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 #41138 – Physics – Mechanics | Kinematics | Dynamics

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

Solution:

Since the initial velocity of the body is zero, it's total energy is:

$$E = - \frac {G m M}{r}$$

where $m$ is mass of the body, $M$ is the mass of the earth and $r$ is distance from the centre of the earth. When the body reaches the earth, let its velocity be $v$ and its distance from the centre of the earth is the earth's radius $R$. Therefore, the energy now is

$$E = \frac {1}{2} m v ^ {2} - \frac {G m M}{R}$$

Equating (1) and (2) we get

$$\frac {1}{2} m v ^ {2} - \frac {G m M}{R} = - \frac {G m M}{r}$$ $$v ^ {2} = 2 G M \left(\frac {1}{R} - \frac {1}{r}\right)$$

Also $g = \frac{GM}{R^2}$. Therefore $GM = gR^2$. Using this in above equation we get

$$v = R \left(2 g \left(\frac {1}{R} - \frac {1}{r}\right)\right) ^ {\frac {1}{2}}$$

Now $r = 2R$ (given). Therefore

$$v = R \left(2 g \left(\frac {1}{R} - \frac {1}{r}\right)\right) ^ {\frac {1}{2}} = \sqrt {g R}$$

Answer: velocity of the body when it strikes the surface of the earth will be $\sqrt{gR}$.

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