JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Gravitation

The value of acceleration due to gravity at Earth's surface is

9.8 ms^{- 2}

. The altitude above its surface at which the acceleration due to gravity decreases to

4.9 ms^{- 2}

, is close to: (Radius of earth

= 6.4 \times 1 0^{6} m

)

(A) $6.4 \times 1 0^{6} m$
(B) $9.0 \times 1 0^{6} m$
(C) $2.6 \times 1 0^{6} m$ ✓ Correct
(D) $1.6 \times 1 0^{6} m$

Answer: (C)

Solution

Gravity at height

h

g^{'} = \frac{GM}{( R + h ) ^{2}}

, against

g = \frac{GM}{R ^{2}}

at the surface. Setting

g^{'} = \frac{g}{2}

gives

\frac{1}{( R + h ) ^{2}} = \frac{1}{2 R ^{2}}

, so

R + h = 2 R

and

h = (2 - 1) R

. With

R \approx 6.4 \times 1 0^{6}

m this is about

0.414 \times 6.4 \times 1 0^{6} \approx 2.6 \times 1 0^{6}

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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