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)

Question

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) (A) 8 km/s (B) 4 km/s (C) 10 km/s (D) 5 km/s

Tardigrade Admin · Accepted Answer

V= √(2gh/(1+(h)R)) = √gR = √gR=√64×106=8km/s

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 = $6.4 \times 1 0^{6}$ m and acceleration due to gravity on earth’s surface g = 10 $m / s^{2}$ )

8 km/s

4 km/s

10 km/s

5 km/s

$V = \frac{2 g h}{( 1 + \frac{h}{R} )}$
$= g R$
$= g R = 64 \times 1 0^{6} = 8 km / s$

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 = 6.4×106 m and acceleration due to gravity on earth’s surface g = 10 m/s2)