Question

A ball at rest is released from a height equal to $R_E$ above Earth's surface where $R_E$ is the radius of Earth. What is the speed of the ball when it is at a distance $\frac{R_E}{2}$ above the Earth's surface? ($M_E$ is mass of earth)

• A. $\sqrt{\frac{GM_E}{3R_E}}$
• B. $\sqrt{\frac{GM_E}{2R_E}}$
• C. $\sqrt{\frac{GM_E}{6R_E}}$
• D. $\sqrt{\frac{GM_E}{R_E}}$

Answer 1

Answer: (A)

$\sqrt{\frac{GM_E}{3R_E}}$

Answer 2

Given:

• Initial height above Earth’s surface = $R_E$
  $\Rightarrow$ Initial distance from Earth’s center, $r_i = R_E + R_E = 2R_E$
• Final height above Earth’s surface = $\frac{R_E}{2}$
  $\Rightarrow$ Final distance from Earth’s center, $r_f = R_E + \frac{R_E}{2} = \frac{3R_E}{2}$

Using conservation of energy:

$$\text{Initial Energy, } E_i = \text{K.E.}_i + \text{P.E.}_i = 0 - \frac{GM_E m}{2R_E} = -\frac{GM_E m}{2R_E}$$ $$\text{Final Energy, } E_f = \frac{1}{2} m v^2 - \frac{GM_E m}{\frac{3R_E}{2}} = \frac{1}{2} m v^2 - \frac{2GM_E m}{3R_E}$$

Setting $E_i = E_f$:

$$-\frac{GM_E m}{2R_E} = \frac{1}{2} m v^2 - \frac{2GM_E m}{3R_E}$$

Cancel $m$ and solve for $v^2$:

$$\frac{1}{2} v^2 = \frac{2GM_E}{3R_E} - \frac{GM_E}{2R_E}$$

Finding a common denominator:

$$\frac{2GM_E}{3R_E} = \frac{4GM_E}{6R_E}, \quad \frac{GM_E}{2R_E} = \frac{3GM_E}{6R_E}$$ $$\frac{1}{2} v^2 = \frac{4GM_E - 3GM_E}{6R_E} = \frac{GM_E}{6R_E}$$ $$v^2 = \frac{2GM_E}{6R_E} = \frac{GM_E}{3R_E}$$ $$v = \sqrt{\frac{GM_E}{3R_E}}$$

Explanation:

1. Determine initial and final distances from Earth’s center.
2. Apply energy conservation between the two points.
3. Rearrange to solve for $v$.