

. Their ratio is

\frac{g _{d}}{g _{h}} = (1 - \frac{r}{R _{E}}) (1 + \frac{r}{R _{E}})^{2} .

Writing

x = r / R_{E}

, this expands as

(1 - x) (1 + 2 x + x^{2}) = 1 + x - x^{2} - x^{3}

, so

\frac{g _{d}}{g _{h}} = 1 + \frac{r}{R _{E}} - \frac{r ^{2}}{R _{E}^{2}} - \frac{r ^{3}}{R _{E}^{3}} .

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option