Question

A body is released from a height equal to the radius R of the earth . The velocity of the body when it strikes the surface of the earth will be

Answer 1

Answer on Question #41142, Physics, Mechanics | Kinematics | Dynamics

Question:

A body is released from a height equal to the radius R of the Earth. The velocity of the body when it strikes the surface of the earth will be

Answer:

The law of conservation of energy:

$$T + U = \text{const}$$

where $T = \frac{mv^2}{2}$ is kinetic energy, $m$ - mass, $v$ - speed

$U = -\frac{GmM}{R + h}$ is potential energy, $h$ is height, $R$ is the radius of the earth

$$-\frac{GmM}{R + h} = -\frac{GmM}{R + 0} + \frac{mv^2}{2}$$ $$v^2 = 2GM \left(\frac{1}{R} - \frac{1}{R + h}\right) = \frac{2GM}{R^2} \frac{Rh}{R + h} = \frac{2ghR}{R + h}$$

If height equal to the radius R of the earth:

$$v = \sqrt{\frac{2gR^2}{R + R}} = \sqrt{gR} = \sqrt{9.81 \frac{m}{s^2} \cdot 6380000 \, m} = 7911 \, \frac{m}{s}.$$

Air resistance is not taken into account here.

Answer: $7911 \, \frac{m}{s}$

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