1. Tardigrade
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  3. Physics
  4. A body is dropped from a height h = R over the Earth surface. The velocity with which it will strike the earth surface is (R = Radius of earth = 6.4 × 106 m and acceleration due to gravity on earth’s surface g = 10 m/s2)

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Q. A body is dropped from a height h = R over the Earth surface. The velocity with which it will strike the earth surface is (R = Radius of earth = $6.4 \times 10^6$ m and acceleration due to gravity on earth’s surface g = 10 $m/s^2$)

Solution:

$V= \sqrt{\frac{2gh}{\left(1+\frac{h}{R}\right)}}$
$= \sqrt{gR}$
$= \sqrt{gR}=\sqrt{64\times10^6}=8km/s$