Gravitation

JEE Main level, single correct, medium

49 of 223

Question 49 of 223

When one moves from a point 1616 km below the earth's surface to a point 1616 km above the earth's surface. The change in gg is approximately α\alpha %. The value of α\alpha is _____.
(Take radius of the earth =6400= 6400 km.)

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Answer: (B) 0.250.25

The acceleration due to gravity at a depth dd below the earth's surface is given by:

gd=g(1−dR)g_d = g\left(1 - \dfrac{d}{R}\right)

The acceleration due to gravity at a height hh above the earth's surface (for h≪Rh \ll R) is given by:

gh=g(1−2hR)g_h = g\left(1 - \dfrac{2h}{R}\right)

Given d=16d = 16 km and h=16h = 16 km, we can find the values of gdg_d and ghg_h:

gd=g(1−166400)=g(1−1400)g_d = g\left(1 - \dfrac{16}{6400}\right) = g\left(1 - \dfrac{1}{400}\right)

gh=g(1−2×166400)=g(1−2400)g_h = g\left(1 - \dfrac{2 \times 16}{6400}\right) = g\left(1 - \dfrac{2}{400}\right)

The magnitude of the change in acceleration due to gravity when moving from the depth dd to the height hh is:

Δg=gd−gh=g(1−1400)−g(1−2400)\Delta g = g_d - g_h = g\left(1 - \dfrac{1}{400}\right) - g\left(1 - \dfrac{2}{400}\right)

Δg=g(2400−1400)=g400\Delta g = g\left(\dfrac{2}{400} - \dfrac{1}{400}\right) = \dfrac{g}{400}

The percentage change in gg is approximately:

Δgg×100=g400g×100=100400=0.25%\dfrac{\Delta g}{g} \times 100 = \dfrac{\dfrac{g}{400}}{g} \times 100 = \dfrac{100}{400} = 0.25\%

Thus, the value of α\alpha is 0.250.25.

Answer: 0.250.25

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