JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 2)
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 × 106 m)
1.6 × 106 m
9.0 × 106 m
6.4 × 106 m
2.6 × 106 m
Explanation
$${{GM} \over {{{\left( {R + h} \right)}^2}}} = {{GM} \over {2{R^2}}}$$
$$R + h = \sqrt 2 R$$
$$h = \left( {\sqrt 2 - 1} \right)R \simeq 2.6 \times {10^6}m$$
$$R + h = \sqrt 2 R$$
$$h = \left( {\sqrt 2 - 1} \right)R \simeq 2.6 \times {10^6}m$$
Comments (0)
